Question:

The potential for the given half cell at 298 K is()×102V.(-) \ldots \times 10^{-2} \, \text{V}.2H+(aq)+2eH2(g)2\text{H}^+ (\text{aq}) + 2e^- \rightarrow \text{H}_2 (\text{g})[H+]=1M,PH2=2atm[\text{H}^+] = 1M, \, P_{\text{H}_2} = 2 \, \text{atm}(Given: 2.303RTF=0.06V,log2=0.3 2.303 \frac{RT}{F} = 0.06 \, \text{V}, \log 2 = 0.3 )

Updated On: Nov 15, 2024
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Correct Answer: 1

Solution and Explanation

The potential is given by the Nernst equation: E=E0.062log(PH2[H+]2) E = E^\circ - \frac{0.06}{2} \log \left( \frac{P_{H_2}}{[H^+]^2} \right) Substituting the values: E=00.062log(212) E = 0 - \frac{0.06}{2} \log \left( \frac{2}{1^2} \right) E=0.03×0.3=0.9×102V E = -0.03 \times 0.3 = -0.9 \times 10^{-2} \, \text{V}
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