The free energy change for the following reaction at 25 °C, when [Pb2+] = 3.82×10-3 M and [Cu2+] = 1.20 M, is 68.1 kJ: Pb2+(3.82×10-3 M) + 2Cu+(aq)Pb(s) + 2Cu2+(1.20 M) ΔG = 68.1 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction?

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