voltaic cell help?

Question

A voltaic cell consists of a Ag/Ag+ electrode (E° = 0.80 V) and a Fe2+/Fe3+ electrode (E° = 0.77 V) with the following initial molar concentrations: [Fe2+] = 0.30 M; [Fe3+] = 0.10 M; [Ag+] = 0.30 M. What is the equilibrium concentration of Fe3+? (Assume the anode and cathode solutions are of equal volume, and a temperature of 25°C.)
0.030 M
0.043 M
0.085 M
0.11 M
0.17 M

Answer 1

Since it's a voltaic cell, the reaction must be spontaneous, therefore the reaction is:

Ag+ + Fe2+ ---> Ag + Fe3+

Eo = 0.80 - 0.77 = 0.03 V

ln K = nFEo/RT = 1*96500*0.03/(8.314*298) = 1.168
K = e^1.168 = 3.22

Q for the initial condition of the cell = [Fe3+]/[Ag+][Fe2+] = 0.10/(0.30)^2 = 1.11

Since Q < K, the reaction will shift to the right, so only the last two answers are possible.

If [Fe3+] = 0.11, then [Ag+]=[Fe2+] = 0.29 (as x = 0.01), but that gives a Q of 1.31 (0.11 / 0.29^2). Using [Fe3+] = 0.17 and [Ag+]=[Fe2+] = 0.23, you get a Q of 3.22,, which equals K, so [Fe3+] = 0.17 (you could have used the quadratic to solve for x, but given only two answers, it was easier to go trial and error).