The standard reduction potentials of and are and respectively. Based on these thermodynamic values, — Electrochemistry Chemistry Question

💡 Solution & Explanation

Standard potentials cannot be added directly, but Gibbs free energy can. $\Delta G^\circ_1 (Cu^{2+} \rightarrow Cu) = -2F(0.337) = -0.674F$. $\Delta G^\circ_2 (Cu^{2+} \rightarrow Cu^+) = -1F(0.153) = -0.153F$. Target reaction: $Cu^+ + e^- \rightarrow Cu$. $\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2 = -0.674F - (-0.153F) = -0.521F$. Using $\Delta G^\circ_3 = -1 \times F \times E^\circ$, we get $E^\circ = 0.521\text{ V}$.