What is $p H$ of the half cell $H^{+} (aq) | H_{2} (g, 1 atm) | P t (s)$
if $E_{H^{+} (aq) | H_{2} (g)} = - 0.25 V$ ?

4.2

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6.4

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1.5

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Solution

The correct option is A $4.2$
$2 H_{(a q)}^{+} + 2 e^{-} \to H_{2 (g)}$
According to nernst equation for electrode potential,
$E_{H^{+} | H_{2}} = E_{H^{+} | H_{2}}^{\circ} - \frac{0.591}{n} \times l o g (\frac{P_{H_{2}}}{[H^{+}]^{2}})$
Here number of electrone involved, $n = 2$
$p H = - log [H^{+}]$
$P_{H_{2}} = 1 atm$
$E_{H^{+} | H_{2}}^{\circ} = 0$
Substituting the values,
$- 0.25 = 0 - \frac{0.0591}{2} \times 2 \times p H$
$\Rightarrow p H = 4.237$

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