For the cell reaction Cu²+ (aq) + Fe (s) -> Fe²+ (aq) + Cu (s), determine the standard Gibbs free energy change. Take E°(Cu²+/Cu) = +0.34 V and E°(Fe²+/Fe) = -0.44 V.

-150540 J
+150540 J
-15054 J
+15054 J

Correct answer: -150540 J

Solution

Fe is oxidized (anode) and Cu²+ is reduced (cathode). E°cell = 0.34 - (-0.44) = 0.78 V. With n = 2, ΔG° = -2 * 96500 * 0.78 = -150540 J, which is negative, confirming a spontaneous reaction.

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