Consider the following electochemical cell.

Question

Pt | Fe3+ (0.0600 M), Fe2+ (6.50e-5 M) || Sn2+ (0.0620 M), Sn4+ (3.90e-4 M) | Pt

(a) Calculate the voltage of the cell, Ecell, including the sign.

Answer 1

Use the Nernst equation to calculate the half cell potential of the Fe+3/Fe+2 electrode.
Use the Nernst equation to calculate the half cell potential of the Sn+4/Sn+2 electrode. Note I used REDUCTIONS for both. Then I would proceed as follows:
Identify the more negative electrode of the two reduction potentials, reverse it and the sign, add the two half equations to find the SPONTANEOUS cell reaction and spontaneous cell potential. Compare the spontaneous reaction with the reaction of the cell, which I've given below. Adjust your equation and potential (the sign) as needed. The reaction of the cell, as written, is:
2Fe^+3 + Sn^+2 ==> 2Fe+2 + Sn^+4

Answer 2

klj