3 Ce4+(aq) + Cr(s) �¨ 3 Ce3+(aq) + Cr3+(aq)

Question

What is the emf of this cell when [Ce4+] = 2.9 M, [Ce3+] = 0.11 M, and [Cr3+] = 0.015 M?

Answer 1

Do it this way.
Ce^4+ + e ==> Ce^3+ Eo = Eo1
Cr ==> Cr^3+ + 3e Eo = Eo2
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Eocell = Eo1 + Eo2

Then Ecell = Eocell - (0.05916/n)log Q where
Q = (Ce^3+)^3x(Cr^3+)/(Ce^4+)^3
Substitute the values and solve for Ecell.

Answer 2

for the E0cell i got 2.35 and for Q i got 8.186 then when i plugged that into the equation i get 2.7101V is that correct? I plugged 1 in for n.