[raafiki]

Use the following cell at 25 oC to answer the questions.
Cr / Cr3+ (0.0460 M) // Ag1+ (0.300 M) / Ag
What is the cell potential?

So I used the Ecell=Eo -.o592/n *log(Q)

So my half reactions were

Cr-> Cr3+ +3e-     Eo=-.74

3Ag+ + 3e- -> 3Ag   Eo=+.8

My combined reaction was
3Ag+ + Cr -> 3Ag + Cr3+


So I know n=3 because thats how many electrons got transfered, and I
know Q= (Ag)^3 * (Cr3+)/ (Ag+)^3 *(Cr) = (1^3)(.046)/(.3)^3(1) = 1.704

so the only problem I'm having is figuring out what I would use for Eo in the equation
 Ecell=Eo - .0592/3 *log(1.704)

Do I use the Eo value of the anode?.

[raafiki]

nevermind, i figured out that the Eo would be Eo(cathode) -Eo(anode) which would be .80-(-.74)=1.54