The potential of hydrogen electrode in contact with a solution whose pH is equal to 10 is

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-0.0591

0.0591

0.591

-0.591

answer is C.

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Detailed Solution

$[H^{+}] = 10^{- 10}$
$2 H^{+} + 2 e^{-} \to H_{2} (1 a t m)$
$E_{r e d} = E °_{r e d} - \frac{0.0591}{1} \log \frac{1}{[H^{+}]}$
$E = - 0.0591 (p H)$
$E_{Re d} = - 0.0591 \times 10 = - 0.591 V$

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