Question

The pressure of $H_2$ required to make the potential of $H_2$-electrode zero in pure water at $298 \,K$ is

• A. $10^{-12} $ atm
• B. $10^{-10} $ atm
• C. $10^{-4} $ atm
• D. $10^{-14} $ atm

Answer 1

Answer: (D)

$2H^{+} + 2e^{-} \rightarrow H_2(g)$
$E_{H^{+}/H_{2}} = - \frac{0.0591}{2} \log \frac{P_{H_2}}{\left[H^{+}\right]^{2}}$
$ \log \frac{P_{H_2}}{\left[H^{+}\right]^{2}} = 0, \frac{P_{H_2}}{\left[H^{+}\right]^{2}} = 10^{0} = 1 $
$ PH_{2} = \left[H^{+}\right]^{2} $
For pure $ H_{2}O; H^{+} = 10^{-7} M$
$ P_{H_2} = \left(10^{-7}\right)^{2} = 10^{-14} $ atm