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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
Explanation
E = 0 requires $P_{H_2} = [H^+]^2$.
Detailed Solution
For water, pH = 7, so $[H^+] = 10^{-7}$
$2H^+(aq) + 2e^- \rightarrow H_2(g)$
$E = E^o - \frac{0.0591}{2}\log\frac{P_{H_2}}{[H^+]^2} \Rightarrow 0 = 0 - \frac{0.0591}{2}\log\frac{P_{H_2}}{[H^+]^2}$
$\frac{P_{H_2}}{(10^{-7})^2} = 1$ (since log 1 = 0)
$P_{H_2} = 10^{-14}$ atm
$2H^+(aq) + 2e^- \rightarrow H_2(g)$
$E = E^o - \frac{0.0591}{2}\log\frac{P_{H_2}}{[H^+]^2} \Rightarrow 0 = 0 - \frac{0.0591}{2}\log\frac{P_{H_2}}{[H^+]^2}$
$\frac{P_{H_2}}{(10^{-7})^2} = 1$ (since log 1 = 0)
$P_{H_2} = 10^{-14}$ atm