[Single-digit Integer] The reduction potential at 25°C for Fe^3+ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Standard Reduction Potential ($E^\circ$) of the $\ce{Fe^3+|Fe^2+}$ Electrode To find the standard reduction potential of the $\ce{Fe^3+|Fe^2+}$ couple, we use standard Gibbs free energy ($\Delta G^\circ$) additivity with the given standard reduction potentials of the iron system: 1. **Reduction of $\ce{Fe^3+}$ to metallic iron ($\ce{Fe}$):** $$\ce{Fe^3+(aq) + 3e^- -> Fe(s)} \quad E^\circ_1 = -0.04\text{ V}, \quad n_1 = 3$$ $$\Delta G^\circ_1 = -n_1 F E^\circ_1 = -3 \times F \times (-0.04\text{ V}) = +0.12 F$$ 2. **Reduction of $\ce{Fe^2+}$ to metallic iron ($\ce{Fe}$):** $$\ce{Fe^2+(aq) + 2e^- -> Fe(s)} \quad E^\circ_2 = -0.44\text{ V}, \quad n_2 = 2$$ $$\Delta G^\circ_2 = -n_2 F E^\circ_2 = -2 \times F \times (-0.44\text{ V}) = +0.88 F$$ We want to find the standard potential ($E^\circ_3$) for the single-electron reduction: 3. **Reduction of $\ce{Fe^3+}$ to $\ce{Fe^2+}$:** $$\ce{Fe^3+(aq) + e^- -> Fe^2+(aq)} \quad E^\circ_3 = ?, \quad n_3 = 1$$ $$\Delta G^\circ_3 = -n_3 F E^\circ_3 = -1 \times F \times E^\circ_3 = -F E^\circ_3$$ Reaction (3) can be obtained by subtracting reaction (2) from reaction (1): $$\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2$$ $$-F E^\circ_3 = +0.12 F - 0.88 F$$ $$-F E^\circ_3 = -0.76 F \implies E^\circ_3 = +0.76\text{ V}$$ Thus, the standard reduction potential $E^\circ(\ce{Fe^3+|Fe^2+}) = +0.76\text{ V}$. Step 2 - Set up the Nernst Equation for the Electrode The reduction half-reaction at the electrode is: $$\ce{Fe^3+(aq) + e^- -> Fe^2+(aq)}$$ At $25^\circ\text{C}$ ($298\text{ K}$), we apply the Nernst equation for this single-electron transfer process ($n = 1$): $$E = E^\circ - \frac{2.303 RT}{F} \log_{10} \left( \frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} \right)$$ Using the textbook approximation $\frac{2.303 RT}{F} = 0.06$: $$E = E^\circ - 0.06 \log_{10} \left( \frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} \right)$$ Step 3 - Solve for the Concentration Ratio *(Note: The textbook question asks for the ratio of $[\ce{Fe^2+}]/[\ce{Fe^3+}]$ when the non-standard potential $E = +0.718\text{ V}$).* Substituting the calculated $E^\circ = 0.76\text{ V}$ and the given electrode potential $E = +0.718\text{ V}$: $$0.718\text{ V} = 0.76\text{ V} - 0.06 \log_{10} \left( \frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} \right)$$ Rearranging the terms to isolate the logarithmic expression: $$0.06 \log_{10} \left( \frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} \right) = 0.76 - 0.718 = 0.042\text{ V}$$ $$\log_{10} \left( \frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} \right) = \frac{0.042}{0.06} = 0.7$$ We are given that $\log_{10} 5 = 0.7$. Therefore: $$\frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} = 5$$ The ratio of the molar concentrations of $\ce{Fe^2+}$ to $\ce{Fe^3+}$ is exactly $5$. Since the question requires a single-digit integer format: $$\boxed{5}$$