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Nernst Equation

I 2 s | I -0....

Question

$I_{2} (s) | I^{-} (0.1 M)$ half cell is connected to a $H^{+} (a q .) | H_{2} (1 bar) | P t$ half cell and emf is found to be $0.7714 V$ . If $E_{I_{2} | I^{-}}^{\circ} = 0.535 V$ . Find the $p H$ of $2 H^{+} | H$ half cell.

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Solution

The correct option is B $3$
The cell reaction is :
$H_{2} (g) + I_{2} (s) ⇌ 2 H^{+} (a q) + 2 I^{-} (a q)$
$E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{2} log \frac{[H^{+}]^{2} [I^{-}]^{2}}{(P_{H_{2}})}$
$0.7714 = 0.535 - \frac{0.0591}{2} log \frac{[H^{+}]^{2} (0.1)^{2}}{1}$
$\Rightarrow 0.2364 = - \frac{0.0591}{2} log \frac{[H^{+}]^{2} (0.1)^{2}}{1}$
$\Rightarrow 8 = - log \frac{[H^{+}]^{2} (0.1)^{2}}{1}$
$\Rightarrow 8 = 2 - 2 l o g [H^{+}]$
$\Rightarrow 3 = - l o g [H^{+}]$
$∴ p H = 3$

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Similar questions

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