Chapter 18Electrochemistry

Definition

The study of the interchange of chemical and electrical energy in oxidation-reduction (redox) reactions

This interchange can occur in both directions:

1. Conversion of Chemical Energy into Electrical Energy: an electrical current is generated from a spontaneous chemical

reaction (galvanic cell)

2. Conversion of Electrical Energy into Chemical Energy: a chemical reaction is induced by an electrical current (electrolysis)

Oxidation-Reduction (Redox) Reactions

Reactions in which electrons are transferred from one substance to another

The substance that loses electrons undergoes oxidation and is said to be

oxidized

The substance that gains electrons undergoes reduction and is said to be

reduced

A simple way to remember:

OIL RIG

Oxidation Is Loss of electrons

Reduction Is Gain of electrons

Redox reactions most commonly occur between metals and non-metals in the

formation of ionic compounds

2Mg(s) + O2(g) → 2MgO(s)

Electrons are transferred from Mg to O2Mg is oxidized to Mg2+

O2 is reduced to O2-

Oxidizing AgentA substance that contains an atom that

accepts electrons (from an atom in another substance that is being oxidized) and is

reduced in the process

Reducing AgentA substance that contains an atom that

donates electrons (to an atom in another substance that is being oxidized) and is

oxidized in the process

2Mg(s) + O2(g) → 2MgO(s)

Mg donates electrons so it is the reducing agent

O2 accepts electrons so it is the oxidizing agent

Summary

Molecules can also undergo redox reactions:

CH4(g) + 2O2(g) → CO2(g)+ 2H2O(g)

6CO2(g) + 6H2O(l) → C6H12O6(aq) + 6O2(g)

However, it is not clear how the electrons are transferred and which species are

oxidized and which are reduced!

Need a general way to follow the transfer of electrons in any reaction!

Oxidation States (Numbers)

A systematic way to keep track of electrons in redox reactions containing covalent

compounds

Numbers are assigned to each atom in a compound which represent the imaginary charge each would have if the electrons in the molecule were distributed according to

their ability to attract electrons (electronegativity)

For binary ionic compounds, the oxidation state of each ion is equal to its charge

For a neutral compound, the sum of oxidations states is zero

For an ion, the sum of oxidation states is equal to the charge on the ion

Application to Redox Reactions

In all redox reactions:

• one atom loses electrons and has its oxidation increase (is oxidized) and another gains electrons and has its oxidation state decrease (is reduced)

• the number of electrons lost by the atom being oxidized must be equal to the number of electrons

gained by the atom being reduced

• if the oxidation states do not change then the reaction is not a redox reaction!

Identify the atoms or ions which are oxidized and reduced and identify the oxidizing and reducing agents:

16H+(aq) + 2Cr2O72-(aq) + C2H5OH(l)

4Cr3+(aq) + 11H2O(l) + 2CO2(g)

Half-ReactionsA redox reaction can be split into two half-reactions, one oxidation,

one reduction:

Example:

Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)

Oxidation half-reaction:

Mg(s) → Mg2+(aq) + 2e-

Reduction half-reaction:

Fe2+(aq) + 2e- → Fe(s)

Luigi Galvani(1737-1798)

Galvanic CellsIf we physically separate one half-reaction from another other and then provide a wire path for the electrons to flow across, the redox

reaction can still proceed

Since the redox reaction is causing electrons to flow it is generating an electric current which can then be passed through a device (e.g. a

light bulb) to produce useful work

Such an arrangement is referred to a galvanic, voltaic or electrochemical cell

Batteries are examples of galvanic cells

Example:

Reaction between MnO4- and Fe2+ in acidic solution:

5Fe2+(aq) + MnO4-(aq) + 8H+(aq) →

5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

Broken down into half-reactions:

Fe2+(aq) → Fe3+(aq) + e-

MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)

Separate the two half-reactions and connect them with electrodes and a wire:

Anode (-)Fe2+(aq) → Fe3+(aq) + e-

Oxidation takes placeElectrons are produced

Cathode (+)

MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)

Reduction takes placeElectrons are used up

Electrons flow along the wire from anode (-) to cathode (+)

To remember the correct polarity think:

anode = anion = negative (-)cathode = cation = positive (+)

Problem: electron flow would create a charge separation which would require an huge amount of energy to

maintain

Solution: solutions must be connected (without mixing) so that ions can also flow to keep the charge neutral

This is accomplished by connecting the two solutions with a salt bridge (a U-shaped tube filled with a strong

electrolyte) or a porous disk in a connecting tube which allows ion flow but prevents mixing

This allows the electrons to flow through the wire from the anode (-) to the cathode (+), while the ions flow from

one compartment to the other to keep the net charge zero

Direction of Electron Flow

The direction of electron flow can be predicted by the activity series

Electrons always flow from the most active species (most powerful reducing agent) to the least active species (least powerful reducing

agent)

Therefore the electrode containing the most active metal becomes the anode (-) and the electrode containing the least active metal

becomes the cathode (+)

Metal Activity Series

Zn is more active than Cu

anode(-) cathode(+)

Mg is more active than Zn

anode(-) cathode(+)

Cell Potential

In a galvanic cell, we can think of the reducing agent at the anode literally “pushing” the electrons through the wire connecting the

electrodes to the oxidizing agent at the cathode

The strength of this “push” on the electrons is referred to as the cell potential (Ecell) or electromotive force (emf) of the cell and is

determined by the difference in activity (potential difference) between the substances at the anode and the cathode

The unit of electrical potential is the volt (V), which measures the number of joules of work produced per coulomb (C) of charge

transferred

Alessandro Volta (1745-1827)

Cell potentials are commonly measured with a digital voltmeter:

Half-reaction Potentials

It is possible to determine the potential of a half-reaction individually by constructing a galvanic cell in which one of the electrodes is a standard hydrogen electrode consisting of a chemically inert

platinum electrode in contact with 1 M H+ ions (e.g. 1 M HCl) bathed in H2 gas at 1 atm pressure

The potential of a half-reaction is measured by arbitrarily assigning a voltage of exactly 0 volts to the standard hydrogen electrode

If the constituents of the half-reaction are also in their standard states (molar concentrations, 1 atm pressures) then:

Ecell0 = EH2

0 + Ehalf-reaction0 = 0 + Ehalf-reaction

0

Ehalf-reaction0 = Ecell

0

In this way the potential (relative to the standard hydrogen electrode) of any half-reaction can be easily measured

2H+(aq) + Zn(s) Zn2+(aq) + H2(g)

Ecell0 = 0.76 V = EH2

0 + EZnZn2+0 = 0.00 + EZnZn2+

0

EZnZn2+0 = 0.76 V

Standard Reduction Potentials (E 0)

By convention, the potentials of half-reactions are always written as reduction processes

The E 0 values corresponding to these reduction reactions under standard conditions (solutes at 1 M and gases at 1 atm) are referred

to as standard reduction potentials

Elements, compounds or ions which are reduced more easily than hydrogen have positive E 0 values while elements, compounds or ions which are reduced less easily than hydrogen have negative E 0 values

The more positive the reduction potential, the easier the species is to reduce and the more powerful an oxidizing agent it is

The more negative the reduction potential, the more difficult the species is to reduce and the more powerful a reducing agent it is

The order of standard reduction potentials is opposite to the activity series which are always written as oxidation processes

Value of Cell Potential

The value of the cell potential in any galvanic cell is equal to the sum of standard reduction potentials of the two half reactions

Ecell0 = EZnZn2+

0 + E Cu2+Cu0 = 0.76 V + 0.34 V = 1.10 V

Calculating Cell Potentials

1. Write the two half-reactions and their standard potentials2. The half-reaction with the largest positive potential is left

unchanged while the other half-reaction is written in reverse (since redox reactions must involve both oxidation and reduction) and the sign of its potential changed. This is done because a cell will always run spontaneously to produce a positive cell potential

3. Balance the number of electrons lost and the number of electrons gained by multiplying by integers (the potentials are not changed

since they are intensive)4. Add the two balanced half-reactions to give the balanced cell

reaction 5. Add the two potentials together to obtain the overall cell potential. Since reduction occurs at the cathode and oxidation at

the anode we can write:

Ecell0 = E 0(cathode) + [-E 0(anode)]

General Rule:

The anode compartment (-) will always contain the species with the smallest standard reduction potential since it is most easily

oxidized and wants to give up electrons

The cathode (+) compartment will always contain the species with the largest standard reduction potential since it is most easily

reduced and wants to accept electrons

This is why electrons flow from the anode to cathode!

Line Diagrams

Double vertical lines indicate the salt bridge or the porous disk while a single vertical line represents a phase boundary

Substitute ElectrodesIn some half-cells there is no component that can be used as an electrode, e.g. if there are two ionic components. In such cases

chemically inert electrodes made from platinum or graphite are used

Example:2MnO4

-(aq) + 6H+(aq) + 5ClO3-(aq) 2Mn2+(aq) + 3H2O(l) + 5ClO4

-(aq)

Half-reactions:MnO4

-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l)

ClO3-(aq) + H2O(l) ClO4

-(aq) + 2H+(aq) + 5e-

Line diagram:

Pt(s)|ClO3-(aq),ClO4

-(aq),H+(aq) || MnO4-(aq),H+(aq),Mn2+(aq)|Pt(s)

or

Pt | ClO3-, ClO4

-, H+ || MnO4-, H+, Mn2+ | Pt

Note: coefficients, water and spectator ions are never written in line diagrams! State symbols are sometimes not included!

A galvanic cell consists of a graphite electrode in an acidified plumbic oxide and plumbous sulfate solution and a graphite electrode in a

solution of ferrous chloride and ferric chloride. Assume all components are in standard states.

Write the half reactions, balanced reaction, line diagram and calculate the potential for the cell

Write the half-reactions and the balanced reaction from the line diagram of the galvanic cell below:

Ag(s) | Cl-(aq), AgCl(s) || NO3-(aq), H+(aq), NO(g) | C(s)

Complete Description of a Galvanic Cell

Given its half-reactions, a complete description of a galvanic cell includes a diagram showing the following information:

1. The cell potential2. The direction of electron flow

3. Designation of the anode and cathode4. The nature of each electrode (an inert conductor must be used if

none of the substances in the half-reaction is a conducting solid) and the ions present in each compartment

5. A balanced equation for the redox reaction6. A line diagram for the cell

Electrical Work from Galvanic Cells

The amount of work that can be extracted from a galvanic cell depends on the on the potential difference between the anode and

cathode and the amount of charge transferred:

w = -qE

where w is the work (in Joules), q the charge transferred (in Coulombs) and E is the cell potential (in Volts: 1 V = JC-1)

For a cell to produce a current, the cell potential, E must be positive and the work, w must be negative (since useful work must flow out

of the system into the surrondings)

The maximum possible amount of work, wmax will always be obtained at the maximum cell potential, E max:

wmax = -qE max

However, the actual work obtained is always less than the maximum due to resistive losses as the current flows through the wire

Michael Faraday(1791-1867)

The Faraday (F)

A constant equal to the charge on 1 mole of electrons

1 F = 96,485 Cmol-1

From this definition, the charge transferred in a galvanic cell can be expressed as follows:

charge transferred = moles of electrons transferred x 1 Faraday

q = nF

Cell Potential and Free Energy

At constant temperature and pressure (Chapter 17):

wmax = ∆G

For a galvanic cell:

wmax = -qE max

so:

∆G = -qE max

since q = nF and omitting the subscript on E max:

∆G = -nFE

Under standard conditions:

∆G0 = -nFE 0

∆G = -nFE

Conclusions:

1. The maximum cell potential is directly related to the free energy difference between the reactants and products in the cell

2. ∆G for a cell can be determined experimentally from the measured cell potential, E

3. Consistent with the fact that cells run in the direction of positive E, since this corresponds to a negative ∆G (the condition for

spontaneity)4. This formula also applies to half-reactions

Cell Potential and Concentration

If the concentration of an ion in a galvanic cell is changed from its standard concentration (1 M), the cell potential will change from its

standard cell potential

Example:

Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)zEcell

0 = 0.78 V

Increase [Cu2+]:

Forward reaction is favored so Ecell > Ecell0

Increase [Fe2+]:

Reverse reaction is favored so Ecell < Ecell0

Walther Hermann Nernst(1864-1941)

Cell Potential and Concentration

The potential of a galvanic cell can be related to the concentrations of their components using the relationship between free energy and

concentration:

∆G = ∆G0 + RTln(Q)

Since ∆G = -nFE and ∆G0 = -nFE 0:

-nFE = -nFE 0 + RTln(Q)

Dividing each side by –nF:

E =E 0 - (RT/nF)ln(Q)

At 25 0C, changing from natural to base 10 logarithms:

E =E 0 - (0.0591/n)log(Q)

The Nernst Equation

For the general cell reaction:

aA + bB cC + dD

The Nernst equation at 25 oC is:

E = E 0 - (0.0591/n)log(Q)

or:

E = E 0 - (0.0591/n)log([C]c[D]d / [A]a[B]b)

Concentration Cells

Galvanic cells in which both compartments have the same components but at different concentrations

The current flows in the direction which will equalize the concentrations of the ions in the two cells - low concentration (anode)

to high concentration (cathode):

Ag | Ag+(0.1 M) || Ag+(1 M) | Ag

Anode (-):Ag Ag+ + e-

[Ag+] increases

Cathode (+):Ag+ + e- Ag

[Ag+] decreases

The Nernst Equation for Concentration Cells

For the general concentration cell:

M(s) + Mn+(aq)(M1) Mn+(aq)(M2) + M(s)

M1 > M2

The Nernst equation for a concentration cell at 25 oC is:

E = E 0 - (0.0591/n)log(Q)

E 0 = 0.00 V

so:

E = - (0.0591/n)log([Mn+(aq)](M2) / [Mn+(aq)](M1))

Note: the smallest concentration is always on top!

Ion Selective Electrodes

Since the cell potential depends on the concentration of reactants and products, measured potentials can be used to measure the

concentration of ions

Example – a pH meter:

The concentrations of other ions can be measured in a similar manner:

Changes in Potential with Time

The potential calculated from the Nernst equation is the maximum potential before any current flow has occurred

As the cell discharges and current flows from anode to cathode, the concentrations will change (products increase, reactants decrease

and Q will decrease) and therefore Ecell will gradually decrease

This will continue until the cell reaches equilibrium:

Q = K and Ecell = 0

A “dead” battery is one in which the cell reaction has reached equilibrium so there is no longer any chemical driving force pushing the electrons through the wire and the cell no longer has the ability

to do work

At equilibrium, the components in the two cell compartments have the same free energy (∆G = 0)

Equilibrium Constants from Cell Potentials

For a cell at equilibrium, Q = K and Ecell = 0

Applying this to the Nernst equation at 25 0C:

E =E 0 - (0.0591/n)log(Q)

gives:

0 =E 0 - (0.0591/n)log(K)

so:

log(K) = nE 0/0.0591

Batteries

A Galvanic cell or a group of galvanic cells connected in series used as a source of direct current

Lead Storage (Car) Battery

Anode: Pb(s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2e-

Cathode: PbO2(s) + HSO4-(aq) + 3H+(aq) + 2e- → PbSO4(aq) + 2H2O(l)

Dry Cell Battery

Anode: Zn(s) → Zn2+(aq)+ 2e-

Cathode:2NH4+(aq) + 2MnO2(s) + 2e- → Mn2O3(s)+ 2NH3(aq) + H2O(l)

Alkaline Battery

Anode: Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2e-

Cathode: 2MnO2(s) + H2O(l) + 2e- → Mn2O3(s) + 2NH3(aq) + 2OH-(aq)

Fuel Cells

A galvanic cell for which the reactants are continuously supplied

Anode: H2(g) + 2OH-(aq) 2H2O(l) + 2e-

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

Corrosion

Involves oxidation of a metal in air (to form oxides and sulfides)

Most metals corrode easily because they have standard reduction potentials which are less positive than oxygen gas so when these half-reactions are reverse and combined with the reduction half-

reaction for oxygen, a positive cell potential results

This process is often slowed by the formation of a thin oxide coating which prevents further corrosion

Metals with large standard positive reduction potentials (for example gold) do not corrode and are referred to as noble metals

Corrosion of Iron

Since steel (an alloy of iron and carbon) has an inhomogeneous surface, regions form where the iron is more easily oxidized (anodic

regions) than at others (cathodic regions)

In the cathodic regions, the ferrous ions react with oxygen to form rust (a hydrated ferric oxide of variable composition):

Prevention of Corrosion

This is achieved by applying a coating (usually paint or a metal plating) to protect the metal from oxygen and moisture

Chromium and tin are often used to plate steel (via electrolysis) which then form a protective oxide coating

Zinc can also be used to coat steel in a process called galvanizingwhich forms a protective mixed oxide-carbonate coating

Since zinc is a more active metal than iron, any oxidation that does occur involves the zinc rather than the iron. Zinc is said to act as a

“sacrificial” metal

Corrosion can also be prevented by alloying

For example, stainless steel contains chromium and nickel which form oxide coatings that increases the steel’s reduction potential so it

is less likely to corrode

Electrolysis

The forcing of a current through a cell to produce a chemical change for which the cell potential is negative (non-spontaneous)

An electrolytic cell therefore uses electrical energy to produce a chemical change

Galvanic Cell

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

Anode: Zn(s) Zn2+(aq) + 2e-

Cathode: Cu2+(aq) + 2e- Cu(s)

Electrolytic Cell

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

Anode: Cu(s) Cu2+(aq) + 2e-

Cathode: Zn2+(aq) + 2e- Zn(s)

Calculating Chemical Change from Current Flow

Given the amount of current flowing (in amperes , A = Cs-1) for a specified time period it is possible to calculate the amount of neutral

metal deposited at the cathode of an electrolytic cell as follows:

1. Multiply the current (in amperes) by time in seconds to obtain the total charge (in coulombs) passed into the solution at the cathode

2. Calculate the number of moles of electrons required to carry this charge by dividing by the Faraday (96,485 Cmol-1)

3. Use the half-reaction for the reduction to calculate the number of moles of neutral metal deposited from the moles of electrons

4. Convert the moles of neutral metal into grams by multiplying by the average atomic mass

Caution!

Standard reduction potentials must be used carefully in predicting the order of oxidation or reduction in electrolytic cells since some species

require a much higher potential then expected (called an overvoltage) due to difficulties in transferring electrons from species

in the solution to atoms on the electrode across the electrode-solution interface

Example:

Electrolysis of NaCl(aq)

Two processes are possible at the anode:

As voltage is increases we would expect to see O2 produced at the anode first since it is easier to oxidize H2O

However, in practice the Cl2 is seen at the anode first!

The Electrolysis of Water2H2O(l) → 2H2(g) + O2(g)

Electroplating

Anode: Ag(s) → Ag+(aq) + e-

Cathode: Ag+(aq) + e- → Ag(s)

Electrolysis of Sodium Chloride

Anode: 2Cl-(l) → Cl2(g) + 2e-

Cathode: Na+(l) + e- → Na(l)