How does concentration affect the cell potential? Do two identical half-cells co

Question

How does concentration affect the cell potential? Do two identical half-cells constitute a galvanic cell? Construct the following cells and measure their cell potentials: Zn(s)|Zn^2+(aq, 0.1M)||Cu^2+(aq, 0.1M)|Cu(s) Zn(s)|Zn^2+(aq, 1 M)|| Cu^2+(aq, 0.1M)|Cu(s) Zn(s)|Zn^2+(aq, 0.1M)||Cu^2+(aq, 1M)|Cu(s) Zn(s)|Zn^2+(aq, 0.1M)||Zn^2+(aq, 0.1M)|Zn(s) Zn(s)|Zn^2+(aq, 0.1M)||Zn^2+(aq, 1M)|Zn(s) Cu(s)|Cu^2+(aq, 0.1M)||Cu^2+(aq, 1M)|Cu(s) Calculate the theoretical cell potentials for these six cells and com with the experimental ones.

Explanation / Answer

Oxidation Cell

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

Reduction Cell

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

Net Reaction

Zn(s) + Cu(+2) ------ Zn(+2) + Cu(s)

Eocell = Eocell(oxidation) + Eocell(reduction) = 0.763+0.337 = 1.100V

a) K = [Zn(+2)]/[Cu(+2)] = 1

Ecell = Eocell - 0.0592/2 log(1) = 1.100V

b) K = [Zn(+2)]/[Cu(+2)] = 10

Ecell = Eocell - 0.0592/2 log(010) = 1.0704V

c) K = [Zn(+2)]/[Cu(+2)] = 0.1

Ecell = Eocell - 0.0592/2 log(0.1) = 1.1296V

d) K = [Zn(+2)]/[Cu(+2)] = 1

Ecell = Eocell - 0.0592/2 log(1) = 1.100V

e) K = [Zn(+2)]/[Cu(+2)] = 0.1

Ecell = Eocell - 0.0592/2 log(0.10) = 1.1296V

c) K = [Cu(+2)]/[Cu(+2)] = 0.1

Ecell = 0- 0.0592/2 log(0.1) = 0.0296V

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