A voltaic cell consists of an Al/Al3+ half-cell and a Cd/Cd2+ half-cell. Calculate {Al3+} when {Cd2+} = 0.401 M and Ecell = 1.299 V.Use reduction potential values of Al3+ = -1.66 V and for Cd2+ = -0.40 V.

Answer 1

2Al(s) ==> 2Al^3+ + 3e Eo ox = +1.66 v
3Cd^2+ + 2e ==> 3Cd(s) Eo red = -0.40
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2Al(s) + 3Cd^2+ ==> 2Al^3+ + 3Cd(s)

So Eo cell = Eo ox + Eo red = +1.66-0.40 = ?
You know Ecell from the problem of 1.299. Plug those values into
Ecell = Eocell + (0.05916/n)log(K)
n = 6 for number of electrons transferred and
K = (Al^3+)^2((Cd)^3/((Al)^2(Cd^2|+)^3
Remember that (Al) and (Cd) in the solid state are 1 by definition. You know (Cd^2+) from the problem and you can calculate (Al^3+).
Post your work if you get stuck.

Answer 2

Thanks for your reply. I was able to get the answer.

Answer 3

Thank you for letting me know.