Chemistry 112

Cell Potentials at Nonstandard Conditions

From thermodynamics, recall

ΔG = ΔG° + RTln Q

If we divide everything by –nF

ΔG        –nF = ΔG°        –nF + RT        –nF ln Q

since ΔG = –nFE, or E = ΔG/(–nF)

E = E° – RT        nF ln Q

R = 8.314 J/mol·K (the gas constant)

F = 96485 coul/mol (Faraday's constant)

T = absolute temperature

n = number of moles of electrons transferred in the balanced equation

Q = reaction quotient

This equation is known as the Nernst Equation and can be used to find a cell potential at any set of conditions.

Example

Consider the Daniell Cell at 25 °C

Zn(s) + Cu²⁺(aq) → Cu(s) + Zn²⁺(aq)

Find the cell potential at the following conditions when [Cu²⁺] = 1.00 M, [Zn²⁺] = 1.0×10^–9 M and when [Cu²⁺] = 0.10 M, [Zn²⁺] = 0.90 M.

Recall that the standard potential for the Daniell cell is E° = +1.10 V

Then, the Nernst equation can be used to find the potentials at the nonstandard conditions:

E = E° – RT        nF ln Q

Q = [Zn²⁺]/[Cu²⁺]

E = E° – RT        nF ln ([Zn²⁺]/[Cu²⁺])

E = +1.10 – (8.314)(298)                       (2)(96485) ln ([Zn²⁺]/[Cu²⁺])

E = 1.10 – 0.0128 ln ([Zn²⁺]/[Cu²⁺])

This is the working equation specific to this problem.

When [Cu²⁺] = 1.00 M, [Zn²⁺] = 1.0×10^–9 M so E = +1.10 – 0.0128 ln(1.0×10^–9/1.00) = 1.37 V

When [Cu²⁺] = 0.10 M, [Zn²⁺] = 0.90 M so E = +1.10 – 0.0128 ln(0.90/0.10) = 1.07 V

A plot of the cell potential versus the reaction progress is shown below:

Consider the cell:

Pt(s)|OH^–(aq)(pH = 13.0)|O₂(g)(0.2 atm)||O₂(g)(0.2 atm)|H⁺(aq)(pH = 1.0)|Pt(s)

What are the cell potentials at 25 , 37, and 50 °C?

Anode (oxidation) reaction:

4 OH^–(aq) → O₂(g) + 2 H₂O(l) + 4 e^–

E°_ox = –0.401 V

Cathode (reduction) reaction:

O₂(g) + 4 H⁺(aq) + 4 e^– → 2 H₂O(l)

E°_red = +1.229 V

Net reaction:

4 H⁺(aq) + 4 OH^–(aq) → 4 H₂O(l)

OR:

H⁺(aq) + OH^–(aq) → H₂O(l)

E° = –0.401 + 1.229 = +0.828 V

E = E° – RT        nF ln Q

Q = Q_anode×Q_cathode

Q = P_O2anode                  [OH^–]⁴_anode × 1                                 P_O2cathode [H⁺]⁴_cathode

P_O2anode = P_O2cathode = 0.2 atm

[H⁺]_cathode = [OH^–]_cathode = 0.1 M

Q = (0.2)/(0.10)⁴(0.2)(0.1)⁴ = 1×10⁸

E = +0.828 – (8.314)T                (4)(96485) ln (1×10⁸)

E = 0.828 – 0.000397T

T (°C)T (K) E (V)

25298 +0.710

37310 +0.705

50323 +0.700

The temperature dependence of cell voltages is typically quite small.

Quiz 7