Question

Given that $E_{ H _{2} O \left| H _{2}\right| P }=0$ at $298\, K$. The pressure of $H _{2}( g )$ would be

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

Answer 1

Answer: (D)

The cell reaction is $: 2 H ^{+}+2 e \rightarrow H _{2}$

$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{p_{ H _{2}}}{\left[ H ^{+}\right]^{2}}$

$E_{ cell }^{\circ} =0,$

$E_{\text {cell }} =0$

$\therefore p_{ H _{2}}=\left[ H ^{+}\right]^{2}$

In pure water, $\left[ H ^{+}\right]=10^{-7}(M)$

$\therefore p_{ H _{2}}=10^{-14}\, bar.$