The standard reduction potential of oxygen in acidic solution is +1.23 V. What is the standard reduc — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Half-Reactions for Oxygen Reduction In acidic medium: $\ce{O2(g) + 4H+(aq) + 4e^- -> 2H2O(l)} \quad E^\circ_{\text{acidic}} = +1.23\text{ V}$ Step 2 - Relate Basic Standard Conditions to pH Under standard basic conditions: $[\ce{OH^-}] = 1\text{ M}$, so $[\ce{H^+}] = 10^{-14}\text{ M}$, giving $\text{pH} = 14$. Step 3 - Apply the Nernst Equation For the acidic half-reaction with $n = 4$ and $p_{\ce{O2}} = 1\text{ atm}$: $$E = E^\circ_{\text{acidic}} - \frac{0.0591}{4} \times 4 \times \text{pH} = E^\circ_{\text{acidic}} - 0.0591 \times \text{pH}$$ Step 4 - Calculate $E^\circ_{\text{basic}}$ At $\text{pH} = 14$ (standard basic conditions, $E = E^\circ_{\text{basic}}$): $$E^\circ_{\text{basic}} = 1.23\text{ V} - 0.0591 \times 14$$ $$E^\circ_{\text{basic}} = 1.23\text{ V} - 0.8274\text{ V} = +0.4026\text{ V} \approx \boxed{+0.404\text{ V}}$$ Step 5 - Explanation of Options * **Option (A)** $+0.404\text{ V}$ — correct. * **Option (B)** $-0.404\text{ V}$ — incorrect sign. * **Option (C)** $+2.056\text{ V}$ — numerically incorrect. * **Option (D)** $-2.056\text{ V}$ — numerically incorrect. $$\text{Correct Answer: } \boxed{\text{A}}$$