A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of $p H = 10$ and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be:

0.059 V

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0.59 V

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0.118 V

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1.18 V

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Solution

The correct option is A $0.59 V$
The solution of HCl has $p H = 10$
Hence, $[H^{+}] = 10^{- p H} = 10^{- 10} M$
The pressure of hydrogen gas is one atm.
The oxidation potential of electrode would be

$E_{c e l l} = \frac{0.059}{n} log \frac{(P_{H_{2}})^{0.5}}{[H^{+}]} = \frac{0.059}{1} log \frac{1}{10^{- 10}}$
$= + 0.59 V$ .
Hence, the oxidation potential of hydrogen electrode is +0.59 V.

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