Use the Nernst equation to calculate the iron (II) concentration for the following

Question

iron/copper electrochemical cell when [Cu2+] = 0.012 M and Ecell = 0.750 V at 35°C.
Cu2+ (aq) + Fe (s) --> Cu (s) + Fe2+ (aq)

Answer 1

I presume you meant 25C and not 35. If you meant 35 then replace the 0.0592 below by (RT/2*F)
Ecell = Eocell -(0.0592/2)log Q
You have Ecell.
Eocell = Eo for Cu^2+ + 2e --> Cu
and add to Eo for Fe(s) ==> Fe^2+ + 2e

That leaves Q. Q in this case is
Q = (Fe^2+)/(Cu^2+)

Substitute and go.

Answer 2

Thank you very much, but the problem indeed says 35 degree C. I was wondering if there is a way to calculate the Fe2+ (aq)?

Answer 3

Yes, if you follow my directions above that will calculate Fe^2+. You have Eocell and Ecell which allows you to calculate Q. Then Q has the definition I gave and that has the concentrations in it. Q = (Fe^2+)/(Cu^2+). You know Q and Cu^2+, solve for Fe^2+.
If you need help in evaluating RT/2F note that I omitted the 2.303 to convert log base e to log base 10.
RT/2*F = (2.303)(8.314*308.15)/(2*96,485) = ?.

Answer 4

Ahhhh, now it's all clear! Thank you!