Chemistry 112

## Cell Potentials at Nonstandard Conditions

From thermodynamics, recall

G = Go + RTlnQ

If we divide everything by –nF

since G = –nFE, or E = G/(–nF)

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 Eo = +1.10 V

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

This is the working equation specific to this problem.

When [Cu2+] = 1.00 M, [Zn2+] = 1.0×10–9 M

When [Cu2+] = 0.10 M, [Zn2+] = 0.90 M

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

Eoox = –0.401 V

Cathode (reduction) reaction:

O2(g) + 4 H+(aq) + 4 e 2 H2O(l)

Eored = +1.229 V

Net reaction:

4 H+(aq) + 4 OH(aq) 4 H2O(l)

OR:

H+(aq) + OH(aq) H2O(l)

Eo = –0.401 + 1.229 = +0.828 V

 T E 25 °C = 298 K +0.710 V 37 °C = 310 K + 0.705 V 50 °C = 323 K +0.700 V

The temperature dependence of cell voltages is typically quite small.