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.
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+] = 1.0×109 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+] = 1.0×109 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
4 H+(aq) + 4 OH(aq) 4 H2O(l)
H+(aq) + OH(aq) H2O(l)
Eo = 0.401 + 1.229 = +0.828 V
25 °C = 298 K
37 °C = 310 K
+ 0.705 V
50 °C = 323 K
The temperature dependence of cell voltages is typically quite small.