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?.