The half-cell reactions for the rusting of iron are given as: with , and with . What is the standard — Electrochemistry Chemistry Question
Question
The half-cell reactions for the rusting of iron are given as: $2H^+(aq) + \frac{1}{2}O_2(g) + 2e^- \rightarrow H_2O(l)$ with $E^\circ = +1.23\text{ V}$, and $Fe^{2+}(aq) + 2e^- \rightarrow Fe(s)$ with $E^\circ = -0.44\text{ V}$. What is the standard Gibbs free energy change ($\Delta G^\circ$) for the overall rusting reaction? (Use $1\text{ F} = 96500\text{ C mol}^{-1}$)
💡 Solution & Explanation
The overall reaction is obtained by reversing the iron half-reaction and adding it to the oxygen reduction: $Fe(s) + 2H^+(aq) + \frac{1}{2}O_2(g) \rightarrow Fe^{2+}(aq) + H_2O(l)$. The standard cell potential is $E^\circ = E^\circ_{cathode} - E^\circ_{anode} = 1.23 - (-0.44) = 1.67\text{ V}$. The number of moles of electrons transferred is $n=2$. Thus, $\Delta G^\circ = -nFE^\circ = -2 \times 96500 \times 1.67 = -322310\text{ J mol}^{-1} \approx -322\text{ kJ mol}^{-1}$.