Calculate [Ag+] in Test Solution with Nernst Eq. (Chem 102 Lab)

[loadsy]

Homework Statement

The question asks: Using the measured cell potential and the Nernst equation, calculate the [Ag+] in the test solution. However, you guys will need a little bit of background information to answer this question since it is a lab. First of all this reaction was:

Ag2CO3(s) <-> 2Ag+(aq) +CO3^2-(aq)

Also the Nernst equation we are using is the following:
Ecell = E^0cell -0.05916V/n logQcell

And, the Ecell potential that I calculated was 0.280V

2. The attempt at a solution

So I've made an attempt at this question, but I'm not too positive if I'm solving it correctly. From the reaction we know that n=1 (1 electron being transferred between the reactants and products). Therefore:

0.280V = 0.800V-0.468V(Standard potentials off CDS)- 0.05916V/1 log[Ag+]^2
0.280V = 0.332V-0.5916V log[Ag+]^2
1.026 = log[Ag+]^2
1.013 = log[Ag+]
[Ag+] 0.00562 = 5.63X10^-3 mol/L

However, I'm not too positive if this is correct or if I should be looking in the Chemistry Data Sheet for the standard potentials of the cathode and anode. Thanks for any help guys. :D

 

[chaoseverlasting]

Actually, here n will be 2 since the total number of electrons traded is 2. Your calculation looks wrong. If 1.013=log[Ag+], then [Ag+] should be $$10^{1.013}$$ which should be greater than 10, not smaller.

 

[loadsy]

Ahhh okay, yeah I'm not sure why I said 1, it clearly is n=2 for the number of electrons being traded. That makes a bit more sense. Thank you for your help.

My only other question is whether or not I got the standard cell potentials right from the chemistry data sheet. Because I know that the cathode is Ag in this case, so looking on the CDS, we find it is 0.800V and then the anode I said was from the equation Ag2CO3(s) + 2e^- -> 2Ag(s) + CO3^2- which is 0.468. I'm assuming those are correct for this case, and that it's just an error in my calculation. Thanks again though.