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The value of $E_{H_{2} O} / H_{2} (1 a t m) / P t$ at 298 K would be

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–0.414 V

–0.207 V

0.207 V

0.414 V

answer is B.

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Detailed Solution

$H_{2} O ⇌ H^{+} + {OH}^{-}$

$H^{+} + e^{-} ⇌ \frac{1}{2} H_{2}$

$E = E^{0} - \frac{0 .059}{n} \log \frac{P_{H_{2}}}{[H^{+}]} E = 0 - \frac{0 .059}{1} \log \frac{1}{10^{- 7}} [[H^{+}] = 10^{- 7}] E = - \frac{0 .059}{1} \log 10^{7} = - 0.414 V$

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