The oxidation potential of hydrogen electrode at pH = 10 and P $_{H_{2}}$ = 1 atm is __________.

0.51 V

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0.00 V

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+ 0.59 V

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- 0.059 V

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Solution

The correct option is B $+ 0.59 V$
Given that,
$p H = 10 \Rightarrow [H^{+}] = 10^{- 10} M$

Oxidation reaction of hydrogen:

$\frac{1}{2} H_{2} \to H^{+} + e^{-}$

Using Nerst equation,

$E_{o x} = E_{o x}^{\circ} - \frac{0.0591}{n} log \frac{[H^{+}]}{P_{H_{2}}}$

$E_{o x} = 0 - \frac{0.0591}{1} log \frac{10^{- 10}}{1} = 0.59 V$ .

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

p H = 10

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0.059 V

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