Question

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

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

Answer 1

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

Cell reaction :-

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

Concentration of $\left[H^{+}\right]$ in pure water $={10}^{-7}$

According to nearest equation :-

$E=E^0-\frac{0.0591}{2}log\frac{P_{H_2}}{{\left[H^{+}\right]}^2}$

$E=0-\frac{0.0591}{2}log\frac{P_{H_2}}{{\left[{10}^{-7}\right]}^2}$

$E=0-\frac{0.0591}{2}log\frac{P_{H_2}}{{10}^{-14}}$ ...... For potential to be zero $P_{H_2}={10}^{-14}$ atm

$\therefore$ $E=0-\frac{0.0591}{2}log\frac{{10}^{-14}}{{10}^{-14}}$

$\therefore$ $E=0-\frac{0.0591}{2}log1$

$E=0$