$I_{2} (S) | I (0.1 M)$ half cell is connected to $H_{(a q)}^{+} | H_{2} (1 a t m) | P t$ half cell and its half cell potential is found to be 0.7714 V. If $E_{I_{2} / I}^{o} = 0.535 V$ , the pH of $H^{+} / H_{2}$ half cell will be:

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Solution

$P t | H_{2} (1 a t m) / H^{+} (a_{0} M) | | I^{(} 0.1 M) | I_{2} (s)$
(C) $I_{2} (s) + 2 e^{-} \to 2 I^{-}$
(A) $H_{2} (g) \to 2 H^{+} + 2 e^{-}$
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$H_{2} (g) + I_{2} (s) \to 2 H^{+} a q + 2 I^{-}$
$Q = \frac{[H^{+}]^{2} [I]^{2}}{p H_{2}} = \frac{10^{- 2} \times a_{0}^{2}}{1}$
$E_{c e l l}^{0} = 0.533 V$
$E_{c e l l} = 0.7714 = 0.533 - \frac{0.059}{2} \times l o g [10^{- 2} \times a_{0}^{2}]$
$0.7714 - 0.533 = - 0.0591 \times l o g [H^{+}] \times 0.1$
$= 0.0591 \times (+ 1) + 0.0591 \times p H$
$p H ≃ 3$
Hence, the pH of $H^{+} / H_{2}$ half cell is 3.

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