The electrode potential of hydrogen electrode in neutral solution and 298 K is — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Half-Cell Reaction for Hydrogen Electrode $$\ce{H+(aq) + e^- <=> \frac{1}{2} H2(g)}$$ Step 2 - Apply the Nernst Equation With $E^\circ_{\ce{H+|H2}} = 0.00\text{ V}$, $n = 1$, $p_{\ce{H2}} = 1\text{ atm}$: $$E = 0.00 - 0.0591 \log\left(\frac{1}{[\ce{H+}]}\right) = -0.0591 \times \text{pH}$$ Step 3 - Calculate the Potential in a Neutral Solution For a neutral aqueous solution at $298\text{ K}$: $\text{pH} = 7$ $$E = -0.0591 \times 7 = -0.4137\text{ V} \approx \boxed{-0.413\text{ V}}$$ Step 4 - Explanation of Options * **Option (A) is correct:** $-0.413\text{ V}$ matches our calculation. * **Option (B) is incorrect:** Zero potential requires $\text{pH} = 0$ ($[\ce{H+}] = 1\text{ M}$). * **Option (C) is incorrect:** $-0.826\text{ V}$ would require $\text{pH} = 14$ (strongly basic, not neutral). * **Option (D) is incorrect:** $+0.413\text{ V}$ would need $\text{pH} \approx -7$ (impossible under normal conditions). $$\text{Correct Answer: } \boxed{\text{A}}$$