Question

The potential for the given half cell at 298 K is\[(-) \ldots \times 10^{-2} \, \text{V}.\]\[2\text{H}^+ (\text{aq}) + 2e^- \rightarrow \text{H}_2 (\text{g})\]\[[\text{H}^+] = 1M, \, P_{\text{H}_2} = 2 \, \text{atm}\](Given: \( 2.303 \frac{RT}{F} = 0.06 \, \text{V}, \log 2 = 0.3 \))

Answer 1

Correct Answer: 1

The potential is given by the Nernst equation: \[ E = E^\circ - \frac{0.06}{2} \log \left( \frac{P_{H_2}}{[H^+]^2} \right) \] Substituting the values: \[ E = 0 - \frac{0.06}{2} \log \left( \frac{2}{1^2} \right) \] \[ E = -0.03 \times 0.3 = -0.9 \times 10^{-2} \, \text{V} \]