Question

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

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

Answer 1

The correct option is A ${10}^{-14}atm$

$2H^{+}+2e^{-}\to H_2$
$E=E^0-\frac{0.0592}{n}log\frac{P_{H_2}}{[H^{+}{]}^2}$
In 1.0 M hydrogen ion solution with hydrogen pressure of 1 atm,
$E=0-\frac{0.0592}{2}log\frac{1}{1^2}$
$E=0-\frac{0.0592}{2}log1=0-0=0$
In pure water, the hydrogen ion concentration is ${10}^{-7}$ M,
with hydrogen pressure of P atm,
$E=E^0-\frac{0.0592}{n}log\frac{P_{H_2}}{[H^{+}{]}^2}$
$0=0-\frac{0.0592}{2}log\frac{P}{({10}^{-7}{)}^2}$
$0=log\frac{P}{({10}^{-7}{)}^2}$
$1=\frac{P}{({10}^{-7}{)}^2}$
$P={10}^{-14}atm$