The rusting of iron can be treated as an electrochemical process. The half-cell reactions are: () an — Electrochemistry Chemistry Question
Question
The rusting of iron can be treated as an electrochemical process. The half-cell reactions are: $O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$ ($E^\circ = 1.23\text{ V}$) and $Fe^{2+}(aq) + 2e^- \rightarrow Fe(s)$ ($E^\circ = -0.44\text{ V}$). An iron pipe is exposed to moisture containing dissolved oxygen at a partial pressure of $0.1\text{ atm}$ and a pH of $3.0$. If the concentration of $Fe^{2+}$ in the moisture layer is $10^{-4}\text{ M}$, calculate the instantaneous operating potential ($E_{cell}$) of this corrosion cell in Volts at $298\text{ K}$. (Take $\frac{2.303RT}{F} = 0.0591\text{ V}$)
💡 Solution & Explanation
The overall reaction is $2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l)$, involving $n=4$ electrons. Standard EMF $E^\circ_{cell} = 1.23 - (-0.44) = 1.67\text{ V}$. At pH = $3.0$, $[H^+] = 10^{-3}\text{ M}$. Reaction quotient $Q = \frac{[Fe^{2+}]^2}{P_{O_2}[H^+]^4} = \frac{(10^{-4})^2}{(0.1)(10^{-3})^4} = \frac{10^{-8}}{10^{-1} \cdot 10^{-12}} = \frac{10^{-8}}{10^{-13}} = 10^5$. By Nernst equation: $E_{cell} = E^\circ_{cell} - \frac{0.0591}{4} \log(10^5) = 1.67 - \frac{0.0591 \times 5}{4} = 1.67 - 0.073875 \approx 1.596\text{ V}$. Rounding slightly gives $1.60\text{ V}$.