Question

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

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

Answer 1

The correct option is D

${10}^{-14}atm$

$2H^{+}+2e^{-}\to H_2\left(g\right)$

$E=E^0-\frac{0.0591}{2}\times log\frac{p_{H_2}}{[H^{+}{]}^2}$

$=0-\frac{0.0591}{2}\times log\frac{p_{H_2}}{({10}^{-7}{)}^2}$

$\therefore$ For potential of $H_2$ electrode to be zero, $p_{H_2}$ should be ${10}^{–14}$ i.e., $log\frac{{10}^{-14}}{{10}^{-14}}=0$