The standard reduction potential data at 25°C: ; ; ; ; ; ; ; . Match the of the redox pair in Column — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Standard Reduction Potential for (A) $E^\circ(\ce{Fe^3+/Fe})$ To find the standard reduction potential of the $\ce{Fe^3+/Fe}$ couple, we use the standard reduction potentials of the individual steps of the iron redox system: 1. $\ce{Fe^3+(aq) + e^- -> Fe^2+(aq)} \quad E^\circ_1 = +0.77\text{ V}, \quad n_1 = 1$ The standard Gibbs free energy change is: $$\Delta G^\circ_1 = -n_1 F E^\circ_1 = -1 \times F \times 0.77 = -0.77 F$$ 2. $\ce{Fe^2+(aq) + 2e^- -> Fe(s)} \quad E^\circ_2 = -0.44\text{ V}, \quad n_2 = 2$ The standard Gibbs free energy change is: $$\Delta G^\circ_2 = -n_2 F E^\circ_2 = -2 \times F \times (-0.44) = +0.88 F$$ The target reduction reaction is: 3. $\ce{Fe^3+(aq) + 3e^- -> Fe(s)} \quad E^\circ_3 = ?, \quad n_3 = 3$ The standard Gibbs free energy change is: $$\Delta G^\circ_3 = -n_3 F E^\circ_3 = -3 F E^\circ_3$$ Since reaction (3) is the direct sum of reactions (1) and (2), we apply the additivity rule of Gibbs free energy: $$\Delta G^\circ_3 = \Delta G^\circ_1 + \Delta G^\circ_2$$ $$-3 F E^\circ_3 = -0.77 F + 0.88 F$$ $$-3 E^\circ_3 = 0.11\text{ V}$$ $$E^\circ_3 = -\frac{0.11}{3}\text{ V} \approx -0.0367\text{ V} \approx -0.04\text{ V}$$ Thus, **(A) matches with (R)**. Step 2 - Calculate the Standard Reduction Potential for (B) $E^\circ(\ce{4H2O + 4e^- -> 2H2 + 4OH^-})$ This reaction represents the reduction of water under basic conditions. We can simplify this reaction on a per-electron basis: $$\ce{2H2O(l) + 2e^- -> H2(g) + 2OH^-(aq)}$$ We can calculate this potential at $25^\circ\text{C}$ ($298.15\text{ K}$) using the Nernst equation relative to the Standard Hydrogen Electrode (SHE): $$\ce{2H^+(aq) + 2e^- -> H2(g)} \quad E^\circ = 0.00\text{ V}$$ For standard basic conditions, the concentration of hydroxide ions is $[\ce{OH^-}] = 1.0\text{ M}$. Using the ion product of water ($K_w = 1.0 \times 10^{-14}$), the hydrogen ion concentration is: $$[\ce{H^+}] = \frac{K_w}{[\ce{OH^-}]} = 1.0 \times 10^{-14}\text{ M} \implies \text{pH} = 14$$ Assuming $P_{\ce{H2}} = 1\text{ atm}$, we apply the Nernst equation for the hydrogen reduction couple at $\text{pH} = 14$: $$E = E^\circ - \frac{2.303 RT}{F} \text{pH}$$ $$E = 0.00\text{ V} - 0.0592\text{ V} \times 14 \approx -0.8288\text{ V} \approx -0.83\text{ V}$$ Thus, **(B) matches with (S)**. Step 3 - Calculate the Potential for (C) $E^\circ(\ce{Cu^2+ + Cu -> 2Cu^+})$ To find the standard cell potential for this comproportionation reaction, we first determine the standard reduction potential of the $\ce{Cu^2+/Cu^+}$ couple ($E^\circ_3$) from the following given data: 1. $\ce{Cu^2+(aq) + 2e^- -> Cu(s)} \quad E^\circ_1 = +0.34\text{ V}, \quad n_1 = 2$ $$\Delta G^\circ_1 = -n_1 F E^\circ_1 = -2 \times F \times 0.34 = -0.68 F$$ 2. $\ce{Cu^+(aq) + e^- -> Cu(s)} \quad E^\circ_2 = +0.52\text{ V}, \quad n_2 = 1$ $$\Delta G^\circ_2 = -n_2 F E^\circ_2 = -1 \times F \times 0.52 = -0.52 F$$ Subtracting reaction (2) from reaction (1) gives the target single-electron reduction: 3. $\ce{Cu^2+(aq) + e^- -> Cu^+(aq)} \quad E^\circ_3 = ?, \quad n_3 = 1$ $$\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2$$ $$-1 F E^\circ_3 = -0.68 F - (-0.52 F)$$ $$-E^\circ_3 = -0.16\text{ V} \implies E^\circ_3 = +0.16\text{ V}$$ For the target overall cell reaction: $$\ce{Cu^2+(aq) + Cu(s) -> 2Cu^+(aq)}$$ We separate this into the cathode and anode half-reactions: * **Cathode (Reduction):** $\ce{Cu^2+(aq) + e^- -> Cu^+(aq)} \quad E^\circ_{\text{cathode}} = +0.16\text{ V}$ * **Anode (Oxidation):** $\ce{Cu(s) -> Cu^+(aq) + e^-} \quad E^\circ_{\text{anode}} = +0.52\text{ V}$ The overall standard potential ($E^\circ_{\text{cell}}$) for this reaction involving two equivalents of $\ce{Cu^+}$ is: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.16\text{ V} - 0.52\text{ V} = -0.36\text{ V}$$ This standard cell potential of $-0.36\text{ V}$ corresponds to a $2\text{-electron}$ transfer process. Normalizing this value on a per-electron basis (or per mole of $\ce{Cu^+}$ formed, representing $\frac{1}{2}\ce{Cu^2+ + 1/2Cu -> Cu^+}$): $$E^\circ = \frac{-0.36\text{ V}}{2} = -0.18\text{ V}$$ Thus, **(C) matches with (P)**. Step 4 - Calculate the Standard Reduction Potential for (D) $E^\circ(\ce{Cr^3+/Cr^2+})$ We are given: 1. $\ce{Cr^3+(aq) + 3e^- -> Cr(s)} \quad E^\circ_1 = -0.74\text{ V}, \quad n_1 = 3$ $$\Delta G^\circ_1 = -n_1 F E^\circ_1 = -3 \times F \times (-0.74) = +2.22 F$$ 2. $\ce{Cr^2+(aq) + 2e^- -> Cr(s)} \quad E^\circ_2 = -0.91\text{ V}, \quad n_2 = 2$ $$\Delta G^\circ_2 = -n_2 F E^\circ_2 = -2 \times F \times (-0.91) = +1.82 F$$ Subtracting reaction (2) from reaction (1) yields the single-electron reduction step: 3. $\ce{Cr^3+(aq) + e^- -> Cr^2+(aq)} \quad E^\circ_3 = ?, \quad n_3 = 1$ $$\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2$$ $$-1 F E^\circ_3 = +2.22 F - 1.82 F$$ $$-E^\circ_3 = 0.40\text{ V} \implies E^\circ_3 = -0.40\text{ V}$$ Thus, **(D) matches with (Q)**. Step 5 - Match Summary Based on the calculations: * (A) $\rightarrow$ (R) * (B) $\rightarrow$ (S) * (C) $\rightarrow$ (P) * (D) $\rightarrow$ (Q) $$\boxed{\text{A}\rightarrow\text{R};\ \text{B}\rightarrow\text{S};\ \text{C}\rightarrow\text{P};\ \text{D}\rightarrow\text{Q}}$$