Chapter 21

Electrochemistry:

Chemical Change and Electrical Work

Key Points About Redox Reactions

•Oxidation (electron loss) always accompanies reduction (electron gain).

•The oxidizing agent is reduced, and the reducing agent is oxidized.

•The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent.

A summary of redox terminology

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

OXIDATION

Zn loses electrons.

Zn is the reducing agent and becomes oxidized.

The oxidation number of Zn increases from x to +2.

REDUCTION

One reactant loses electrons.

Reducing agent is oxidized.

Oxidation number increases.

Hydrogen ion gains electrons.

Hydrogen ion is the oxidizing agent and becomes reduced.

The oxidation number of H decreases from +1 to 0.

Other reactant gains electrons.

Oxidizing agent is reduced.

Oxidation number decreases.

Half-Reaction Method for Balancing Redox Reactions

Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions.

•Each reaction is balanced for mass (atoms) and charge.

•One or both are multiplied by some integer to make the number of electrons gained and lost equal.

•The half-reactions are then recombined to give the balanced redox equation.

•The separation of half-reactions reflects actual physical separations in electrochemical cells.

Balancing Redox Reactions in Acidic Solution

Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq)

1. Divide the reaction into half-reactions -

Determine the O.N.s for the species undergoing redox.

Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq)

+6 -1 +3 0

2. Balance atoms and charges in each half-reaction -

I - I2

2 + 7H2O(l)

Cr2O72- Cr3+

14H+(aq) + Cr2O72- Cr3+

Cr is going from +6 to +3

I is going from -1 to 0

net: +12 net: +6 Add 6e- to left.

2Cr2O72- Cr3+ + 7H2O(l)14H+(aq) +6e- +

Balancing Redox Reactions in Acidic Solution continued

+ 2e-I - I22

Cr2O72- Cr3+ + 7H2O(l)14H+(aq) +6e- +

3. Multiply each half-reaction by an integer, if necessary -

Cr(+6) is the oxidizing agent and I(-1) is the reducing agent.

X 3

4. Add the half-reactions together -

+ 2e-I - I22

3I - I26 + 6e-

Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l)14H+(aq) +

2

Cr2O72- Cr3+ + 7H2O(l)14H+ +6e- + 2

Do a final check on atoms and charges.

Balancing Redox Reactions in Basic Solution

Balance the reaction in acid and then add OH- so as to neutralize the H+ ions.

Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l)14H+(aq) +

+ 14OH-(aq) + 14OH-(aq)

Cr2O72- + 6 I- 2Cr3+ + 3I2 + 7H2O + 14OH-14H2O +

Reconcile the number of water molecules.

+ 14OH-Cr2O72- + 6 I- 2Cr3+ + 3I27H2O +

Do a final check on atoms and charges.

The redox reaction between dichromate ion and iodide ion.

Cr2O72- I-

Cr3+ + I2

Practice 1 Balancing Redox Reactions by the Half-Reaction Method

PROBLEM: Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate ion to form carbonate ion and solid mangaese dioxide. Balance the skeleton ionic reaction that occurs between NaMnO4 and Na2C2O4 in basic solution:

MnO4-(aq) + C2O4

2-(aq) MnO2(s) + CO32-(aq)

Energy is absorbed to drive a nonspontaneous redox reaction

General characteristics of voltaic and electrolytic cells

VOLTAIC CELL ELECTROLYTIC CELLEnergy is released from

spontaneous redox reaction

Reduction half-reactionY++ e- Y

Oxidation half-reactionX X+ + e-

System does work on its surroundings

Reduction half-reactionB++ e- B

Oxidation half-reactionA- A + e-

Surroundings(power supply)do work on system(cell)

Overall (cell) reactionX + Y+ X+ + Y; G < 0

Overall (cell) reactionA- + B+ A + B; G > 0

The spontaneous reaction between zinc and copper(II) ion

Zn(s) + Cu2+(aq)

Zn2+(aq) + Cu(s)

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A voltaic cell based on the zinc-copper reaction

Oxidation half-reactionZn(s) Zn2+(aq) + 2e-

Reduction half-reactionCu2+(aq) + 2e- Cu(s)

Overall (cell) reactionZn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

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Notation for a Voltaic Cell

components of anode compartment

(oxidation half-cell)

components of cathode compartment

(reduction half-cell)

phase of lower oxidation state

phase of higher oxidation state

phase of higher oxidation state

phase of lower oxidation state

phase boundary between half-cells

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

Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s)

graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite

inert electrode

A voltaic cell using inactive electrodes

Reduction half-reactionMnO4

-(aq) + 8H+(aq) + 5e-

Mn2+(aq) + 4H2O(l)

Oxidation half-reaction2I-(aq) I2(s) + 2e-

Overall (cell) reaction2MnO4

-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l)

Practice 2 Diagramming Voltaic Cells

PROBLEM: Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode.

Why Does a Voltaic Cell Work?

The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit.

Ecell > 0 for a spontaneous reaction

1 Volt (V) = 1 Joule (J)/ Coulomb (C)

Voltages of Some Voltaic Cells

Voltaic Cell Voltage (V)

Common alkaline battery

Lead-acid car battery (6 cells = 12V)

Calculator battery (mercury)

Electric eel (~5000 cells in 6-ft eel = 750V)

Nerve of giant squid (across cell membrane)

2.0

1.5

1.3

0.15

0.070

Determining an unknown E0half-cell with the standard

reference (hydrogen) electrode

Oxidation half-reactionZn(s) Zn2+(aq) + 2e-

Reduction half-reaction2H3O+(aq) + 2e- H2(g) + 2H2O(l)Overall (cell) reaction

Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l)

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Practice 3 Calculating an Unknown E0half-cell from E0

cell

PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal:

Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) E0cell = 1.83V

Calculate E0bromine given E0

zinc = 0.76V

Selected Standard Electrode Potentials (298K)

Half-Reaction E0(V)

2H+(aq) + 2e- H2(g)

F2(g) + 2e- 2F-(aq)

Cl2(g) + 2e- 2Cl-(aq)

MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)

Ag+(aq) + e- Ag(s)

Fe3+(g) + e- Fe2+(aq)

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

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

N2(g) + 5H+(aq) + 4e- N2H5+(aq)

Fe2+(aq) + 2e- Fe(s)

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

Na+(aq) + e- Na(s)

Li+(aq) + e- Li(s)

+2.87

-3.05

+1.36

+1.23

+0.96

+0.80

+0.77

+0.40

+0.34

0.00

-0.23

-0.44

-0.83

-2.71

•By convention, electrode potentials are written as reductions.

•When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential.

•The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E0

cell.

•When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents.

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

stronger reducing agent

weaker oxidizing agent

stronger oxidizing agent

weaker reducing agent

Practice 4 Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength

PROBLEM: (a) Combine the following three half-reactions into three spontaneous, balanced equations (A, B, and C), and calculate E0

cell for each.

(b) Rank the relative strengths of the oxidizing and reducing agents:

E0 = 0.96V(1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)

E0 = -0.23V(2) N2(g) + 5H+(aq) + 4e- N2H5+(aq)

E0 = 1.23V(3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

Free Energy and Electrical Work

G -Ecell

-Ecell =-wmax

charge

charge = n F

n = #mols e-

F = Faraday constant

F = 96,485 C/mol e-

1V = 1J/C

F = 9.65x104J/V*mol e-

G = wmax = charge x (-Ecell)

G = -n F Ecell

In the standard state -

G0 = -n F E0cell

G0 = - RT ln K

E0cell = - (RT/n F) ln K

G0

E0cell K

G0 KReaction at

standard-state conditions

E0cell

The interrelationship of G0, E0, and K

< 0 spontaneous

at equilibrium

nonspontaneous

0

> 0

> 0

0

< 0

> 1

1

< 1

G0 = -RT lnKG0 = -nFEo

cell

E0cell = -RT lnK

nF

By substituting standard state values into E0

cell, we get

E0cell = (0.0592V/n) log K (at 250C)

Practice 5 Calculating K and G0 from E0cell

PROBLEM: Lead can displace silver from solution:

As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and G0 at 250C for this reaction.

Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s)

Practice 6 Using the Nernst Equation to Calculate Ecell

PROBLEM: In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions:

[Zn2+] = 0.010M [H+] = 2.5M P = 0.30atmH2

Calculate Ecell at 250C.

The Effect of Concentration on Cell Potential

G = G0 + RT ln Q

-nF Ecell = -nF Ecell + RT ln Q

Ecell = E0cell - ln Q

RT

nF

•When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell

•When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell

•When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell

Ecell = E0cell -

log Q0.0592

n

The relation between Ecell and log Q for the zinc-copper cell

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A concentration cell based on the Cu/Cu2+ half-reaction

Overall (cell) reactionCu2+(aq,1.0M) Cu2+(aq, 0.1M)

Oxidation half-reactionCu(s) Cu2+(aq, 0.1M) + 2e-

Reduction half-reactionCu2+(aq, 1.0M) + 2e- Cu(s)

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Practice 7 Calculating the Potential of a Concentration Cell

PROBLEM: A concentration cell consists of two Ag/Ag+ half-cells. In half-cell A, electrode A dips into 0.0100M AgNO3; in half-cell B, electrode B dips into 4.0x10-4M AgNO3. What is the cell potential at 298K? Which electrode has a positive charge?

The corrosion of iron

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Enhanced corrosion at sea

The effect of metal-metal contact on the corrosion of iron

faster corrosion cathodic protection

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The use of sacrificial anodes to prevent iron corrosion

The tin-copper reaction as the basis of a voltaic and an electrolytic cell

Oxidation half-reactionSn(s) Sn2+(aq) + 2e-

Reduction half-reactionCu2+(aq) + 2e- Cu(s)

Oxidation half-reactionCu(s) Cu2+(aq) + 2e-

Reduction half-reactionSn2+(aq) + 2e- Sn(s)

Overall (cell) reactionSn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)

Overall (cell) reactionSn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)

voltaic cell electrolytic cell

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The processes occurring during the discharge and recharge of a lead-acid battery

VOLTAIC(discharge)

ELECTROLYTIC(recharge)

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Comparison of Voltaic and Electrolytic Cells

Cell Type G Ecell

Electrode

Name Process Sign

Voltaic

Voltaic

Electrolytic

Electrolytic

< 0

< 0

> 0

> 0

> 0

> 0

< 0

< 0

Anode

Anode

Cathode

Cathode

Oxidation

Oxidation

Reduction

Reduction

-

-

+

+

Practice 8 Predicting the Electrolysis Products of a Molten Salt Mixture

PROBLEM: A chemical engineer melts a naturally occurring mixture of NaBr and MgCl2 and decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and write balanced half-reactions and the overall cell reaction.

The electrolysis of water

Oxidation half-reaction2H2O(l) 4H+(aq) + O2(g) + 4e-

Reduction half-reaction2H2O(l) + 4e- 2H2(g) + 2OH-(aq)

Overall (cell) reaction2H2O(l) H2(g) + O2(g)

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Practice 9 Predicting the Electrolysis Products of Aqueous Ionic Solutions

PROBLEM: What products form during electrolysis of aqueous solution of the following salts: (a) KBr; (b) AgNO3; (c) MgSO4?

A summary diagram for the stoichiometry of electrolysis

MASS (g)of substance oxidized or

reduced

MASS (g)of substance oxidized or

reduced

AMOUNT (MOL)of electrons transferred

AMOUNT (MOL)of electrons transferred

AMOUNT (MOL)of substance oxidized or

reduced

AMOUNT (MOL)of substance oxidized or

reduced

CHARGE (C)CHARGE (C)

CURRENT (A)CURRENT (A)

balanced half-reaction

Faraday constant (C/mol e-)

M(g/mol)

time(s)

Practice 10 Applying the Relationship Among Current, Time, and Amount of Substance

PROBLEM: A technician is plating a faucet with 0.86g of Cr from an electrolytic bath containing aqueous Cr2(SO4)3. If 12.5 min is allowed for the plating, what current is needed?