For the reaction, 2Cu+→Cu2++Cu the cathode is Cu+∕Cu and anode is Cu+∕Cu2+. Given, Cu2++2e−→Cu;E°1=0.34V...(1) Cu2++e−→Cu+;E°2=0.15V...(2) Cu++e−→Cu;E°3=?...(3) Now ΔG°1=−nFE°1=−2×0.34×F ΔG°2=−1×0.15×F, ΔG°3=−1×E°3×F Again, ΔG°1=ΔG°2+ΔG°3 ⇒−0.68F=−0.15F−E°3×F ⇒E°3=0.68−0.15=0.53V As, E°cell=E°cathode(Cu+∕Cu)−E°anode(Cu2+∕Cu+) =0.53−0.15=0.38V