The hydrogen electrode is dipped in a solution of $p H = 3$ at 25 $^{o}$ C. The potential would be: (the value of $\frac{2.303 R T}{F}$ is $0.059 V$ )

0.177 V

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0.087 V

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0.059 V

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- 0.177 V

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Solution

The correct option is B $- 0.177 V$
$p H = 3$
$[H_{3} O^{+}] = 10^{- p H} = 10^{- 3} M$
$E_{e l} = E_{e l}^{0} - \frac{0.059}{n} l o g \frac{P_{H 2}}{[H^{+}]^{2}}$
$E_{e l} = 0 - \frac{0.059}{2} l o g \frac{1}{(10^{- 3})^{2}}$
$E_{e l} = - 0.177 V$

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