- We have the following two half-reactions:
- Oxidation: Sn(s)→Sn2+(aq, 1.50 M )+2e−
- Reduction: ClO2(g, 0.180 atm )+e−→ClO−2(aq, 1.65 M )
- To calculate the cell potential, we use the formula: Ecell = E°cell - (RT/nF)ln(Q)
E°cell = Standard state cell potential
R = 0.0821 Lkmol^-1K^-1 (gas constant)
T = 298 K
n = Number of electrons transferred in balanced redox reaction = 2 (from the half-reactions)
F = 96485 C/mol (Faraday's constant)
Q = Reaction quotient = [Sn^2+] [ClO2^-] / [Sn] [ClO2]
1. Standard state cell potential (E°cell): Since we have Sn/Sn^2+ and ClO2/ClO2^- half-cells, E°cell = E°Sn/Sn^2+ - E°ClO2/ClO2^-
= -0.76 V - 0.94 V = -1.7 V
2. Reaction quotient (Q):
[Sn^2+] = 1.50 M
[ClO2^-] = 1.65 M
[Sn] = 1 M (assumed, since Sn is solid)
[ClO2] = 0.180 atm = 0.180 M
So Q = (1.50 M) (1.65 M) / (1 M) (0.180 M) = 9:1
3. Substitute into cell potential formula:
Ecell = -1.7 V - (0.0821 Lkmol^-1K^-1 * 298 K) * ln(9)
Ecell = -1.7 V - 0.0613 * ln(9)
Ecell = -1.76 V
So the cell potential at 25°C is -1.76 V
Let me know if you have any other questions!