The standard reduction potential of oxygen in acidic solution is 1.23 V ( + 4H3O^+ + 4e^- -> 6). The — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Given Parameters and the Half-Cell Reactions We are given the standard reduction potential of oxygen in an acidic solution at $25^\circ\text{C}$: $$\ce{O2(g) + 4H^+(aq) + 4e^- -> 2H2O(l)} \quad E^\circ_{\text{acid}} = 1.23\text{ V}$$ We need to calculate the standard reduction potential of oxygen in a basic solution: $$\ce{O2(g) + 2H2O(l) + 4e^- -> 4OH^-(aq)} \quad E^\circ_{\text{basic}} = ?$$ In a standard basic solution, the concentration of hydroxide ions ($\ce{OH^-}$) is defined as: $$[\ce{OH^-}] = 1.0\text{ M}$$ We can solve this problem using two elegant pedagogical methods: using the Nernst equation under basic conditions, or utilizing thermodynamic relationships. Step 2 - Method 1: Using the Nernst Equation at pH = 14 Under standard basic conditions, we determine the concentration of hydrogen ions ($\ce{H^+}$) using the ionic product of water ($K_w = 1.0 \times 10^{-14}$ at $25^\circ\text{C}$): $$K_w = [\ce{H^+}][\ce{OH^-}] = 10^{-14}$$ $$[\ce{H^+}] = \frac{10^{-14}}{1.0\text{ M}} = 10^{-14}\text{ M}$$ This hydrogen ion concentration corresponds to a pH of: $$\text{pH} = -\log[\ce{H^+}] = 14$$ The potential ($E$) of the acidic reduction half-reaction under these specific concentrations represents the standard reduction potential of oxygen in a basic medium ($E^\circ_{\text{basic}}$). We apply the Nernst equation for the reaction $\ce{O2(g) + 4H^+(aq) + 4e^- -> 2H2O(l)}$: $$E = E^\circ_{\text{acid}} - \frac{2.303RT}{nF} \log Q$$ Where: * $n = 4$ (number of electrons transferred) * $Q = \frac{1}{P_{\ce{O2}} [\ce{H^+}]^4}$ (reaction quotient) * $P_{\ce{O2}} = 1\text{ atm}$ (standard state pressure) * $\frac{2.303RT}{F} = 0.06\text{ V}$ (given) Substitute these values into the Nernst equation: $$E = E^\circ_{\text{acid}} - \frac{0.06\text{ V}}{4} \log\left(\frac{1}{1 \times (10^{-14})^4}\right)$$ $$E = 1.23\text{ V} - 0.015\text{ V} \log\left(10^{56}\right)$$ $$E = 1.23\text{ V} - 0.015\text{ V} \times 56$$ $$E = 1.23\text{ V} - 0.84\text{ V}$$ $$E = \mathbf{+0.39\text{ V}}$$ Thus, the standard reduction potential of oxygen in a basic solution is $+0.39\text{ V}$. Step 3 - Method 2: Thermodynamic Relationship via Water Autoionization Let us couple the two reduction half-reactions thermodynamically: 1. **Reduction in basic medium:** $$\ce{O2(g) + 2H2O(l) + 4e^- -> 4OH^-(aq)} \quad \Delta G^\circ_1 = -4FE^\circ_{\text{basic}}$$ 2. **Oxidation in acidic medium (reverse of acidic reduction):** $$\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-} \quad \Delta G^\circ_2 = +4FE^\circ_{\text{acid}}$$ Adding these two reactions yields the net chemical reaction: $$\ce{4H2O(l) <=> 4H^+(aq) + 4OH^-(aq)} \quad \Delta G^\circ_3 = \Delta G^\circ_1 + \Delta G^\circ_2$$ $$\Delta G^\circ_3 = 4F\left(E^\circ_{\text{acid}} - E^\circ_{\text{basic}}\right)$$ This combined reaction is exactly $4$ times the autoionization reaction of water ($\ce{H2O(l) <=> H^+(aq) + OH^-(aq)}$), which has an equilibrium constant of $K_w = 10^{-14}$. Therefore, the equilibrium constant ($K$) for the combined reaction is: $$K = (K_w)^4 = 10^{-56}$$ Using the Gibbs free energy relationship ($\Delta G^\circ = -2.303RT \log K$) and substituting $2.303RT = 0.06F$: $$\Delta G^\circ_3 = -2.303RT \log(10^{-56})$$ $$\Delta G^\circ_3 = -0.06F \times (-56) = +3.36F$$ Equating the two expressions for $\Delta G^\circ_3$: $$4F\left(E^\circ_{\text{acid}} - E^\circ_{\text{basic}}\right) = 3.36F$$ $$E^\circ_{\text{acid}} - E^\circ_{\text{basic}} = \frac{3.36}{4} = 0.84\text{ V}$$ $$E^\circ_{\text{basic}} = E^\circ_{\text{acid}} - 0.84\text{ V}$$ $$E^\circ_{\text{basic}} = 1.23\text{ V} - 0.84\text{ V} = \mathbf{+0.39\text{ V}}$$ Both methods yield the identical, robust result of $+0.39\text{ V}$. Step 4 - Evaluate the Options * **Option (A) is incorrect:** $-1.23\text{ V}$ is obtained by simply changing the sign of the acidic reduction potential, which is thermodynamically incorrect. * **Option (B) is incorrect:** $-0.39\text{ V}$ is obtained by an incorrect sign change in the final step of the calculation. * **Option (C) is correct:** As mathematically demonstrated by both methods, the standard reduction potential of oxygen in basic solution is exactly $+0.39\text{ V}$. * **Option (D) is incorrect:** $+2.07\text{ V}$ is obtained if the potential change of $0.84\text{ V}$ is incorrectly added to $1.23\text{ V}$ instead of subtracted. $$\text{Correct Option: } \boxed{\text{C}}$$