The Oxidation potential of hydrogen electrode at $P^{H} = 10$ and $P_{H_{2}}$ at 1atm is :

0.51 V

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0.00V

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+0.59 V

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0.059 V

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Solution

The correct option is C
+0.59 V

$E_{o x} = E_{o x}^{0} - \frac{0.0591}{n} l o g \frac{[H^{+}]}{P H_{2}}$
$= 0 - \frac{0.0591}{1} l o g \frac{[10^{- 10}]}{1} = 0.59 v$

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