Answer to Question #29891 in Physical Chemistry for Branden
I2(s) + 2e− → 2I−(aq)
Eanode = 0.53V
Fe3+(aq) + e− → Fe2+(aq)
Ecathode = 0.77 V
Since Fe3+/Fe2+ is more positive, it should oxidize I– to I2. This makes the I2/I– half-reaction the anode. The standard emf can be found
Ecell = Ecathode − Eanode = 0.77 V − 0.53 V = 0.24 V
Dear Bodacious Bonds! Thank you for adding new information.
The question was never explicitly answered. It should be stated that because E(cell) is positive this DOES indeed favor the forward reaction of Fe3+ oxidizing I2 to I-.