Calculate the \of the galvanic cell containing the copper electrode and the zinc electrode where one is immersed in1.5 M CuSO4 solution and the other in 0.01 M ZnSO4 solution (E(Zn=-0.76 V, E(Cu=+0.34 V)

Answer 1

Zn ==> Zn^2+ + 2e .........Eo ox = +0.76
Cu^2+ + 2e ==> Cu..........Eo red = +0.34
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Zn(s) + Cu^2+(1M) ==> Zn^2+(1M) + Cu(s) .......Eo cell (std) = 1.10 v
Ecell = Eocell std - (0.059/n)*log[(Zn^2)(Cu)/(Zn)(Cu^2+)]
So substitute and solve for Ecell.
Eo cell std is the 1.10 v
n = total electrons exchanged = 2
(Zn) = 1.0 and (Cu) = 1.0 by definition
(Zn^2+) = 0.01 M from the problem.
(Cu^2+) = 1.5 M from the problem.
Post your work if you get stuck.