Question

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

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

Answer 1

Answer: (D)

$2H^{+}+2e^{-}\to H_2(g)$
$E_{H^{+}/H_2}=-\frac{0.0591}{2}\log \frac{P_{H_2}}{{\left[H^{+}\right]}^2}$
$\log \frac{P_{H_2}}{{\left[H^{+}\right]}^2}=0,\frac{P_{H_2}}{{\left[H^{+}\right]}^2}=10^0=1$
$P{H}_2={\left[H^{+}\right]}^2$
For pure $H_{2}O;H^{+}=10^{-7}M$
$P_{H_2}={\left(10^{-7}\right)}^2=10^{-14}$ atm