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

Correct Option : (C)

Explanation :

${pH}=7$ for water.

$-\log \left[{H}^{+}\right]=7 \Rightarrow\left[{H}^{+}\right]=10^{-7}$

$2 {H}_{(a q)}^{+}+2 e^{-} \longrightarrow {H}_{2({~g})}$

$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{p_{{H}_2}}{\left[{H}^{+}\right]^2}$

$0=0-\frac{0.0591}{2} \log \frac{p_{{H}_2}}{\left(10^{-7}\right)^2}$

$\log \frac{p_{{H}_2}}{\left(10^{-7}\right)^2}=0 \Rightarrow \frac{p_{{H}_2}}{\left(10^{-7}\right)^2}=1 \quad[\because \log 1=0]$

$p_{{H}_2}=10^{-14} {atm}$