electrochemistry.doc

Electrolytic and Non-electrolytic conductors

Introduction

Electrochemistry deals with the interactions of electrical energy with chemical species. It is broadly divided into two categories, namely (i) production of chemical change by electrical energy (phenomenon of electrolysis) and (ii) conversation of chemical energy into electrical energy, i.e., generation of electricity by spontaneous redox reactions. In this chapter both of these aspects will be described. All electrochemical reactions involve transfer of electrons and are, therefore, oxidation-reduction (redox) reactions.

Substances which allow the passage of electric current through then are called electrical conductors or simply conductors. Those which do not allow the flow of electric current through them are termed insulators. Electrical conductors are of two types:

(i) Metallic or electronic conductors:

Conductors which transfer electric current by transfer of electrons, without transfer of any matter, are known as metallic or electronic conductors. Metals such as copper, silver, aluminum, etc., non-metals like carbon (graphite - an allotropic form of carbon) and various alloys belong to this class. These materials contain electrons which are relatively free to move. The passage of current through these materials has no observable effect other than a rise in their temperature.

(ii) Electrolytic conductors:

Conductors like aqueous solutions of acids, bases and salts in which the flow of electric current is accompanied by chemical decomposition are known as electrolytic conductors. The substances whose aqueous solutions allow the passage of electric current and are chemically decomposed, are termed electrolytes.

The substances whose aqueous solutions do not conduct electric current are called non-electrolytes. Solutions of cane sugar, glycerine, alcohol, etc., are examples of non-electrolytes.

In order to pass the current through an electrolytic conductor (aqueous solution or fused electrolyte), two rods or plates (metallic conductors) are always needed which are connected with the terminals of a battery. These rods or plates are known as electrodes. The electrode through which the current enters the electrolytic solution is called the anode (positive electrode) with the electrode through which the current leaves the electrolytic solution is known as cathode (negative electrode). The electrolytic solution conducts electricity by the movement of charged particles called ions towards the respective oppositely charged electrodes. The ions which carry positive charge and move towards cathode are termed cations while ions carrying negative charge which move towards anode are called anions. When these ions reach the boundary between a metallic and an electrolytic conductor, electrons are being either attached to or removed from the ions. Removal of electrons is termed oxidation (de-electronation) which occurs at anode while addition of electrons is called reduction (electronation) that takes place at cathode. Hence, flow of electrons through the outer circuit from anode to cathode across the boundary is accompanied by oxidation and reduction.

Distinction between metallic and electrolytic conduction

Metallic conduction Electrolytic conduction

1. Electric current flows by movement of electrons.

1. Electric current flows by movement of ions.

2. No chemical change occurs. 2. Ions are oxidized or reduced at the electrodes.

3. It does not involve the transfer of any matter.

3. It involves transfer of matter in the form of ions.

4. Ohm's law is followed. 4. Ohm's low is followed.

5. Resistance increases with increase of temperature.

5. Resistance decreases with increase of temperature.

6. Faraday law is not followed. 6. Faraday law is followed.

The process of chemical decomposition of an electrolyte by passage of electric current through its solution is called electrolysis.

Molecules of an electrolyte when dissolved in water split up into ions, i.e., into cations and anions. On passing current, these ions move towards oppositely charged electrodes. On reaching the electrodes the ions lose their charge either by accepting electrons or losing electrons and thereby deposited at the respective electrodes or undergo a secondary change. For example, when electric current is passed through a solution of hydrochloric acid, the H+ ions move towards cathode and CI- ions move towards anode.

The decomposition of HCl into H2 and Cl2 as a result of passage of current is termed electrolysis of HCl. It is, thus, a process in which electric current brings the chemical change.

The device in which electrolysis (chemical reaction involving oxidation and reduction) is carried out by using electricity or in which convertion of electrical

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energy into chemical change is done is known as electrolytic cell. An electrolytic cell consists of a vessel for the electrolytic solution or fused electrolyte and two metallic electrodes immersed in the reaction material which are connected to a source of electric current. The metallic electrodes which do not react with ions or final products are called inert electrodes. Inert electrodes are usually used in an electrolytic cell.

Electrolysis and Electrolytic cell

Preferential Discharge Theory

If an electrolytic solution consists of more than two ions and the electrolysis is done, it is observed that all the ions are not discharged at the electrodes simultaneously but certain ions are liberated at the electrodes in preference to others. This is explained by preferential discharge theory. It states that if more than one type of ions are attracted towards a particular electrode, then the one discharged is the ion which requires least energy. The potential at which the ion is discharged is called discharge or deposition potential. The values of discharge potential are different for different ions. For example, the discharge potential of H+ ions is lower than Na+ ions when platinum or most of the other metals* are used as cathodes. Similarly, discharge potential of Cl - ions is lower than that of OH - ions . My note: As a result, the electrolysis of a solytion of ClNa yields hydrogen at the cathod and chlorium at the anode. The “less preferred” ions Na+ and OH- form the base NaOH. This can be explained by some examples given below:

(i) Electrolysis of sodium chloride solution:

The solution of sodium chloride besides Na+ and Cl- ions possesses H+ and OH- ions due to ionization of water. However, their number is small as water is a weak electrolyte. When potential difference is established across the two electrodes, Na+ and H+ ions move towards cathode and Cl- and OH- ions move towards anode. At cathode H+ ions are discharged in preference to Na+ ions as the discharge potential of H+ ions is lower than Na+ ions. Similarly at anode, Cl- ions are discharged in preference to OH- ions.

NaCl <--> Na+ + Cl-

H2O <--> H+ + OH-

At cathode At Anode

H+ + e- --> H Cl- --> Cl + e-

2H --> H2 2Cl --> Cl2

Thus, Na+ and OH- ions remain in solution and the solution when evaporated yields crystals of sodium hydroxide.

(ii) Electrolysis of copper sulphate solution using platinum electrodes:

CuSO4 <--> Cu2+ + SO42-

H2O <--> H+ + OH-

At cathode At Anode

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Cu2+ + 2e- --> Cu 2OH- --> H2O + O + 2e-

O + O --> O2

Copper is discharged at cathode as Cu2+ ions have lower discharge potential than H+ ions. OH- ions are discharged at anode as these have lower discharge potential than SO4

2- ions. Thus, copper is deposited at cathode and oxygen gas is evolved at anode. My note: The “less prefered” ions H+ (from H2O) and SO4

2- (from CuSO4) form sulfuric acid H2SO4.

Electrolysis and Electrolytic cell

(iii) Electrolysis of sodium sulphate solution using inert electrodes:

Na2SO4 <--> 2Na+ + SO42-

H2O <--> H+ + OH-

At cathode At Anode

H+ + e- --> H 2OH- --> H2O + O + 2e-

2H --> H2 O + O --> O2

Hydrogen is discharged at cathode as H+ ions have lower discharge potential than Na+ ions. OH- ions are discharged at anode as these have lower discharge potential than SO4

2- ions. Thus, hydrogen is evolved at cathode and oxygen is evolved at anode, i.e., the net reaction describes the electrolysis of water. The ions of Na2SO4 conduct the current through the solution and take no part in the overall chemical reaction.

The decreasing order of discharge potential or the increasing order of deposition of some of the ions is given below:

For cations: K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, H+, Cu2+, Hg2+, Ag+

For anions: SO42-, NO3

-, OH-, Cl-, Br-, I-

(iv) Electrolysis of copper sulphate solution using copper electrodes:

CuSO4 <--> Cu2+ + SO42-

At cathode, copper is deposited.

Cu2+ + 2e- Cu

At anode, the copper of the electrode is oxidised to Cu2+ ions or ions solution dissolve equivalent amount of copper of the anode.

Cu Cu2+ + 2e-

or Cu + CuSO4 + 2e-

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Thus, during electrolysis, copper is transferred from anode to cathode.

(v) Electrolysis of silver nitrate solution using silver electrodes:

AgNO2 <--> Ag+ + NO42-

At cathode, silver is deposited.

Ag+ + e- --> Ag

At anode, the silver of the electrode is oxidised to Ag+ ions which go into the solution or ions dissolve equivalent amount of silver of the electrode.

Ag --> Ag+ + e-

Ag + NO3- --> AgNO3 + e-

Table 12.1 Some more examples of electrolysis

Electrolyte Electrode Cathodic reaction Anodic reaction

Aqueous acidified CuCl2 solution

Molten PbBr2

Sodium chloride solution

Silver nitrate solution

Sodium nitrate solution

Pt

Pt

Hg

Pt

Pt

Cu2+ + 2e---> Cu

Pb2+ + 2e- --> Pb

2Na+ + 2e- --> 2Na

Ag+ + e- --> Ag

2H+ + 2e- --> H2

2Cl- --> Cl2 + 2e-

2br- --> Br2 + 2e-

2Cl- --> Cl2 + 2e-

2OH- --> 1/2 O2 + H2O + 2e-

2OH- -->1/2 O2 + H2O + 2e-

Faraday’s Laws of Electrolysis

The relationship between the quantity of electric charge passed through an electrolyte and the amount of the substance deposited at the electrodes was presented by Faraday in 1834, in the form of laws of electrolysis.

(i) Faraday's First Law

When an electric current is passed through an electrolyte, the amount of substance deposited is proportional to the quantity of electric charge passed through the electrolyte.

If W be the mass of the substance deposited by passing Q coulomb of charge, then according to the law, we have the relation:

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W Q

A coulomb is the quantity of charge when a current of one ampere is passed for one second. Thus, amount of charge in coulombs,

Q = current in amperes × time in seconds

= i × t

So W i × t

or W = z × i × t

where z is a constant, known as electro-chemical equivalent, and is characteristic of the substance deposited.

When a current of one ampere is passed for one second, i.e., one coulomb (Q = 1), then

W = Z

Thus, electrochemical equivalent can be defined as the mass of the substance deposited by one coulomb of charge or by one ampere of current passed for one second. For example, when a charge of one coulomb is passed through silver nitrate solution, the amount of silver deposited is 0.001118 g. this is the value of electrochemical equivalent of silver.

(i) Faraday's Second Law

When the same quantity of charge is passed through different electrolytes, then the masses of different substances deposited at the respective electrodes will be in the ratio of their equivalent masses.

The law can be illustrated by passing same quantity of electric current through three voltametres containing solutions of H2SO4, CuSO4 and AgNO3 respectively as shown in Fig.12.1. In the first voltameter, hydrogen and oxygen will be liberated; in the second, copper will be deposited and in the third, silver will be deposited.

(Mass of hydrogen)/(Mass of copper) = (Equivalent mass of hydrogen)/ (Equivalent mass of copper)

or (Mass of copper)/(Mass of silver) = (Equivalent mass of copper)/ (Equivalent mass of silver)

or (Mass of silver)/(Mass of hydrogen) = (Equivalent mass of silver)/ (Equivalent mass of hydrogen)

It is observed that by passing one coulomb of electric charge:

Hydrogen evolved = 0.00001036 g

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Copper deposited = 0.0003292 g

and Silver deposited = 0.001118 g

These masses are in the ratio of their equivalent masses. From these masses, the amount of electric charge required to deposit one equivalent of hydrogen or copper or silver can be calculated.

For hydrogen = 1/0.0001036= 96500 coulomb

For copper = 31.78/0.0003292= 96500 coulomb

For silver = 107.88/0.001118 = 96500 coulomb

Thus it follows that 96500 coulomb electric charge will deposit one g equivalent of any substance. 96500 coulomb us termed as one Faraday and is denoted by F.

Again according to first law,

W = Z×Q

Then Q = 96500 coulomb, W becomes gram equivalent mass (E).

Thus, E = Z×96500

or Z = E/96500

z1/z2 = E1/E2

Fundamental unit of charge:

As one g-equivalent of an ion is liberated by 96500 coulomb, it follows that charge carried by one g-equivalent of an ion is 96500 coulomb. If the valency of an ion is 'n', then one mole of these ions will carry a charge of nF coulomb. One g-mole of an ion contains 6.02 × 10 23 ions .

Then,

The charge carried by an ion = nF/(6.02 × 1023 ) coulomb

For n = 1,

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The fundamental unit of charge = F/(6.02 × 1023 )

i.e., 96500/(6.02 × 1023 ) = 1.6 × 10-19 coulomb

or 1 coulomb*=6.24 × 1018 electrons

The rate of flowing of electric charge through a conductor is called the electric current.

Coulomb:

It is the unit of electric charge. It is the amount of charge that moves past a given point in a circuit when a current of 1 ampere is supplied for one second.

1 coulomb = 1 ampere - second

It is also defined as the amount of charge which is required to deposit by electrolysis 0.001118 g of silver from a solution of silver nitrate.

An electron has 1.6×10-19 coulomb of negative charge. Hence, one coulomb of charge is carried by 6.24 × 1018 electrons. 1 mole of electrons carry a charge of 96500 coulomb. This quantity of charge is called Faraday.

Charge carried by 1 mole of electrons

= (6.023×1023)(1.6×10-19)

= 96368 coulomb

= 96500 coulomb

Electric current = (Electric charge)/Time

1 ampere = (1 coulomb)/(1 second)

Volt is a unit of electrical potential difference. It is defined as potential energy per unit charge.

1 volt = (1 joule)/(1 coulomb) = (1 newton×1 metre)/(1 ampere×1 second)

Electrical energy = Potential difference × Quantity of charge

= V × Q

= V × 1 × 1 (1 = ampere; 1 = second)

= Watt-second

Faraday's law of gaseous electrolytic product

We know W = ZQ

= Zlt

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W = (It E)/96500 ....... (i)

where Z = E/96500

Equation, V = Volume of gas evolved at S.T.P. at an electrode

Ve = Equivalent volume

= Volume of gas evolved at an electrode at S.T.P. by 1 Faraday charge

Illustration

O2: M = 32, E = 8

32 g O2 = 22.4 L at S.T.P. [M = Molecular mass]

8 g O2 = 5.6 L at S..T.P. [E = Equivalent mass]

Thus, Ve of O2 = 5.6 L

H2: M = 2, E = 1

2 g H2 = 22.4 L at S.T.P.

1 g H2 = 11.2 : at S.T.P.

Thus Ve of Cl2 = 11.2L

Applications of Electrolysis

The phenomenon of electrolysis has wide application. The important ones are:

(1) Determination of equivalent masses of elements:

According to second law of electrolysis when the same quantity of electronic current is passed through solutions of salts of two different cells, the amounts of the metals deposited on the cathodes of the two cells are proportional to their equivalent masses of the respective metals. If the amounts of the metals deposited on the cathodes be WA and WB respectively, then

WA/WB = (Equivalent mass of A)/(Equivalent mass of B)

Knowing the equivalent mass of one metal, the equivalent mass of the other metal can be calculated from the above relationship. The equivalent masses of those non-metals which are evolved at anodes can also be determined by this method.

(2) Electronmetallurgy:

The metals like sodium, potassium, magnesium, calcium aluminum, etc., are obtained by electrolysis of fused electrolytes.

Fused electrolyte Metal isolated

NaCl + CaCl2 + KF Na

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CaCl2 + CaF2

Ca

Al2O3 + cryolite Al

MgCi2 (35%) + NaCl (50%) + CaCl2 (15%) Mg

NaOH Na

KCl + CaCl2 K

(3) Manufacture of non-metals:

Non-metals like hydrogen, fluorine, chlorine are obtained by electrolysis.

(4) Electro-refining of metals:

The metals like copper, silver, gold, aluminum, tin, etc., are refined by electrolysis.

(5) Manufacture of compounds:

Compounds like NaOH, KOH, Na2CO3 KCIO3, white lead, KMnO4, etc., are manufactured by electrolysis.

(6) Electroplating:

The process of coating an inferior metal with a superior metal by electrolysis is known as electroplating.

The aims of electroplating are:

(i) To prevent the inferior metal from corrosion.

(ii) To make it more attractive in appearance.

The object to be electroplated is made the cathode and block of the metal to be deposited is made the anode in an electrolytic bath containing a solution of a salt of the anodic metal. On passing electric current in the cell, the metal of the anode dissolves out and is deposited on the cathode-article in the form of a thin film. The following are the requirements for fine coating:

(i) The surface of the article should be free from greasy matter and its oxide layer. The surface is cleaned with chromic acid or detergents.

(ii) The surface of the article should be rough so that the metal deposited sticks permanently.

(iii) The concentration of the electrolyte should be so adjusted as to get smooth coating.

(iv) Current density must be the same throughout.

For Anode Cathode Electrolyte

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electroplating

With copper

With silver

With nickel

With gold

With zinc

With thin

Cu

Ag

Ni

Au

Zn

Sn

Object

Object

Object

Object

Iron objects

Iron objects

CuSo4 + dilute H2So4

KAg(CN)2

Nickel ammonium sulphate

KAu(CN)2

ZnSO4

SnSO4

Thickness of coated layer

Let the dimensions of metal sheet to be coated be (a cm × b cm).

Thickness of coated layer = c cm

Volume of coated layer = (a × b × c) cm3

Mass of the deposited substance = Volume × density

= (a × b × c) × d g

=> (a × b × c) × d = (I×t×E)/96500

Using above relation we may calculate the thickness of coated layer.

Note: Sometimes radius of atom of deposited metal is given instead of density, e.g.,

Radius of silver atom = 10-8 cm

Atomic mass of Ag = 108

Mass of single silver atom = 108/(6.023× 1023 ) g

Volume of single atom = 4/3× πR3

= 4/3×3.14×(10-8)3 cm3

Density of Ag = (Mass of single atom )/(Volume of single atom)

= (108/6.023 × 1023)/(4/3×3.14×10-8 )3

Current efficiency

Sometimes the ammeter shows false current due to mechanical fault. In this case,

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% Current efficiency = (Actual current)/(Ammeter current)×100

Solved Examples on Faraday’s Laws of Electrolysis

Some Solved Examples

Example 1. Find the charge in coulomb on 1 g-ion of

Solution: Charge on one ion of N3-

= 3 × 1.6 × 10-19 coulomb

Thus, charge on one g-ion of N3-

= 3 × 1.6 10-19 × 6.02 × 1023

= 2.89 × 105 coulomb

Example 2. How much charge is required to reduce (a) 1 mole of Al3+ to Al and (b)1 mole of to Mn2+?

Solution: (a) The reduction reaction is

Al3+ + 3e- --> Al

1 mole 3 mole

Thus, 3 mole of electrons are needed to reduce 1 mole of Al3+.

Q = 3 × F

= 3 × 96500 = 289500 coulomb

(b) The reduction is

Mn4- + 8H + 5e- --> MN2+ + 4H2O

1 mole 5 mole

Q = 5 × F

= 5 × 96500 = 48500 coulomb

Example 3. How much electric charge is required to oxidise (a) 1 mole of H2O to O2 and (b)1 mole of FeO to Fe2O3?

Solution: (a) The oxidation reaction is

H2O --> 1/2 O2 + 2H+ + 2e-

1 mole 2 mole

Q = 2 × F

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= 2 × 96500=193000 coulomb

(b) The oxidation reaction is

FeO + 1/2 H2O --> Fe2O3 + H++ e-

Q = F = 96500 coulomb

Example 4. Exactly 0.4 faraday electric charge is passed through three electrolytic cells in series, first containing AgNO3, second CuSO4 and third FeCl3 solution. How many gram of rach metal will be deposited assuming only cathodic reaction in each cell?

Solution: The cathodic reactions in the cells are respectively.

Ag+ + e- --> Ag

1 mole 1 mole

108 g 1 F

Cu2+ + 2e- --> Cu

1 mole 2 mole

63.5 g 2 F

and Fe3+ + 3e- --> Fe

1 mole 3 mole

56 g 3 F

Hence, Ag deposited = 108 × 0.4 = 43.2 g

Cu deposited = 63.5/2×0.4 = 12.7 g

and Fe deposited = 56/3×0.4 = 7.47 g

Example 5. An electric current of 100 ampere is passed through a molten liquid of sodium chloride for 5 hours. Calculate the volume of chlorine gas liberated at the electrode at NTP.

Solution: The reaction taking place at anode is

2Cl- --> Cl2 + 2e-

71.0 g 71.0 g 2 × 96500 coulomb

1 mole

Q = I × t = 100 × 5 × 600 coulomb

The amount of chlorine liberated by passing 100 × 5 × 60 × 60 coulomb of electric charge.

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= 1/(2×96500)×100×5×60×60 = 9.3264 mole

Volume of Cl2 liberated at NTP = 9.3264 × 22.4 = 208.91 L

Example 6. A 100 watt, 100 volt incandescent lamp is connected in series with an electrolytic cell containing cadmium sulphate solution. What mass of cadmium will be deposited by the current flowing for 10 hours?

Solution: We know that

Watt = ampere × volt

100 = ampere × 110

Ampere = 100/110

Quantity of charge = ampere × second

= 100/110×10×60×60 coulomb

The cathodic reaction is

Cd2+ + 2e- --> Cd

112.4 g 2 × 96500 C

Mass of cadmium deposited by passing 100/110×10×60×60

Coulomb charge = 112.4/(2×96500)×100/110×10×60×60 = 19.0598 g

Example 7. In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution gold salt and the second cell contains copper sulphate solution. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, find the amount of copper deposited on the cathode in the second cell. Also calculate the magnitude of the current in ampere.


(Mass of Au deposited)/(Mass f Cu deposited)=(Eq.mass of Au)/(Eq.Mass of Cu)

Eq. mass of Au = 197/3 ; Eq. mass of Cu 63.5/2

Mass of copper deposited

= 9.85 × 63.5/2×3/197 g = 4.7625 g

Let Z be the electrochemical equivalent of Cu.

E = Z × 96500

or Z = E/96500 = 63.5/(2×96500)

Applying W = Z × I × t

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T = 5 hour = 5 × 3600 second

4.7625 = 63.5/(2×96500) × I × 5 × 3600

or I = (4.7625 × 2 × 96500)/(63.5 × 5 × 3600) = 0.0804 ampere

Example 8. How long has a current of 3 ampere to be applied through a solution of silver nitrate to coat a metal surface of 80 cm2 with 0.005 cm thick layer? Density of silver is 10.5 g/cm3.

Solution: Mass of silver to be deposited

= Volume × density

= Area ×thickness × density

Given: Area = 80 cm2, thickness = 0.0005 cm and density = 10.5 g/cm3

Mass of silver to be deposited = 80 × 0.0005 × 10.5

= 0.42 g

Applying to silver E = Z × 96500

Z = 108/96500 g

Let the current be passed for r seconds.

We know that

W = Z × I × t

So, 0.42 = 108/96500×3×t

or t = (0.42 × 96500)/(108×3) = 125.09 second

ARRHENIUS THEORY OF ELECTOLYTIC DISSOCIATION

In order to explain the properties of electrolytic solutions, Arrhenius put forth, in 1884, a comprehensive theory which is known as theory of electrolytic dissociation or ionic theory. The main points of the theory are:

(i) An electrolyte, when dissolved in water, breaks up into two types of charged particles, one carrying a positive charge and the other a negative charge. These charged particles are called ions. Positively charged ions are termed cations and negatively charged as anions.

AB --> A+ + B-

NaCl --> Na+ + CL-

K2SO4 --> 2K++ SO42-

Electrolyte Ions

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In its modern form, the theory assumes that solid electrolytes are composed of ions which are held together by electrostatic forces of attraction. When an electrolyte is issolved in a solvent, these forces are weakened and the electrolyte undergoes dissociation into ions. The ions are solvated.

A+B- --> A+ + B-

or A+B-+ aq --> A+(aq)+B- (aq)

(ii) The process of splitting of the molecules into ions of an electrolyte is called ionization. The fraction of the total number of molecules present in solution as ions is known as degree of ionization or degree of dissociation. It is denoted by

α= (Number of molecules dissociated into ions)/(Total number of molecules)

It has been observed that all electrolytes do not ionize to the same extent. Some are almost completely ionized while others are feebly ionized. The degree of ionization depends on a number of factors (see 12.6).

(iii) Ions present in solution constantly re-unite to form neutral molecules and, thus, there is a state of dynamic equilibrium between the ionized the ionized and non-ionised molecules, i.e.,

AB <--> A+ + B-

Applying the law of mass action to above equilibrium

[A+ ][B- ] /[AB] =K

K is known as ionization constant. The electrolytes having high value of K are termed strong electrolytes and those having low value of K as weak electrolytes.

(iv) When an electric current is passed through the electrolytic solution, the positive ions (cations) move towards cathode and the negative ions (anions) move towards anode and get discharged, i.e., electrolysis occurs.

The ions are discharged always in equivalent amounts, no matter what their relative speeds are.

(v) The electrolytic solutions is always neutral in nature as the total charge on one set of ions is always equal to the total charge on the other set of ions. However, it is not necessary that the number of two sets of ions must be equal always.

AB <--> A+ + B- (Both ions are equal)

NaCl <--> Na+ + Cl- (Both ions are equal)

AB2 <--> A2+ + 2B- (Anions are double that of cations)

BaCl2 <--> Ba2+ + 2Cl- (Anions are double that of cations)

A2B <--> 2a+ + B2- (Anions are double that of cations)

Na2SO4 <--> 2Na+ + (Anions are double that of cations)

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(vi) The properties of electrolytes in solution are the properties of ions present in solution. For example, acidic solution always contains H+ ions while basic solution contains OH- ions and characteristic properties of solutions are those of H- ions and OH- ions respectively.

(vii) The ions act like molecules towards depressing the freezing point, elevating the boiling point, lowering the vapour pressure and establishing the osmotic pressure.

(viii) The conductivity of the electrolytic solution depends on the nature and number of ions as the current is carried through solution by the movement of ions.

Evidences in favour of ionic theory

A large number of experimental observations are available which support Arrhenius theory. A few of them are given below:

(i) Ions present in solid electrolytes:

X-ray diffraction studies have shown that electrolytes are composed of ions. For example, a crystal of NaCl does not contain NaCl units but Na+ and Cl- ions. Each Na+ ion is surrounded by six Cl- ions and each Cl- ion in turn is surrounded by six Na+ and Cl- ions. The ionic compounds behave as good conductors in fused state. It can only be possible if ions are already present in ionic solids.

(ii) Ohm's law applicability:

The electrolytic solutions like metallic conductors obey Ohm's law, i.e., the strength of the current flowing conductor is directly proportional to potential difference (E) applied across the conductor and is inversely proportional to the resistance of the conductor. Mathematically,

I = E/R

This can only be possible if ions are already present in the solution and no part of the current has only directive effect on the ions.

(iii) Ionic reaction:

Evidence for the existence of ions in aqueous solutions of electrolytes is furnished by well known reactions in inorganic chemistry. A white precipitate of silver chloride is obtained whenever Ag + ions come in contact with chloride ions .

Ag+ + + Na+ Cl- AgCl + Na+ + NO3-

But no precipitation occurs when AgNO3 solution is added to CCl4, CHCl3 or C2H5Cl as these substances being non-electrolytes do not furnish Cl - ions in solution .

An acid which gives all tests of H+ ions in aqueous solution, does not give the same tests when dissolved in any organic solvent because no ionization of the acid occurs in the common reaction.

(iv) Heat of neutralization:

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When one gram equivalent of a strong acid is neutralized by one gram equivalent of a strong base, the heat evolved is always the same, i.e., 13.7 kcal. This can be explained on the basis of Arrhenius theory that an acid furnished H + ions and base OH - ions when dissolved in water and the process of neutralization involves the common reaction.

H+ + OH- <--> H2O + 13.7 kcal.

Thus, heat of neutralization is actually the heat of formation of H2O from H+ and OH- ions.

(v) Abnormal colligative properties:

The abnormal behavior towards colligative properties as observed in the case of electrolytes can be explained on the basis of ionic theory. When an electrolyte is dissolved in water, the number of molecules actually dissolved due to ionization. The can't Hoff factor,

i= (Observed colligative property)/(Calculated colligative property)

is always more than one, i.e., i = 1 + (n-1) where 'n' is the number of ions produced by the ionization of one molecule of the electrolyte and ' is the degree of ionization.

(vi) Colour of solution:

The color of the electrolytes in solution. If any, is due to their ions, the CuSO4 is blue in solution due to the presence of Cu 2+ ions . Potassium permanganate (KMnO4) is purple in solution due to the presence of ions .

(vii) Explanation of some other phenomena:

Ionic theory provides satisfactory explanations regarding various phenomena such as electrolysis, conductivity, salt hydrolysis, solubility product, etc.

Limitations of Arrhenius theory

(i) Ostwald's dilution law which is based on Arrhenius theory is not applicable to strong electrolytes.

(ii) Strong electrolytes conduct electricity infused state, i.e., in absence of water. this is in contradiction of Arrhenius theory according to which the presence of solvent is a must for ionization.

(iii) Arrhenius theory assumes independent existence of ions but fails to account for the factors which influence the mobility of the ions.

Factors pertaining to degree of ionization

The degree of ionization of an electrolyte in solution depends upon the following factors:

(i) Nature of solute:

When the ionisable parts of a molecule of a substance are held more by covalent bonding than by electrovalent bonding, less ions are furnished in solution. Such

18

substances are termed weak electrolytes. H2S, HCN, NH4OH, CH3COOH are examples of this class. NaCl, Ba(NO3)2, KOH, etc., are strong electrolytes, in which the transfer of electrons seems to be more or less complete, furnish ions immediately when dissolved. Strong electrolytes are almost completely ionized in solution.

(ii) Nature of solvent:

The main function of the solvent is to weaken the electrostatic forces of attraction between the two ions and separate them. The force of attraction holding the ions together in any medium is expressed as

F= 1/(K ) (q1 q2)/r2 Coulomb’s law

where K is the dielectric constant of medium.

Any solvent which has high value of dielectric constant has the capacity of separating ions. Water is considered to be the best solvent as it has the highest dielectric constant. The dielectric constants of some of the solvents are given below at 25 0 C .

Water Methyl alcohol Ethyl alcohol Acetone

81 35 27 21

(iii) Dilution:

The extent of ionization of an electrolyte is inversely proportional to the concentration of its solution. Thus, degree of ionization increases with the increase of dilution of the solution, i.e., decreasing the concentration of the solution.

(iv) Temperature:

The degree of ionization increases with the increase of temperature. This is due to the fact that at higher temperature molecular speed is increased which overcomes the forces of attraction between the ions.

Conductance, Specific conductance and Molar conductance

Electrolytic Conductance

The conductance is the property of the conductor (metallic as well as electrolytic) which facilitates the flow of electricity through it. It is equal to the reciprocal of resistance i.e.,

Conductance = 1/Resistance = 1/R ..... (i)

It is expressed on the unit called reciprocal ohm (ohm-1 or mho) or siemens.

Specific conductance or conductivity:

The resistance of any conductor varies directly as its length (l) and inversely as its cross-sectional area (a), i.e.,

19

R ∝ l/α or R = ρ l/α ..... (ii)

where ρ is called the specific resistance.

If l = 1 cm and a = 1 cm2, then

R = ρ ..... (iii)

The specific resistance is, thus, defines as the resistance of one centimeter cube of a conductor.

The reciprocal of specific resistance is termed the specific conductance or it is the conductance of one centimeter cube of a conductor.

It is denoted by the symbol Κ. Thus,

Κ= 1/ρ, Κ = kappa - the specific conductance ...... (iv)

Specific conductance is also called conductivity.

From Eq. (ii), we have

ρ = (a/l)R or 1/ρ = (l/a)(1/R)

K = (l/a)×C (l/a = cell constant)

or Specific conductance = Conductance × cell constant

In the case of electrolytic solutions, the specific conductance is defined as the conductance of a solution of definite dilution enclosed in a cell having two electrodes of unit area separated y one centimeter apart as shown in Fig. 12.2.

The unit of specific conductance is ohm-1 cm-1.

Equivalent conductance:

One of the factors on which the conductance of an electrolytic solution depends is the concentration of the solution. In order to obtain comparable results for different electrolytes, it is necessary to take equivalent conductance's.

20

Equivalent conductance is defined as the conductance of all the ions produced by one gram equivalent of an electrolyte in a given solution. It is denoted by A.

To understand the meaning of equivalent conductance, imagine a rectangular trough with two opposite sides made of metallic conductor (acting as electrodes) exactly 1 cm apart, If 1 cm3 (1 mL) solution containing 1 gram equivalent of an electrolyte is places in this container is measured.

According to definitions,

Conductance = Specific conductance (K)

= Equivalent conductance (A)

If the solution is diluted to say (9 cm3) (9 mL), the conductance of the solution will be the same but specific conductance becomes 1/9th as it contains nine cubes. The conductance is also equal to the equivalent because the solution still has 1 g equivalent of the electrolyte. This is shown in Fig. 12.3. Thus,

Equivalent conductance (A) = 9 × k

In general,

A = k × V ..... (v)

where V is the volume in mL containing 1 g equivalent of the electrolyte.

In case, if the concentration of the solution is c g equivalent per litre, then the volume containing 1 g equivalent of the electrolyte will be 1000/e.

So equivalent conductance

A = k × 1000/c ..... (vi)

A = k × 1000/N

where N = normality

The unit of equivalent conductance is ohm-1 cm-2 equi-1.

21

Molar conductance

The molar conductance is defined as the conductance of all the ions produced by ionization of 1 g mole of an electrolyte when present in V mL of solution. It is denoted by .

Molar conductance μ = k ×V .... (vii)

where V is the volume in mL containing 1 g mole of the electrolyte. If c is the concentration of the solution in g mole per litre, then

μ = k × 1000/c

It units are ohm-1 cm2 mol-1.

Equivalent conductance = (MOlar conductance)/n

where n = (Molecular mass)/(Equivalent mass)

Measurement of conductance

It is now known to us that when the solution of an electrolyte is taken between two parallel electrodes of cross-sectional area 'a' and 'l' cm apart, then the specific conductance, k, should be:

k= (l/a)(1/R)

Thus, knowing the values of R, l and a, the specific conductance can be measured. The resistance of the solution between two parallel electrodes is determined by using the Wheatstone bridge method. The diagram of the apparatus is shown in Fig. 12.4 AB is a uniform wire and X is a sliding contact which moves over it. C is the conductivity cell containing the solution of the electrolyte and S represents the source of alternating current. R is the resistance box and T is a headphone to detect the flow of current. A suitable resistance is taken out from the resistance box and the sliding contact X is moves on the wire to search a point of minimum sound in the headphone. At this point, the bridge is balanced.

(Resistance of solution)/(Resistance from resistance box)

=(Resistance XB)/(Resistance XA)

=(Length XB)/(Length XA)

Thus, resistance of solution can be used because it produce two complications.

22

(i) Change in the concentration of the solution occurs due to electrolysis which will change the resistance.

(ii) Polarisation at the electrodes sets in which also changes the resistance.

Thus, an alternating current (A.C.) is used to overcome the above complications.

The solution whose conductance is to be measured is taken in a special type of cell known as conductivity cell. Various types of cells are shown in Fig. 12.5. The electrodes consist of platinum discs coated with finely divided platinum black and welded to platinum wires fused in two glass tubes. The glass tubes contain mercury and are finely fixed in the cover of cells. Contact with the platinum is made by dipping the copper wires of the circuit in the mercury contained in the tubes. As the conductivity changes with temperature, the cell is usually placed in a constant temperature bath during the experiment. Cells with long paths are used for concentrated solution and cells with short paths and large electrodes are used for dilute solutions.

Cell constant

Since the electrodes are not exactly 1 unit apart and may not possess a surface area of 1 square unit, the measured resistance does not give the specific conductance of the solution. Actual measurements of l and a being inconvenient, an indirect method is employed to determine the value of which is a constant quantity for a particular cell and is known as cell constant. We know that

23

(Specific conductance)/Conductance=1/a= cell constant

The resistance of a cell, i.e., conductance is measured when filled with a standard solution (say N/10 KCl solution) at a given temperature. The standard values of specific conductance of KCl solutions of various concentrations at different temperature are known. Thus, the cell constant is calculated by using the above equation. The same cell constant applies to a measurement with any other solution.

The determination of specific conductance of an electrolytic solution, thus, consists of two steps:

(i) Determination of cell constant by using a standard KCl solution of known concentration in the conductivity cell.

(ii) Determination of resistance of the given solution using the same cell. The reciprocal of this gives the value of conductance.

Multiplication of conductance and cell constant gives the value of specific conductance of the solution.

In order t determine equivalent conductance or molar conductance, the concentration of the experimental solution should be known. In conductance measurements, the solutions are always prepared in conductivity water which has no conductance due to dissolved impurities. It is prepared by distilling a number of times the distilled water to which a little KMnO4 and KOH have been added in a hard glass distillation assembly. Such water has very low conductance of the order of 4.3 × 10-8 ohm-1. For ordinary purposes, double distilled water may be used.

Effect of dilution on equivalent conductance

The value of equivalent conductance increases with dilution. This is due to the fact that degree of ionization increases with dilution thereby increasing the total number of ions in solution. Solution which contains large number of ions compared to another solution of the same concentration at the same temperature has more conductance and is said to be stronger electrolyte. The one which has relatively small number of ions

24

is called a weak electrolyte. The number of ions from an electrolyte depends on the degree of dissociation. The curve (Fig.12.6) shows the variation of the equivalent conductance of some electrolytes with dilution. It shows that electrolytes behave in two ways o dilution.

(i) Electrolytes like KCl have high value of conductance even at low concentration and there is no rapid increase in their equivalent conductance on dilution. Such electrolytes are termed strong electrolytes. In the case of strong electrolytes, there is a tendency for equivalent conductance to approach a limiting value when the concentration approaches zero. When the whole of the electrolyte has ionized, further addition of the water does not bring any change in the value of equivalent conductance. This stage is called infinite dilution. The equivalent conductance has a limiting value at infinite dilution and is represented by /\∞ .

(ii) Electrolytes like acetic acid have a low value at high concentration and there is a rapid increase in the value of equivalent conductance with dilution. Such electrolytes are termed weak electrolytes. There is no indication that a limiting value of equivalent conductance can be attained even when the concentration approaches zero. Thus, graphically, /\∞ of weak electrolytes cannot be obtained.

It is thus concluded that equivalent conductance of electrolytes whether strong or weak increases with dilution and reaches to a maximum or limiting value which is termed /\∞ (equivalent conductance at infinite dilution.) /\∞ in the case of strong electrolytes can be obtained by extrapolation of the graph of equivalent conductance to zero concentration but in the case of weak electrolytes it cannot be obtained accurately. An indirect method for obtaining /\∞ for weak electrolyte has been given by Kohlrusch.

KOHLRAUSCH'S LAW

"At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards equivalent conductance of the electrolyte irrespective of the nature of the ion with which it is associated and the value of equivalent conductance at infinite dilution for any electrolyte is the sum of contribution of its constituent ions", i.e., anions and cations. Thus,

/\∞ = λa + λc

The and are called the ionic conductance of cation and anion at infinite dilution respectively. The ionic conductances are proportional to their ionic mobilities. Thus, at infinite dilution,

λc = kuc

and λa = kua

where uc and ua are ionic mobilities of cation and anion respectively at infinite dilution. The value of k is equal to 96500 c, i.e., one Faraday.

Thus, assuming that increase in equivalent conductance with dilution is due to increase in the degree of dissociation of the electrolyte; it is evident that the electrolyte achieves the degree of dissociation as unity when it is completely ionized at infinite dilution. Therefore, at any other dilution, the equivalent conductance is proportional to the degree of dissociation. Thus,

Degree of dissociation

25

α = /\/(/\∞ )

=(Equivalent conductance at a given concentration)/(Equivalent conductance at infinite dilution)

Calculation of absolute ionic mobilities:

It has been experimentally found that ionic conductance is directly proportional to ionic mobilities.

λ+ ∝ u+

λ- ∝ u-

where u+ and u- are ionic mobilities of cations and anions.

λ+ = Fu+ where F = Faraday

λ- = Fu- = 96500 coulomb

Ionic mobility= (Ionic velocity)/(Potential gradient)

=(Ionic velocity (cm/sec))/(Potential gradient (volt)/elctrode seperation)

Relation between equivalent and molar conductance at infinite dilution

Molar conductance at infinite dilution

26

THEORY OF WEAK ELECTROLYTES

(i) Weak electrolytes are not completely ionized when dissolved in a polar medium like water. There exists equilibrium between ions and unionized molecules.

AB ⇌ A+ + B-

(ii) Concept of chemical equilibrium and law of mass action can be applied to ionic equilibrium also.

AB ⇌ A+ + B-

t =0 C 0 0

teq C-Cα Cα Cα

K = ([A+ ][B-])/[AB] = (Cα×Cα)/C(1-α)

K = (Cα2)/(1-α) ....... (i)

For weak electrolyte, α<<1

(1- α) ≈1

Thus equation (i) can be written as

K = Cα2

α= √(K/C) ....... (ii)

From (ii) it is clear that on dilution concentration decreases, as a result of which degree of ionization ' ' increases. At high degree of ionization bot equivalent and molar conductance increase.

(iii) Degree of ionization can be calculated as

α= (/\eC)/(/\e

∞ )=(/\mC)/(/\m

∞ ) ....... (iii)

/\eC, /\m

C = Equivalent and molar conductance at concentration 'C'.

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/\e∞, /\m

∞ = Equivalent and molar conductance at infinite dilution.

Subtracting the values of ' ' from (ii) in (i), we get

Equations (iv) and (v) are called Ostwald equations.


Example 24: 1.0 N solution of a salt surrounding two platinum electrodes 2.1 cm apart and 4.2 sq.cm in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductivity of the solution.

Solution: Given, l = 2.1 cm, a = 4.2 sq xm, R = 5- ohm

Specific conductance, k = l/a.1/R

or k = 2.1/4.2×1/50 = 0.01 ohm-1 cm-1

Equivalent conductivity = k = V

V = the volume containing 1 g equivalent = 1000 mL

So Equivalent conductivity = 0.01 × 1000

= 10 ohm-1 cm-2 equiv-1

Example 25: Specific conductance of a decinormal solution of KCl is 0.0112 ohm-1 cm-1. The resistance of a cell containing the solution was found to be 56. What is the cell constant?


Sp. conductance = Cell constant × conductance

or Cell constant = (Sp.conductance)/Conductance

= Sp. conductance × Resistance

= 0.0112 × 56

= 0.06272 cm-1

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Example 26: The specific conductivity of 0.02 M KCl solution at 250C is 2.768 × 10-3 ohm-1 cm-1. The resistance of this solution at 250C when measured with a particular cell was 250.2 ohms. The resistance of 0.01 M CuSO4 solution at 250C measured with the same cell was 8331 ohms. Calculate the molar conductivity of the copper sulphate solution.

Solution: Cell constant = (Sp.cond.of KCl)/(Conductane of KCl)

= (2.768× 10-3)/(I/250.2)

= 2.768 × 10-3 × 250.2

For 0.01 M CuSO3 solution

Sp. conductivity = Cell constant × conductance

= 2.768 × 10-3 × 250.2 × 1/8331

Molar conductance = Sp. cond. × 1000/c

= (2.768×10-3 × 25.2)/8331 × 1000/(1/100)




or k = 2.1/4.2×1/50 = 0.01 ohm-1 cm-1










= 0.0112 × 56

= 0.06272 cm-1

29



= (2.768× 10-3)/(I/250.2)

= 2.768 × 10-3 × 250.2



= 2.768 × 10-3 × 250.2 × 1/8331


= (2.768×10-3 × 25.2)/8331 × 1000/(1/100)

ELECTROCHEMICAL CELL

Electrochemical cell is a system or arrangement in which two electrodes are fitted in the same electrolyte or in two different electrolytes which are joined by a salt bridge. Electrochemical cells are of two types:

(a) Electrolytic cell

(b) Galvanic or voltaic cell

Electrolytic cell

It is a device in which electrolysis (chemical reaction involving oxidation and reduction) is carried out by using electricity or in which conversion of electrical energy into chemical energy is done.

Galvanic or voltaic cell

It is a device in which a redox reaction is used to convert chemical energy into electrical energy, i.e., electricity can be obtained with the help of oxidation and reduction reaction. The chemical reaction responsible for production of electricity takes place in two separate compartments. Each compartment consists of a suitable electrolyte solution and a metallic conductor. The metallic conductor acts as an electrode. The compartments containing the electrode and the solution of the electrolyte are called half-cells. When the two compartments are connected by a salt bridge and electrodes are joined by a wire through galvanometer the electricity begins to flow. This is the simple form of voltaic cell.

DANIELL CELL

30

It is designed to make use of the spontaneous redox reaction between zinc and cupric ions to produce an electric current (Fig.12.7). It consists of two half-cells. The half-cells on the left contains a zinc metal electrode dipped in ZnSO4 solution.

The half-cell on the right consists of copper metal electrode in a solution CuSO4. The half-cells are joined by a salt bridge that prevents the mechanical mixing of the solution.

When the zinc and copper electrodes are joined by wire, the following observations are made:

(i) There is a flow of electric current through the external circuit.

(ii) The zinc rod loses its mass while the copper rod gains in mass.

(iii) The concentration of ZnSO4 solution increases while the concentration of copper sulphate solution decreases.

(iv) The solutions in both the compartments remain electrically neutral.

During the passage if electric current through external circuit, electrons flow from the zinc electrode to the copper electrode. At the zinc electrode, the zinc metal is oxidized to zinc ions which go into the solution. The electrons released at the electrode travel through the external circuit to the copper electrode where they are used in the reduction of Cu2+ ions to metallic copper which is deposited on the electrode. Thus, the overall redox reaction is:

Zn(s) + Cu2+ Cu(s) + Zn2+(aq)

Thus, indirect redox reaction leads to the production of electrical energy. At the zinc rod, oxidation occurs. It is the anode of the cell and is negatively charged while at copper electrode, reduction takes, place; it is the cathode of the cell and is positively charged.

Thus, the above points can be summed up as:

(i) Voltaic or Galvanic cell consists of two half-cells. The reactions occurring in half-cells are called half-cell reactions. The half-cell in which oxidation taking place in it is called oxidation half-cell and the reaction taking place in it is called

31

oxidation half-cell reaction. Similarly, the half-cell occurs is called reduction half-cell and the reaction taking place in it is called reduction half-cell reaction.

(ii) The electrode where oxidation occurs is called anode and the electrode where reduction occurs is termed cathode.

(iii) Electrons flow from anode to cathode in the external circuit.

(iv) Chemical energy is converted into electrical energy.

(v) The net reaction is the sum of two half-cell reactions. The reaction is Daniel cell can be represented as

Oxidation half reaction, Zn(s) --> Zn2+(aq) + 2e-

Reduction half reaction, Cu2+(aq) + 2e- --> Cu (s)

---------------------------------------------------------------

Net reaction Zn(s) + Cu2+ (aq) --> Zn2+(aq) + Cu(s)

Electrode Signs

The signs of the anode and cathode in the voltaic or galvanic cells are opposite to those in the electrolytic cells (Fig. 12.8).

ELECTROLYTIC CELL VOLTAIC OR GALVANIC CELL

(e.m.f. is applied to cell) (e.m.f. is generated by cell)

Electrolytic cell

Anode Cathode

Voltaic or Galvanic cell

Anode Cathode

Sign + - - +

32

Electron flow

Half-reaction

Out in

Oxidation reductionOut in

Oxidation reduction

Difference in electrolytic cell and galvanic cell

Electrolytic cell Galvanic cell

1. Electrical energy is converted into chemical energy.

2. Anode positive electrode. Cathode negative electrode.

3. Ions are discharged on both the electrodes.

4. If the electrodes are inert, concentration of the electrolyte decreases when the electric current is circulated.

5. Both the electrodes can be fitted in the same compartment.

Chemical energy is converted into electrical energy.

Anode negative electrode. Cathode positive electrode.

Ions are discharged only on the cathode.

Concentration of the anodic half-cell increases while that of cathodic half-cell decreases when the two electrodes are joined by a wire.

The electrodes are fitted in different compartments.





---------------------------------------------------------------


SALT BRIDGE AND ITS SIGNIFICANCE

Salt bridge is usually an inverted U-tube filled with concentrated solution of inert electrolytes. An inert electrolyte is one whose ions are neither involved in any electrochemical change nor do they react chemically with the electrolytes in the two half-cells. Generally salts like KCl, KN03, NH4N03, etc., are used. For the preparation of salt bridge, gelatin or agar-agar is dissolved in a hot concentrated aqueous solution of an inert electrolyte and the solution thus formed is filled in the U-tube. On cooling the solution sets in the form of a gel in the U-tube. The ends of the U-tube are plugged with cotton wool as to minimize diffusion effects. This is used as a salt bridge.

33

Significance of salt bridge:

The following are the functions of the salt bridge:

(i) It connects the solutions of two half-cells and completes the cell circuit.

(ii) It prevents transference or diffusion of the solutions from one half-cell to the other.

(iii) It keeps the solutions in two half-cells electrically neutral. In anodic half cell, positive ions pass into the solution and there shall be accumulation of extra positive charge in the solution around the anode which will prevent the flow of electrons from anode. This does not happen because negative ions are provided by salt bridge. Similarly, in cathodic half-cell negative ions will accumulate around cathode due to deposition of positive ions by reduction. To neutralize these negative ions, sufficient number of positive ions is provided by salt bridge. Thus, salt bridge maintains electrical neutrality.

(iv) It prevents liquid-liquid junction-potential, i.e., the potential difference which arises between two solutions when in contact with each other.

A broken vertical line or two parallel vertical lines in a cell reaction indicates the salt bridge.

Zn|Zn2+||Cu2+|Cu

Salt bridge can be replaced by a porous partition which allows the migration of ions without allowing the solutions to intermix.





---------------------------------------------------------------


REPRESENTATION OF AN ELECTROCHEMICAL CELL (GALVANIC CELL)

The following universally accepted conventions are followed in representing an electrochemical cell:

(i) The anode (negative electrode) is written on the left hand side and cathode (positive electrode) on the right hand side.

(ii) A vertical line or semicolon (;) indicates a contact between two phases. The anode of the cell is represented by writing metal first and then the metal ion present in the electrolytic solution. Both are separated by a vertical line or a semicolon. For example,

34

Zn|Zn2+ or Zn;Zn2+

The molar concentration or activity of the solution is written in brackets after the formula of the ion. For example:

Zn|Zn2+(1 M) or Zn | Zn2+(0.1 M)

(iii) The cathode of the cell is represented by writing the cation of the electrolyte first and then metal. Both are separated by a vertical line or semicolon. For example,

Cu2+|Cu or Cu2+;Cu or Cu2+(1 M)|Cu

(iv) The salt bridge which separates the two half-cells is indicated by two parallel vertical lines.

(v) Sometimes negative and positive signs are also put on the electrodes.

The Daniell cell can be represented as:

Zn|ZnS04(aq)||CuS04(aq)|C+u

Anode Salt bridge Cathode

Oxidation half-cell Reduction half-cell

or Zn|Zn2+||Cu2+|Cu

or Zn|Zn2+(1 M)||Cu2+(1 M)|Cu





---------------------------------------------------------------


ELECTRODE POTENTIAL

When a metal is placed in a solution of its ions, the metal acquires either a positive or negative charge with respect to the solution. On account of this, a definite potential difference is developed between the metal and the solution. This potential difference is called electrode potential. For example, when a plate of zinc is placed in a solution having Zn2+ ions, it becomes negatively charged with respect to the solution and thus a potential difference is set up between zinc plate and the solution. This potential difference is termed the electrode potential of zinc. Similarly, when copper is placed in a solution having Cu2+ ions, it becomes positively charged with respect to solution. A potential difference is set up between the copper plate

35

and the solution. The potential difference thus developed is termed as electrode potential of copper. The potential difference is established due to the formation of electrical double layer at the

M(s) --> Mn + ne-

(a) Oxidation:

Metal ions pass from the electrode into solution leaving an excess of electrons and thus a negative charge on the electrode.

(b) Reduction:

Metal ions in solution gain electrons from the electrode leaving a positive charge on the electrode.

interface of metal and the solution. The development of negative charge (as on zinc plate) or positive charge (as on copper plate) can be explained in the following manner. When a metal rod is dipped in its salt solution, two changes occur:

(i) The conversion of metal atoms into metal ions by the attractive force of polar water molecules.

M --> Mn + ne-

The metal ions go into the solution and the electrons remain on the metal making it negatively charged. The tendency of the metal to change into ions is known as electrolytic solution pressure.

(ii) Metal ions start depositing on the metal surface leading to a positive charge on the metal.

36

Mn+ + ne- --> M

This tendency of the ions is termed osmotic pressure.

In the beginning, both these changes occur with different speeds but soon an equilibrium is established.

M <--> Mn+ + ne-

In practice, one effect is greater than the other, if first effect is greater than the second, the metal acquires a negative charge with respect to solution and if the second is greater than the first, it acquires positive charge with respect to solution, thus in both the cases a potential difference is set up.

The magnitude of the electrode potential of a metal is a measure of its relative tendency to lose or gain electrons, i.e., it is a measure of the relative tendency to undergo oxidation (loss of electrons) or reduction (gain of electrons). |The magnitude of potential depends on the following factors:

(i) Nature of the electrode,

(ii) Concentration of the ions in solution,

(iii) Temperature.

Depending on the nature of the metal electrode to lose or gain electrons, the electrode potential may be of two types:

(i) Oxidation potential:

When electrode is negatively charged with respect to solution, i.e., it acts as anode. Oxidation occurs.

M --> Mn+ + ne-

(ii) Reduction potential:

When electrode is positively charged with respect to solution, i.e., it acts as cathode. Reduction occurs.

Mn+ + ne- --> M

It is not possible to measure the absolute value of the single electrode potential directly. Only the difference in potential between two electrodes can be measured experimentally. It is, therefore, necessary to couple the electrode with another electrode whose potential is known. This electrode is termed as reference electrode. The emf of the resulting cell is measured experimentally. The emf of the cell is equal to the sum of potentials on the two electrodes.

Emf of the cell = EAnode + ECathode

= Oxidation potential of anode

37

+ Reduction potential of cathode

Knowing the value of reference electrode, the value of other electrode can be determined.

STANDARD ELECTRODE POTENTIAL

In order to compare the electrode potentials of various electrodes, it is necessary to specify the concentration of the ions present in solution in which the electrode is dipped and the temperature of the half-cell. The potential difference developed between metal electrode and the solution of its ions of unit molarity (1M) at 25°C (298 K) is called standard electrode potential.

According to the IUPAC convention, the reduction potential alone be called as the electrode potential (EO), i.e., the given value of electrode potential be regarded as reduction potential unless it is specifically mentioned that it is oxidation potential. Standard reduction potential of an electrode means that reduction reaction is taking place at the electrode. If the reaction is reversed and written as oxidation reaction, the numerical value of electrode potential will remain same but the sign of standard potential will have to be reversed. Thus

Standard reduction potential = - (Standard oxidation potential)

or Standard oxidation potential = - (Standard reduction potential)

REFERENCE ELECTRODE

(STANDARD HYDROGEN ELECTRODE, SHE OR NHE)

Hydrogen electrode is the primary standard electrode. It consists of a small platinum strip coated with platinum black as to adsorb hydrogen gas. A platinum wire is welded to the platinum strip and sealed in a glass tube as to make contact with the outer circuit through mercury. The platinum strip and glass tube is surrounded by an outer glass tube which has an inlet for hydrogen gas at the top and a number of holes at the base for the escape of excess of hydrogen gas. The platinum strip is placed in an acid solution which has H+ ion concentration 1 M. Pure hydrogen gas is circulated at one atmospheric pressure. A part of the gas is adsorbed and the rest escapes through holes. This gives an equilibrium between the adsorbed hydrogen and hydrogen ions in the solution.

H2 <--> 2H+ + 2e-

38

The temperature of the cell is maintained at 250C. By international agreement the standard hydrogen electrode is arbitrarily assigned a potential of exactly ± 0.000 . .. volt.

The hydrogen electrode thus obtained forms one of two half-cells of a voltaic cell. When this half-cell is connected with any other half-cell, a voltaic cell is constituted. The hydrogen electrode can act as cathode or anode with respect to other electrode.

SHE half reaction Electrode potential

H2 --> 2H+ + 2e- 0.0 V (Anode)

2H+ + 2e- --> H2 0.0 V (Cathode)

MEASUREMENT OF ELECTRODE POTENTIAL

The measurement of electrode potential of a given electrode is made by constituting a voltaic cell, i.e., by connecting it with a standard hydrogen electrode (SHE) through a salt bridge. 1 M solution is used in hydrogen half-cell and the temperature is

39

maintained at 25oC. The emf of the cell is measured either by a calibrated potentiometer or by a high resistance voltmeter, i.e., a valve voltmeter. The reading of the voltmeter gives the electrode potential of the electrode in question with respect to the hydrogen electrode. The standard electrode potential of a metal may be determined as it is the potential difference in volt developed in a cell consisting of two electrodes: the pure metal is contact with a molar solution of one of its ions and the standard hydrogen electrode.

(i) Determination of standard electrode potential of Zn/Zn2+ electrode:

A zinc rod is dipped in 1 M zinc sulphate solution. This half-cell is combined with a standard hydrogen electrode through a salt bridge. Both the electrodes are con-nected with a voltmeter as shown in Fig. 12.12. The deflection of the voltmeter indicates that current is flowing from hydrogen electrode to metal electrode or the electrons are moving from zinc rod to hydrogen electrode. The zinc electrode acts as an anode and the hydrogen electrode as cathode and the cell can be represented as

Zn|Zn2+ (aq)/Anode(-) || 2H(aq)| H2 (g)/Cathode (+)

Zn --> Zn2+ + 2e-; 2H+ + 2e- --> H2↑

(Oxidation) (Reduction)

The emf of the cell is 0.76 volt

ECell = EoAnode + Eo

Cathode

0.76 = EoAnode + 0 or Eo

Anode = +0.76 V

As the reaction on the anode is oxidation, i.e.,

Zn --> Zn2+ + 2e,

EoAnode is the standard oxidation potential of zinc. This potential is given the

positive sign

40

Eoox (Zn/Zn2+) = +0.76 volt

So standard reduction potential of Zn, i.e., Eo (Zn/Zn2+)

= Eoox = -(+0.76)

= -0.76 volt

The emf of such a cell gives the positive value of standard oxidation potential of metal M. The standard reduction potential (Eo) is obtained by reversing the sign of standard oxidation potential.

(i) Determination of standard electrode potential of Cu2+/Cu, electrode:

A copper rod is dipped in 1 M solution of CuSO4. It is combined with hydrogen electrode through a salt bridge. Both the electrodes joined through a voltmeter. The deflection of the voltmeter indicates that current is flowing from copper electrode towards hydrogen electrode, i.e., the electrons ate moving from hydrogen to copper electrode. The hydrogen electrode acts as an anode and the copper electrode as a cathode. The cell can be represented as

41

H2 (g)|2H+ (aq) ||Cu2+ (aq) |Cu

Anode (-) Cathode (+)

H2 --> 2H+ + 2e- ; Cu2+ 2e- --> Cu

Oxidation Reduction

The emf of the cell is 0.34 volt.

EoCell = Eo

Anode + EoCathode

0.34 = 0 + EoCathode

Since the reaction on the cathode is reduction, i.e.,Cu2+ + 2e- --> Cu, Eo

Cathode is the standard reduction potential of copper. This is given the +ve sign.

E°, i.e., standard reduction potential of Cu2+/Cu = 0.34 volt

So E°x (standard oxidation potential of copper) = -0.34 volt

The emf of such a cell gives positive value of reduction potential of metal electrode. The standard oxidation potential of this electrode is obtained by reversing the sign of standard reduction potential.

The emf of such a cell gives the positive value of standard oxidation potential of metal M. The standard reduction potential (E°) is obtained by reversing the sign of standard oxidation potential.

It is thus concluded that at the metal electrode which acts as anode with respect to hydrogen electrode (cathode), the reduction potential is given the minus sign and at the metal electrode which acts as cathode with respect to hydrogen electrode (anode), the reduction potential is given the positive sign.

The standard electrode potentials (oxidation or reduction) of various elements can be measured by combining the electrode in question with a standard hydrogen electrode and measuring the emf of the cell constituted.

EMF OF A GALVANIC CELL

42

Every galvanic or voltaic cell is made up of two half-cells, the oxidation half-cell (anode) and the reduction half-cell (cathode). The potentials of these half-cells are always different. On account of this difference in electrode potentials, the electric current moves from the electrode at higher potential to the electrode at lower potential, i.e., from cathode to anode. The direction of the flow of electrons is from anode to cathode.

Flow of electrons

Anode <==============> Cathode

Flow of current

The difference in potentials of the two half-cells is known as the electromotive force (emf) of the cell or cell potential.

The emf of the cell or cell potential can be calculated from the values of electrode potentials of the two half-cells constituting the cell. The following three methods are in use:

(i) When oxidation potential of anode and reduction potential of cathode are taken into account:

ECello = Oxidation potential of anode

+ Reduction potential of cathode

= Eoxo (anode) + Ered

o (cathode)

(ii) When reduction potentials of both electrodes are taken into account:

ECello = Reduction potential of cathode - Reduction potential of anode

= ECathodeo - EAnode

o

= Erighto - Eleft

o

(iii) When oxidation potentials of both electrodes are taken into account:

= Oxidation potential of anode - Oxidation potential of cathode

ECello = Eox

o (anode) - Eredo (cathode)

Difference between Emf and potential difference:

The potential difference is the difference between the electrode potentials of the two electrodes of the cell under any condition while emf is the potential generated by a cell when there is zero electron flow, i.e., it draws no current. The points of difference are given below:

Emf Potential difference

1. It is the potential difference 1. It is the difference of the

43

between two electrodes when no current is flowing in the circuit.

2. It is the maximum voltage that the cell can deliver.

3. It is responsible for the steady flow of current in the cell.

electrode potentials of the two electrodes when the cell is under operation.

2. It is always less than the maximum the cell can deliver.

3. It is not responsible for the steady flow of current in the cell.

REVERSIBLE AND IRREVERSIBLE CELLS

Daniell cell has the emf value 1.09 volt. If an opposing emf exactly equal to 1.09 volt is applied to the cell, the cell reaction,

Zn + Cu2+ --> Cu + Zn2+

stops but if it is increased infinitesimally beyond 1.09 volt, the cell reaction is reversed.

Cu + Zn2+ --> Zn + Cu2+

Such a cell is termed a reversible cell. Thus, the following are the two main conditions of reversibility:

(i) The chemical reaction of the cell stops when an exactly equal opposing emf is applied.

(ii) The chemical reaction of the cell is reversed and the current flows in opposite direction when the opposing emf is slightly greater than that of the cell.

Any other cell which does not obey the above two conditions is termed as irreversible. A cell consisting of zinc and copper electrodes dipped into the solution of sulphuric acid is irreversible. Similarly, the cell

Zn|H2S04(aq)|Ag

is also irreversible because when the external emf is greater than the emf of the cell, the cell reaction,

Zn + 2H+ --> Zn2+ + H2

is not reversed but the cell reaction becomes

2Ag + 2H+ --> 2Ag+ + H2

SOME OTHER REFERENCE ELECTRODES

Since a standard hydrogen electrode is difficult to prepare and maintain, it is usually replaced by other reference electrodes, which are known as secondary reference electrodes. These are convenient to handle and are prepared easily. Two important secondary reference electrodes are described here.

44

(i) Calomel electrode: It consists of mercury at the bottom over which a paste of mercury-mercurous chloride is placed. A solution of potassium chloride is then placed over the paste. A platinum wire sealed in a glass tube helps in making the electrical contact. The electrode is connected with the help of the side tube on the left through a salt bridge with the other electrode to make a complete cell.

The potential of the calomel electrode depends upon the concentration of the potassium chloride solution. If potassium

chloride solution is saturated, the electrode is known as saturated calomel electrode (SCE) and if the potassium chloride solution is 1 N, the electrode is known as normal calomel electrode (NCE) while for 0.1 N potassium chloride solution, the electrode is referred to as decinormal calomel electrode (DNCE). The electrode reaction when the electrode acts as cathode is:

1/2 Hg2Cl2 + e- <---> Hg + Cl-

The reduction potentials of the calomel electrodes on hydrogen scale at 298K are as follows:

Saturated KC1 0.2415 V

1.0NKC1 0.2800 V

0.1NKC1 0.3338 V

The electrode potential of any other electrode on hydrogen scale can be measured when it is combined with calomel electrode. The emf of such a cell is measured. From the value of electrode potential of calomel electrode, the electrode potential of the other electrode can be evaluated.

(ii) Silver-silver chloride electrode: This is another widely used reference electrode. It is reversible and stable and can be combined with cells containing chlorides without insetting liquid junctions.

45

Silver chloride is deposited electrolytically on a silver or platinum wire and it is then immersed in a solution containing chloride ions. Its standard electrode potential with respect to the standard hydrogen electrode is 0.2224 V at 298 K. the electrode is represented as:

Ag|AgCl|Cl-

and the electrode reaction is,

AgCl+e- --> Ag + Cl-

PREDICTION FOR OCCURENCE OF A REDOX REACTION

Any redox reaction would occur spontaneously if the free energy change (∆G) is negative. The free energy is related to cell emf in the following manner:

∆Go = - nFEo

where n is the number of electrons involved, F is the value of Faraday and Eo is the cel emf. ∆G can be negative if Eo is positive.

When Eo is positive, the cell reaction is spontaneous and serves as a source of electrical energy.

To predict whether a particular redox reaction will occur or not, write down the redox reaction into two half reactions, one involving oxidation reaction and the other involving reduction reaction. Write the oxidation reaction and reduction potential value for reduction reaction. Add these two values, if the algebraic summation gives a positive value, the reaction will occur, otherwise not.

ELECTRODE AND CELL POTENTIALS /NERNST EQUATION

The electrode potential and the emf of the cell depend upon the nature of the electrode, temperature and the activities (concentrations) of the ions in solution. The variation of electrode and cell potentials with concentration of ions in solution can be obtained from thermodynamic considerations. For a general reaction such as

M1A + m2B ..... n1X + n2Y + .... .......(i)

occurring in the cell, the Gibbs free energy change is given by the equation

G = ∆Go + 2.303RT log10 (axn

1 × ayn

2)/(aAm

1 × aBm

2) ....... (ii)

where 'a' represents the activities of reactants and products under a given set of conditions and ∆Go refers to free energy change for the reaction when the various reactants and products are present at standard conditions. The free energy change of a cell reaction is related to the electrical work that can be obtained from the cell, i.e., ∆Go = -nFEcell and ∆Go = -nFEo. On substituting these values in Eq. (ii) we get

-nFEcell = -nFEo + 2.30eRT log10 (axn

1 × ayn

2)/(aAm

1 × aBm

2) ....... (iii)

or Ecell = Ecello - 2.303RT/nF log10 (ax

n1 × ay

n2)/(aA

m1 × aB

m2) ....... (iv)

46

This equation is known as Nearnst equation.

Putting the values of R=8.314 JK-1 mol-1, T = 298 K and F=96500 C, Eq. (iv) reduces to

E = Eo - 0.0591/n log10 (axn

1 × ayn

2)/(aAm

1 × aBm

2) ....... (v)

= Eo - 0.0591/n log10 ([Products])/([Reactants]) ....... (vi)

Potential of single electrode (Anode): Consider the general oxidation reaction,

M --> Mn+ + ne-

Applying Nernst equation,

Eox = Eoxo - 0.0591/n log10 [Mn+]/[M]

where Eox, the oxidation potential of the electrode (anode), is the standard oxidation potential of the electrode.

[Note: The concentration of pure solids and liquids are taken as unity.]

Eox = Eoxo - 0.0591/n log10 [Mn+]

Let us consider a Daniell cell to explain the above equations. The concentrations of the electrolytes are not 1 M.

Zn(s)+Cu2+(aq) <=> Zn2+(aq) + Cu(s)

Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu

Potential at zinc electrode (Anode)

Eox = Eoxo - 0.0591/n log10 [Zn3+]

Potential at copper electrode (Cathode)

Ered = Eredo - 0.0591/n log10 [Cu2+]

Emf of the cell

Ecell = Eox + Ered

= (Eoxo + Ered

o )- 0.0591/n [Zn2+/Cu2+]

The value of n = 2 for both zinc and copper.

Let us consider an example, in which the values of n for the two ions in the two half-cells are not same. For example, in the cell

Cu|Cu2+||Ag+|Ag

The cell reaction is

47

Cu(s) + 2Ag+ ---> Cu2+ + 2Ag

The two half-cell reaction are:

Cu --> Cu2+ + 2e-

Ag+ + e- --> Ag

The second equation is multiplied by 2 to balance the number of electrons.

2Ag+ + 2e- --> 2 Ag

Eox = Eoxo - 0.0591/2 log10[Cu2+]

Ered = Eredo - 0.0591/2 log10[Ag+]2

Ecell = Eox + Ered

= Eoxo - 0.0591/2 log10 [Cu2+]/[Ag+]2

= Ecell = 0.0591/2 log10 [Cu2+]/[Ag+]2

Example 31: Construct the cells in which the following reactions are taking place. Which of the electrodes shall act as anode (negative electrode) and which one as cathode (positive electrode)?

(a) Zn + CuSO4 = ZnSO4 + Cu

(b) Cu + 2AgNO3 = Cu(NO3)2 + 2 Ag

(c) Zn + H2SO4 = ZnSO4 + H2

(d) Fe + SnCl2 = FeCl2 + Sn

Solution: It should always be kept in mind that the metal which goes into solution in the form of its ions undergoes oxidation and thus acts as negative electrode (anode) and the element which comes into the free state undergoes reduction and acts as positive electrode (cathode):

(a) In this case Zn is oxidized to Zn2+ and thus acts as anode (negative electrode) while Cu2+ is reduced to copper and thus acts as cathode (positive electrode). The cell can be represented

as Zn|ZnSO4||CuSO4|Cu

or Zn|Zn2+||Cu2+|Cu


(b) In this case Cu is oxidized to Cu2+ and Ag+ is reduced to Ag. The cell can be represented as

Cu|Cu(NO3)2||AgNO3|Ag

or Cu|Cu2+||Ag+|Ag

48


(c) In this case Zn is oxidized to Zn2+ and H+ is reduced to H2. The cell can be represented as

Zn|ZnSO4||H2SO4|Cu

or Zn|Zn2+||2H+|H2(Pt)


(d) Here Fe is oxidized to Fe2+ and Sn2+ is reduced to Sn. The cell can be represented as

Fe|FeCl2||SnCl2|Sn

or Fe|Fe2+||Sn2+|Sn


Example 32: Consider the reaction,

2Ag+ + Cd --> 2Ag + Cd2+

The standard electrode potentials for Ag+ --> Ag and Cd2+ --> Cd couples are 0.80 volt and -0.40 volt, respectively.

(i) What is the standard potential Eo for this reaction?

(ii) For the electrochemical cell in which this reaction takes place which electrode is negative electrode?

Solution: (i) The half reactions are:

2Ag+ + 2e- --> 2Ag.

Reduction

(Cathode)

EoAg

+/Ag =0.80 volt (Reduction potential)

Cd --> Cd2+ + 2e-,

Oxidation

(Anode)

EoCd

+/Cd = -0.40 volt (Reduction potential)

or EoCd

+/Cd

2 = +0.40 volt

Eo = EoCd

+/Cd

2 + Eo

Ag+

/Ag = 0.40+0.80 = 1.20 volt

49

(ii) The negative electrode is always the electrode whose reduction potential has smaller value or the electrode where oxidation occurs. Thus, Cd electrode is the negative electrode.

Example 33: Consider the cell,

Zn|Zn2+(aq)(1.0M)||Cu2+(aq)(1.0M)|Cu

The standard (reduction) electrode potentials are

Cu2+ + 2e- --> Cu(aq) Eo = 0.350 volt

Zn2+ + 2e- --> Zn(aq) Eo = -0.763 volt

(i) Write down the cell reaction.

(ii) Calculate the emf of the cell

Solution: (i) Reduction potential of Zn is less than copper, hence Zn acts as anode and copper as cathode.

At anode Zn --> Zn2+ + 2e- (Oxidation)

At cathode Cu2+ + 2e- --> Cu (Reduction)

Σημείωση: Το θετικό reduction potential του χαλκού σημαίνει ότι ο χαλκός δίνει αρνητικό ∆Go, σύμφωνα με την εξίσωση ∆Go = - nFEo, άρα η ημιαντίδραση του χαλκού είναι αυθόρμητη. Τα ιόντα χαλκού θέλουν να πάρουν ηλεκτρόνια από το ηλεκτρόδιο χαλκού, το οποίο θα είναι κάθοδος. Αντίθετα, η ημιαντίδραση του ψευδάργυρου δεν είναι αυθόρμητη, αυθόρμητη είναι η αντίστροφη ημιαντίδραση, δηλαδή άτομα ψευδάργυρου θέλουν να αφήσουν ηλεκτρόνια στο ηλεκτρόδιο ψευδάργυρου, που θα είναι άνοδος και να γίνουν ιόντα ψευδάργυρου.

--------------------------------------------------------

Cell reaction Zn + Cu2+ --> Zn2+ + Cu

(ii) EoCell = Eo

Zn/Zn2+ + Eo

Cu2+

/Cu

= Oxi. Potential of zinc + Red. Potential of copper

EoZn

2+/Zn = -0.763 (Reduction potential)

EoZn/Zn

2+ = +0.763 (Oxidation potential)

and EoCu

2+/Cu = 0.350 (Reduction potential)

So Ecello= 0.763+0.350 = 1.113 volt

Oxidation potential is EoM/M

a+ while reduction potential is represented as EoM

a+/M. The value of Eo

Zn/Zn2+ (oxidation potential of Zn) is +0.76 volt and the value of

EoCu

2+/Cu (reduction potential of copper) is +0.34 volt. The electrode having

lower value of reduction potential acts as an anode while that having higher value of reduction potential acts as cathode.

50

Σημείωση: Αυτό συμβαίνει γιατί θετικό reduction potential σημαίνει αυθόρμητη αντίδραση και όσο πιο μεγάλο είναι τόσο περισσότερη ενέργεια απελευθερώνεται.

Example 34: Write the electrode reactions and the net cell reactions for the following cells. Which electrode would be the positive terminal in each cell?

(a) Zn|Zn2+||Br-, Br2|Pt

(b) Cr|Cr3+||I-, I2|Pt

(c) Pt |H2, H+||Cu2+|Cu

(d) Cd|Cd2+||Cl-, AgCl|Ag

Solution: (a) Oxidation half reaction, Zn --> Zn2++2e-

Reduction half reaction, Br2 + 2e- --> 2Br-

-------------------------------------------------

Net cell reaction Zn + Br2 --> Zn2+ + 2Br-

Positive terminal-Cathode Pt

(b) Oxidation half reaction, [Cr --> Cr3+ + 3e-]× 2

Reduction half reaction, [I2 + 2e- --> 2Ir-] × 3

----------------------------------------------------

Net cell reaction 2Cr + 3I2 --> 2Cr3+ + 6I-

Positive terminal-Cathode Pt

(c) Oxidation half reaction, H2 --> 2H+ + 2e-

Reduction half reaction, Cu2+ + 2e---> Cu

-------------------------------------------------

Net cell reaction H2 + Cu2+ --> Cu + 2H+

Positive terminal-Cathode Cu

(d) Oxidation half reaction, Cd --> Cd2+ + 2e-

Reduction half reaction, [AgCl+e- --> Ag+Cl- ]×2

------------------------------------------------------

Net cell reaction Cd+2AgCl --> Cd2++2Ag+2Cl-

51

Positive terminal-Cathode Ag

Example 35: Will Fe be oxidiesed to Fe2+ by reaction with 1.0 M HCl? Eo for Fe/Fe2+ = +0.44 volt.

Solution: The reaction will occur if Fr is oxidized to Fe2+.

Fe + 2HCl --> FeCl2 + H2

Writing two half reaction,

Fe --> Fe2+ + 2e- Oxidation EoFe/Fe

2+ = 0.44 volt

2H++ 2e- --> H2 Reduction EoH

+/H = 0.0 volt

--------------------------------------

Adding, emf = 0.44 volt

Since emf is positive, the reaction shall occur.

Example 36: The values of Eo of some of the reactions are given below:

I2 + 2e- --> 2I-; Eo = +0.54 volt

Cl2 + 2e- --> 2Cl-; Eo = +1.36 volt

Fe3+ + e- --> Fe2+; Eo = +0.76 volt

Ce4+ + e- --> Ce3+; Eo = +1.60 volt

Sn4+ + 2e- --> Sn2+; Eo = +0.15 volt

On the basis of the above data, answer the following questions:

(a) Whether Fe3+ oxidizes Ce3+ or not ?

(b) Whether I2 displaces chlorine form KCl ?

(c) Whether the reaction between FeCl3 and SnCl2 occurs or not ?

Solution: (a) Chemical reaction,

Fe3+ + Ce3+ --> Ce4+ + Fe2+

Two half reactions,

Fe3+ + e --> Fe2+ Reduction Eo = 0.76 volt

Ce3+ --> Ce4+ + e- Oxidation Eoox = -1.60 volt

---------------------------

Adding = -0.84 volt

52

Since, emf is negative the reaction does not occur, i.e., Fe3+ does not oxidise Ce3+.

(b) Chemical reaction

I2 + 2KCl = 2Kl + Cl2

Half reactions

I2 + 2e- --> 2I- Reduction Eo = 0.54 volt

2Cl- --> Cl2 + 2e- Oxidation Eoox = -1.36 volt

---------------------------

Adding = -0.82 volt

Since, emf is negative, the reaction does not occur, i.e., I2 does not displace Cl2

from KCl.

(c) Chemical reaction

SnCl2 + 2FeCl3 --> SnCl4 + 2FeCl2

Half reactions

Fe3+ + e Fe2+ Reduction Eo = 0.76 volt

Ce2+ Sn4+ + 2e- Oxidation Eo = -0.15 volt

-------------------------

Adding = +0.61 volt

Since, emf is positive, the reaction will occur.

Example 37: Calculate the electrode potential at a copper electrode dipped in a 0.1 M solution of copper sulphate at 25o C. The standard electrode potential of Cu2+/Cu system is 0.34 volt at 298 K.

Solution: We know that Ered = Eored + 0.0591/n log10[ion]

Putting the values of Eored =0.34 V, n = 2 and [Cu2+]= 0.1 M

Eored = 0.34+0.0591/2 log10[0.1]

= 0.34 + 0.02955 × (-1)

= 0.34 - 0.02955 = 0.31045 volt

Example 38: What is the single electrode potential of a half-cell foe zinc electrode dipping in 0.01 M ZnSO4 solution at 25o C? The standard electrode potential of Zn/Zn2+ system is 0.763 volt at 25o C.

Solution: We know that Eox = Eored - 0.0591/n log10[ion]

53

Putting the value of Eoox=0.763 V,n=2 and

[Zn2+]=0.01 M

Eoox = 0.763-0.0591/2 log_10 [0.01]

= 0.763 - 0.02955 × (-2)

= (0.763 + 0.0591) volt = 0.8221 volt

Example 38: The standard oxidation potential of zinc is 0.76 volt and of silver is -0.80 volt. Calculate the emf of the cell:

Zn|Zn(NO3)2||AgNO3|Ag

0.25 M 0.1 M

at 250C.

Solution: The cell reaction is

Zn + 2 Ag+ --> 2Ag + Zn2+

Eoox of Zn = 0.76 volt

Eoox of Ag = 0.80 volt

Eocell = Eoox of Zn + of Ag = 0.76 + 0.80

= 1.56 volt

= 1.56 - 0.0591/2×1.3979

= (1.56-0.0413) volt

= 1.5187 volt

Alternative method:

First of all, the single electrode potentials of both the electrodes are determined on the basis of given concentrations.

Eox(Zinc) = Eoox -0.0591/2 log 0.25

= 0.76 + 0.0177 = 0.7777 volt

Ered(Silver) = Eored -0.0591/2 log 0.1

= 0.80 + 0.0591 = 0.7409 volt

Ecell = Eox(Zinc) + Ered(Silver)

Example 40: The emf(E°) of the following cells are:

54

Ag|Ag|(1 M)||Cu2+(1 M)|Cu; E° = -0.46 volt

Zn|Zn2(1 M)||Cu2|(1 M)|Cu; E° = +1.10 volt

Calculate the emf of the cell:

Zn|Zn2+(1 M)||Ag+(1 M)|Ag

Solution: Zn|Zn2+(1 M)||Ag+(1 M)|Ag

Ecell = Eox(Zn/Zn2+) + Ered (Ag+/Ag)

With the help of the following two cells, the above equation can be obtained.

Ag|Ag+(1 M)||Cu2+(1 M)|Cu, E° = -0.46 volt

or Cu|Cu2+(1 M)||Ag+(1 M)|Ag, E° will be +0.46 volt

or +0.46 = Eox(cwcu2+

) + Ered (Ag+/Ag) .... (i)

Zn|Zn2+(1 M)||Cu2+1|Cu, E° = +1.10 volt

+ 1.10 = Eox(Zn/Zn+

) + Ered(Cu2+/Cu) ....... (ii)

Adding Eqs. (i) and (ii),

+ 1.56 = Eox(Cu/Cu2+

) + Ered(Ag+

/Ag) + Eox(Zn/Zn2+

) + Ered(Cu2+/Cu)

Since Eox(Cu/Cu2+) - Ered(Cu2+/Cu)

So +1.56 = Em(Zn/Zn+

) + Ered(Ag+

/Ag)

Thus, the emf of the following cell is

Zn|Zn2+(1 M)||Ag+(1 M)|Ag is +1.56 volt.

Example 41: Calculate the e.m.f of the cell.

Mg(s)|Mg2+(0.2M)||Ag+(1×10-3)|Ag

EoAg+/Ag = +0.8 volt, Eo

Mg2+/Mg = -2.37 volt

What will be the effect on e.m.f. if concentration

of Mg2+ ion is decreased to 0.1 M?

Solution: Eocell = Eo

Cathode - Eoanode

= 0.80-(-2.37) = 3.17 volt

Cell reaction

Mg + 2Ag+ --> 2Ag + Mg2+

55

Ecell = Ecello - 0.0591/n log(Mg2+)/[Ag+]2

= 3.17 -0.0591/2 log 0.2/[1× 10-3 ]2

= 3.17 - 0.1566 = 3.0134 volt

when Mg2+ = 0.1 M

Ecell = Eocell - 0.0591/n log(0.1)/[1 x 10-3]2

= (3.17 - 0.1477) volt

= 3.0223 volt

Example 42: To find the standard potential of M3+/M electrode, the following cell is constituted:

Pt|M|M3+(0.0018 mol-1L)||Ag+(0.01 mol-1L)|Ag

The emf of this cell is found to be 0.42 volt. Calculate the standard potential of the half reaction M3+ + 3e- M3+. = 0.80 volt.

Solution: The cell reaction is

M + 3Ag+ ---> 3Ag + M3+

Applying Nernst equation,

Ecell = Ecello - 0.0591/n log(Mg2+)/[Ag+]3

0.42 = Ecello - 0.0591/n log (0.0018)/(0.01)3 = Ecell

o - 0.064

Ecello =(0.042+0.064)= 0.484 volt

Eocell = Eo

cathode - Eoanode

or Eoanode = Eo

cathode - Eocell

= (0.80-0.484) = 0.32 volt

ELECTROCHEMICAL SERIES

By measuring the potentials of various electrodes versus standard hydrogen electrode (SHE), a series of standard electrode potentials has been established. When the electrodes (metals and non-metals) in contact with their ions are arranged oh the basis of the values of their standard reduction potentials or standard oxidation potentials, the resulting series is called the electrochemical or electromotive or activity series of the elements.

By international convention, the standard potentials of electrodes are tabulated for reduction half reactions, indicating the tendencies of the electrodes to behave as cathodes towards SHE. Those with positive E° values for reduction half reactions do in fact act as cathodes versus SHE, while those with negative E° values of reduction half reactions behave instead as anodes versus SHE. The electrochemical series is shown in the following table.

56

Standard Aqueous Electrode Potentials at 25°C 'The Electrochemical Series'

ElementElectrode Reaction

(Reduction)

Standard Electrode Reduction potential

Eo, volt

Li

K

Ca

Na

Mg

Al

Zn

Cr

Fe

Cd

Ni

Sn

H2

Cu

I2

Ag

Hg

Br2

Cl2

Au

F2

Li+ + e- = Li

K+ + e- = K

Ca2+ + 2e- = Ca

Na+ + e- = Na

Mg2+ + 2e- = Mg

Al3+ + 3e- = Al

Zn2+ + 2e- = Zn

Cr3+ + 3e- = Cr

Fe2+ + 2e- = Fe

Cd2+ + 2e- = Cd

Ni2+ + 2e- = Ni

Sn2+ + 2e- = Sn

2H+ + 2e- = H2

Cu2+ + 2e- = Cu

I2 + 2e- = 2I-

Ag+ + e- = Ag

Hg2+ + 2e- = Hg

Br2 + 2e- = 2Br-

Cl2 + 2e- = 2Cl-

Au3+ + 3e- = Au

F2 + 2e- = 2F-

Characteristics of Electrochemical series

(i) The negative sign of standard reduction potential indicates that an electrode when joined with SHE acts as anode and

57

oxidation occurs on this electrode. For example, standard reduction potential of zinc is -0.76 volt. When zinc electrode is joined with SHE, it acts as anode (-ve electrode) i.e., oxidation occurs on this electrode. Similarly, the +ve sign of standard reduction potential indicates that the electrode when joined with SHE acts as cathode and reduction occurs on this electrode.

(ii) The substances which are stronger reducing agents than hydrogen are placed above hydrogen in the series and have negative values of standard reduction potentials. All those substances which have positive values of reduction potentials and placed below hydrogen in the series are weaker reducing agents than hydrogen.

(iii) The substances which are stronger oxidising agents than H+ ion are placed below hydrogen in the series.

(iv) The metals on the top (having high negative values of standard reduction potentials) have the tendency to lose electrons readily. These are active metals. The activity of metals decreases from top to bottom. The non-metals on the bottom (having high positive values of standard reduction potentials) have the tendency to accept electrons readily. These are active non-metals. The activity of non-metals increases from top to bottom.

Applications of Electrochemical series

(i) Reactivity of metals:

The activity of the metal depends on its tendency to lose electron, i.e., tendency to form cation (M"+). This tendency depends on the magnitude of standard reduction potential. The metal which has high negative value (or smaller positive value) of standard reduction potential readily loses the electron or electrons and is converted into cation. Such a metal is said to be chemically active.

The chemical reactivity of metals decreases from top to bottom in the series. The metal higher in the series is more active than the metal lower in the series. For example,

(a) Alkali metals and alkaline earth metals having high negative values of standard reduction potentials are chemically active. These react with cold water and evolve hydrogen. These readily dissolve in acids forming corresponding salts and combine with those substances which accept electrons.

(b) Metals like Fe, Pb, Sn, Ni, Co, etc., which lie a little down in the series do not react with cold water but react with steam to evolve hydrogen.

(c) Metals like Cu, Ag and Au which lie below hydrogen are less reactive and do not evolve hydrogen from water.

(ii) Electropositive character of metals:

The electropositive character also depends on the tendency to lose electron or electrons. Like reactivity, the electropositive character of metals decreases from top to bottom in the electrochemical series. On the basis of standard reduction potential values, metals are divided into three groups:

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(a) Strongly electropositive metals: Metals having standard reduction potential near about -2.0 volt or more negative like alkali metals, alkaline earth metals are strongly electropositive in nature.

(b) Moderately electropositive metals: Metals having values of reduction potentials between 0.0 and about -2.0 volt are moderately electropositive. Al, Zn, Fe, Ni, Co, etc., belong to this group.

(c) Weakly electropositive metals: The metals which are below hydrogen and possess positive values of reduction potentials are weakly electropositive metals. Cu, Hg, Ag, etc., belong to this group.

(iii) Displacement reactions:

(a) To predict whether a given metal will displace another, from its salt solution. A metal higher in the series will displace the metal from its solution which is lower in the series, i.e., the metal having low standard reduction potential will displace the metal from its salt's solution which has higher value of standard reduction potential. A metal higher in the series has greater tendency to provide electrons to the cations of the metal to be precipitated.

(b) Displacement of one nonmetal from its salt solution by another nonmetal: A nonmetal higher in the series (towards bottom side), i.e., having high value of reduction potential will displace another nonmetal with lower reduction potential i.e., occupying position above in the series. The nonmetal's which possess high positive reduction potentials have the tendency to accept electrons readily. These electrons are provided by the ions of the nonmetal having low value of reduction potential. Thus, Cl2 can displace bromine and iodine from bromides and iodides.

Cl2 + 2KI --> 2KC1 + I2

21- --> I2 + 2e- (Oxidation)

Cl2 + 2e- --> 2C1- (Reduction)

[The activity or electronegative character or oxidising nature of the nonmetal increases as the value of reduction potential increases.]

(c) Displacement of hydrogen from dilute acids by metals: The metal which can provide electrons to H+ ions present in dilute acids for reduction, evolve hydrogen from dilute acids.

Mn --> Mn"+ + ne- (Oxidation)

2H+ + 2e- --> H2 (Reduction)

The metal having negative values of reduction potential possess the property of losing electron or electrons.

Thus, the metals occupying top positions in the electrochemical series readily liberate hydrogen from dilute acids and on descending in the series tendency to liberate hydrogen gas from dilute acids decreases.

The metals which are below hydrogen in electrochemical series like Cu, Hg, Au, Pt, etc., do not evolve hydrogen from dilute acids.

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(d) Displacement of hydrogen from water: Iron and the metals above iron are capable of liberating hydrogen from water. The tendency decreases from top to bottom in electrochemical series.

Alkali and alkaline earth metals liberate hydrogen from cold water but Mg, Zn and Fe liberate hydrogen from hot water or steam.

(iv) Reducing power of metals:

Reducing nature depends on the tendency of losing electron or electrons. The more the negative reduction potential, the more is the tendency to lose electron or electrons. Thus, reducing nature decreases from top to bottom in the electrochemical series. The power of the reducing agent increases as the standard reduction potential becomes more and more negative.

Sodium is a stronger reducing agent than zinc and zinc is a stronger reducing agent than iron.

Element Na Zn Fe

Reduction potential -2.71 -0.76 -0.44

-------------------------------------------->

Reducing nature decreases

Alkali and alkaline earth metals are strong reducing agents.

(v) Oxidising nature of nonmetals:

Oxidising nature depends on the tendency to accept electron or electrons. More the value of reduction potential, higher is the tendency to accept electron or electrons. Thus, oxidising nature increases from top to bottom in the electrochemical series. The strength of an oxidising agent increases as the value of reduction potential becomes more and more positive.

F2 (Fluorine) is a stronger oxidant than Cl2, Br2 and I2.

Cl2 (Chlorine) is a stronger oxidant than Br2 and I2.

Element I2 Br2 Cl2 F2

Reduction potential +0.53 +1.06 +1.36 +2.85

--------------------------------------->

Oxidising nature increases

Thus, in electrochemical series

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(vi) Thermal stability of metallic oxides:

The thermal stability of the metal oxide depends on its electropositive nature. As the electropositivity decreases from top to bottom, the thermal stability of the oxide also decreases from top to bottom. The oxides of metals having high positive reduction potentials are not stable towards heat. The metals which come below copper form unstable oxides, i.e., these are decomposed on heating.

Heat

Ag2O ---------> 1/2 O2

2 Ag

Heat

2HgO ------------> 1/2 O2

2 Hg

(vii) Products of electrolysis:

In case two or more types of positive and negative ions are present in solution, during electrolysis certain ions are discharged or liberated at the electrodes in preference to others. In general, in such competition the ion which is stronger oxidising agent (high value of standard reduction potential) is discharged first at the cathode. The increasing order of deposition of few cations is:

K+, Ca2+, Na+, Mg2+, Al3+, Zn2+, Fe2+, H+, Cu2+, Ag+, Au3+

------------------------------------------------------------>

Increasing order of deposition (same as in electrochemical series)

Similarly, the anion which is stronger reducing agent (low value of standard reduction potential) is liberated first at the anode.

The increasing order of discharge of few anions is:

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SO42-, NO3

-, OH-, Cl-, Br-, I-

------------------------------------->

Increasing order of discharge

Thus, when an aqueous solution of NaCl containing Na+, Cl-, H+ and OH" ions is electrolysed, H+ ions are discharged at cathode and Cl- ions at the anode, i.e., H2 is liberated at cathode and chlorine at anode.

When an aqueous solution of CuS04 containing Cu2+, , H+ and OH- ions is electrolysed, Cu2+ ions are discharged at cathode and OH- ions at the anode.

Cu2+ + 2e- --> Cu (Cathodic reaction)

4OH- --> O2 + 2H2O + 4e- (Anodic reaction)

Cu is deposited on cathode while 02 is liberated at anode.

(viii) Corrosion of metals:

Corrosion is defined as the deterioration of a substance because of its reaction with its environment. This is also defined as the process by which metals have the tendency to go back to their combined state, i.e., reverse of extraction of metals.

Ordinary corrosion is a redox reaction by which metals are oxidised by oxygen in presence of moisture. Oxidation of metals occurs more readily at points of strain. Thus, a steel nail first corrodes at the tip and head. The end of a steel nail acts as an anode where iron is oxidised to Fe 2+ ions .

Fe --> Fe2 + 2e- (Anode reaction)

The electrons flow along the nail to areas containing impurities which act as cathodes where oxygen is reduced to hydroxyl ions.

O2 + 2H2O + 4e- --> 4OH- (Cathode reaction)

The overall reaction is

2Fe + Oz + 2H2O = 2Fe(OH)2

Fe(OH)2 may be dehydrated to iron oxide, FeO, or further oxidised to Fe(OH)3 and then dehydrated to iron rust, Fe203

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Several methods for protection of metals against corrosion have been developed. The most widely used are (i) plating the metal with a thin layer of a less easily oxidised metal (ii) allowing a protective film such as metal oxide (iii) galvanising-steel is coated with zinc (a more active metal).

(ix) Extraction of metals:

A more electropositive metal can displace a less electropositive metal from its salt's solution. This principle is applied for the extraction of Ag and Au by cyanide process. Silver from the solution containing sodium argento cyanide, NaAg(CN)2, can be obtained by the addition of zinc as it is more electro-positive than Ag.

2NaAg(CN)2 + Zn --> Na2Zn(CN)4 + 2Ag

Relation between Equilibrium constant, Gibbs free energy and EMF of the cell

Concept of equilibrium in electrochemical cell

In an electrochemical cell a reversible redox process takes place, e.g., in Daniell cell:

Zn(s)+Cu2+(aq) <==> Zn2+(aq)+Cu(s)

(1) At equilibrium mass action ratio becomes equal to equilibrium constant,

i.e., Q = Ke

(2) Oxidation potential of anode = -Reaction potential of cathode

i.e., emf = oxidation potential of anode + Reduction potential of cathode

= 0

Cell is fully discharged

According to Nernst equation:

E = Eo - 0.0591/n log10 Q at 25o C

At equilibrium, E = 0, Q = K

0 = Eo 0.0591/n log10 K

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K = Antilog [(nEo)/0.0591]

Work done by the cell

Let n faraday charge be taken out of a cell of emf E; then work done by the cell will be calculated as:

Work = Charge x Potential

= nFE

Work done by the cell is equal to decrease in free energy.

-∆G = nFE

Similarly, maximum obtainable work from the cell will be

Wmax = nFE°

where, Eo = standard emf or standard cell potential.

-∆G = nFE

The relationship among K, ∆Go and Eo cell

Heat of reaction in an electrochemical cell

Let n Faraday charge flows out of a cell of emf E,

Then -∆G = nFE ...... (i)

Gibbs-Helmholtz equation from thermodynamics may be given as

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∆G = ∆H + T (∂∆G/∂T)P ...... (ii)

From equation (i) and (ii) we get

-nFE = ∆H + T (∂(-nFE)/∂T)P = ∆H-nFT(∂E/∂T)P

∆H = -nFE + nFT(∂E/∂T)P

Here (∂E/∂T)P = Temperature coefficient of cell

Case I: When (∂E/∂T)P = 0, then ∆H=-nFE

Case II: When (∂E/∂T)>0, then nFE>∆H, i.e., process inside the cell is endothermic.

Case III: When (∂E/∂T)<0 , then nFE<∆H, i.e., process inside the cell is exothermic.

PRIMARY VOLTAIC CELL (THE DRY CELL)

In this cell, once the chemicals have been consumed, further reaction is not possible. It cannot be regenerated by reversing the current flow through the cell using an external direct current source of electrical energy. The most common example of this type is dry cell.

The container of the dry cell is made of zinc which also serves as one of the electrodes. The other electrode is a carbon rod in the centre of the cell. The zinc container is lined with a porous paper. A moist mixture of ammonium chloride, manganese dioxide, zinc chloride and a porous inert filler occupy the space between the paper lined zinc container and the carbon rod. The cell is sealed with a material like wax.

As the cell operates, the zinc is oxidised to Zn2+

Zn ---> Zn2+ + 2e- (Anode reaction)

The electrons are utilized at carbon rod (cathode) as the ammonium ions are reduced.

2NH4++2e- --> 2NH3 + H2 (Cathode reaction)

The cell reaction is

Zn+ 2 NH4+ ---> Zn2+ + 2NH3 + H2

Hydrogen is oxidized by MnO2 in the cell.

2MnO2 + H2 ---> 2MnO(OH)

Ammonia produced at cathode combines with zinc ions to form complex ion.

Zn2+ + 4NH3 ---> [Zn(NH3)4]2+

Ecell is 1.6 volt

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Alkaline dry cell is similar to ordinary dry cell. It contains potassium hydroxide. The reaction in alkaline dry cell are:

Zn + 2OH- ---> Zn(OH)2 + 2e- (Anode reaction)

2MnO2 + 2H2O + 2e- ---> 2MnO(OH) + 2OH- (Cathode reaction)

Zn + 2MnO2 + 2H2O ---> Zn(OH)2 + 2MnO(OH) (Overall)

Ecell is 1.5 volt.

SECODARY VOLTAIC CELL (LEAD STORAGE BATTERY)

The cell in which original reactants are regenerated by passing direct current from external source, i.e., it is re-charged, is called secondary cell. Lead storage battery is the example of this type.

It consists of a group of lead plates bearing compressed spongy lead, alternating with a group of lead plates bearing leaf dioxide, PbO2. These plates are immersed in a solution of about 30% H2SO4. When the cell discharge; it operates as a voltaic cell. The spongy lead is oxidized to Pb2+ ions and lead plates acquire a negative charge.

Pb --> Pb2+ + 2e- (Anode reaction)

Pb2+ ions combine with sulphate ions to form insoluble lead sulphate, PbSO4, which begins to coat lead electrode.

Pb2+ + SO42- ---> PbSO4 (Precipitation)

The electrons are utilized at PbO2 electrode.

PbO2 + 4H+ + 2e- ---> Pb2+ 2H2O (Cathode reaction)

Pb2+ + SO42- ---> PbSO4 (Precipitation)

Overall cell reaction is:

Pb + PbO2 + 4H+ + 2 SO42- ---> 2PbSO4 + 2H2O

Ecell is 2.041 volt.

When a potential slightly greater than the potential of battery is applied, the battery can be re-charged.

2PbSO4 + 2H2O ---> Pb + PbO2 + 2H2SO4

After many repeated charge-discharge cycles, some of the lead sulphate falls to the bottom of the container, the sulphuric acid concentration remains low and the battery cannot be recharged fully.

FUEL CELL

Fuel cells are another means by which chemical energy may be converted into electrical energy. The main disadvantage of a primary

66

cell is that it can deliver current for a short period only. This is due to the fact that the quantity of oxidising agent and reducing agent is limited. But the energy can be obtained indefinitely from a fuel cell as long as the outside supply of fuel is maintained. One of the examples is the hydrogen-oxygen fuel cell. The cell consists of three compartments separated by a porous electrode. Hydrogen gas is introduced into one compartment and oxygen gas is fed into another compartment. These gases then diffuse slowly through the electrodes and react with an electrolyte that is in the central compartment. The electrodes are made of porous carbon and the electrolyte is a resin containing concentrated aqueous sodium hydroxide solution. Hydrogen is oxidised at anode and oxygen is reduced at cathode. The overall cell reaction produces water. The reactions which occur are:

Anode [H2(g) + 2OH-(aq) ---> 2H2O(l) + 2e-] × 2

Cathode O2(g) + 2H2O(l) + 4e- ---> 4OH-(aq)

-------------------------------------------------------------

Overall 2H2(g) + O2(g) ---> 2H20(l)

This type of cells are used in space-crafts. Fuel cells are efficient and pollution free.

CONCENTRATIONS CELLS

If two plates of the same metal are dipped separately into two solutions of the same electrolyte and are connected with a salt bridge, the whole arrangement is found to act as a galvanic cell. In general, there are two types of concentration cells:

(i) Electrode concentration cells:

In these cells, the potential difference is developed between two like electrodes at different concentrations dipped in the same solution of the electrolyte. For example, two hydrogen electrodes at different has pressure in the same solution of hydrogen ions constitute a cell of this type.

(Pt,H2 (Pressure p1))/Anode |H+ | (H2 (Pressure p2)Pt)/Cathode

If p1, p2 oxidation occurs at L.H.S. electrode and reduction occurs at R.H.S. electrode.

Ecell = 0.0591/2 log(p1/p2) at 25o C

In the amalgam cells, two amalgams of the same metal at two different concentrations are interested in the same electrolyte solution.

(ii) Electrolyte concentration cells:

In these cells, electrodes are identical but these are immersed in solutions of the same electrolyte of different concentrations. The source of electrical energy in the cell is the tendency of the electrolyte to diffuse from a solution of higher concentration to that of lower concentration. With the expiry of time, the two

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concentrations tend to become equal. Thus, at the start the emf of the cell is maximum and it gradually falls to zero. Such a cell is represented in the following manner:

(C2 is greater than C1).

M|Mn+(C1)||Mn+(C2)|M

or (Zn|Zn2+ (C1))/Anode || (Zn2+ (C2 )|Zn)/Cathode

The emf of the cell is given by the following expression:

Ecell = 0.0591/n log C(2(R.H.S.))/C(1(L.H.S.)) at 25o C

The concentration cells are used to determine the solubility of sparingly soluble salts, valency of the cation of the electrolyte and transition point of the two allotropic forms of a metal used as electrodes, etc.

Example 43. A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 10-6 M hydrogen ions. The emf of the cell is 0.118 volt at 25° C. Calculate the concentration of hydrogen ions at the positive electrode.

Solution: The cell may be represented as

Pt|H2(1 atm)|H+||H+|H2(1 atm)|Pt

10-6 M CM

Anode Cathode

(-ve) (+ve)

H2 ---> 2H+ + 2e- 2H+ + 2 ---> H2

Ecell = 0.0591/2 log([H+ ]cathode2)/[10-6 ]2

0.081 = (0.0591) log ([H+])/10-6

log[H+ ]cathode/10-6 =0.118/0.0591=2

[H+ ]cathode/10-6 = 102

[H+]cathode = 10-6 = 10-4 M

Example 44. The emf of the cell Ag|Agl in 0.05 MK\Sol. NH4NO3|10.05 M AgNO3\Ag is 0.788 volt at 25°C. The activity coefficient of KI and silver nitrate in the above solution is 0.90 each. Calculate (i) the solubility product of Agl, and (ii) the solubility of Agl in pure water at 25°C.

Solution: Ag+ ion concentration on AgN03 side

= 0.9 × 0.5 = 0.045 M

Similarly I- ion concentration in 0.05 M KI solution

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= 0.05 × 0.9 - 0.045 M

Ecell = 0.0591/1 log[Ag+ ](R.H.S.)/[Ag+ ](L.H.S.) = 0.0591 log 0.045/[Ag+ ](L.H.S.)

or log 0.045/[Ag+ ](L.H.S.) = 0.788/0.0591 = 13.33

[Ag+]L.H.S. = 0.045/(2.138× 1013 )

= 2.105 × 10-15 M

Solubility product of Agl = [Ag+][I-]

= 2.105 × 10-15 × 0.045

= 9.427 × 10-17

Solubility of Agl = √(Solubility product of Agl)

= √(9.472×〖10〗^(-17) )

= 9.732 × 10-9 g mol L-1

= 9.732 × 10-9 × 143.5 g L-1

= 1.396 × 10-6 g L-1

Example 45: The observed emf of the cell

Pt|H2(1 atm)|H+(3×10-4 M)||H+(M1)|H2(1 atm)|Pt

is 0.154 V. Calculate the value of M1 and pH of cathodic solution.

Solution: Ecell = 0.0591 log M1/(3×10-4 )

or log M1/(3×10-4 ) = 0.0154/0.0591 = 2.6058

M1/(3×10-4) = 4.034 × 102

M1 = 4.034 × 102 × 3 × 10-4 M

= 0.121 M

pH = -log [H+]=-log 0.121 = 0.917

Example 46: Calculate the emf of the following cell at 25oC.

Pt H2|HCl|H2 Pt

2 atm 10 atm

Solution: Ecell =0.0591/2 log P1/P2

= 0.0591/2 log 2/10

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= -0.0206 volt

COMMERCIAL PRODUCTION OF CHEMICALS

The wide applications of electrolysis have been listed in section 12.4 of this chapter. A large number of chemicals are produced by electrolysis. A few of these are described below:

1. Manufacture of sodium

Sodium is obtained on large scale by two processes:

(i) Castner's process:

In this process, electrolysis of fused sodium hydroxide is carried out at 330°C using iron as cathode and nickel as anode.

2NaOH <==> 2Na+ + 2OH

At cathode 2Na+ + 2e --> 2Na

At anode 4OH- --> 2H2O + O2 + 4e

During electrolysis, oxygen and water are produced. Water formed at the anode gets partly evaporated and is partly broken down and hydrogen is discharged at cathode.

H2O <==> H+ OH-

At cathode 2H+ + 2e --> 2H H2

(ii) Down's process: Now-a-days sodium metal is manufactured by this process. It involves the electrolysis of fused sodium chloride containing calcium chloride and potassium fluoride using iron as cathode and graphite as anode at about 600oC (Fig. 12.18)

NaCl <==> Na+ + Cl-

At cathode Na+ + e -->Na

At anode 2Cl- --> Cl2 + 2e

The electrolysis of pure NaCl presents the following difficulties:

(a) The fusion temperature of NaCl is high, i.e., 803°C. At this temperature both sodium and chlorine are corrosive.

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(b) Sodium forms a metallic fog at this temperature.

To remove above difficulties, the fusion temperature is reduced to 600°C by adding CaCl2 and KF. This is a cheaper method and chlorine is obtained as a by-product. The sodium obtained is of high purity (about 99.5%).

2. Sodium hydroxide (Caustic soda), NaOH

Caustic soda is manufactured by the electrolysis of aqueous solution of sodium chloride in an electrolytic cell.

Principle: A sodium chloride solution contains Na+, H+, Cl- and OH- ions.

NaCl <==> Na+ + Cl-

H20 <==> H+ + OH-

On passing electricity, Na+ and IF ions move towards cathode and CI- and OH" ions move towards anode. The discharge potential of H+ ions is less than Na+ ions, thus hydrogen ions get discharged easily and hydrogen is liberated. Similarly, at anode CI- ions are easily discharges as their discharge potential is less than that of OH- ions. Cl2 gas is, therefore, liberated at anode.

The solution on electrolysis becomes richer in Na+ and OH- ions.

Since chlorine reacts with sodium hydroxide solution even in the cold forming sodium chloride and sodium hypochlorite, it is necessary that chlorine should not come in contact with sodium hydroxide during electrolysis.

2NaOH + Cl2 --> NaCl + NaCIO + H20

To overcome this problem, the anode is separated from the cathode in the electrolytic cell either by using a porous diaphragm or by using a mercury cathode.

(i) Porous diaphragm process (Nelson cell process): Nelson cell consists of a perforated steel tube lined inside with asbestos. The tube acts as a cathode (Fig. 12.19).

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It is suspended in a steel tank. A graphic rod dipped in sodium chloride solution serves as anode. On passing electric current, chlorine is liberated at the anode and let out through the outlet. Sodium ions penetrate through the asbestos and reach the cathode where hydrogen and OH- ions are formed by reduction of water. Sodium ions combine with OH- ions to form NaOH which is collected in the outer tank while hydrogen is drawn off through the outlet. The steam blown during the process keeps the electrolyte warm and helps to keep perforation clear.

NaCl <==> Na+ + Cl- (in solution)

At Cathode 2H2O+2e <==>H2 + 2OH-

Na+ OH- <==> NaOH

At anode 2Cl- --> Cl2 + 2e

The solution containing NaOH and Nacl as impurity is taken out and evaporated to dryness.

(ii) Castner-Kellner cell:

This is the common cell in which mercury is used as cathode. The advantage to using Hg as a cathode is that the discharge potential of Na+ ions is less than that of H+ ions. Na+ ions get discharged on mercury and the sodium so deposited combines with mercury to form sodium amalgam. The cell consists of a large rectangular trough divided into three compartments by slate partitions which do not touch the bottom of the cell but dipping in mercury as shown in the Fig. 12.20.

The mercury can flow from one compartment into other but the solution kept in one compartment cannot flow into other. Sodium chloride solution is placed in the two outer compartments and a dilute solution of sodium hydroxide in the inner compartment. Two graphite electrodes which act as anodes are fixed in the outer compartments and a series of iron rods fitted in the inner compartment acts as cathode. Mercury in the outer compartments acts as cathode while in the inner

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compartment it acts as anode by induction. The cell is kept rocking with the help of an eccentric wheel.

When electricity is circulated, sodium chloride in the outer compartments is electrolysed. Chlorine is evolved at the graphite anode while Na+ ions are discharged at the Hg cathode. The liberated sodium forms amalgam with mercury.

NaCl <==> Na+ + Cl-

At anode 2Cl- 2Cl + 2e Cl2

At cathode Na+ + e --> Na

Na + Hg --> Amalgam

The sodium amalgam thus formed comes in the inner compartment due to rocking. Here, the sodium amalgam acts as the anode and iron rods acts as cathode.

At anode Na-amalgam --> Na+ + Hg + e

At cathode 2H2O + 2e --> H2 + 2OH-

The concentrated solution of sodium hydroxide (about 20%) is taken out from the inner compartment and evaporated to dryness to get solid NaOH.

(i) Kellner-Solvay cell:

This is the modified cell. This cell has no compartments. The flowing mercury as shown in Fig.12.21 acts as cathode. A number of graphite rods dipping in sodium chloride solution acts as anode. A constant level of sodium chloride solution is maintained in the cell. On electrolysis chlorine gas is librated and Na+ ions are discharged ay cathode (mercury). Sodium discharged dissolves in Hg and

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forms amalgam. This amalgam flows out in a vessel containing water. sodium hydroxide is formed with evolution of hydrogen.

Preparation of pure sodium hydroxide:

Commercial sodium hydroxide is purified with the help of alcohol. Sodium hydroxide dissolves in alcohol while impurities like NaCl, Na2CO3, Na2SO4, etc., remain insoluble. The alcoholic filtrate is distilled. The alcohol distills off while pure solid sodium hydroxide is left behind.

3. Manufacture of aluminium

Aluminium is manufactured from pure bauxite ore by electrolysis. The bauxite ore usually contains impurities such as iron oxide, silica, etc. These impurities are first removed by the application of the following methods in order to get pure alumina, i.e., pure bauxite ore.

(a) Hall's process

(b) Baeyer's process

(c) Serpeck's process.

Electrolytic reduction of pure alumina:

The electrolysis of pure alumina faces two difficulties: (i) Pure alumina is a bad conductor of electricity, (ii) The fusion temperature of pure alumina is about 2000°C and at this temperature when the electrolysis is carried of the fused mass, the metal formed vapourises as the boiling point of aluminium is 1800°C.

The above difficulties are overcome by using a mixture containing alumina, cryolite (Na3AlF6) and fluorspar (CaF2) in the ratio of 20 : 60 : 20. The fusion temperature of this mixture is 900°C and it is a good conductor of electricity.

The electrolysis is carried out in an iron box lined inside with gas carbon which acts as cathode. The anode consists of carbon rods which dip in the fused mixture of the electrolyte from above. The fused electrolyte is covered with a layer of coke (Fig. 12.22).

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The current passed through the cell serves two purposes: (i) Heating of the electrolyte: The temperature of the cell is automatically maintained at 900-950°C. (ii) Electrolysis: On passing current, aluminium is discharged at cathode.

Aluminium being heavier than the electrolyte sinks to the bottom and is tapped out periodically from a tapping hole. Oxygen is liberated at anode. It attacks the carbon rods forming CO and C02. The process is continuous. When the concentra-tion of the electrolyte decreases, the resistance of the cell increases. This is indicated by the glowing of a lamp placed in parallel. At this stage more of alumina is added.

The exact mechanism of the electrolysis is not yet known.

Two concepts have been proposed.

First concept: AlF3 from cryolite ionizes as

AlF3 <==> Al3+ + 3F-

Al3+ ions are discharged at cathode and F- ions at anode.

Al3+ + 3e -->Al (at cathode)

2F- --> F2 + 2e (at anode)

The liberated fluorine reacts with alumina to form AlF3 and O2. The oxygen attacks the carbon anodes to form CO and CO2.

Al2O3 + 3F3 --> 2AlF3 + O2

2C + O2 --> 2CO

C + O2 --> CO2

Anodes are replaced frequently.

Second concept: Alumina (Al2O3) ionizes as

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Al2O3 --> Al3+ + AlO33-

Cathode Anode

Al3+ + 3e Al (at cathode)

At anode AlO3- --> 2Al2O3 + 3O2 + 12e (at anode)

Thus, the overall chemical reaction taking place during electrolysis is

2Al2O3 --> 4Al + 3O2

Aluminum of 99.8% purity is obtained from this process.

Refining of aluminum by Hoope's electrolytic method:

Aluminum is further purified by Hoope's process. The electrolytic cell consists of an iron box lined inside with carbon.

The cell consists of three layers which differ in specific gravities. The upper layer is of pure aluminium which acts as cathode. The middle layer consists of a mixture of the fluorides of Al, Ba and Na. The lowest layer consists of impure aluminium which acts as anode. The middle layer works as electrolyte.

The graphite rods are dipped in pure aluminium and Cu-Al alloy rods at the bottom of impure aluminium work as conductors. On electrolysis, aluminium is deposited at the cathode from the middle layer and an equivalent amount of aluminium is taken up by the middle layer from the bottom layer (impure aluminium). Therefore, aluminium is transferred from bottom to the top layer through middle layer while impurities are left behind. Aluminium thus obtained is 99.98%.

4. Isolation of fluorine

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Fluorine presented many difficulties in its isolation. It remained a difficult problem in chemistry for many years and after a hard labour of many chemists for about 75 years it could be isolated finally by Moissan in 1886. The reasons for its late discovery were its high reactivity and non-conducting nature of hydrofluoric acid. Fluorine attacked the material of the vessels used for its isolation. Carbon vessel was attacked with formation of CF4 and platinum vessel was reduced to chocolate powder. The vessels of other metals were also affected. Platinum and carbon could not be used as electrodes. Another difficulty experienced was that when the electrolysis of aqueous hydrofluoric acid was carried out, hydrogen and oxygen (ozone) were obtained and when anhydrous hydrofluoric acid was tried it was found to be a bad conductor of electricity.

Moissan finally solved the problem and isolated fluorine by the electrolysis of anhydrous hydrofluoric acid in the presence of potassium hydrogen fluoride using Pt-Ir alloy vessel at -23°C. The electrodes used were also of Pt-Ir alloy.

Modern methods of isolation:

In modern methods, fluorine is prepared by electrolysis of a fused fluoride (usually potassium hydrogen fluoride, KHF2). The electrolytic cells are made of copper, nickel or monel metal. The anode is generally of graphite and the fluorine set free contains some carbon tetrafluoride.

Reactions in the electrolytic cell

The following precautions should be taken in the preparation of fluorine:

(i) The electrolyte must be completely dry. In presence of moisture, the evolved fluorine reacts with moisture to form 02 and 03.

(ii) The parts of the apparatus which come in contact with fluorine must be free from oil and grease.

(iii) The vessel in which fluorine is collected should also be absolutely dry.

(iv) The gas must be made free from HF before storing by passing through sodium fluoride (NaF), otherwise HF will attack the vessel.

[Note: HF is more corrosive and reactive than fluorine.]

Dennis method:

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The electrolytic cell used in this method consists of a V-shaped copper tube (5 cm in diameter) fitted with copper caps. Graphite electrodes through these caps are sealed and insulated in the tube by bakelite cement which is not. affected by fluorine. The cen is covered with an insulating layer of asbestos cement over which is wound a resistance wire for electrical heating. The tube is thickly lagged to prevent the loss of heat.

The electrolyte consists of fused potassium hydrogen fluoride which has already been dried for 48 hours at 130°C. The electrolyte is kept in fused state by electrical heating externally. For electrolysis, a current of 5 amperes and 12 volts is used. On electrolysis fluorine is liberated at anode. To make the liberated fluorine free from HF vapours, it is passed through copper U-tubes containing sodium fluoride.

NaF + HF ---> NaHF2

The following difficulty is experienced in this method: The liberated fluorine at anode does not escape fast enough due to narrow exit. The escape is further hindered due to frothing in the electrolyte. There are, thus, chances of mixing of H2 and F2 which may result in explosion. To avoid this, a modified apparatus has been devised by Whytlaw-Gray.

Whytlaw-Gray method:

It consists of a copper cell wound with resistance wire for electrical heating. The pure graphite anode is enclosed in a copper cylinder which is perforated at the bottom (Fig. 12.25). The electrolysis of fused KHF2 is carried out in this cell. The escape of fluorine is fast enough and thus no frothing in the electrolyte occurs. There are no chances of mixing of H2 and F2 in this cell.

Solved Examples On Electrochemistry

Example 1. Find the charge in coulomb on 1 g-ion of

Solution: Charge on one ion of N3-

= 3 × 1.6 × 10-19 coulomb

Thus, charge on one g-ion of N3-

= 3 × 1.6 10-19 × 6.02 × 1023

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= 2.89 × 105 coulomb

Example 2. How much charge is required to reduce (a) 1 mole of Al3+ to Al and (b)1 mole of to Mn2+?

Solution: (a) The reduction reaction is

Al3+ + 3e- --> Al

1 mole 3 mole

Thus, 3 mole of electrons are needed to reduce 1 mole of Al3+.

Q = 3 × F

= 3 × 96500 = 289500 coulomb

(b) The reduction is

Mn4- + 8H+ 5e- --> MN2+ + 4H2O

1 mole 5 mole

Q = 5 × F

= 5 × 96500 = 48500 coulomb

Example 3. How much electric charge is required to oxidise (a) 1 mole of H2O to O2 and (b)1 mole of FeO to Fe2O3?

Solution: (a) The oxidation reaction is

H2O --> 1/2 O2 + 2H+ + 2e-

1 mole 2 mole

Q = 2 × F

= 2 × 96500=193000 coulomb

(b) The oxidation reaction is

FeO + 1/2 H2O --> 1/2 Fe2O3 + H++e-

Q = F = 96500 coulomb

Example 4. Exactly 0.4 faraday electric charge is passed through three electrolytic cells in series, first containing AgNO3, second CuSO4 and third FeCl3 solution. How many gram of rach metal will be deposited assuming only cathodic reaction in each cell?

Solution: The cathodic reactions in the cells are respectively.

Ag+ + e- --> Ag

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1 mole 1 mole

108 g 1 F

Cu2+ + 2e- -->Cu

1 mole 2 mole

63.5 g 2 F

and Fe3+ + 3e- --> Fe

1 mole 3 mole

56 g 3 F

Hence, Ag deposited = 108 × 0.4 = 43.2 g

Cu deposited = 63.5/2×0.4=12.7 g

and Fe deposited = 56/3×0.4=7.47 g

Example 5. An electric current of 100 ampere is passed through a molten liquid of sodium chloride for 5 hours. Calculate the volume of chlorine gas liberated at the electrode at NTP.

Solution: The reaction taking place at anode is

2Cl- --> Cl2 + 2e-

71.0 g 71.0 g 2 × 96500 coulomb

1 mole

Q = I × t = 100 × 5 × 600 coulomb

The amount of chlorine liberated by passing 100 × 5 × 60 × 60 coulomb of electric charge.

=1/(2×96500)×100×5×60×60=9.3264 mole

Volume of Cl2 liberated at NTP = 9.3264 × 22.4 = 208.91 L

Example 6. A 100 watt, 100 volt incandescent lamp is connected in series with an electrolytic cell containing cadmium sulphate solution. What mass of cadmium will be deposited by the current flowing for 10 hours?


Watt = ampere × volt

100 = ampere × 110

Ampere = 100/110

80

Quantity of charge = ampere × second

= 100/110×10×60×60 coulomb

The cathodic reaction is

Cd2+ + 2e- --> Cd

112.4 g 2 × 96500 C

Mass of cadmium deposited by passing 100/110×10×60×60

Coulomb charge = 112.4/(2×96500)×100/110×10×60×60=19.0598 g

Example 7. In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution gold salt and the second cell contains copper sulphate solution. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, find the amount of copper deposited on the cathode in the second cell. Also calculate the magnitude of the current in ampere.


(Mass of Au deposited)/(Mass f Cu deposited)=(Eq.mass of Au)/(Eq.Mass of Cu)

Eq. mass of Au = 197/3; Eq. mass of Cu 63.5/2

Mass of copper deposited

= 9.85 × 63.5/2 x 3/197 g = 4.7625 g

Let Z be the electrochemical equivalent of Cu.

E = Z × 96500

or Z =E/96500=63.5/(2×96500)

Applying W = Z × I × t

T = 5 hour = 5 × 3600 second

4.7625 = 63.5/(2×96500) × I × 5 × 3600

or I = (4.7625 × 2 × 96500)/(63.5 × 5 × 3600)=0.0804 ampere

Example 8. How long has a current of 3 ampere to be applied through a solution of silver nitrate to coat a metal surface of 80 cm2 with 0.005 cm thick layer? Density of silver is 10.5 g/cm3.

Solution: Mass of silver to be deposited

= Volume × density

= Area ×thickness × density

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Given: Area = 80 cm2, thickness = 0.0005 cm and density = 10.5 g/cm3

Mass of silver to be deposited = 80 × 0.0005 × 10.5

= 0.42 g

Applying to silver E = Z × 96500

Z = 108/96500 g

Let the current be passed for r seconds.

We know that

W = Z × I × t

So, 0.42 = 108/96500 x 3 x t

or t = (0.42 × 96500)/(108×3)=125.09 second

Example 9. What current strength in ampere will be required to liberate 10 g of chlorine from sodium chloride solution in one hour?

Solution: Applying E = Z × 96500 (E for chlorine = 35.5)

35.5 = Z × 96500

or Z = 35.5/96500 g

Now, applying the formula

W = Z × I × t

Where W = 10 g, Z= 35.5/96500 t=60×60=3600 second

I = 10x96500/35.5x96500 = 7.55 ampere

Example 10. 0.2964 g of copper was deposited on passage of a current of 0.5 ampere for 30 minutes through a solution of copper sulphate. Calculate the atomic mass of copper. (1 faraday = 96500 coulomb)

Solution: Quantity of charge passed

0.5 × 30 × 60 = 900 coulomb

900 coulomb deposit copper = 0.2964 g

96500 coulomb deposit copper = 0.2964/900×96500=31.78 g

Thus, 31.78 is the equivalent mass of copper.

At. mass = Eq. mass × Valency

= 31.78 × 2 = 63.56

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Example 11. 19 g of molten SnCI2 is electrolysed for some time using inert electrodes until 0.119 g of Sn is deposited at the cathode. No substance is lost during electrolysis. Find the ratio of the masses of SnCI2 : SnCI4 after electrolysis.

Solution: The chemical reaction occurring during electrolysis is

2SnCl2 --> SnCl4 + Sn

2×190 g 261 g 119 g

119 g of Sn is deposited by the decomposition of 380 g of SnCl2

So, 0.119 g of SnCl2 of Sn is deposited by the decomposition of

380/119×0.119=0.380 g of SnCl2

Remaining amount of SnCl2 = (19-0.380) = 18.62 g

380 g of SnCl2 produce = 261 g of SnCl4

So 0.380 g of SnCl2 produce = 261/380×0.380=0.261 g of SnCl

Thus, the ratio SnCl2 : SnCl4 =18.2/0.261 , i.e., 71.34 : 1

Example 12. A current of 2.68 ampere is passed for one hour through an aqueous solution of copper sulphate using copper electrodes. Calculate the change in mass of cathode and that of the anode. (At. mass of copper = 63.5).

Solution: The electrode reactions are:

Cu2+ + 2e- --> Cu (Cathode)

1 mole 2 × 96500 C

Cu --> Cu2+ + 2e- (Anode)

Thus, cathode increases in mass as copper is deposited on it and the anode decreases in mass as copper from it dissolves.

Charge passed through cell = 2.68 × 60 × 60 coulomb

Copper deposited or dissolved = 63,5/(2×96500)×2.68×60×60=3.174 g

Increase in mass of cathode = Decrease in mass of anode = 3.174 g

Example 13. An ammeter and a copper voltameter are connected in series through which a constant current flows. The ammeter shows 0.52 ampere. If 0.635 g of copper is deposited in one hour, what is the percentage error of the ammeter? (At. mass of copper = 63.5)

Solution : The electrode reaction is:

Cu2+ + 2e- --> Cu

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1 mole 2 × 96500 C

63.5 g of copper deposited by passing charge = 2 × 96500 Coulomb

0.635 g of copper deposited by passing charge

=(2×96500)/63.5×0.653 coulomb

= 2 × 965 coulomb

= 1930 coulomb

We know that

Q = l × t

1930 = I × 60 × 60

I= 1930/3600=0.536 ampere

Percentage error = ((0.536-0.52))/0.536×100=2.985

Example 14. A current of 3.7 ampere is passed for 6 hours between platinum electrodes in 0.5 litre of a 2 M solution of Ni(NO3)2. What will be the molarity of the solution at the end of electrolysis? What will be the molarity of solution if nickel electrodes are used? (1 F = 96500 coulomb; Ni = 58.7)

Solution: The electrode reaction is

Ni2+ + 2e- --> Ni

1 mole 2 × 96500 C

Quantity of electric charge passed

= 3.7 × 6 × 60 × 60 coulomb = 79920 coulomb

Number of moles of Ni(NO3)2 decomposed or nickel deposited = (1.0 - 0.4140) = 0.586

Since 0.586 moles are present in 0.5 litre,

Molarity of the solution = 2 × 0.586 = 1.72 M

When nickel electrodes are used, anodic nickel will dissolve and get deposited at the cathode. The molarity of the solution will, thus, remain unaffected.

Example 15: An acidic solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with volume of solution kept at 100 mL and the current at 1.2 amp. Calculate the gases evolved at NTP during the entire electrolysis.

Solution: 0.4 g of Cu2+ = 0.4/31.75 = 0.0126 g equivalent

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At the same time, the oxygen deposited at anode

= 0.0126 g equivalent

= 8/32 × 0.0126 = 0.00315 g mol

After the complete deposited of copper, the electrolysis will discharge hydrogen at cathode and oxygen at anode.

The amount of charge passed = 1.2 × 7 × 60 = 504 coulomb

So, Oxygen liberated = 1/96500 × 504 = 0.00523 g equivalent

= 8/32 × 0.00523 = 0.001307 g mole

Hydrogen liberated = 0.00523 g equivalent

= 1/2 × 0.00523 = 0.00261 g mole

Total gases evolved = (0.00315 + 0.001307 + 0.00261) g mole

= 0.007067 g mole

Volume of gases evolved at NTP

= 22400 × 0.007067 mL

= 158.3 mL

Example 16: An acidic solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with volume of solution kept at 100 mL and the current at 1.2 amp. Calculate the gases evolved at NTP during the entire electrolysis.

Solution: Amount of charge passed = 1.70 × 230 coulomb

Amount of actual charge passed = 90/100 × 1.70 × 230

= 351.9 coulomb

No. of moles of Zn deposited by passing 351.9 coulomb of charge

= 1/(2x96500) × 351.9 = 0.000182

Molarity of Zn2+ ions after deposition of zinc

= [0.160-(0.000182×1000)/300]M

= 0.154 M

Example 18: Calculate the electricity that would be required to reduce 12.3 g of nitrobenzene to aniline, if the current efficiency for the process is 50 per cent. If the potential drop across the cell is 3.0 volt, how much energy will be consumed?

Solution: The reduction reaction is

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C6H5NO2 + 3H2 C6H5NH2 + 2H2O

123 g 6g

1 mole 3 mole

Hydrogen required for reduction of 12.3/123 or 0.1 mole of nitrobenzene = 0.1 × 3 = 0.3 mole

Amount of charge required for liberation of 0.3 mole of hydrogen = 2 × 96500 × 0.3 = 57900 coulomb

Actual amount of charge required as efficiency is 50%

= 2 × 57900 = 115800 coulomb

Energy consumed = 115800 × 3.0 = 347400 J

= 347.4 kJ

Example 18: An aqueous solution of sodium chloride on electrolysis gives H2(g), CI2(g) and NaOH according to the reaction:

2Cl-(aq) + 2H2O ® 2OH-(aq) + H2(g) + Cl2(g)

A direct current of 25 ampere with a current efficiency 62% is passed through 20 L of NaCl solution (20% by mass). Write down the reactions taking place at the anode and cathode. How long will it take to produce 1 kg of Cl2? What will be the molarity of the solution with respect to hydroxide ion? Assume no loss due to evaporation.

Solution: Reactions at anode and cathode are

2Cl- ---> Cl2 + 2e- (at anode)

2H2O + 2e- ---> H2 + 2OH- (at cathode)

1 kg of Cl2 = 1000/71.0 = 14.08 mole

Charge to produce one mole of Cl2 = 2 × 96500 coulomb

Charge to produce 14.08 mole of Cl2 = 2 × 96500 × 14.08 coulomb

Effective current = 62/100 × 25.0 = 15.5 ampere

Time = charge/current = (2×96500×14.08)/15.5

= 175318.7 second = 48.699 hour

OH- ions produced= 2 × mole of Cl2

= 2 × 14.08 = 28.16 mole

Molarity Mole/Volume = 28.16/20 = 1.408 M

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Example 19: Chromium metal can be plated out from an acidic solution containing CrO3 according to the following reaction:

CrO3 + 6H+ + 6e- --> Cr + 3H2O

Solution: CrO3 + 6H+ + 6e- --> Cr + 3H2O

6 × 96500C 1 mole

52 g.

Mass of chromium plated out by 24000 coulomb charge

= 52/6*96500 × 24000 = 2.155 g

Charge required for plating out 1.5 g of chromium

= 6*96500/52 × 1.5 = 16701.92 coulomb

Time = charge/current = 1336.15 second

= 22.27 minute

Example 20: After electrolysis of a sodium chloride solution with inert electrodes for a certain period of time, 600 mL of the solution was left which was found to be 1 N in NaOH. During the same period 31.75 g of copper was deposited in the copper voltameter in series with the electrolytic cell. Calculate the percentage theoretical yield of NaOH obtained.

Soution: Equivalent mass of NaOH = 40/1000 × 600 = 24 g

Amount of NaOH formed = 40/1000 × 600 = 24 g

31.75 g of Cu = 1 g equivalent of Cu.

During the same period, 1 g equivalent of NaOH should have been formed.

1 g equivalent of NaOH = 40 g

% yield = 24/40 × 100 = 60

Example 21: Peroxy disulphuric acid (H2S2O8) can be prepared by electrolytic oxidation of H2SO4 as:

2H2SO4 --> H2S2O8 + 2H+ +2e-

Oxygen and hydrogen are by products. In such an electrolysis 9.72 litre of H2 and 2.35 litre of O2 were generated at NTP. What is the mass of peroxy disulphuric acid formed?

Solution:

Anodic reaction 2H2SO4 --> H2S2O8 + 2H+ + 2e-

2H2O --> 4H+ + O2 + 2e-

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Cathodic reaction 2H2O + 2e- --> 2OH- + H2

Total equivalent of H2S2O8 + equivalent of oxygen = Equivalent of H2

9.72 litre H2 = 9.72/11.2 = 0.868 equivalent

2.35 litre O2 = 2.35/5.6 = 0.42 equivalent

Equivalent of H2S2O8 = (0.868 - 0.420)

= 0.448

Mass of H2S2O8 = 0.448 × 194/2 = 43.456 g.

Example 22: Cadmium amalgam is prepared by electrolysis of a solution of CdCl2 using a mercury cathode. Find how long a current of 5 ampere should be passed in order to prepare 12% Cd-Hg amalgam on a cathode of 2 g mercury. At mass of Cd = 112.40.

Solution: 2 g Hg require Cd to prepare 12% amalgam

= 12/88 × 2 = 0.273 g

Cd2+ + 2e- --> Cd

1 mole 2 × 96500C

112.40g

Charge required to deposit 0.273 g of Cd

= 2*96500/112.40 × 0.273 coulomb

Charge = ampere × second

Second = 2*96500*0.273/112.40*5 = 93.75

Example 23: Assume that impure copper contains irons, gold and silver as impurities. After passing a current of 140 ampere for 482. Second, the mass of anode decreased by 22.260 g and the cathode increased in mass by 22.011 g. Estimate the percentage of iron and copper originally present.

Solution: The increase at the cathode is due to copper only. Hence, there is 22.011 g of copper and rest impurities of iron, gold and silver.

Mass of impurities = (22.260 - 22.011) = 0.249 g)

At anode, only copper and iron are oxidised; the gold and silver collect below anode in the form of anodic mud.

M --> M2+ + 2e-

(copper and iron)

No. of moles of metal oxidised = (140×482.5)/(2×96500) = 0.35

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No. of moles of copper = 22.011/63.5 = 0.3466

No. of moles of iron = (0.35 - 0.3466) = 0.0034





or k= 2.1/4.2×1/50=0.01 ohm-1 cm-1










= 0.0112 × 56

= 0.06272 cm-1



= (2.768×10-3)/(I/250.2)

= 2.768 × 10-3 × 250.2


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= 2.768 × 10-3 × 250.2 × 1/8331


= (2.768×10-3×25.2)/8331×1000/(1/100)

Example 27: The equivalent conductances of sodium chloride, hydrochloric acid and sodium acetate at infinite dilution are 126.45, 426.16 and 91.0 ohm-1 cm2

equiv-1, respectively, at 250C. Calculate the equivalent conductance of acetic acid at infinite dilution.

Solution: According to Kohlrausch's law,⋀_(∞CH_3 COONa)=λ_(CH_3 COO^- )+λ_(Na^+ )=91.0 ......(i)

⋀∞HCl = λH+ + λCl

-=426.16 ...(ii)

⋀∞NaCl = λNa+ + λCl

-=126.45 ...(iii)

Adding Eqs. (i) and (ii) and subtracting (iii),

λCH3 COO- + λNa

+ + λH+ + λCl

- - λNa+ + λCl

-

= 91.0 + 426.16 - 126.45

λCH3 COO- + λH

+ =⋀∞CH3 COOH = 390.7 ohm-1 cm-2 equiv-1

Example 28: The equivalent conductivity of N/10 solution of acetic acid at 250C is 14.3 ohm-1 cm2 equiv-1. Calculate the degree of dissociation of CH3COOH if is 390.71.

Solution: ⋀∞CH3 COOH =390.7 ohm-1 cm(-2) equiv(-1)

⋀(∞CH3 COOH)=143.3 ohm(-1) cm(-2) equiv(-1)

Degree of dissociation, α= ⋀v/⋀∞ =14.3/390.71

= 0.0366 i.e., 3.66% dissociated

Example 29: A decinormal solution of NaCI has specific conductivity equal to 0.0092. If ionic conductances of Na+ and Cl- ions at the same temperature are 43.0 and 65.0 ohm-1 respectively, calculate the degree of dissociation of NaCl solution.

Solution: Equivalent conductance of N/10 NaCl solution

90

⋀v = Sp. conductivity × dilution

= 0.0092 × 10,000

= 92 ohm-1

⋀∞ = λ(Na+

)+ λCl-

= 43.0 + 65.0

= 108 ohm-1

Degree of dissociation,

Connecting rural India to the world

It is an ambitious vision: all villages in the country, connected to the world via technology. Village knowledge centres (VKCs) -- a dream that former President A P J Abdul Kalam shared with IIT alumni -- are gaining steam. While the country has 50 VKCs, only one has been functioning in Karnataka. The number is now all set to jump.

A knowledge centre -- which provides access to books, internet and training in different streams to villagers -- has been in place at Ittamadu, a small village 50 km from Bangalore, for three years. Now, neighbouring villages are demanding VKCs of their own. IIT-Bombay alumni, along with local gram panchayats, are planning to expand the service to 10 VKCs by March 2009. A concert by alumni from across the country will be held in Bangalore, on November 8, to raise funds for this.

Typically, a VKC operates in a small room provided by the gram panchayat. A library and 3-4 computers are set up. Services include internet banking and reservation of tickets. Children can access education-related software and the library. Information useful to farmers -- like new methods, best practices, employment opportunities and prices of agricultural products -- is also provided.

The Ittamadu VKC was a pilot project set up in 2005. The computers and library were provided by IIT-Bombay alumni. Principal secretary V P Baligar, an alumnus, believes VKCs are a successful model. "I was involved in setting up the first VKC at Ittamadu. There is a great demand for them in other villages as well. We plan to set up nine more centres. Alumni are trying their best to raise funds,' he says.

Eight hundred students of five primary shools in the area have benefited from the Ittamadu centre. Many children visit the VKC to use computers.

The centre was proposed as a joint venture between the panchayat, villagers and alumni. "There had to be a sense of ownership amongst villagers. So the panchayat provided us with a small hut that has now transformed into a proper room. They provided the salary for the keeper and paid for maintenance. We involved the villagers to ensure we had everything they wanted. To set up more VKCs, we need money. Funding has to be relatively continuous so the project is sustained in many places,' says Srikanth Rao, treasurer of the IIT Bombay Alumni Association, Bangalore chapter.

The nine proposed VKCs, four of which will be in Ittamadu gram panchayat and five in the neighbouring Melgopahalli gram panchayat, will require a capital expenditure of Rs 12.5 lakh and revenue expenditure of Rs 7.5 lakh. The alumni also plan to expand computer-training centres. Rao said the Nemmadi common services centre will help raise funds for the VKC.

Electrolysis refers to the decomposition of a substance by an electric current. The electrolysis of sodium and potassium hydroxides, first carried out in 1808 by Sir Humphrey Davey, led to

91

http://www.corrosion-doctors.org/Biographies/DavyBio.htm

the discovery of these two metallic elements and showed that these two hydroxides which had previously been considered un-decomposable and thus elements, were in fact compounds:

"By means of a flame which was thrown on a spoon containing potash, this alkali was kept for some minutes at a strong red heat, and in a state of perfect fluidity." One pole of a battery of copper-zinc cells was connected to the spoon, and the other was connected to platinum wire which dipped into the melt. "By this arrangement some brilliant phenomena were produced. The potash appeared to be a conductor in a high degree, and as long as the communication was preserved, a most intense light was exhibited at the negative wire, and a column of flame, which seemed to be owing to the development of combustible matter, arose from the point of contact." The flame was due to the combustion in the air of metallic potassium. In another experiment, Davey observed "small globules having a high metallic lustre, precisely similar in visible characters to quicksilver, some of which burnt with explosion and bright flame, as soon as they were formed, and others remained, and were merely tarnished, and finally covered by a white film which formed on their surfaces."

Electrolysis of molten alkali halides is the usual industrial method of preparing the alkali metals:

cathode: Na+ + e– → Na(l) E° = –2.71 v

anode: Cl– → ½ Cl2(g) + e– E° = –1.36 v

net: Na+ + Cl– → Na(l) + ½ Cl2(g) E° = –4.1 v

Ions in aqueous solutions can undergo similar reactions. Thus if a solution of nickel chloride undergoes electrolysis at platinum electrodes, the reactions are

cathode: Ni2+ + 2 e– → Ni(s) E° = –0.24 v

anode: 2 Cl– → Cl2(g) + 2 e– E° = –1.36 v

net: Ni2+ + 2 Cl– → Ni(s) + Cl2(g) E° = –1.60 v

Both of these processes are carried out in electrochemical cells which are forced to operate in the "reverse", or non-spontaneous direction, as indicated by the negative for the above cell reaction. The free energy is supplied in the form of electrical work done on the system by the outside world (the surroundings). This is the only fundamental difference between an electrolytic cell and the galvanic cell in which the free energy supplied by the cell reaction is extracted as work done on the surroundings.

A common misconception about electrolysis is that "ions are attracted to the oppositely-charged electrode." This is true only in the very thin interfacial region near the electrode surface. Ionic motion throughout the bulk of the solution occurs mostly by diffusion, which is the transport of molecules in response to a concentration gradient. Migration— the motion of a charged particle due to an applied electric field, is only a minor player,

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producing only about one non-random jump out of around 100,000 random ones for a 1 volt cm–1 electric field. Only those ions that are near the interfacial region are likely to undergo migration.

Electrolysis in aqueous solutions

Water is capable of undergoing both oxidation

H2O → O2(g) + 4 H+ + 2 e– E° = -1.23 v

and reduction

2 H2O + 2 e– → H2(g) + 2 OH– E° = -0.83 v

Thus if an aqueous solution is subjected to electrolysis, one or both of the above reactions may be able to compete with the electrolysis of the solute.

For example, if we try to electrolyze a solution of sodium chloride, hydrogen is produced at the cathode instead of sodium:

cathode: H2O + 2 e– → H2(g) + 2 OH– E =+0.41 v ([OH–] = 10-7 M)

anode: Cl– → ½ Cl2(g) + e– E° = –1.36 v

net: Cl– + H2O → 2 H2(g) + ½ Cl2(g) + 2 OH– E = –0.95 v

[This illustration is taken from the excellentPurdue University Chemistry site]

Reduction of Na+ (E° = –2.7 v) is energetically more difficult than the reduction of water (–1.23 v), so in aqueous solution the latter will prevail.

Electrolysis of salt ("brine") is carried out on a huge scale and is the basis of the chloralkali industry .

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http://electrochem.cwru.edu/ed/encycl/art-b01-brine.htm


http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php


Electrolysis of water

Pure water is an insulator and cannot undergo signifigant electrolysis without adding an electrolyte. If the object is to produce hydrogen and oxygen, the electrolyte must be energetically more difficult to oxidize or reduce than water itself. Electrolysis of a solution of sulfuric acid or of a salt such as NaNO3 results in the decomposition of water at both electrodes:

cathode: H2O + 2 e– → H2(g) + 2 OH– E =+0.41 v ([OH–] = 10-7 M)

anode: 2 H2O → O2(g) + 4 H+ + 2 e– E° = -0.82 v

net: 2 H2O(l) → 2 H2(g) + O2(g) E = -1.23 v

Electrolytic production of hydrogen is usually carried out with a dilute solution of sulfuric acid; see here for a concise summary of the chemistry and energetics. This process is generally too expensive for industrial production unless highly pure hydrogen is required. However, it becomes more efficient at higher temperatures, where thermal energy reduces the amount of electrical energy required, so there is now some interest in developing high-temperature electrolytic processes. Most hydrogen gas is manufactured by the steam reforming of natural gas.

Faraday's laws of electrolysis

One mole of electric charge (96,500 coulombs), when passed through a cell, will discharge half a mole of a divalent metal ion such as Cu2+. This relation was first formulated by Faraday in 1832 in the form of two laws of electrolysis:

1. The weights of substances formed at an electrode during electrolysis are directly proportional to the quantity of electricity that passes through the electrolyte.

2. The weights of different substances formed by the passage of the same quantity of electricity are proportional to the equivalent weight of each substance.

The equivalent weight of a substance is defined as the molar mass, divided by the number of electrons required to oxidize or reduce each unit of the substance. Thus one mole of V3+ corresponds to three equivalents of this species, and will require three faradays of charge to deposit it as metallic vanadium.

Most stoichiometric problems involving electrolysis can be solved without explicit use of Faraday's laws. The "chemistry" in these problems is usually very elementary; the major difficulties usually stem from unfamiliarity with the basic electrical units:

current (amperes) is the rate of charge transport; 1 amp = 1 c/sec.

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http://www.chemheritage.org/EducationalServices/chemach/eei/mf.html

http://en.wikipedia.org/wiki/Steam_reforming

http://en.wikipedia.org/wiki/High-temperature_electrolysis

http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/electrol.html

power (watts) is the rate of energy production or consumption;1 w = 1 J/sec = 1 volt-amp; 1 watt-sec = 1 J, 1 kw-h = 3600 J.

Problem Example 1

A metallic object to be plated with copper is placed in a solution of CuSO4. a) To which electrode of a direct current power supply should the object be connected?b) What mass of copper will be deposited if a current of 0.22 amp flows through the cell for 1.5 hours?

Solution:

a) Since Cu2+ ions are being reduced, the object acts as a cathode and must be connected to the negative terminal (where the electrons come from!)

b) The amount of charge passing through the cell is

(0.22 amp) × (5400 sec) = 1200 cor

(1200 c) ÷ (96500 c F–1) = 0.012 F

Since the reduction of one mole of Cu2+ ion requires the addition of two moles of electrons, the mass of Cu deposited will be

(63.54 g mol–1) (0.5 mol Cu/F) (.012 F) = 0.39 g of copper

Problem Example 2

How much electric power is required to produce 1 metric ton (1000 kg) of chlorine from brine, assuming the cells operate at 2.0 volts and assuming 100 % efficiency?

Solution:

moles of Cl2 produced: (106 g) ÷ 70 g mol–1 = 14300 mol Cl2

faradays of charge: (2 F/mol) × (14300 mol) = 28600 F

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charge in coulombs: (96500 c/F) × (28600 F) = 2.76 × 109 c

duration of electrolysis: (3600 s/h) x (24 h) = 86400 s

current (rate of charge delivery): (2.76 × 109 amp-sec) ÷ (86400 sec) = 32300 amps

power (volt-amps): (2.0 v) × (32300 a) = 64.6 kw

energy in kW-h: (64.6 kw) × (24 h) = 1550 kw-h

energy in joules: (1550 kw-h) × (3.6Mj/kw-h) = 5580 Mj (megajoules)

(In the last step, recall that 1 w = 1 j/s, so 1 kw-h = 3.6 Mj)

Industrial electrolytic processes

For many industrial-scale operations involving the oxidation or reduction of both inorganic and organic substances, and especially for the production of the more active metals such as sodium, calcium, magnesium, and aluminum, the most cost-effective reducing agent is electrons supplied by an external power source. The two most economically important of these processes are described below.

The chloralkali industry

The electrolysis of brine is carried out on a huge scale for the industrial production of chlorine and caustic soda (sodium hydroxide). Because the reduction potential of Na+ is much higher than that of water, the latter substance undergoes decomposition at the cathode, yielding hydrogen gas and OH–.

anodereactions

2 Cl– → Cl2(g) + 2 e– -1.36 v i

4 OH– → O2(g) + 2 H2O + 4 e–-0.40 v ii

cathodereactions

Na+ + e– → Na(s) -2.7 v iii

H2O + 2 e– → H2(g) + 2 OH– +.41 v iv

A comparison of the E°s would lead us to predict that the reduction (ii) would be favored over that of (i). This is certainly the case from a purely energetic standpoint, but as was mentioned in the section on fuel cells, electrode reactions involving O2 are notoriously slow (that

96

is, they are kinetically hindered), so the anodic process here is under kinetic rather than thermodynamic control. The reduction of water (iv) is energetically favored over that of Na+ (iii), so the net result of the electrolysis of brine is the production of Cl2 and NaOH ("caustic"), both of which are of immense industrial importance:

2 NaCl + 2 H2O → 2 NaOH + Cl2(g) + H2(g)

A modern industrial chloralkali plant

Schematic diagram of a cell for the production of chlorine

Since chlorine reacts with both OH– and H2, it is necessary to physically separate the anode and cathode compartments. In modern plants this is accomplished by means of an ion-selective polymer membrane, but prior to 1970 a more complicated cell was used that employed a pool of mercury as the cathode. A small amount of this mercury would normally find its way into the plant's waste stream, and this has resulted in serious pollution of many major river systems and estuaries and devastation of their fisheries.

Here is an excellent survey of industrial brine electrolysis from Case Western Reserve University

Electrolytic refining of aluminum

Aluminum is present in most rocks and is the most abundant metallic element in the earth's crust (eight percent by weight.) However, its isolation is very difficult and expensive to accomplish by purely chemical means, as evidenced by the high E° (–1.66 v) of the Al3+/Al couple. For the same reason, aluminum cannot be

isolated by electrolysis of aqueous solutions of its compounds, since the water would be electrolyzed preferentially. And if you have ever

97


tried to melt a rock, you will appreciate the difficulty of electrolyzing a molten aluminum ore! Aluminum was in fact considered an exotic and costly metal until 1886, when Charles Hall (U.S.A) and Paul Hérault (France) independently developed a practical electrolytic reduction process.

The Hall-Hérault process takes advantage of the principle that the melting point of a substance is reduced by admixture with another substance with which it forms a homogeneous phase. Instead of using the pure alumina ore Al2O3 which melts at 2050°C, it is mixed with cryolite, which is a natural mixture of NaF and AlF3, thus reducing the temperature required to a more manageable 1000°C. The anodes of the cell are made of carbon (actually a mixture of pitch and coal), and this plays a direct role in the process; the carbon gets oxidized (by the oxide ions left over from the reduction of Al3+ to CO, and the free energy of this reaction helps drive the aluminum reduction, lowering the voltage that must be applied and thus reducing the power consumption. This is important, because aluminum refining is the largest consumer of industrial electricity, accounting for about 5% of all electricity generated in North America. Since aluminum cells commonly operated at about 100,000 amperes, even a slight reduction in voltage can result in a large saving of power.

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http://www.fact-index.com/p/pa/paul_heroult.html

http://www.fact-index.com/p/pa/paul_heroult.html

http://www.oberlin.edu/archive/WWW_files/hall_cm_b.html

The net reaction is

2 Al2O3 + 3 C → 4 Al + 3 CO2

However, large quantities of CO and of HF (from the cryolite), and hydrocarbons (from the electrodes) are formed in various side reactions, and these can be serious sources of environmental pollution.

Brief biographies of Charles and Julia Hall, Paul Hérault

Excellent Wikipedia article on the Hall process

Aluminum production - Electrochemistry Encyclopedia

Details of Hall's development of his refining process

More on aluminum ore processing and refining

Above: a modern aluminum cell [image]Below: Cell pot line in Høyanger, Norway [Norsk Hydro]

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http://www.hydro.com/en/Press-room/Image-gallery/Aluminium/Aluminium-facilities/

http://www.qatalum.com/en/Media-room/News/2004/Plans-for-world-class-aluminium-project-in-Qatar/

http://www.industrialcenter.org/HeatTreat/MetalsAdvisor/aluminum/mining_and_primary_processing/mining_and_primary_process_description.htm

http://www.oberlin.edu/external/EOG/OYTT-images/CMHall.html

http://electrochem.cwru.edu/encycl/art-a01-al-prod.htm

http://en.wikipedia.org/wiki/Hall%E2%80%93H%CE%B9roult_process

http://www.chemheritage.org/classroom/chemach/electrochem/heroult-hall.html

What you need to know

Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic.

In electrolysis, an external power source supplies the free energy required to drive a cell reaction in its non-spontaneous direction. An electrolytic cell is in this sense the opposite of a galvanic cell. In practice, the products of electrolysis are usually simpler than the reactants, hence the term electro-lysis.

Transport of ions in the electrolyte in response to the potential difference between the electrodes ("drift") is largely restricted to the regions very close to the electrodes. Ionic transport in the greater part of the electrolyte is by ordinary thermal diffusion— the statistical tendency of concentrations to become uniform.

When an aqueous solution is subjected to electrolysis, the oxidation or reduction of water can be a competing process and may dominate if the applied voltage is sufficiently great. Thus an attempt to electrolyze a solution of NaNO3 will produce only H2 and O2.

A large number of electrolysis processes are employed by industry to refine metals and to produce both inorganic and organic products. The largest of these are the chloralkali industry (chlorine and "caustic"), and the refining of aluminum; the latter consumes approximately 5% of the electrical power generated in North America.

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Concept Map

References

Electrolysis tutorial (with practice problems) from Purdue University

Electrolytic Cells

The Electrolysis of Molten NaCl

The Electrolysis of Aqueous NaCl

The Electrolysis of Water Faraday's Law

Voltaic cells use a spontaneous chemical reaction to drive an electric current through an external circuit. These cells are important because they are the basis for the batteries that fuel

101

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php#law%23law

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php#water%23water

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php#aq%23aq


http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php#molten%23molten



modern society. But they aren't the only kind of electrochemical cell. It is also possible to construct a cell that does work on a chemical system by driving an electric current through the system. These cells are called electrolytic cells. Electrolysis is used to drive an oxidation-reduction reaction in a direction in which it does not occur spontaneously.


An idealized cell for the electrolysis of sodium chloride is shown in the figure below. A source of direct current is connected to a pair of inert electrodes immersed in molten sodium chloride. Because the salt has been heated until it melts, the Na+ ions flow toward the negative electrode and the Cl- ions flow toward the positive electrode.

When Na+ ions collide with the negative electrode, the battery carries a large

enough potential to force these ions to pick up electrons to form sodium metal.

Negative electrode (cathode): Na+ + e- Na

Cl- ions that collide with the positive electrode are oxidized to Cl2 gas,

which bubbles off at this electrode.

Positive electrode (anode): 2 Cl- Cl2 + 2 e-

The net effect of passing an electric current through the molten salt in this cell is to

decompose sodium chloride into its elements, sodium metal and chlorine gas.

Electrolysis of NaCl: Cathode (-): Na+ + e- Na

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Anode (+): 2 Cl- Cl2 + 2 e-

The potential required to oxidize Cl- ions to Cl2 is -1.36 volts and the potential needed to reduce Na+ ions to sodium metal is -2.71 volts. The battery used to drive this reaction must therefore have a potential of at least 4.07 volts.

This example explains why the process is called electrolysis. The suffix -lysis comes from the Greek stem meaning to loosen or split up. Electrolysis literally uses an electric current to split a compound into its elements.

electrolysis 2 NaCl(l) 2 Na(l) + Cl2(g)

This example also illustrates the difference between voltaic cells and electrolytic cells. Voltaic cells use the energy given off in a spontaneous reaction to do electrical work. Electrolytic cells use electrical work as source of energy to drive the reaction in the opposite direction.

The dotted vertical line in the center of the above figure represents a diaphragm that keeps the Cl2 gas produced at the anode from coming into contact with the sodium metal generated at the cathode. The function of this diaphragm can be understood by turning to a more realistic drawing of the commercial Downs cell used to electrolyze sodium chloride shown in the figure below.

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Chlorine gas that forms on the graphite anode inserted into the bottom of this cell bubbles through the molten sodium chloride into a funnel at the top of the cell. Sodium metal that forms at the cathode floats up through the molten sodium chloride into a sodium-collecting ring, from which it is periodically drained. The diaphragm that separates the two electrodes is a screen of iron gauze, which prevents the explosive reaction that would occur if the products of the electrolysis reaction came in contact.

The feed-stock for the Downs cell is a 3:2 mixture by mass of CaCl2 and NaCl. This

mixture is used because it has a melting point of 580oC, whereas pure sodium chloride

has to be heated to more than 800oC before it melts.


The figure below shows an idealized drawing of a cell in which an aqueous solution of sodium chloride is electrolyzed.

Once again, the Na+ ions migrate toward the negative electrode and the Cl- ions migrate toward the positive electrode. But, now there are two substances that can be reduced at the cathode: Na+ ions and

water molecules.

Cathode (-): Na+ + e- Na Eo

red = -2.71 V 2 H2O + 2 e- H2 + 2 OH- Eo

red = -0.83 V

Because it is much easier to reduce water than Na+ ions, the only

product formed at the cathode is hydrogen gas.

Cathode (-): 2 H2O(l) + 2 e- H2(g) + 2 OH-(aq)

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There are also two substances that can be oxidized at the anode: Cl-

ions and water molecules.

Anode (+): 2 Cl- Cl2 + 2 e- Eo

ox = -1.36 V 2 H2O O2 + 4 H+ + 4 e- Eo

ox = -1.23 V

The standard-state potentials for these half-reactions are so close to each other that we might expect to see a mixture of Cl2 and O2 gas collect at the anode. In practice, the only product of this reaction is Cl2.

Anode (+): 2 Cl- Cl2 + 2 e-

At first glance, it would seem easier to oxidize water (Eoox = -1.23

volts) than Cl- ions (Eoox = -1.36 volts). It is worth noting, however,

that the cell is never allowed to reach standard-state conditions. The solution is typically 25% NaCl by mass, which significantly decreases the potential required to oxidize the Cl- ion. The pH of the cell is also kept very high, which decreases the oxidation potential for water. The deciding factor is a phenomenon known as overvoltage, which is the extra voltage that must be applied to a reaction to get it to occur at the rate at which it would occur in an ideal system.

Under ideal conditions, a potential of 1.23 volts is large enough to oxidize water to O2 gas. Under real conditions, however, it can take a much larger voltage to initiate this reaction. (The overvoltage for the oxidation of water can be as large as 1 volt.) By carefully choosing the electrode to maximize the overvoltage for the oxidation of water and then carefully controlling the potential at which the cell operates, we can ensure that only chlorine is produced in this reaction.

In summary, electrolysis of aqueous solutions of sodium chloride doesn't give the same products as electrolysis of molten sodium chloride. Electrolysis of molten NaCl decomposes this compound into its elements.


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Electrolysis of aqueous NaCl solutions gives a mixture of hydrogen and chlorine gas and an aqueous sodium hydroxide solution.

electrolysis 2 NaCl(aq) + 2 H2O(l)

2 Na+(aq) + 2 OH-(aq) + H2(g) + Cl2(g)

Because the demand for chlorine is much larger than the demand for sodium, electrolysis of aqueous sodium chloride is a more important process commercially. Electrolysis of an aqueous NaCl solution has two other advantages. It produces H2 gas at the cathode, which can be collected and sold. It also produces NaOH, which can be drained from the bottom of the electrolytic cell and sold.

The dotted vertical line in the above figure represents a diaphragm that prevents the Cl2 produced at the anode in this cell from coming into contact with the NaOH that accumulates at the cathode. When this diaphragm is removed from the cell, the products of the electrolysis of aqueous sodium chloride react to form sodium hypo-chlorite, which is the first step in the preparation of hypochlorite bleaches, such as Chlorox.

Cl2(g) + 2 OH-(aq) Cl-(aq) + OCl-(aq) + H2O(l)

Electrolysis of Water

A standard apparatus for the electrolysis of water is shown in the figure below.

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electrolysis 2 H2O(l) 2 H2(g) + O2(g)

A pair of inert electrodes are sealed in opposite ends of a container designed to collect the H2 and O2 gas given off in this reaction. The electrodes are then connected to a battery or another source of electric current.

By itself, water is a very poor conductor of electricity. We therefore add an electrolyte

to water to provide ions that can flow through the solution, thereby completing the

electric circuit. The electrolyte must be soluble in water. It should also be relatively

inexpensive. Most importantly, it must contain ions that are harder to oxidize or

reduce than water.

2 H2O + 2 e- H2 + 2 OH- Eored = -0.83 V

2 H2O O2 + 4 H+ + 4 e- Eoox = -1.23 V

The following cations are harder to reduce than water: Li+, Rb+, K+, Cs+, Ba2+, Sr2+, Ca2+, Na+, and Mg2+. Two of these cations are more likely candidates than the others because they form inexpensive, soluble salts: Na+ and K+.

The SO42- ion might be the best anion to use because it is the most difficult anion to

oxidize. The potential for oxidation of this ion to the peroxydisulfate ion is -2.05

volts.

2 SO42- S2O8

2- + 2 e- Eoox = -2.05 V

107

When an aqueous solution of either Na2SO4 or K2SO4 is electrolyzed in the apparatus shown in the above figure, H2 gas collects at one electrode and O2 gas collects at the other.

What would happen if we added an indicator such as bromothymol blue to this

apparatus? Bromothymol blue turns yellow in acidic solutions (pH < 6) and blue in

basic solutions (pH > 7.6). According to the equations for the two half-reactions, the

indicator should turn yellow at the anode and blue at the cathode.

Cathode (-): 2 H2O + 2 e- H2 + 2 OH-

Anode (+): 2 H2O O2 + 4 H+ + 4 e-

Faraday's Law

Faraday's law of electrolysis can be stated as follows. The amount of a substance consumed or produced at one of the electrodes in an electrolytic cell is directly proportional to the amount of electricity that passes through the cell.

In order to use Faraday's law we need to recognize the relationship between current, time, and the amount of electric charge that flows through a circuit. By definition, one coulomb of charge is transferred when a 1-amp current flows for 1 second.

1 C = 1 amp-s

Example: To illustrate how Faraday's law can be used, let's calculate the number of grams of sodium metal that will form at the cathode when a 10.0-amp current is passed through molten sodium chloride for a period of 4.00 hours.

We start by calculating the amount of electric charge that flows through the cell.

Before we can use this information, we need a bridge between this macroscopic quantity and the phenomenon that occurs on the atomic scale. This bridge is represented by Faraday's constant, which describes the number of coulombs of charge carried by a mole of electrons.

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Thus, the number of moles of electrons transferred when 144,000 coulombs of electric charge flow through the cell can be calculated as follows.

According to the balanced equation for the reaction that occurs at the cathode of this

cell, we get one mole of sodium for every mole of electrons.

Cathode (-): Na+ + e- Na

Thus, we get 1.49 moles, or 34.3 grams, of sodium in 4.00 hours.

The consequences of this calculation are interesting. We would have to run this

electrolysis for more than two days to prepare a pound of sodium.

Practice Problem 13:

Calculate the volume of H2 gas at 25oC and 1.00 atm that will collect at the cathode when an aqueous solution of Na2SO4 is electrolyzed for 2.00 hours with a 10.0-amp current.

Click here to check your answer to Practice Problem 13

Click here to see a solution to Practice Problem 13

We can extend the general pattern outlined in this section to answer questions that

might seem impossible at first glance.


Determine the oxidation number of the chromium in an unknown salt if electrolysis of a molten sample of this salt for 1.50 hours with a 10.0-amp current deposits 9.71 grams of chromium metal

109

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/problems/ex20_14s.html

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/problems/ex20_14a.html

at the cathode.



Purdue University’s chemical site

Electrolytic Cells



The Electrolysis of Water Faraday's Law

Voltaic cells use a spontaneous chemical reaction to drive an electric current through an external circuit. These cells are important because they are the basis for the batteries that fuel modern society. But they aren't the only kind of electrochemical cell. It is also possible to construct a cell that does work on a chemical system by driving an electric current through the system. These cells are called electrolytic cells. Electrolysis is used to drive an oxidation-reduction reaction in a direction in which it does not occur spontaneously.


An idealized cell for the electrolysis of sodium chloride is shown in the figure below. A source of direct current is connected to a pair of inert electrodes immersed in molten sodium chloride. Because the salt has been heated until it melts, the Na+ ions flow toward the negative electrode and the Cl- ions flow toward the positive electrode.

110

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php#law%23law

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php#water%23water







When Na+ ions collide with the negative electrode, the battery carries a large

enough potential to force these ions to pick up electrons to form sodium metal.

Negative electrode (cathode): Na+ + e- Na

Cl- ions that collide with the positive electrode are oxidized to Cl2 gas,

which bubbles off at this electrode.

Positive electrode (anode): 2 Cl- Cl2 + 2 e-

The net effect of passing an electric current through the molten salt in this cell is to

decompose sodium chloride into its elements, sodium metal and chlorine gas.

Electrolysis of NaCl: Cathode (-): Na+ + e- Na Anode (+): 2 Cl- Cl2 + 2 e-

The potential required to oxidize Cl- ions to Cl2 is -1.36 volts and the potential needed to reduce Na+ ions to sodium metal is -2.71 volts. The battery used to drive this reaction must therefore have a potential of at least 4.07 volts.

This example explains why the process is called electrolysis. The suffix -lysis comes from the Greek stem meaning to loosen or split up. Electrolysis literally uses an electric current to split a compound into its elements.


This example also illustrates the difference between voltaic cells and electrolytic cells. Voltaic cells use the energy given off in a spontaneous reaction to do electrical work. Electrolytic cells use

111

electrical work as source of energy to drive the reaction in the opposite direction.

The dotted vertical line in the center of the above figure represents a diaphragm that keeps the Cl2 gas produced at the anode from coming into contact with the sodium metal generated at the cathode. The function of this diaphragm can be understood by turning to a more realistic drawing of the commercial Downs cell used to electrolyze sodium chloride shown in the figure below.

Chlorine gas that forms on the graphite anode inserted into the bottom of this cell bubbles through the molten sodium chloride into a funnel at the top of the cell. Sodium metal that forms at the cathode floats up through the molten sodium chloride into a sodium-collecting ring, from which it is periodically drained. The diaphragm that separates the two electrodes is a screen of iron gauze, which prevents the explosive reaction that would occur if the products of the electrolysis reaction came in contact.

The feed-stock for the Downs cell is a 3:2 mixture by mass of CaCl2 and NaCl. This

mixture is used because it has a melting point of 580oC, whereas pure sodium chloride

has to be heated to more than 800oC before it melts.


112

The figure below shows an idealized drawing of a cell in which an aqueous solution of sodium chloride is electrolyzed.

Once again, the Na+ ions migrate toward the negative electrode and the Cl- ions migrate toward the positive electrode. But, now there are two substances that can be reduced at the cathode: Na+ ions and

water molecules.

Cathode (-): Na+ + e- Na Eo

red = -2.71 V 2 H2O + 2 e- H2 + 2 OH- Eo

red = -0.83 V

Because it is much easier to reduce water than Na+ ions, the only

product formed at the cathode is hydrogen gas.

Cathode (-): 2 H2O(l) + 2 e- H2(g) + 2 OH-(aq)

There are also two substances that can be oxidized at the anode: Cl-

ions and water molecules.

Anode (+): 2 Cl- Cl2 + 2 e- Eo

ox = -1.36 V 2 H2O O2 + 4 H+ + 4 e- Eo

ox = -1.23 V

The standard-state potentials for these half-reactions are so close to each other that we might expect to see a mixture of Cl2 and O2 gas collect at the anode. In practice, the only product of this reaction is Cl2.

Anode (+): 2 Cl- Cl2 + 2 e-

At first glance, it would seem easier to oxidize water (Eoox = -1.23

volts) than Cl- ions (Eoox = -1.36 volts). It is worth noting, however,

113

that the cell is never allowed to reach standard-state conditions. The solution is typically 25% NaCl by mass, which significantly decreases the potential required to oxidize the Cl- ion. The pH of the cell is also kept very high, which decreases the oxidation potential for water. The deciding factor is a phenomenon known as overvoltage, which is the extra voltage that must be applied to a reaction to get it to occur at the rate at which it would occur in an ideal system.

Under ideal conditions, a potential of 1.23 volts is large enough to oxidize water to O2 gas. Under real conditions, however, it can take a much larger voltage to initiate this reaction. (The overvoltage for the oxidation of water can be as large as 1 volt.) By carefully choosing the electrode to maximize the overvoltage for the oxidation of water and then carefully controlling the potential at which the cell operates, we can ensure that only chlorine is produced in this reaction.

In summary, electrolysis of aqueous solutions of sodium chloride doesn't give the same products as electrolysis of molten sodium chloride. Electrolysis of molten NaCl decomposes this compound into its elements.


Electrolysis of aqueous NaCl solutions gives a mixture of hydrogen and chlorine gas and an aqueous sodium hydroxide solution.

electrolysis 2 NaCl(aq) + 2 H2O(l)

2 Na+(aq) + 2 OH-(aq) + H2(g) + Cl2(g)

Because the demand for chlorine is much larger than the demand for sodium, electrolysis of aqueous sodium chloride is a more important process commercially. Electrolysis of an aqueous NaCl solution has two other advantages. It produces H2 gas at the cathode, which can be collected and sold. It also produces NaOH, which can be drained from the bottom of the electrolytic cell and sold.

114

The dotted vertical line in the above figure represents a diaphragm that prevents the Cl2 produced at the anode in this cell from coming into contact with the NaOH that accumulates at the cathode. When this diaphragm is removed from the cell, the products of the electrolysis of aqueous sodium chloride react to form sodium hypo-chlorite, which is the first step in the preparation of hypochlorite bleaches, such as Chlorox.

Cl2(g) + 2 OH-(aq) Cl-(aq) + OCl-(aq) + H2O(l)

Electrolysis of Water

A standard apparatus for the electrolysis of water is shown in the figure below.

electrolysis 2 H2O(l) 2 H2(g) + O2(g)

115

A pair of inert electrodes are sealed in opposite ends of a container designed to collect the H2 and O2 gas given off in this reaction. The electrodes are then connected to a battery or another source of electric current.

By itself, water is a very poor conductor of electricity. We therefore add an electrolyte

to water to provide ions that can flow through the solution, thereby completing the

electric circuit. The electrolyte must be soluble in water. It should also be relatively

inexpensive. Most importantly, it must contain ions that are harder to oxidize or

reduce than water.

2 H2O + 2 e- H2 + 2 OH- Eored = -0.83 V

2 H2O O2 + 4 H+ + 4 e- Eoox = -1.23 V

The following cations are harder to reduce than water: Li+, Rb+, K+, Cs+, Ba2+, Sr2+, Ca2+, Na+, and Mg2+. Two of these cations are more likely candidates than the others because they form inexpensive, soluble salts: Na+ and K+.

The SO42- ion might be the best anion to use because it is the most difficult anion to

oxidize. The potential for oxidation of this ion to the peroxydisulfate ion is -2.05

volts.

2 SO42- S2O8

2- + 2 e- Eoox = -2.05 V

When an aqueous solution of either Na2SO4 or K2SO4 is electrolyzed in the apparatus shown in the above figure, H2 gas collects at one electrode and O2 gas collects at the other.

What would happen if we added an indicator such as bromothymol blue to this

apparatus? Bromothymol blue turns yellow in acidic solutions (pH < 6) and blue in

basic solutions (pH > 7.6). According to the equations for the two half-reactions, the

indicator should turn yellow at the anode and blue at the cathode.

Cathode (-): 2 H2O + 2 e- H2 + 2 OH-

Anode (+): 2 H2O O2 + 4 H+ + 4 e-

Faraday's Law

Faraday's law of electrolysis can be stated as follows. The amount of a substance consumed or produced at one of the

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electrodes in an electrolytic cell is directly proportional to the amount of electricity that passes through the cell.

In order to use Faraday's law we need to recognize the relationship between current, time, and the amount of electric charge that flows through a circuit. By definition, one coulomb of charge is transferred when a 1-amp current flows for 1 second.

1 C = 1 amp-s

Example: To illustrate how Faraday's law can be used, let's calculate the number of grams of sodium metal that will form at the cathode when a 10.0-amp current is passed through molten sodium chloride for a period of 4.00 hours.

We start by calculating the amount of electric charge that flows through the cell.

Before we can use this information, we need a bridge between this macroscopic quantity and the phenomenon that occurs on the atomic scale. This bridge is represented by Faraday's constant, which describes the number of coulombs of charge carried by a mole of electrons.

Thus, the number of moles of electrons transferred when 144,000 coulombs of electric charge flow through the cell can be calculated as follows.

According to the balanced equation for the reaction that occurs at the cathode of this

cell, we get one mole of sodium for every mole of electrons.

Cathode (-): Na+ + e- Na

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Thus, we get 1.49 moles, or 34.3 grams, of sodium in 4.00 hours.

The consequences of this calculation are interesting. We would have to run this

electrolysis for more than two days to prepare a pound of sodium.





We can extend the general pattern outlined in this section to answer questions that

might seem impossible at first glance.


Determine the oxidation number of the chromium in an unknown salt if electrolysis of a molten sample of this salt for 1.50 hours with a 10.0-amp current deposits 9.71 grams of chromium metal at the cathode.



Practice Problem 13


Solution

We start by calculating the amount of electrical charge that passes through the solution.

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We then calculate the number of moles of electrons that carry this charge.

The balanced equation for the reaction that produces H2 gas at the cathode indicates that we get a mole of H2 gas for every two moles of electrons.

Cathode (-): 2 H2O + 2 e- H2 + 2 OH-

We therefore get one mole of H2 gas at the cathode for every two moles of electrons that flow through the cell.

We now have the information we need to calculate the volume of the gas produced in this reaction.

Practice Problem 14

Determine the oxidation number of the chromium in an unknown salt if electrolysis of a molten sample of this salt for 1.50 hours with a 10.0-amp current deposits 9.71 grams of chromium metal at the cathode.

Solution

We start, as before, by calculating the number of moles of electrons that passed through the cell during electrolysis.

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Since we don't know the balanced equation for the reaction at the cathode in this cell, it doesn't seem obvious how we are going to use this information. We therefore write down this result in a conspicuous location, and go back to the original statement of the question, to see what else can be done.

The problem tells us the mass of chromium deposited at the cathode. By now it should be obvious that we need to transform this information into the number of moles of chromium metal generated.

We now know the number of moles of chromium metal produced and the number of molus of electrons it took to produce this metal. We might therefore look at the relationship between the moles of electrons consumed in this reaction and the moles of chromium produced.

Three moles of electrons are consumed for every mole of chromium metal produced. The only way to explain this is to assume that the net reaction at the cathode involves reduction of Cr3+ ions to chromium metal.

Cathode (-): Cr3+ + 3 e- Cr

Thus, the oxidation number of chromium in the unknown salt must be +3.

Electrochemical Reactions

Electrical Work From Spontaneous

Oxidation-Reduction Reactions

Voltaic CellsStandard-State Cell

Potentials for Voltaic Cells

Predicting Spontaneous Redox Reactions from the

Sign of E

Standard-State Reduction Half-Cell

Potentials

Predicting Standard-State Cell

Potentials

Line Notation for Voltaic Cells

The Nernst Equation

Using the Nernst Equation to

Measure Equilibrium Constants

Table: Standard State Reduction Potentials

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Στη χημεία, η ένωση ενός στοιχείου με οξυγόνο ονομάζεται καύση ή οξείδωση (oxidation) της στοιχείου. Δοθέντος ότι στις ενώσεις του το οξυγόνο αφαιρεί δυο ηλεκτρόνια από το μοριακό υπόλοιπο για τη συμπλήρωση της στοιβάδας σθένους του, στις χημικές αντιδράσεις οξείδωση υποδηλώνει γενικότερα την αφαίρεση ηλεκτρονίων σθένους από μια χημική ουσία (Σχ. α). Λέμε τότε ότι η ουσία «οξειδώνεται». Αντίστοιχα, η ένωση ενός στοιχείου με υδρογόνο ονομάζεται αναγωγή (reduction) του στοιχείου και επειδή το υδρογόνο παραχωρεί το ηλεκτρόνιο σθένους του, αναγωγή υποδηλώνει γενικότερα την παραχώρηση ηλεκτρονίων σθένους σε μια ουσία, η οποία «ανάγεται».

Οι έννοιες οξείδωση/αναγωγή υποδηλώνουν γενικά την απώλεια/πρόκτηση ηλεκτρονίων στις χημικές αντιδράσεις ενός στοιχείου ή μιας ουσίας. Συναφώς με αυτούς χρησιμοποιούνται και οι όροι αναγωγικό/οξειδωτικό μέσο για να δηλώσουν τη δράση ενός

Σχ. Χρήση των όρων οξείδωση/αναγωγή όταν αναφέρονται α) στην αλλαγή

μιας ουσίας και β) στη δράση μιας ουσίας σε άλλες. στοιχείου ή μιας ουσίας σε άλλες (Σχ. β). Π.χ. το υδρογόνο και τα μέταλλα σε όλες τις αντιδράσεις τους χάνουν ηλεκτρόνια και μετατρέπονται σε θετικά ιόντα, δηλαδή οξειδώνονται. Αναφορικά με τις άλλες ουσίες δίνουν ηλεκτρόνια και αποτελούν αναγωγικά μέσα.

Π.χ. στη δημιουργία νερού (H2O) με καύση υδρογόνου, το υδρογόνο οξειδώνεται και το οξυγόνο ανάγεται. Στα ηλεκτρολυτικά διαλύματα, το νερό συμπεριφέρεται σαν ασθενής ηλεκτρολύτης, ένα μέρος των μορίων του διίστανται ηλεκτρολυτικά σε ιόντα, H2O

<==> 2Η+ + O . Αν τα κατιόντα Η+ μετατρέπονται στο ηλεκτρόδιο της καθόδου σε αέριο H2 αφαιρώντας από αυτό ηλεκτρόνια, η

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αντίδραση αυτή του υδρογόνου είναι αναγωγή. Ενώ, δηλαδή, κατά την καύση του το υδρογόνο οξειδώνεται, στην ηλεκτρόλυση τα κατιόντα Η+ ανάγονται σε H2.

Με την παραχώρηση ηλεκτρονίων, η κάθοδος λειτουργεί γενικά σαν μέσο αναγωγής ηλεκτροθετικών ιόντων και ταυτόχρονα, με την αφαίρεση ηλεκτρονίων, η άνοδος λειτουργεί σαν μέσο οξείδωσης ηλεκτραρνητικών ιόντων. Συμπεραίνουμε ότι η ροή ηλεκτρικού ρεύματος στα ηλεκτρολυτικά υγρά συνδέεται πάντα με παράλληλες αντιδράσεις αναγωγής (παραχώρησης ηλεκτρονίων) από την κάθοδο και οξείδωσης (απόθεσης ηλεκτρονίων) στην άνοδο (redox reactions).

Electrical Work From Spontaneous Oxidation-Reduction Reactions

The following rule can be used to predict whether an oxidation-reduction reaction should occur. Oxidation-reduction reactions should occur when they convert the stronger of a pair of oxidizing agents and the stronger of a pair of reducing agents into a weaker oxidizing agent and a weaker reducing agent.

Practice Problem 1:

Predict whether zinc metal should dissolve in acid.


Practice Problem 1

Predict whether zinc metal should dissolve in acid.

Answer

Zinc will dissolve in acid if the metal is a strong enough reducing agent to reduce H+ ions to H2 gas.

Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g)

Zinc is a stronger reducing agent than H2, and the H+ ion is a stronger oxidizing agent than the Zn2+ ion.

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Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g) strongerreducingagent

strongeroxidizing agent

weakeroxidizingagent

weakerreducingagent

Zinc therefore should dissolve in acid.

We can test this prediction by adding a few chunks of mossy zinc to a beaker of concentrated hydrochloric acid. Within a few minutes, the zinc metal dissolves, and significant amounts of hydrogen gas are liberated.

The reaction in Practice Problem 1 has some of the characteristic features of oxidation-reduction reactions.

It is exothermic, in this case giving off 153.89 kilojoules per mole of zinc consumed.

The equilibrium constant for the reaction is very large (Kc = 6 x 1025), and chemists often write the equation for this reaction as if essentially all of the reactants were converted to products.

Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g)

It can be formally divided into separate oxidation

and reduction half-reactions.

Oxidation: Zn Zn2+ + 2 e-

Reduction: 2 H+ + 2 e- H2

By separating the two half-reactions, the energy given off by this reaction can be used to do work.

According to the first law of thermodynamics, the energy given off in a chemical reaction can be converted into heat, work, or a mixture of heat and work. By running the half-reactions in separate containers, we can force the electrons to flow from the oxidation to the reduction half-reaction through an external wire, which allows us to capture as much as possible of the energy given off in the reaction as electrical work.

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We can start by immersing a strip of zinc metal into a 1 M Zn2+ ion solution, as shown in the figure below. We then immerse a piece of platinum wire in a second beaker filled with 1 M HCl and bubble H2 gas over the Pt wire. Finally, we connect the zinc metal and platinum wire to form an electric circuit.

We've now made a system in which electrons can flow from one half-reaction, or half-cell, to another. The same driving force that makes zinc metal react with acid when the two are in contact should operate in this system. Zinc atoms on the metal surface lose electrons to form Zn2+ ions, which go into solution.


The electrons given off in this half-reaction flow through the circuit and eventually accumulate on the platinum wire to give this wire a net negative charge. The H+ ions from the hydrochloric acid are attracted to this negative charge and migrate toward the platinum wire. When the H+ ions touch the platinum wire, they pick up electrons to form hydrogen atoms, which immediately combine to form H2 molecules.


The oxidation of zinc metal releases Zn2+ ions into the Zn/Zn2+ half-cell. This half-cell therefore picks up a positive charge that interferes with the transfer of more electrons. The reduction of H+ ions in the H2/H+ half-cell leads to a net negative charge as these H+ ions are

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removed from the solution. This negative charge also interferes with the transfer of more electrons.

To overcome this problem, we complete the circuit by adding a U-tube filled with a saturated solution of a soluble salt such as KCl. Negatively charged Cl- ions flow out of one end of the U-tube to balance the positive charge on the Zn2+ ions created in one half-cell. Positively charged K+ ions flow out of the other end of the tube to replace the H+ ions consumed in the other half cell. The U-tube is called a salt bridge, because it contains a solution of a salt that literally serves as a bridge to complete the electric circuit.

Voltaic Cells

Electrochemical cells that use an oxidation-reduction reaction to generate an electric current are known as galvanic or voltaic cells. Because the potential of these cells to do work by driving an electric current through a wire is measured in units of volts, we will refer to the cells that generate this potential from now on as voltaic cells.

Let's take another look at the voltaic cell in the figure below.

Within each half-cell, reaction occurs on the surface of the metal electrode. At the zinc electrode, zinc atoms are oxidized to form Zn2+ ions, which go into solution. The electrons liberated in this reaction flow through the zinc metal until they reach the wire that connects the zinc electrode to the platinum wire. They then flow through the platinum wire, where they eventually reduce an H+ ion in the neighboring solution to a hydrogen atom, which combines with another hydrogen atom to form an H2 molecule.

The electrode at which oxidation takes place in a electrochemical cell is called the anode. The electrode at which reduction occurs is called the cathode. The identity of the cathode and anode can be remembered by recognizing that positive ions, or cations, flow toward the cathode, while negative ions, or anions, flow toward the anode. In the voltaic cell shown above, H+ ions flow toward the

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cathode, where they are reduced to H2 gas. On the other side of the cell, Cl- ions are released from the salt bridge and flow toward the anode, where the zinc metal is oxidized.

Standard-State Cell Potentials for Voltaic Cells

The cell potential for a voltaic cell is literally the potential of the cell to do work on its surroundings by driving an electric current through a wire. By definition, one joule of energy is produced when one coulomb of electrical charge is transported across a potential of one volt.

The potential of a voltaic cell depends on the concentrations of any species present in solution, the partial pressures of any gases involved in the reaction, and the temperature at which the reaction is run. To provide a basis for comparing the results of one experiment with another, the following set of standard-state conditions for electrochemical measurements has been defined.

All solutions are 1 M. All gases have a partial pressure of 0.1

MPa (0.9869 atm).

Although standard-state measurements can be made at any temperature, they are often taken at 25oC.

Cell potentials measured under standard-state conditions are represented by the symbol Eo. The standard-state cell potential, E o , measures the strength of the driving force behind the chemical reaction. The larger the difference between the oxidizing and reducing strengths of the reactants and products, the larger the cell potential. To obtain a relatively large cell potential, we have to react a strong reducing agent with a strong oxidizing agent.

Example: The experimental value for the standard-state cell potential for the reaction between zinc metal and acid is 0.76 volts.

Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g) Eo = 0.76 V

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The cell potential for this reaction measures the relative reducing power of zinc metal compared with hydrogen gas. But it doesn't tell us anything about the absolute value of the reducing power for either zinc metal or H2.

We therefore arbitrarily define the standard-state potential for the reduction of H+ ions to H2 gas as exactly zero volts.

2 H+ + 2 e- H2 Eo = 0.000... V

We will then use this reference point to calibrate the potential of any other half-reaction.

The key to using this reference point is recognizing that the overall cell potential for a reaction must be the sum of the potentials for the oxidation and reduction half-reactions.

Eooverall = Eo

ox + Eored

If the overall potential for the reaction between zinc and acid is 0.76 volts, and the half-cell potential for the reduction of H+ ions is 0 volts,

then the half-cell potential for the oxidation of zinc metal must be 0.76 volts.

Zn Zn2+ + 2 e- Eoox = 0.76 V

+ 2 H+ + 2 e- H2 Eored = 0.00 V

Zn + 2 H+ Zn2+ + H2 Eo = Eoox + Eo

red = 0.76 V

Predicting Spontaneous Redox Reactions From the Sign of Eo

The magnitude of the cell potential is a measure of the driving force behind a reaction. The larger the value of the cell potential, the further the reaction is from equilibrium. The sign of the cell potential tells us the direction in which the reaction must shift to reach equilibrium.

Consider the reaction between zinc and acid, for example.


The fact that E o is positive tells us that when this system is present at standard-state conditions, it has to shift to the right to reach

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equilibrium. Reactions for which E o is positive therefore have equilibrium constants that favor the products of the reaction. It is tempting to describe these reactions as "spontaneous."

What happens to the cell potential when we reverse the direction in which a reaction is written? Turning the reaction around doesn't change the relative strengths of the oxidizing or reducing agents. The magnitude of the potential must remain the same . But turning the equation around changes the sign of the cell potential , and can therefore turn an unfavorable reaction into one that is spontaneous, or vice versa.

Practice Problem 2:

Use the overall cell potentials to predict which of the following reactions are spontaneous.

(a) Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) Eo = 0.46 V

(b) 2 Fe3+(aq) + 2 Cl-(aq) 2 Fe2+(aq) + Cl2(g) Eo = -0.59 V

(c) 2 Fe3+(aq) + 2 I-(aq) 2 Fe2+(aq) + I2(aq) Eo = 0.24 V

(d) 2 H2O2(aq) 2 H2O(l) + O2(aq) Eo = 1.09 V

(e) Cu(s) + 2 H+(aq) Cu2+(aq) + H2(g) Eo = -0.34 V


Practice Problem 3:

Use the standard-state cell potential for the following reaction

Cu(s) + 2 H+(aq) Cu2+(aq) + H2(g) Eo = -0.34 V

to predict the standard-state cell potential for the opposite reaction.

Cu2+(aq) + H2(g) Cu(s) + 2 H+(aq) Eo = ?


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Standard-State Reduction Half-Cell Potentials

The standard-state cell potentials for some common half-reactions are given in the table below.

Standard-State Reduction Potentials, Eored

Half-Reaction Eored

K+ + e- K (best reducing agent) -2.924 BestBa2+ + 2 e- Ba -2.90 reducingCa2+ + 2 e- Ca -2.76 agentsNa+ + e- Na -2.7109Mg2+ + 2 e- Mg -2.375H2 + 2 e- 2 H- -2.23Al3+ + 3 e- Al -1.706Mn2+ + 2 e- Mn -1.04Zn2+ + 2 e- Zn -0.7628Cr3+ + 3 e- Cr -0.74S + 2 e- S2- -0.5082 CO2 + 2 H+ + 2 e- H2C2O4 -0.49Cr3+ + e- Cr2+ -0.41Fe2+ + 2 e- Fe -0.409Co2+ + 2 e- Co -0.28Ni2+ + 2 e- Ni -0.23Sn2+ + 2 e- Sn -0.1364Pb2+ + 2 e- Pb -0.1263Fe3+ + 3 e- Fe -0.0362 H+ + 2 e- H2 0.0000...S4O6

2- + 2 e- 2 S2O32- 0.0895

Oxidizing Sn4+ + 2 e- Sn2+ 0.15

power Cu2+ + e- Cu+ 0.158 Reducingincreases Cu2+ + 2 e- Cu 0.3402 power

O2 + 2 H2O + 4 e- 4 OH- 0.401 increases

Cu+ + e- Cu 0.522I3- + 2 e- 3 I- 0.5338 MnO4

- + 2 H2O + 3 e- MnO2 + 4 OH- 0.588

O2 + 2 H+ + 2 e- H2O2 0.682Fe3+ + e- Fe2+ 0.770

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Hg22+ + 2 e- Hg 0.7961

Ag+ + e- Ag 0.7996Hg2+ + 2 e- Hg 0.851H2O2 + 2 e- 2 OH- 0.88HNO3 + 3 H+ + 3 e- NO + 2 H2O 0.96Br2(aq) + 2 e- 2 Br- 1.0872 IO3

- + 12 H+ + 10 e- I2 + 6 H2O 1.19CrO4

2- + 8 H+ + 3 e- Cr3+ + 4 H2O 1.195Pt2+ + 2 e- Pt 1.2MnO2 + 4 H+ + 2 e- Mn2+ + 2 H2O 1.208O2 + 4 H+ + 4 e- 2 H2O 1.229Cr2O7

2- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O

1.33

Cl2(g) + 2 e- 2 Cl- 1.3583PbO2 + 4 H+ + 2 e- Pb2+ + 2 H2O 1.467MnO4

- + 8 H+ + 5 e- Mn2+ + 4 H2O 1.491Au+ + e- Au 1.68H2O2 + 2 H+ + 2 e- 2 H2O 1.776Co3+ + e- Co2+ 1.842

Best S2O82- + 2 e- 2 SO4

2- 2.05oxidizing O3(g) + 2 H+ + 2 e- O2(g) + H2O 2.07agents F2(g) + 2 H+ + 2 e- 2 HF(aq) 3.03

There is no need to remember that reducing agents become stronger toward the upper right corner of this table, or that the strength of the oxidizing agents increases toward the bottom left corner. All you have to do is remember some of the chemistry of the elements at the top and bottom of this table.

Take a look at the half-reaction at the top of the table.

K+ + e- K Eored = -2.924 V

What do we know about potassium metal? Potassium is one of the most reactive metals it bursts into flame when added to water, for example. Furthermore, we know that metals are reducing agents in all of their chemical reactions. When we find potassium in this table, we can therefore conclude that it is listed among the strongest reducing agents.

Conversely, look at the last reaction in the table.

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F2 + 2 e- 2 F- Eored = 3.03 V

Fluorine is the most electronegative element in the periodic table. It shouldn't be surprising to find that F2 is the strongest oxidizing agent in the above table.

Referring to either end of this table can also help you remember the sign convention for cell potentials. The previous section introduced the following rule: Oxidation-reduction reactions that have a positive overall cell potential are spontaneous. This is consistent

with the data in the above table. We know that fluorine wants to gain electrons to

form fluoride ions, and the half-cell potential for this reaction is positive.

F2 + 2 e- 2 F- Eored = 3.03 V

We also know that potassium is an excellent reducing agent. Thus, the potential for the reduction of K+ ions to potassium metal is negative

K+ + e- K Eored = -2.924 V

but the potential for the oxidation of potassium metal to K+ ions is

positive.

K K+ + e- Eoox = -(-2.924 V) = 2.924 V

Practice Problem 4:

Which of the following is the strongest oxidizing agent?

(a) H2O2 in acid

(b) H2O2 in base

(c) MnO4- in acid

(d) MnO4- in base

(e) CrO42- in acid


Practice Problem 4

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Which of the following is the strongest oxidizing agent?

(a) H2O2 in acid

(b) H2O2 in base

(c) MnO4- in acid

(d) MnO4- in base

(e) CrO42- in acid

Answer

We start by searching the table of standard-state reduction potentials for the reduction half-cell potential for the oxidizing agent in each of these solutions.

(a) H2O2 + 2 H+ + 2 e- 2 H2O Eored = 1.776 V

(b) H2O2 + 2 e- 2 OH- Eored = 0.88 V

(c) MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O Eo

red = 1.491 V

(d) MnO4- + 2 H2O + 3 e- MnO2 + 4 OH- Eo

red = 0.588 V

(e) CrO42- + 8 H+ + 3 e- Cr3+ + 4 H2O Eo

red = 1.195

According to these data, the strongest oxidizing agent is H2O2 in acid.

Predicting Standard-State Cell Potentials

A voltaic cell stable enough to be used as a battery is called a Daniell cell. For our purposes, we will work with the idealized Daniell cell in the figure below.

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We can use the known values of the standard-state reduction potentials for the Cu/Cu2+ and Zn/Zn2+ half-cells to predict the overall potential for the Daniell cell and to determine which electrode is the anode and which is the cathode.

We start by writing a balanced chemical equation for the reaction that occurs in this cell. The table of standard-state reduction potentials suggests that zinc is a better reducing agent than copper and that the Cu2+ ion is a better oxidizing agent than the Zn2+ ion. The overall reaction therefore involves the reduction of Cu2+ ions by

zinc metal.

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)stronger

reducingagent

strongeroxidizingagent

weaker oxidizingagent

weakerreducingagent

We then divide the reaction into separate oxidation and reduction half-reactions.

Reduction: Cu2+ + 2 e- CuOxidation: Zn Zn2+ + 2 e-

The potential for the reduction of Cu2+ ions to copper metal can be found in the table of standard-state reduction potentials. To find the potential for the oxidation of zinc metal, we have to reverse the sign on the potential for the Zn/Zn2+ couple in this table.

Reduction: Cu2+ + 2 e- Cu Eored = 0.34 V

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Oxidation: Zn Zn2+ + 2 e- Eoox = -(-0.76 V) = 0.76 V

The overall potential for this cell is the sum of the potentials for the two half-cells.

Cu2+ + 2 e- Cu Eored = 0.34 V

+ Zn Zn2+ + 2 e- Eoox = 0.76 V

Zn + Cu2+ Zn2+ + Cu Eo = Eored + Eo

ox = 1.10 V

Oxidation always occurs at the anode and reduction always occurs at the cathode of an electrochemical cell. The Zn/Zn2+ half-cell is therefore the anode, and the Cu2+/Cu half-cell is the cathode, as shown in the

figure of the Daniell cell.

Practice Problem 5:

Use cell potential data to explain why copper metal does not dissolve in a typical strong acid, such as hydrochloric acid,

Cu(s) + 2 H+(aq)

but will dissolve in 1 M nitric acid.

3 Cu(s) + 2 HNO3(aq) + 6 H+(aq) 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l)


Practice Problem 5

Use cell potential data to explain why copper metal does not dissolve in a typical strong acid, such as hydrochloric acid,

Cu(s) + 2 H+(aq)

but will dissolve in 1 M nitric acid.

3 Cu(s) + 2 HNO3(aq) + 6 H+(aq) 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l)

Answer

Copper does not dissolve in a typical strong acid because the overall cell potential for the oxidation of copper metal to Cu2+ ions coupled with the reduction of H+ ions to H2 is negative.

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Cu Cu2+ + 2 e- Eoox = -(0.34 V)

+ 2 H+ + 2 e- H2 Eored = 0.000... V

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Cu + 2 H+ Cu2+ + H2 Eo = Eoox + Eo

red = -0.34 V

Copper dissolves in nitric acid because the reaction at the cathode now involves the reduction of nitric acid to NO gas, and the potential for this half-reaction is strong enough to overcome the half-cell potential for oxidation of copper metal to Cu2+ ions.

3 (Cu Cu2+ + 2 e-) Eoox = -(0.34 V)

+ 2(HNO3 + 3 H+ + 3 e- NO + 2 H2O) Eored = 0.96 V

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

3 Cu + 2 HNO3 + 6 H+ 3 Cu2+ + 2 NO + 4 H2O Eo = Eoox + Eo

red = 0.62 V

The units of half-cell potentials are volts, not volts per mole or volts per electron. All

we do when combining half-reactions is add the two half-cell potentials. We do not

multiply these potentials by the integers used to balance the number of electrons

transferred in the reaction.

Practice Problem 6:

Use cell potentials to explain why hydrogen peroxide disproportionates to form oxygen and water.

2 H2O2(aq) 2 H2O(l) + O2(g)


Practice Problem 6

Use cell potentials to explain why hydrogen peroxide disproportionates to form oxygen and water.

2 H2O2(aq) 2 H2O(l) + O2(g)

Answer

Because the oxygen atoms in hydrogen peroxide are in the -1 oxidation state, H2O2 can be either oxidized to O2 or reduced to water.

H2O2 + 2 H+ + 2 e- 2 H2O Eored = 1.776 V

H2O2 O2 + 2 H+ + 2 e- +Eoox = -(0.682 V)

135


_________________________________________

2 H2O2 2 H2O + O2 Eo = Eored + Eo

ox = 1.094 V

The overall cell potential for the disproportionation of H2O2 is positive, which means this should be a spontaneous oxidation-reduction reaction.

Line Notation For Voltaic Cells

Voltaic cells can be described by a line notation based on the following conventions.

A single vertical line indicates a change in state or phase.

Within a half-cell, the reactants are listed before the products.

Concentrations of aqueous solutions are written in parentheses after the symbol for the ion or molecule.

A double vertical line is used to indicate the junction between the half-cells.

The line notation for the anode (oxidation) is written before the line notation for the cathode (reduction).

The line notation for a standard-state Daniell cell is written as follows.

Zn | Zn2+(1.0 M) || Cu2+(1.0 M) | Cuanode

(oxidation)cathode

(reduction)

Electrons flow from the anode to the cathode in a voltaic cell. (They flow from the

electrode at which they are given off to the electrode at which they are consumed.)

Reading from left to right, this line notation therefore corresponds to the direction in

which electrons flow.

Practice Problem 7:

Write the line notation for the cell shown in the figure below.

136


Practice Problem 7

Write the line notation for the cell shown in the figure below.

Answer

The cell is based on the following half-reactions.


137



Since we list the reactants before the products in a half-cell, we write the following line notation for the anode.

anode: Zn | Zn2+(1.0 M)

The line notation for the cathode has to indicate that H+ ions are reduced to H2 gas on a platinum metal surface. This can be done as follows.

cathode: H+(1.0 M) | H2(1 atm) | Pt

Since line notation is read in the direction in which electrons flow, we place the notation for the anode before the cathode, as follows.

Zn | Zn2+(1.0 M) || H+(1.0 M) | H2(1 atm) | Pt

anode cathode

The Nernst Equation

What happens when the Daniell cell is used to do work?

The zinc electrode becomes lighter as zinc atoms are oxidized to Zn2+ ions, which go into solution.

The copper electrode becomes heavier as Cu2+ ions in the solution are reduced to copper metal.

The concentration of Zn2+ ions at the anode increases and the concentration of the Cu2+ ions at the cathode decreases.

Negative ions flow from the salt bridge toward the anode to balance the charge on the Zn2+ ions produced at this electrode.

Positive ions flow from the salt bridge toward the cathode to compensate for the Cu2+ ions consumed in the reaction.

138

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/electro.php#daniell%23daniell

An important property of the cell is missing from this list. Over a period of time, the cell runs down, and eventually has to be replaced. Let's assume that our cell is initially a standard-state cell in which the concentrations of the Zn2+ and Cu2+ ions are both 1 molar.

Zn | Zn2+(1.0 M) || Cu2+(1.0 M) | Cu

As the reaction goes forward as zinc metal is consumed and copper metal is produced the driving force behind the reaction must become weaker. Therefore, the cell potential must become smaller.

This raises an interesting question: When does the cell potential become zero?

The cell potential is zero if and only if the reaction is at equilibrium.

When the reaction is at equilibrium, there is no net change in the amount of zinc metal or copper ions in the system, so no electrons flow from the anode to the cathode. If there is no longer a net flow of electrons, the cell can no longer do electrical work. Its potential for doing work must therefore be zero.

In 1889 Hermann Walther Nernst showed that the potential for an electrochemical reaction is described by the following equation.

In the Nernst equation, E is the cell potential at some moment in time, Eo is the cell potential when the reaction is at standard-state conditions, R is the ideal gas constant in units of joules per mole, T is the temperature in kelvin, n is the number of moles of electrons transferred in the balanced equation for the reaction, F is the charge on a mole of electrons, and Qc is the reaction quotient at that moment in time. The symbol ln indicates a natural logarithm, which is the log to the base e, where e is an irrational number equal to 2.71828...

Three terms in the Nernst equation are constants: R, T, and F. The ideal gas constant is 8.314 J/mol-K. The temperature is usually 25oC.

139

The charge on a mole of electrons can be calculated from Avogadro's number and the charge on a single electron.

Substituting this information into the Nernst equation gives the following equation.

Three of the remaining terms in this equation are characteristics of a particular reaction: n, Eo, and Qc.

Example: The standard-state potential for the Daniell cell is 1.10 V. Two moles of electrons are transferred from zinc metal to Cu2+ ions in the balanced equation for this reaction, so n is 2 for this cell. Because we never include the concentrations of solids in either reaction quotient or equilibrium constant expressions, Qc for this reaction is equal to the concentration of the Zn2+ ion divided by the concentration of the Cu2+ ion.

Substituting what we know about the Daniell cell into the Nernst equation gives the following result, which represents the cell potential for the Daniell cell at 25oC at any moment in time.

The figure below shows a plot of the potential for the Daniell cell as a function of the natural logarithm of the reaction quotient.

140

When the reaction quotient is very small, the cell potential is positive and relatively

large. This isn't surprising, because the reaction is far from equilibrium and the

driving force behind the reaction should be relatively large. When the reaction

quotient is very large, the cell potential is negative. This means that the reaction

would have to shift back toward the reactants to reach equilibrium.

Practice Problem 8:

Calculate the potential in the following cell when 99.99% of the Cu2+ ions have been consumed.

Zn | Zn2+(1.00 M) || Cu2+(1.00 M) | Cu



Practice Problem 8 above raises an important point. The cell potential depends on the logarithm of the ratio of the concentrations of the products and the reactants. As a result, the potential of a cell or battery is more or less constant until virtually all of the reactants have been converted into products.

The Nernst equation can be used to calculate the potential of a cell that operates at

non-standard-state conditions.

Practice Problem 9:

Calculate the potential at 25oC for the following cell.

141

http://chemed.chem.purdue.edu/~genchem/topicreview/bp/ch20/problems/ex20_9s.html


Cu | Cu2+(0.024 M) || Ag+(0.0048 M) | Ag



Practice Problem 9

Calculate the potential at 25oC for the following cell.

Cu | Cu2+(0.024 M) || Ag+(0.0048 M) | Ag

Solution

We start by translating the line notation into an equation for the reaction.

Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)

We then look up the standard-state potentials for the two half-reactions and calculate the standard-state cell potential.

Oxidation: Cu Cu2+ + 2 e- Eoox = -(0.3402 V)

Reduction: Ag+ + e- Ag Eored = 0.7996 V

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Eo = Eored + Eo

ox = 0.4594 V

We now set up the Nernst equation for this cell, noting that n is 2 because two electrons are transferred in the balanced equation for the reaction.

142



Substituting the concentrations of the Cu2+ and Ag+ ions into this equation gives the value for the cell potential.

The Nernst equation can help us understand a popular demonstration that uses glucose, C6H12O6, to reduce Ag+ ions to silver metal, thereby forming a silver mirror on the inner surface of the container. The standard-state half-cell potential for the reduction of Ag+ ions is approximately 0.800 volt.

Ag+ + e- Ag Eored = 0.800 V

The standard-state half-cell potential for the oxidation of glucose is -0.050 V.

C6H12O6 + H2O C6H12O7 + 2 H+ + 2 e- Eoox = -0.050 V

The overall standard-state cell potential for this reaction is therefore favorable.

2 (Ag+ + e- Ag) Eored = 0.800 V

+ C6H12O6 + H2O C6H12O7 + 2 H+ + 2 e- Eoox = -0.050 V

2 Ag+ + C6H12O6 + H2O 2 Ag + C6H12O7 + 2 H+ Eo = 0.750 V

The reaction isn't run under standard-state conditions, however. It takes place in a solution to which aqueous ammonia has been added. Most of the silver is therefore present as the Ag(NH3)2

+ complex ion. This is important, because the half-cell potential for the reduction of this complex is considerably smaller than the potential for the reduction of the Ag+ ion.

Ag(NH3)2+ + e- Ag + 2 NH3 Eo

red = 0.373 V

This leads to a significant decrease in the overall cell potential for the reaction because the Ag(NH3)2

+ ion is a much weaker oxidizing agent than the Ag+ ion.

143

The fact that this reaction is run in an aqueous ammonia solution also has an effect on the potential for the oxidation of glucose, because this half-reaction contains a pair of H+ ions.

C6H12O6 + H2O C6H12O7 + 2 H+ + 2 e-

The half-cell potential for this reaction therefore depends on the pH of the solution. Because two H+ ions are given off when glucose is oxidized, the reaction quotient for this reaction depends on the square of the H+ ion concentration. A change in this solution from standard-state conditions (pH = 0) to the pH of an aqueous ammonia solution (pH 11) therefore results in an increase of 0.650 volts in the half-cell potential for this reaction.

The increase in the reducing strength of glucose when the reaction is run at pH 11 more than compensates for the decrease in oxidizing strength that results from the formation of the Ag(NH3)2

+ complex ion.

Thus, the overall cell potential for the reduction of silver ions to silver metal is

actually more favorable in aqueous ammonia than under standard-state conditions.

2 (Ag(NH3)2+ + e- Ag + 2 NH3)

Eored = 0.373

V+ C6H12O6 + H2O C6H12O7 + 2 H+ + 2 e- Eo

ox = 0.600 V

2 Ag(NH3)2+ + C6H12O6 + H2O 2 Ag + C6H12O7 + 2

NH4+ Eo = 0.973 V

Using the Nernst Equation to Measure Equilibrium Constants

The Nernst equation can be used to measure the equilibrium constant for a reaction.

To understand how this is done, we have to recognize what happens to the cell

potential as an oxidation-reduction reaction comes to equilibrium. As the reaction

approaches equilibrium, the driving force behind the reaction decreases, and the cell

potential approaches zero.

At equilibrium: E = 0

What implications does this have for the Nernst equation?

144

At equilibrium, the reaction quotient is equal to the equilibrium constant (Qc = Kc) and the overall cell potential for the reaction is zero (E = 0).

At equilibrium:

Rearranging this equation gives the following result.

At equilibrium:

According to this equation, we can calculate the equilibrium constant for any oxidation-reduction reaction from its standard-state cell potential:

or, in a more useful arrangement:


Calculate the equilibrium constant at 25oC for the reaction between zinc metal and acid.

Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g)



Practice Problem 10

Calculate the equilibrium constant at 25oC for the reaction between zinc metal and acid.

Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g)

Solution

145



We start by calculating the overall standard-state cell potential for this reaction.

Zn Zn2+ + 2 e- Eoox = -(-0.7628 V)

+ 2 H+ + 2 e- H2 Eored = 0.0000 V

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯


We then substitute into the Nernst equation the implications of the fact that the reaction is at equilibrium: Qc = Kc and E = 0.

We then rearrange this equation

and solve for the natural logarithm of the equilibrium constant.

Substituting what we know about the reaction into this equation gives the following result.

This is a very large equilibrium constant, which means that equilibrium lies heavily on the side of the products. The equilibrium constant is so large that the equation for this reaction is written as if it proceeds to completion.

Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g)

This technique used in Practice Problem 10 can even be used to calculate equilibrium

constants for reactions that don't seem to involve oxidation-reduction.


Use the following standard-state cell potentials to calculate the complex formation equilibrium constant for the Zn(NH3)4

2+

146

complex ion.

Zn(NH3)42+ + 2 e- Zn + 4 NH3 Eo

red = -1.04 VZn2+ + 2 e- Zn Eo

red = -0.7628 V




Use the following standard-state cell potentials to calculate the solubility product at 25oC for Mg(OH)2.

Mg(OH)2 + 2 e- Mg + 2 OH- Eored = -2.69 V

Mg2+ + 2 e- Mg Eored = -2.375 V



Practice Problem 11

Use the following standard-state cell potentials to calculate the complex formation equilibrium constant for the Zn(NH3)4

2+ complex ion.

Zn(NH3)42+ + 2 e- Zn + 4 NH3 Eo

red = -1.04 V

Zn2+ + 2 e- Zn Eored = -0.7628 V

Solution

By reversing the first half-reaction and adding it to the second, we can obtain an overall equation that corresponds to the complex formation equilibrium.

Zn + 4 NH3 Zn(NH3)42+ + 2 e- Eo

ox = 1.04 V

+ Zn2+ + 2 e- Zn Eored = -0.7628 V

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Zn2+ + 4 NH3 Zn(NH3)42+ Eo = 0.28 V

When this reaction comes to equilibrium, the cell potential is zero and the reaction quotient is equal to the complex formation equilibrium constant for the Zn(NH3)4

2+ complex ion.

147





We now solve this equation for the natural logarithm of the equilibrium constant and calculate the value of Kf for this reaction.

Practice Problem 12

Use the following standard-state cell potentials to calculate the solubility product at 25oC for Mg(OH)2.


Mg2+ + 2 e- Mg Eored = -2.375 V

Solution

This time we have to reverse the second half-reaction to obtain an overall equation that corresponds to the appropriate equilibrium expression.


+ Mg Mg2+ + 2 e- Eoox = 2.375 V

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯

Mg(OH)2 Mg2+ + 2 OH- Eo = -0.32 V

Once again, we start with the Nernst equation and assume that the reaction is at equilibrium.

We then rearrange this equation, substitute what we know about the reaction into this equation, and calculate the value of Ksp for this compound.

148

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