Electrochemistry

Electrochemistry 

Reading: from Petrucci, Harwood and Herring (8th edition): 

Required for Part 1: Sections 21-1 through 21-4. 

Examples for Part 1: 21-1 through 21-10. 

Problem Set for Part 1: 

Review: Chapter 5 questions 21-26. 

Chapter 21 questions: 15-17, 32, 34, 43, 53 

Additional problems from Chapter 21: 

York University CHEM 1001 3.0 Electrochemistry - 1

Applications of Electrochemistry 

Spontaneous chemical reactions can be used to produce 

an electric current and do work. (batteries, fuel-cells) 

An electric current can be used to force non-spontaneous 

chemical reactions to occur. (electrolysis) 

Reactions can be made to occur in a specific place. 

(electroplating, electropolishing) 

The voltage produced by a reaction can be used as an 

analytical tool. (pH electrodes) 

The current produced by a reaction can be used as an 

analytical tool. 


G and non-PV Work 

For a reversible process at constant T and P: 

H = q P + w E 

(definition of enthalpy) 

w E = non-PV work done on the system 

q P = TS (reversible, constant T) 

Let w max = maximum non-PV work done by the system 

w max = -w E 

(reversible) 

So 

w max = -H + TS = -G 

Conclusion: The maximum non-PV work that can be 

obtained from a process is equal to -G. 


Work from Chemical Reactions 

Spontaneous chemical reactions can be used to do work. 

How? 

One possibility: 

burn fuel to release heat 

boil water 

use the expanding steam to do work 

Disadvantages: 

inefficient (only part of heat can be turned into work) 

can not readily carry out the reverse process 


Work from Redox Reactions 

Another method: Use redox reactions. 

Cu(s) + 2Ag + Cu 2+ + 2Ag(s) r G° = -88.43 kJ mol -1 

The Cu(s) is oxidized (gives up electrons). 

Half-reaction: Cu(s) Cu 2+ + 2e - 

The Ag + (aq) is reduced (receives electrons). 

Half-reaction: Ag + + e - 2Ag(s) 

This reaction is spontaneous. If we can transfer the electrons 

through an external circuit, we can use it do electrical work. 


Cu is oxidized at 

one electrode: 

Cu(s) Cu 2+ + 2e - 

Ag + is reduced at 

the other electrode: 

e - + Ag + Ag(s) 

Electrons travel 

through the wire. 

Ions travel through 

the salt bridge. 

Electrochemical Cells 


Atomic View of an Electrochemical Cell 


Electrochemical Cells - Terminology 

The anode is the electrode at which oxidation occurs. 

The cathode is the electrode at which reduction occurs. 

The cell potential is the voltage difference when no 

current flows between the electrodes. 

Cell potential is also called cell voltage or EMF 

(electromotive force). 

Cells in which spontaneous reactions produce a current 

are called voltaic cells or galvanic cells. 

In electrolytic cells electricity is used to force a nonspontaneous 

reaction to occur. 


Cell Diagrams 

Anode (oxidation) is placed on left side of diagram. 

Cathode (reduction) is placed on right side of diagram. 

Single vertical line, | , indicates a boundary between 

different phases (i.e., solution | solid). 

Double vertical line, || , indicates a boundary (salt bridge) 

between different half-cell compartments. 

Example: Cu(s) Cu 2+ Ag + Ag(s) 

At the anode: Cu(s) Cu 2+ + 2e - 

At the cathode: 

Overall: 

Ag + + e - Ag(s) 

Cu(s) + 2Ag + Cu 2+ + 2Ag(s) 


Cell Diagrams - examples 

Diagram the cell in which the following overall reaction 

occurs: 

Pb(s) + 2AgCl(s) PbCl 2 (s) + 2Ag(s) 

Answer: Pb(s)Cl - (aq)PbCl 2 (s))AgCl(s)Ag(s)Cl - (aq) 

Write the half-cell reactions for the following cell: 

Ag(s)Ag + (aq)Cl - (aq)AgCl(s)Ag(s) 

Answer: 

Anode (oxidation): Ag(s) Ag + (aq) + e - 

Cathode (reduction): AgCl(s) + e - Ag(s) + Cl - (aq) 


Balancing Redox Reactions - Review 

Example: SO 3 2- + MnO 4- SO 4 2- + Mn 2+ (unbalanced) 

S from +4 to +6 (oxidized). Mn from +7 to +2 (reduced). 

(1) Write balanced half-reactions for oxidation and 

reduction. 

Oxidation half-reaction 

Skeleton reaction: SO 3 

2- 

SO 4 2- + 2e - 

Balanced: SO 3 2- + H 2 O(l) SO 4 2- + 2H + + 2e - 

Reduction half-reaction 

Skeleton reaction: MnO 4 - + 5e - Mn 2+ 

Balanced: 

MnO 4 - + 8H + + 5e - Mn 2+ + 4H 2 O(l) 


Balancing Redox Reactions - continued 

(2) Adjust coefficients so the two half-reactions have the 

same numbers of electrons. 

5SO 3 2- + 5H 2 O(l) 5SO 4 2- + 10H + + 10e - 

2MnO 4 - + 16H + + 10e - 2Mn 2+ + 8H 2 O(l) 

(3) Add the two half-reactions. 

5SO 3 2- + 2MnO 4 - + 6H + 5SO 4 2- + 2Mn 2+ + 3H 2 O(l) 

10 electrons are transferred (important for later). 

(4) Check that the reaction is balanced for both atoms and 

charge. 


Redox Reactions in Basic Solution 

5SO 3 2- + 2MnO 4 - + 6H + 5SO 4 2- + 2Mn 2+ + 3H 2 O(l) 

Since this balanced reaction involves H + , it is appropriate 

for acidic solution 

For the reaction in basic solution, add 

This gives: 

6 × {H 2 O(l) H + + OH - } 

5SO 3 2- + 2MnO 4 - + 3H 2 O(l) 5SO 4 2- + 2Mn 2+ + 6OH - 

This method can also be used to get half-reactions in basic 

solution. 


Current and Charge 

Current is the amount of charge transferred per unit time. 

The amount of charge transferred is determined by the 

stoichiometry of the cell reaction. 

The charge on one mole of electrons is 96,485 coulombs. 

This is called the Faraday constant, F. 

F = 96,485 C mol -1 = 96,485 J V -1 mol -1 

Example: Cu(s) + 2Ag + Cu 2+ + 2Ag(s) 

1.93×10 5 C are transferred per mole of Cu oxidized. 

The current produced by a cell is determined by the reaction 

kinetics and the resistance of the circuit. 


Electrical Work 

The work done in an electrical circuit is 

Units: 

w elec = charge × (potential difference) 

coulombs × volts = joules 

The maximum possible non-PV work equals -G. Let 

n moles of electrons transferred per mole of reaction 

nF = total charge (coulombs) transferred 

E cell maximum possible cell potential (requires zero 

current) 

Then: 

r G = -nFE cell 


Electrical Work - continued 


There is a fundamental connection between r G and cell 

potential. Because of this: 

Cell potentials depend on concentrations. 

Electrochemical cells can be used to measure 

concentrations. (pH electrodes, for example) 

Electrochemical cells can be used to measure r G (and 

to determine r G°). 

Tabulated thermodynamic data can be used to determine 

cell potentials. 


Cell Potential - Example 

A cell is constructed in which the half-cell reactions are: 

Anode: H 2 (g) 2H + (aq) + 2e - 

Cathode: 

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

When P H2 = P Cl2 = 1.000 atm and [H + ] = [Cl - ] = 0.0100 M, 

the cell potential is found to be 1.4813 volts. Find r G 

under these conditions for 

H 2 (g) + Cl 2 (g) 2H + (aq) + 2Cl - (aq) 

Solution: r G = -nFE cell and n = 2 for the overall reaction. 

r G = -2(96,485 J V -1 mol -1 )(1.4813 V) = -285.85 kJ mol -1 


Cell Potentials and Spontaneity 


If r G < 0, the reaction is spontaneous as written. 

If r G > 0, the reaction is non-spontaneous as written 

(reverse reaction is spontaneous). 

If the reaction proceeds as written, then n > 0. 

Therefore: 

If E cell > 0, the reaction is spontaneous as written. 

If E cell < 0, the reaction is non-spontaneous as written. 


Standard Cell Potentials 

Definition: The standard cell potential, E° cell , is the cell 

potential that would obtain if all reactants and products were 

in their standard states. 

Therefore: 

r G° = -nFE° cell 

Standard states may be hypothetical. The standard cell 

potential is used in calculations of actual cell potentials. 

The standard cell potential is the sum of standard potentials 

for the individual half-cells. 


Standard Cell Potential - example 

The cell with the overall reaction 

H 2 (g) + Cl 2 (g) 2H + (aq) + 2Cl - (aq) 

has a standard cell potential of 1.3604 V. Determine f G° 

for Cl - (aq). 

Solution: 

r G° = -nFE° cell 

r G° = -2(96,485 C mol -1 )(1.3604 V) = -262.52 kJ mol -1 

For H 2 (g), Cl 2 (g), and H + (aq); f G° = 0. 

r G° = 2 f G°(Cl - (aq)) 

f G°(Cl - (aq)) = -131.26 kJ mol -1 


Standard Hydrogen Electrode 

To create a voltage, we need two half-cells. So we 

can't measure individual half-cell potentials. 

Convention: The standard hydrogen electrode is 

assigned a half-cell potential of zero. 

2H + (aq) + 2e - H 2 (g) E° = 0 volts 

Standard states (a=1): [H + ] 1 M, P H2 = 1 bar 1 atm 

The half-cell potential will differ from zero if H + and/or 

H 2 are not in their standard states. 


Standard Hydrogen Electrode - continued 

H 2 (g) at one bar 

bubbled over a 

platinum electrode. 

Pt acts a catalyst for 

the reaction. 

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

Used as a basis for 

calculations. Not 

really very practical. 


Standard Electrode Potentials 

The standard electrode potential for a half-cell is the 

potential when all species are in their standard states. 

refers to reduction at the electrode (these days) 

measured relative to a standard hydrogen electrode as 

the anode 

Example: Cell for measuring E° for Cu 2+ /Cu. 

Anode: H 2 (g) 2H + + 2e - (oxidation) 

Cathode: Cu 2+ (1 M) + 2e - Cu(s) (reduction) 

Cell Diagram: PtH 2 (1 bar)H + (1 M)Cu 2+ (1 M)Cu(s) 

Cell potential is 0.340 V. So E° = 0.340 V for Cu 2+ /Cu. 


Standard Electrode Potentials - continued 

Reduction Half-Reaction E° (volts) 

F 2 (g) + 2e - 2F - (aq) 2.866 

O 2 (g) + 4H + (aq) + 4e - 2H 2 O(l) 1.229 

2H + (aq) + 2e - H 2 (g) 0.000 

Zn 2+ (aq) + 2e - Zn(s) -0.763 

Li + (aq) + e - Li(s) -3.040 

F 2 is easiest to reduce (largest E°). F - is hardest to oxidize. 

Li + is hardest to reduce. Li is easiest to oxidize. 

F 2 is best oxidizing agent; Li is best reducing agent. 


Using Standard Electrode Potentials 

A standard cell potential, E° cell , may be calculated from the 

standard electrode potentials for the cathode, E° cathode , and 

anode, E° anode : 

E° cell = E° cathode - E° anode 

The anode potential is subtracted since the potential is for 

reduction and the anode reaction is oxidation. 

Standard electrode potentials are listed in tables. 

Standard electrode potentials do not depend on how a 

reaction is written since they are related to r G° per mole 

of electrons. ( r G° = -nFE° cell ) 


Standard Electrode Potentials - example 

Find the standard cell potential for the reaction: 

Zn(s) + Cl 2 (g) Zn 2+ (aq) + 2Cl - (aq) 

Solution: Write half-cell reactions and find E° values. 

Oxidation: Zn(s) Zn 2+ (aq) + 2e - 

Reduction: Cl 2 (g) + 2e - 2Cl - (aq) 

From Table 21.1: 

Zn 2+ (aq) + 2e - Zn(s) E° = -0.763 V 

Cl 2 (g) + 2e - 2Cl - (aq) E° = 1.358 V 

E° cell = E° cathode - E° anode = 1.358 - (-0.763) = 2.121 V 


Cell Potential and Equilibrium Constant 

We have derived the following two equations: 

r G° = -nFE° cell and r G° = -RTln K eq 

Combining these gives 

Uses of this equation: 

E° cell = lnK eq 

Calculating K eq from standard half-cell potentials (see 

example 21-7 in text). 

Relating E° cell for different reactions. 


Cell Potential and Equilibrium - Example 

At 298.15 K, the standard reduction potential for O 2 (g) in 

acidic solution is 1.229 V: 

O 2 (g) + 4H + (aq) + 4e - 2H 2 O(l) E 1 ° = 1.229 V 

Find the standard reduction potential for O 2 (g) in basic 

solution: 

O 2 (g) + 2H 2 O(l) + 4e - 4OH - (aq) E 2 ° = ? 

Solution: The second reaction is equal to the first plus 

4H 2 O(l) 4H + (aq) + 4OH - (aq) K eq = K W 

4 

So K 2 = K 1 K W 4 , E 2 ° = (RT/nF)ln(K 1 K W 4 ), n = 4 

E 2 ° = E 1 ° + (RT/F)lnK W = 1.229 + 0.02569 ln(1.0×10 -14 ) 

E 2 ° = 0.401 V 


Dissolving Metals with Acids 

Many metals are dissolved by acids with the evolution 

of H 2 (g). 

Oxidation: M(s) M n+ (aq) + ne - E° = E° M 

Reduction: 2H + (aq) + 2e - H 2 (g) E° = 0 

Overall: M(s) + nH + (aq) M n+ (aq) + (n/2)H 2 (g) E° = -E° M 

Conclusions: 

The more negative the standard reduction potential of the 

metal ion, the easier the metal is to dissolve. 

Lowering the pH promotes the dissolution of metals. 


Dissolving Metals with Acids - example 

Determine the concentrations of each of the following 

metals that will dissolve at pH = 7.00 and pH = 0.00. 

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

Pb 2+ (aq) + 2e - Pb(s) 

Zn 2+ (aq) + 2e - Zn(s) 

Solution: M(s) + 2H + (aq) M 2+ (aq) + H 2 (g) 

E° = 0.340 V 

E° = -0.125 V 

E° = -0.763 V 

n = 2 E° cell = -E° T = 298.15 K P H2 1 bar 

E° cell = (RT/nF)lnK eq K eq = exp(-(77.85 V -1 )E°) 

[M 2+ ] [H + ] 2 K eq = [H + ] 2 exp(-(77.85 V -1 )E°) 


Dissolving Metals with Acids - continued 

[M 2+ ] [H + ] 2 exp(-(77.85 V -1 )E°) 

pH = 7 pH = 0 E° (V) 

[Cu 2+ ] eq = 3.2×10 -26 M 3.2×10 -12 M 0.340 

[Pb 2+ ] eq = 1.7×10 -10 M 1.7×10 4 M -0.125 

[Zn 2+ ] eq = 6.3×10 11 M 6.3×10 25 M -0.763 

Metals with E° 0 are difficult to dissolve even in 

strong acids. 

Metals with E° 0 will dissolve in strong acids. 

Metals with E° 0 will dissolve in water. 


Enhancing Dissolution of Metals 

Concentrated HNO 3 will dissolve Cu(s): 

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

NO 3 - + 4H + + 4e - NO(g) + H 2 O(l) 

E° = 0.340 V 

E° = 0.956 V 

For the overall reaction, E° = (0.956 - 0.340) V = 0.616 V. 

This is very favorable. 

Gold can be dissolved using aqua regia (1 part HNO 3 to 

3 parts HCl): 

Au 3+ (aq) + 3e - Au(s) 

E° = 1.52 V 

Au 3+ (aq) + 4Cl - (aq) [AuCl 4 ] - (aq) 


We have shown that 

The Nernst Equation 


But r G depends on the concentrations of reactants and 

products: 

r G = r G° + RTlnQ 

Therefore, E cell also depends on concentrations. Combining 

these gives nFE cell = nFE° cell - RTlnQ. So 

This is known as the Nernst Equation. 


The Nernst Equation - continued 

Here: R = 8.314 J mol -1 K -1 

F = 96,485 C mol -1 = 96,485 J V -1 mol -1 

lnQ = ln(10)×logQ = 2.303 logQ 

So 

E cell = E° cell - (1.984×10 -4 V K -1 )(T/n)logQ 

If T = 298.15 K (25 °C), then 

E cell = E° cell - (1/n)(0.05916 V)logQ 


Using the Nernst Equation 

Find E cell at 298 K for the cell 

PtFe 2+ (0.10 M),Fe 3+ (0.20 M)Ag + (1.0 M)Ag(s) 

Solution: First find E° cell , then use the Nernst Equation. 

Anode: Fe 2+ Fe 3+ + e - E° = 0.771 V 

Cathode: Ag + + e - Ag(s) E° = 0.800 V 

Cell: Fe 2+ + Ag + Fe 3+ + Ag(s) 

E° cell = E° cathode - E° anode = 0.029 V n = 1 

Q = [Fe 3+ ] / [Fe 2+ ][Ag + ] = (0.20) / (0.10)(1.0) = 2.0 

E cell = E° cell - (1/n)(0.05916 V)logQ = 0.011 V 


We can make a cell 

with the same 

reaction occuring at 

both electrodes. 

E° cell = 0 

The cell voltage is 

due to the difference 

in concentration. 

Concentration Cells 


Determining K sp 

With saturated AgI(aq) at the anode and [Ag + ] = 0.100 M at 

the cathode, E cell = 0.417 V. Use this to find K sp . 

Anode: Ag(s) Ag + (aq, sat. AgI) + e - 

Cathode: 

Cell: 

Ag + (aq, 0.1M) + e - Ag(s) 

Ag + (aq, 0.1M) Ag + (aq, sat. AgI) 

E° cell = 0. Q = [Ag + ] sat,KI / [0.1 M]. n = 1. 

The Nernst equation becomes (at 298.15 K): 

0.417 V = E cell = - (0.05916 V) log([Ag + ] sat,KI / [0.1 M]) 

[Ag + ] sat,KI = [0.1 M] 10 -7.049 = 8.94×10 -9 M 

K sp = [I - ][Ag + ] = (8.94×10 -9 ) 2 = 7.99×10 -17 


Electrochemistry Basics - Summary 

Electrochemical cells permit us to couple electrical 

work to chemical reactions. 

Cell potential and r G are directly related: 

r G = -nFE cell . 

Standard reduction potentials are tabulated and may 

be used to compute E° cell and r G°. 

The effect of concentration on cell potential is given 

by the Nernst equation:

Electrochemistry

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