electrochemistry

Electrochemistry• Importance of Electrochemistry

– starting your car– use a calculator– listen to a radio– corrosion of iron– preparation of important industrial

materials

• Electrochemistry - the study of the interchange of chemical and electrical energy

Galvanic Cells

• Redox reactions– electron transfer reactions– oxidation - loss of electrons, i.e., an

increase in oxidation number– reduction - gain electrons, i.e., a

decrease in oxidation number

Galvanic Cells

• Ex: 8 H+ + MnO4- + 5 Fe+2 --> Mn+2 + 5 Fe+3 + 4

H2O

• If MnO4- and Fe+2 are present in the same solution,

the electrons are transferred directly when the reactants collide.– No useful work is done.– Chemical energy is released as heat.

• How can the energy be harnessed?– Separate the half reactions.– Have the electron transfer go through a wire, which

can then be directed through a motor or other useful device

Galvanic Cells

• Galvanic cell– Uses a spontaneous redox

reaction to produce current to do work

– A device in which chemical energy is changed to electrical energy

– Also known as a voltaic cell

Galvanic Cells

• Requirements for a Galvanic Cell– Half reactions are in separate

compartments– Compartments are connected by a

salt bridge• a porous disk connection• a salt bridge (U bridge) containing a

strong electrolyte held in a jello-like matrix

Galvanic Cells• Why a salt bridge?

– Without a salt bridge, the reaction will not go

– Current starts, but stops due to charge buildup in both compartments• one side becomes negatively charged as

electrons are added• one side becomes positively charged as

electrons are lost• Creating a charge separation requires a

large amount of energy

Galvanic Cells

• Reaction in a galvanic cell occurs at each of the electrodes

• Be able to label the anode, cathode, direction of electron flow, and direction of ion flow

Galvanic Cells

• Cell Potential– The oxidizing agent “pulls” electrons

towards itself from the reducing agent– The “pull” or driving force on the

electrons is the cell potential, Ecell, aka, the electromotive force (emf) of the cell.

– Measured in volts, V– 1 V = 1 Joule/Coulomb

Galvanic Cell

• Cell Potential– Measure with a voltmeter

Standard Reduction Potentials

• Predict cell potentials if the half cell potentials are known– add half cell potentials to get cell

potential

– Eox + Ered = Ecell

– There is no way to measure half cell potentials


• So how do we get these half cell potentials if they can’t be measured directly?

• Measure a cell potential, assigning one reaction to be the “zero”– 2H+ + 2 e-

--> H2 Ered = 0.00 V


• So for 2H+ + Zn --> H2 + Zn+2 the voltage is found to be 0.76 V

• If 2 H+ + 2 e- --> H2 Ered = 0.00 V

and Ecell = Eox + Ered,

then Eox = 0.76 - 0.00 V = 0.76 V

• So we can sayZn --> Zn+2 + 2 e- Eox = 0.76 V


• Combine other half reactions with half reactions with known half cell potentials to complete the table of Standard Reduction Potentials

Standard Reduction Potential

• Things to know about standard reduction potentials:– Eo values correspond to solutes at 1 M

and all gases at 1 atm– When a half reaction is reversed, the

sign of Eo is reversed– When a half reaction is multiplied by

an integer, Eo remains the same!! Do not multiply Eo

by any number!


• The cell potential is always positive for a galvanic cellGiven Fe+2 + 2e- --> Fe Ered = -

0.44 V MnO4

- + 5 e- + 8 H+--> Mn+2 + 4H2O Ered = 1.51 V

• Reverse the reaction that will result in an overall cell potential that is positive, i.e., reverse the rxn involving iron


• Find the Eo for the galvanic cell based on the reactionAl+3 + Mg --> Al + Mg+2

• Sketch this cell and label the anode, cathode, direction of electron and ion flow, and indicate what reaction occurs at each electrode


• Calculate Eo and sketch the galvanic cell based on the (unbalanced) reaction:

MnO4- + H+ + ClO3

- --> ClO4

- + Mn+2 + H2O

• Balance the reaction

Galvanic Cells

• Line Notation– Used to describe electrochemical cells– Anode components are listed on the left– Cathode components are listed on the

right– Separate half cells with double vertical

lines: ll– Indicate a phase difference with a

single vertical line: l

Galvanic Cells

• Ex: Mg(s) l Mg+2(aq) ll Al+3(aq) l Al(s)is the line notation for the example in slide # 15

• For slide #16, the reactants and products are all ions, and so cannot act as an electrode– an inert electrode is needed - use Pt

• Pt(s) l ClO3-(aq),ClO4

- ll MnO4-(aq),Mn+2(aq) l

Pt(s)

Cell Potential, Electrical Work, and Free Energy

• Relationship between free energy and cell potential– after some serious derivations:

Go = -nFE o

• n = number of moles of electrons• F = the charge on 1 mole of electrons

= Faraday = 96,485 Coulombs/1 mole e-

– this equation provides an experimental means to obtain Go.

– Cell potential is directly related to Go.– Confirms that a galvanic cell runs in a direction

that results in a positive E o, because a positive E o means a negative Go, which means the reaction is spontaneous

Cell Potential and Free Energy

• Calculate Go for the reactionCu+2 + Fe --> Cu + Fe+2

Is this reaction spontaneous?

Cell Potential and Free Energy

• Predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au+3 solution.

Dependence of Cell Potential on Concentration

• The standard reduction potentials can be used to tell us the cell potential for cells under standard conditions (all concentrations = 1 M)

• What will happen to the cell potential when the concentration is not 1 M?


• Cu(s) + 2Ce+4(aq) --> Cu+2(aq) + 2 Ce+3(aq)

• Under standard conditions, Eo = 1.36 V. What will the cell potential be if [Ce+4] is greater than 1.0 M?

• Use Le Chatelier’s principle.– Increasing the [Ce+4] (when the [Ce+4] > 1 M),

then the forward reaction is favored, so the driving force on the electrons is greater, so the cell potential is greater


2 Al(s) + 3Mn+2(aq) --> 2 Al+3(aq) + 3 Mn(s) Eo = 0.48 V

• Predict whether Ecell is larger or smaller than Eo

cell

a. [Al+3] = 2.0 M, [Mn+2] = 1.0 Mb. [Al+3] = 1.0 M, [Mn+2] = 3.0 M

Concentration Cells

• Cell potential depends on concentration– Make a cell (a concentration cell) in

which both compartments contain the same components, but at different concentrations

– Usually has a small cell potential– The difference in concentration

produces the cell potential

Concentration Cells

• Given a cell in which both compartments contain aquesous AgNO3

• Ag+ + e- ---> Ag Eo = 0.80 V• Left side: 0.10 M Ag+

• Right side: 1.0 M Ag+

• What will happen?

Concentration Cells

• There will be a positive cell potential due to the difference in Ag+ concentrations

• To reach equilibrium, the driving force is to equalize the Ag+ concentration.– This can be done if the 1.0 M Ag+ could

be reduced and the 0.10 M Ag+ could be increased.

– The Ag in the 0.10 M Ag+ compartment will dissolve while the Ag in the 1.0 M Ag+ compartment will increase in mass.

The Nernst Equation

G = Go + RT ln QGo= -nFEo

• -nFE = -nFE o + RT lnQ• E = Eo - RT ln Q Nernst equation nF• The Nernst equation gives the

relationship between cell potential and the concentrations of the cell components

The Nernst Equation

• @ 25oC, the Nernst equation can be written as:E = E o - 0.0592 log Q n

• Use the Nernst equation to calculate the cell potential when one or more of the components are not in their standard states

The Nernst Equation

• For 2 Al(s) + 3 Mn+2 --> 2 Al+3 + 3 Mn(s)– Calculate E when [Mn+2] = 0.50 M and

[Al+3] = 1.50 M– Applying Le Chatelier’s principle, we

would predict that the reverse reaction would be favored, so E should be less than Eo.

The Nernst Equation

• The cell potential calculated by the Nernst equation is the maximum cell potential before any current flows

• As an galvanic cell discharges, the concentrations of the components will change, so E will change over time

• A cell will spontaneously discharge until equilibrium has been reached

The Nernst Equation

• Equilibrium has been reached when Q = K, so Ecell = 0

• A dead battery is one in which the cell reaction has reached equilibrium– There is no driving force for electrons to

be pushed through the wireG = 0 at equilibrium, meaning the cell no

longer has the ability to do work

The Nernst Equation

• Describe the cell based on the following half reactions and concentrations:VO2

+ + 2 H+ + e- --> VO+2 + H2O Eo = 1.00 V

Zn+2 + 2 e- --> Zn Eo = -0.76 VT = 25o C[VO2

+] = 2.0 M

[H+] = 0.50 M[VO+2] = 1.0 x 10-2M[Zn+2] = 1.0 x 10-1 M

Calculating the Keq for Redox

Reactions • For a cell at equilibrium,

Ecell = 0, and Q = K

Ecell = Eocell - 0.0592 log Q

n0 = Eo

cell - 0.0592 log Keq

nEo

cell = 0.0592 log Keq

nlog Keq = nEo

cell @ 25oC

0.0592

Batteries

• A battery is – a galvanic cell– a group of galvanic cells connected in

series• cell potentials of the individual cells add

up to give the total battery cell potential

– a source of direct current

Batteries

• Lead Storage Battery– can function for several years under

temperature extremes from -30oF to 120oF

– 12 Volt storage battery made of six cells– anode = Pb

– cathode = Pb coated with PbO2

– electrolyte solution = H2SO4 solution

Batteries

Pb + HSO4- --> PbSO4 + H+ + 2 e-

PbO2 + HSO4- + 3 H+ + 2 e- --> PbSO4 + 2 H2O

___________________________________________________________

Pb + PbO2 + 2HSO4- + 2 H+ --> 2 PbSO4 + 2 H2O

• PbSO4 adheres to the electrodes

• As the cell discharges, the sulfuric acid is consumed. The condition of the battery can be determined by measuring the density of the solution. As the sulfuric acid concentration decreases, the density decreases.

Batteries

• The lead storage battery can be recharged because the products adhere to the electrodes, so the alternator can force current through the battery in the opposite direction and reverse the reaction

• Even though the battery can be recharged, physical damage from road shock and chemical side reactions (e.g. electrolysis of water)eventually cause battery failure

Batteries

• Dry Cell Battery– invented more than 100 years ago by

George Leclanche• Acid Version

– anode: Zn – cathode: Carbon rod in contact with a moist

paste of solid MnO2, solid NH4Cl, and carbon– 2NH4

+ + 2 MnO2 + Zn --> Zn+2 + Mn2O3 + 2 NH3 + H2O

– produces 1.5 V

Batteries

• Alkaline dry cell– anode: Zn– cathode: carbon rod in contact with a moist

paste of solid MnO2, KOH or NaOH, and carbon

– 2MnO2 + H2O + Zn --> Zn+2 + Mn2O3 + 2 NH3 + H2O

– the alkaline dry cell lasts longer because the zinc doesn’t corrode as fast under basic conditions

Batteries

• Nickel-Cadmium battery• Cd + NiO2 + 2 OH- + 2 H2O --> Cd(OH)2 + Ni(OH)2 + 2

OH-

• This battery can be recharged because, like the lead storage battery, the products adhere to the electrodes.

Fuel Cells• Fuel Cell

– a galvanic cell for which the reactants are continuously supplied

– used in the U.S. space program– based on the reaction of hydrogen

and oxygen to form water:• 2 H2(g) + O2(g) --> 2H2O(l)

– not exactly a portable power source as this cell weighs about 500 pounds

Corrosion

• Corrosion is the return of metals to their natural states, i.e., the ores from which they are obtained.

• Corrosion involves oxidation of the metal• Corroded metal loses its structural integrity

and attractiveness• Metals corrode because they oxidize easily• Metals commonly used for decorative and

structural purposes have less positive reduction potentials than oxygen gas

Corrosion

• Noble Metals– copper, gold, silver, and platinum are

relatively difficult to oxidize, hence the term noble metals

Corrosion

• Some metals form a protective oxide coating, preventing the complete corrosion of the metal– Aluminum is the best example, forming Al2O3,

which adheres to, and protects the aluminum– Copper forms an external layer of copper

carbonate, known as patina– Silver forms silver tarnish which is silver sulfide– Gold does not corrode in air

Corrosion of Iron

• Important to control the corrosion of iron because it is so important as a structural material

• The corrosion of iron is not a direct oxidation process (iron reacting with oxygen), but is actually an electrochemical reaction.

Corrosion of Iron• Steel has a nonuniform surface

– These nonuniform areas are where iron can be more easily oxidized (the anode regions) than at other regions (the cathode regions)

– anode: Fe--> Fe+2 + 2 e- (the electrons flow through the steel to the cathode)

– cathode: O2 + 2 H2O + 4 e- --> 4OH-

– The Fe+2 formed in the anodic regions travel through the moisture to the cathodic regions. There, the Fe+2 reacts with oxygen to form rust, which is hydrated iron (III) oxide, Fe2O3

.nH2O

Corrosion of Iron

• For iron to corrode, moisture must be present to act as a salt bridge between the anode and cathode regions.– Steel does not rust in dry air

• Salt accelerates the rusting process– Cars rust faster where salt is used on roads

to melt ice and snow– Salt on the moist surfaces increases the

conductivity of the aqueous solution formed

Corrosion Prevention

• Apply a protective coating– apply paint or metal plating

• Chromium or tin are often used to plate steel because they form protective oxides

• Zinc can be used to coat steel, a process called galvanizing

– Fe--> Fe+2 + 2 e- Eo = 0.44 V– Zn --> Zn+2 + 2 e- Eo = 0.76 V– Zinc, then, is more likely to be oxidized than

iron. It acts as a protective coating because it will react with the oxygen preferentially, so the oxygen and the iron don’t come into contact. The zinc is sacrificed.

Corrosion Prevention

• Alloying is used to prevent corrosion– Stainless steel does not corrode like

iron• carbon, chromium and nickel have been

added to iron to form stainless steel, an alloy.

• These additions to iron form a protective oxide coating that changes steel’s reduction potential to that of a noble metal.

Electrolysis• Electrolytic Cell

– Uses electrical energy to force a nonspontaneous redox reaction to go

– Current is forced through a cell for which the Eo

cell is negative

– Importance• charging a battery• producing aluminum metal• chrome plating an object

Electrolysis

• Compare an electrolytic cell to a galvanic cell. Be able to label the anode, cathode, direction of ion and electron flow, which half reactions take place at each electrode, and where oxidation or reduction are taking place:

• Zn+2 + Cu --> Cu+2 + Zn vs.• Zn + Cu+2 --> Zn+2 + Cu

Electrolysis

• Stoichiometry of electrolytic process– 1 Ampere = 1 coulomb of charge /sec or– 1 A.sec = 1 C – 96485 C = the charge on 1 mole of electrons

• To plate something is to deposit the neutral metal on the electrode by reducing the metal ions in solution, e.g.

Cu+2 + 2 e---> Cu

Electrolysis

• How long must a currentof 5.00 A be applied to a solution of Ag+ to produce 10.5 g of silver metal?

Electrolysis

• Electrolysis of a mixture of ions– Which metal will plate out first?– Look at the standard reduction

potentials• The more positive the Eo, the more likely

the reaction will proceed

– Given a solution with Cu+2, Ag+, and Zn+2, which metal will plate out first?

Electrolysis

• Given a solution containing Ce+4, VO2

+, and Fe+3, give the order of oxidizing ability of these species and predict which one will be reduced at the cathode of an electrolytic cell at the lowest voltage.

Electrolysis• Overvoltage

– a complex phenomenon which results in more voltage needing to be applied to cause a reaction to occur than predicted from the Eo’s, i.e., there will be exceptions!

• Ex: Electrolysis of a solution of NaCl

– Expect Na+, Cl-, and H2O to be the major species

2Cl- --> Cl2 + 2 e- Eo = -1.36 V

2H2O --> O2 + 4 H+ + 4 e- Eo = -1.23 V

It would appear that H2O would be easier to oxidize because Eo is more positive, but in reality, Cl2 is produced. No good explanation, sorry!

Commercial Electrolytic Processes

• Metals are typically good reducing agents– typically found combined with other

substances in ores, mixtures of ionic substances containing oxides, sulfides, and silicate anions

– nobles metals, Cu, Ag, Pt, and Au, can be found as pure metals


• Production of aluminum– Al is the third most abundant element on

earth– Al is a very active metal found in nature as

the oxide in an ore called bauxite (for Les Baux, France where it was discovered)

– Charles Hall (U.S.) and Paul Heroult (France) discovered an electrolytic process to produce pure aluminum almost simultaneously - Hall-Heroult process

Commercial Electrolytic Processes• Production of Aluminum

– Uses molten cryolite, Na3AlF6 as the solvent for aluminum oxide

• water is more easily reduced than Al+3, so aluminum oxide could not be dissolved in water to produce aluminum

• melting an ionic substance allows for mobility of the ions, but the m.p. of Al2O3 is too high at 2050oC to make melting it practical

• Mixing Al2O3 and Na3AlF6 lowers the m.p. to a mere 1000oC, so electrolysis became economically feasible

• Price of Aluminum dropped from $100,000/lb to $0.74 /lb!


• Aluminum produced in the Hall-Heroult process is 99.5% pure

• To be useful as a structural material, aluminum is alloyed with metals like zinc and manganese.

• Production of aluminum uses about 5% of all the electricity used in the U.S.

Electrolysis of Sodium Chloride

• Electrolysis of sodium chloride– produces pure sodium metal– melt solid sodium chloride (after

mixing with calcium chloride to lower the m.p. from 800oC to 600oC), apply electricity, and sodium is produced

Electrolysis of Sodium Chloride

• Electrolysis of aqueous sodium chloride– aqueous sodium chloride, aka, brine– process is the second largest consumer of

electricity in the U.S.– produces chlorine gas and sodium

hydroxide– not a source of sodium metal because

water is more easily reduced than Na+

• Na+ + e- --> Na Eo = -2.71 V

• 2 H2O + 2 e- --> H2 + 2 OH- Eo = -0.83V

electrochemistry

Documents

known half cell potentials

cell potentialeox ered

h2 ered

h2 zn

h zn

chemical energy

e eox

h mno4