# Electrochemistry question!

1. Nov 26, 2006

### leaf345

Hey guys, I just did a lab on electrochem but we haven't started learning this in our lectures so I'm not too confident on it. One of the lab question asks us to calculate the expected cell voltage. From reading the text book, this is what I did:
The reaction was 2Ag+ + Cu -> 2Ag + Cu2+
I added reduction potentials of Ag and Cu to get 0.46V for the standard voltage. Then I used the nersnt eq'n
E= 0.46 -(0.05915/n)*logQ
What I'm not sure about is...
We used 0.1 M AgNO3 and Cu(NO3)2, so would n be 2 or 0.2?
And would Q= [0.1 Cu2+][1]/[0.1 Ag+]^2*[1]=10?
Help would be appreciated!

2. Nov 26, 2006

### leaf345

Also, we did another cell that is giving me trouble too.
One 1/2 cell was:
Ag electrode
0.1 M AgNO3 solution

The other was:
Pt electrode
0.1 M Fe2+ and 0.1 M Fe3+

I think the overall redox rx'n was Ag+ + Fe2+ --> Fe3+ + Ag
so I calculated the expected cell voltage. But I get a value that is opposite to the sign of the voltage we observed.
What I did was
E=Eo -(0.05915/n)*logQ
where Eo= 0.8-0.77 because that is the difference between the redox potentials of Ag and Fe.
n=1 electron
Q= [Fe3+]/[Fe2+]*[Ag+] = 0.1/(0.1*0.1)=10
so E= negative something

Am I doing this one correct?