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

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Solution

The correct option is B 3
The cell reaction is
$H_{2} + I_{2} (s) ⇌ 2 H^{+} + 2 I^{-} (a q)$
$0.7714 = 0.535 - \frac{0.0591}{2} l o g \frac{[H^{+}]^{2} [I^{-}]^{2}}{P_{H_{2}}}$
$p H = 3$ {Option B}

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Similar questions

I_{2} (s) | I^{-} (0.1 M)

H^{+} (a q .) | H_{2} (1 bar) | P t

0.7714 V

E_{I_{2} | I^{-}}^{\circ} = 0.535 V

p H

2 H^{+} | H

I_{2} (S) | I (0.1 M)

H_{(a q)}^{+} | H_{2} (1 a t m) | P t

E_{I_{2} / I}^{o} = 0.535 V

H^{+} / H_{2}

P t | H_{2} (g) | H^{+}

E_{H^{+} | H_{2}}^{\circ} = - 0.0295 V

P t (H_{2}) | H^{+} (p H = 2) | H^{+} (p H = 3) | P t (H_{2})

X | X^{2 +} (0.1 M)

Y | Y^{2 +} (1.0 M)

25^{\circ} C

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