Question

Given that $E_{H_{2}O|H_2|Pt}=0$ at 298 K. The pressure of $H_2\left(g\right)$ would be -

• A. ${10}^{-7}$ bar
• B. ${10}^{-10}$ bar
• C. ${10}^{-12}$ bar
• D. ${10}^{-14}$ bar

Answer 1

The correct option is D ${10}^{-14}$ bar

The cell reaction is: $2H^{+}+2e\to H_2$
$E_{cell}=E_{cell}^0-\frac{0.059}{2}log\frac{P{H}_2}{[H^{+}{]}^2}$
$E_{cell}^0=0$, $E_{cell}=0$
$\therefore P{H}_2=[H^{+}{]}^2$
In pure water, $\left[H^{+}\right]={10}^{-7}\left(M\right)$
$\therefore P{H}_2={10}^{-14}$ bar