Electrode potential will be more for hydrogen electrode at pH (at the same temperature) — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Half-Cell Reaction of Hydrogen Electrode $$\ce{H+(aq) + e^- -> \frac{1}{2} H2(g)}$$ Step 2 - Derivation of Nernst Equation for Hydrogen Electrode With $E^\circ_{\ce{H+|H2}} = 0.00\text{ V}$, $n = 1$, and $p_{\ce{H2}} = 1\text{ atm}$: $$E = 0.00 - 0.0591 \log\left(\frac{1}{[\ce{H+}]}\right) = -0.0591 \times \text{pH}$$ Step 3 - Evaluating the Options at 298 K Using $E = -0.0591 \times \text{pH}$: * **Option (A) - pH = 4:** $E = -0.0591 \times 4 = -0.2364\text{ V}$ * **Option (B) - pH = 3:** $E = -0.0591 \times 3 = -0.1773\text{ V}$ * **Option (C) - pH = 2:** $E = -0.0591 \times 2 = -0.1182\text{ V}$ * **Option (D) - pH = 5:** $E = -0.0591 \times 5 = -0.2955\text{ V}$ Comparing: $-0.1182\text{ V} > -0.1773\text{ V} > -0.2364\text{ V} > -0.2955\text{ V}$ The electrode potential is highest (least negative) at $\text{pH} = 2$. $$\text{Correct Option: } \boxed{\text{C}}$$