If the $E _ { \text {cell } } ^ { \circ }$ = -0.362 V for a given electrochemical cell at 25 °C, calculate the pH of the solution at the cathode. Pt | H₂(g, 1.0 atm) | H⁺(aq, 1.00 M) || H⁺(aq) | H₂(g, 1.0 atm) | Pt

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Quizplus · Accepted Answer

The cell is a concentration cell, and the Nernst equation is used: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log \frac{[H^+]_{\text{cathode}}}{[H^+]_{\text{anode}}}$. Given $E^\circ_{\text{cell}} = 0$ for a hydrogen cell, solve $-0.362 = -\frac{0.0592}{1} \log \frac{[H^+]_{\text{cathode}}}{1}$ to find $[H^+]_{\text{cathode}} = 7.76 \times 10^{-7}$, giving pH = 6.11.

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