What is the potential of the cell containing two hydrogen electrodes as represented below ?
$p t; \frac{1}{2} H_{2} (g) | H_{2} O | | H^{\oplus} (0.001 M) | 1 / 2 H_{2} (g) P t$

- 0.236 V

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- 0.0591 V

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0.236 V

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0.0591 V

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Solution

The correct option is C $0.236 V$
$E_{c e l l} = \frac{0.059}{1} l o g \frac{[H^{\oplus}]_{R H S}}{[H^{\oplus}]_{L H S}}$
[For $H_{2} O, [H^{\oplus}] = [⊖ O H] = 10^{- 7} M]$
$= 0.059 l o g \frac{10^{- 3}}{10^{- 7}} = 0.059 \times 4 = 0.236 V$
or
$E_{c e l l} = - 0.059 (p H_{c} - p H_{a}) = - 0.059 (3 - 7) = 0.236 V$

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