A voltaic cell operates with 0.40 M Zn 2+ at the solid zinc anode and 1.45 M Zn2+ at the solid zinc cathode at 25 C. Is the reaction thermodynamically favored?

Question

DrBob222 · Accepted Answer

I would write the reaction as
Zn|Zn^2+(0.4M)||Zn^2+(1.45M)|Zn

Zn ==> Zn^2+(0.40) + 2e at the anode
Zn^2+(1.45) + 2e ==> Zn at the cathode
-----------------------------
Zn + Zn^2+(1.45M) ==> Zn + Zn^2+(0.4)

Then Ecell = Eocell - [0.05916/2]*log Q
where Q is 0.40/1.45. Eocell is zero of course. Evaluate -0.05916/2]*log Q. A + voltage means favorable. A - voltage means not favorable.