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Q.

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

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a

$10^{- 10} atm$

b

$10^{- 4} atm$

c

$10^{- 14} atm$

d

$10^{- 12} atm$

answer is C.

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Detailed Solution

$E_{H^{+} / H_{2}} = \frac{- 0.059}{2} \log \frac{{pH}_{2}}{{[H^{+}]}^{2}} = 0$

${∴pH}_{2} = {[H^{+}]}^{2} = 10^{- 14} atm$

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