Some standard electrode potentials are given: Fe^2+ + 2e^- -> Fe; E° = -0.440 V, Fe^3+ + 3e^- -> Fe; — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write down the Given Half-Reactions and Standard Potentials We are given the standard reduction potentials ($E^\circ$) at $25^\circ\text{C}$ for two iron redox couples: 1. For the reduction of $\ce{Fe^{2+}}$ to metallic iron ($\ce{Fe}$): $$\ce{Fe^{2+}(aq) + 2e^{-} -> Fe(s)} \quad E^\circ_1 = -0.440\text{ V}$$ The number of electrons transferred in this reaction is $n_1 = 2$. 2. For the reduction of $\ce{Fe^{3+}}$ to metallic iron ($\ce{Fe}$): $$\ce{Fe^{3+}(aq) + 3e^{-} -> Fe(s)} \quad E^\circ_2 = -0.036\text{ V}$$ The number of electrons transferred in this reaction is $n_2 = 3$. Step 2 - Identify the Target Half-Reaction Our objective is to calculate the standard reduction potential ($E^\circ_3$) for the target half-reaction: $$\ce{Fe^{3+}(aq) + e^{-} -> Fe^{2+}(aq)} \quad E^\circ_3 = ?$$ The number of electrons transferred in this reaction is $n_3 = 1$. Step 3 - Establish the Thermodynamic Relationship Standard electrode potentials ($E^\circ$) are intensive thermodynamic properties and cannot be added or subtracted directly. Instead, we must convert them into standard Gibbs free energy changes ($\Delta G^\circ$), which are extensive properties and are additive: $$\Delta G^\circ = -n F E^\circ$$ Let us express the standard Gibbs free energy changes for the three reactions: * For Reaction 1: $\Delta G^\circ_1 = -n_1 F E^\circ_1 = -2 \times F \times (-0.440\text{ V}) = +0.880 F$ * For Reaction 2: $\Delta G^\circ_2 = -n_2 F E^\circ_2 = -3 \times F \times (-0.036\text{ V}) = +0.108 F$ * For Reaction 3 (target): $\Delta G^\circ_3 = -n_3 F E^\circ_3 = -1 \times F \times E^\circ_3$ We can obtain the target reaction (Reaction 3) by subtracting the first half-reaction (Reaction 1) from the second half-reaction (Reaction 2): $$\text{Reaction 3} = \text{Reaction 2} - \text{Reaction 1}$$ $$\ce{(Fe^{3+} + 3e^{-} -> Fe) - (Fe^{2+} + 2e^{-} -> Fe) \implies Fe^{3+} + e^{-} -> Fe^{2+}}$$ Correspondingly, their standard Gibbs free energies are related as: $$\Delta G^\circ_3 = \Delta G^\circ_2 - \Delta G^\circ_1$$ Step 4 - Calculate the Standard Potential of the Target Half-Cell Substituting the Gibbs free energy expressions into the thermodynamic relation: $$-n_3 F E^\circ_3 = -n_2 F E^\circ_2 - (-n_1 F E^\circ_1)$$ Dividing both sides of the equation by $-F$ gives the general formula: $$n_3 E^\circ_3 = n_2 E^\circ_2 - n_1 E^\circ_1$$ Now, substitute the values of $n_1, E^\circ_1, n_2, E^\circ_2,$ and $n_3$ into the equation: $$1 \times E^\circ_3 = 3 \times (-0.036\text{ V}) - 2 \times (-0.440\text{ V})$$ $$E^\circ_3 = -0.108\text{ V} - (-0.880\text{ V})$$ $$E^\circ_3 = -0.108\text{ V} + 0.880\text{ V}$$ $$E^\circ_3 = \boxed{+0.772\text{ V}}$$ Step 5 - Explanation of Each Option * **Option (A) is incorrect:** A potential of $-0.476\text{ V}$ is obtained by directly adding the standard potentials: $-0.440\text{ V} + (-0.036\text{ V}) = -0.476\text{ V}$. This is thermodynamically incorrect because intensive properties cannot be added directly. * **Option (B) is incorrect:** A potential of $-0.404\text{ V}$ is obtained by directly subtracting the standard potentials: $-0.440\text{ V} - (-0.036\text{ V}) = -0.404\text{ V}$. This is incorrect for the same reason. * **Option (C) is incorrect:** A potential of $+0.988\text{ V}$ is numerically incorrect and does not satisfy the free energy relations. * **Option (D) is correct:** Our calculation using the Gibbs free energy relationship yields exactly $+0.772\text{ V}$. $$\text{Correct Option: } \boxed{\text{D}}$$