The standard potentials of MnO4– \| Mn2+ and MnO2 \| Mn2+ electrodes in acid solution are 1.51 and 1 — Electrochemistry Chemistry Question

💡 Solution & Explanation

(1) $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$; $\Delta G_1^\circ = -5F(1.51) = -7.55F$.<br>(2) $MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{2+} + 2H_2O$; $\Delta G_2^\circ = -2F(1.23) = -2.46F$.<br>Subtracting (2) from (1) gives: $MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O$. $\Delta G^\circ = -7.55F - (-2.46F) = -5.09F$. $E^\circ = -(-5.09F) / 3F = +1.697$ V.