Q72 / 180 · NEET 2016 (Phase 1) · Chemistry

The pressure of $H_{2}$ required to make the potential of hydrogen electrode zero in pure water at 298 K is:

Correct answer
A
$10^{- 14} a t m$
B
$10^{- 12} a t m$
C
$10^{- 10} a t m$
D
$10^{- 4} a t m$

Solution

P t, \frac{H_{2 (g)}}{H^{+}}

Hydrogen electrode

E_{H_{2} / H^{+}}^{o} = 0.0 V o l t

H_{2 (g)} \to 2 H_{(a q)}^{+} + 2 e^{-}

[H^{+}] = 10^{- 7} M

at

25^{o} C

(for Pure water)

E = \frac{- 0.0591}{2} \log (\frac{{[H^{+}]}^{2}}{P_{H_{2}}})

E = 0 = \log \frac{[H^{+}]}{P_{H_{2}}} = 0

∴ {[H^{+}]}^{2} = P_{H_{2}} = {(10^{- 7})}^{2}

∴ P_{H_{2}} = 10^{- 14} a t m

Hence, the correct answer is option: 'A'.

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