Question

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

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

Answer 1

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

$pH=7forwater$
$-log\left[H^{+}\right]=7\Rightarrow \left[H^{+}\right]={10}^{-7}$
$2H_{\left(aq\right)}^{+}+2e^{-}\to H_{2\left(g\right)}$
$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{p_{H_2}}{[H^{+}{]}^2}$
$0=0-\frac{0.0591}{2}log\frac{}{p_{H_2}}({10}^{-7}{)}^2$
$log\frac{p_{H_2}}{{10}^{-7}{)}^2}=0\Rightarrow \frac{p_{H_2}}{({10}^{-7}{)}^2}=1[\because log1=0]$
$p_{H_2}={10}^{-14}atm$