The potential for the given half cell at 298 K is

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Solution

Q.
The potential for the given half cell at 298 K is (-) ______ $\times 10^{- 2} V .$

$2 H_{(a q)}^{+} + 2 e^{-} \to H_{2} (g)$

$[H^{+}] = 1 M, P_{H_{2}} = 2 atm$

(Given: 2.303RT/F = 0.06 V, log2 = 0.3) [2024]

Ans.
(1)

$2 H^{+} (aq) + 2 e^{-} \to H_{2} (g)$

By Nernst equation:

$E_{H^{+} / H_{2}} = E_{H^{+} / H_{2}}^{0} - \frac{2.303 R T}{n F} \log \frac{p_{H_{2}}}{{[H^{+}]}^{2}}$

$E_{H^{+} / H_{2}} = (0 - \frac{0.06}{2} \log \frac{2}{1^{2}}) V$

$= (- 0.03 \times 0.3) V = - 0.9 \times 10^{- 2} V = - 1 \times 10^{- 2} V$

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