What is the potential of the cell containing two hydrogen electrodes as represented below?
$P t; 1 / 2 H_{2} (g) | H^{+} (10^{- 8} M) | | H^{+} (0.001 M) | 1 / 2 H_{2} (g) P t$
1 atm 1 atm

- 0.295 V

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- 0.0591 V

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0.295 V

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0.0591 V

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Solution

The correct option is C $0.295 V$
According to Nernst equation,
$E = E^{o} - 0.059 l o g Q / n$
Here,
$E_{c e l l} = E^{o} + \frac{0.059}{1} l o g \frac{[H^{+}]_{R H S}}{[H^{+}]_{L H S}}$
$= 0 + 0.059 l o g \frac{10^{- 3}}{10^{- 8}} = 0.059 \times 5 = 0.295 V$ (as $E^{o} = 0$ for concentration cell)

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