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The potential...

Question

The potential of hydrogen electrode at $25^{o} C$ will be:
$(P_{H_{2}} = 1$ atm, concentration of $H^{+} = 0.1 M)$

0.00 V

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- 0.059 V

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0.118 V

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0.059 V

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Solution

The correct option is A $0.00 V$
$2 H^{+} + 2 e^{-} \to H_{2}$
$E_{c e l l} = E^{0} - \frac{0.059}{2} l o g ⎡ ⎣ \frac{P_{H_{2}}}{{(H^{+})}^{2}} ⎤ ⎦$
$E_{c e l l} = 0 - 0.059 l o g [\frac{1}{(0.1)}]$
$E_{c e l l} = 0.05 g l o g (8.1)$
$= - 0.05 g V$

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