First calculate E°cell based on the two half-reactions:
Oxidation Zn (s) ––�?gt; Zn2+ (aq) + 2 e�?/SUP> E°ox = +0.76 V
Half-Rxn
Reduction Sn2+ (aq) + 2 e�?/SUP> ––�?gt; Sn (s) E°red = �?.14 V
Half-Rxn
Overall Zn (s) + Sn2+ (aq) ––�?gt; Sn (s) + Zn2+ (aq) E°cell = +0.62 V
Verify the half-reaction potentials from your table of standard reduction potentials. Now you can plug into ΔG° = –n F E°cell. The moles of electrons transferred is 2 (2 electrons lost, 2 electrons gained). Also notice, when E°cell is positive, ΔG° will be negative, meaning that the overall reaction will occur spontaneously, and the cell will therefore produce electricity (the flow of electrons from the anode to the cathode) as the reaction occurs. 
Steve
Free Forum Hosting
Important Announcement
Zn2+(aq) + Sn(s)
First 