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Chemistry Question on Electrochemistry

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: 2.303RTF=0.06 V,log⁡2=0.32.303 \frac{RT}{F} = 0.06 \, \text{V}, \log 2 = 0.3)

Answer

The potential is given by the Nernst equation: E=E∘−0.062log⁡(PH2[H+]2)E = E^\circ - \frac{0.06}{2} \log \left( \frac{P_{H_2}}{[H^+]^2} \right) Substituting the values: E=0−0.062log⁡(212)E = 0 - \frac{0.06}{2} \log \left( \frac{2}{1^2} \right) E=−0.03×0.3=−0.9×10−2 VE = -0.03 \times 0.3 = -0.9 \times 10^{-2} \, \text{V}