The oxidation potential of a hydrogen electrode at $p H = 1$ is ( $T = 298 K$ ):-

0.059 V

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0 V

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- 0.059 V

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0.118 V

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Solution

The correct option is A $0.059 V$
$\frac{1}{2} H_{2} ⟶ H^{+} + e^{-}$ oxidation
$[H^{+}] = 10^{- p H} = 10^{- 1} E = E^{o} - \frac{0.0591}{1} l o g \frac{[H^{+}]}{1} E = 0 - 0.0591 \times l o g 10^{- 1} E = 0.0591 V$

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