The standard reduction potential of oxygen in acidic solution is 1.23 V (O2 + 4H3O+ + 4e– 6H2O). The — Electrochemistry Chemistry Question

💡 Solution & Explanation

Acidic reduction: $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$, $\Delta G^\circ = -4F(1.23) = -4.92F$. Water dissociation: $4H_2O \rightarrow 4H^+ + 4OH^-$, $\Delta G^\circ = -RT \ln(10^{-56}) = -2.303RT \times (-56) = 56 \times 0.06F = 3.36F$. Adding these reactions gives basic reduction: $O_2 + 2H_2O + 4e^- \rightarrow 4OH^-$, $\Delta G^\circ = -4.92F + 3.36F = -1.56F$. Since $\Delta G^\circ = -nFE^\circ$, $E^\circ = -(-1.56F) / 4F = +0.39$ V.