Question

Using the half reaction potentials calculate the E^o cell for the Ni-Al cell.
Ni (aq) + 2e- <-> Ni (s) -0.25v
Al3+ (aq) + 3e- <-> Al (s) -1.66v

Using that answer determine Ecell for Ni-Al cell when the concentration of Al3+ is 4.00M and the concentration of Ni2+ (aq) is 2.00*10^-5 M.

Answer 1

I am unsure about the nomenclature for this. I have set it up for the cell to be spontaneous; the Al is the anode and Ni is the cathode. If the term "Ni-Al cell" means that the cell is non-spontaneous, you will need to reverse everything. Hope this helps.
Al ==> Al^3+ + 3e Eo = 1.66v
Ni^2+ + 2e ==> Ni Eo = -0.25
----------------------------
2Al + 3Ni^2+ ==> 2Al^3+ + 3Ni
Eocell = EAl above + ENi above = 1.66+(-0.25) = ?.

For the second part:
Ecell = Eocell - (0.0592/6)log Q
where Q = (Al^3+)^2(Ni)^3/(Al)^2(Ni^2+)^3 and solve for Ecell.