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The potential of the cell containing two hydrogen electrodes as represented below Pt, $H_{2 (g)} ∣ H^{+}$ $(10^{- 6} M) ∥ H^{+} (10^{- 4} M) ∣ H_{2 (g)}, Pt at 298 K is$

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– 0.118 V

0.0591 V

– 0.0591 V

0.118 V

answer is C.

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Detailed Solution

$E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{{[H^{+}]}_{anode}}{{[H^{+}]}_{cathode}} = 0 - \frac{0.0591}{1} \log \frac{10^{- 6}}{10^{- 4}} on solving E_{cell} = 0.1182 V$

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