Question

Given that $E_{H_{2}O|H_2|Pt}=0$ at K . The pressure of $H_2$ gas would be:

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

Answer 1

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

For water, $\left[H^{+}\right]=1\times {10}^{-7}M.$
The Nernst equation for $E_{H_2}$ for reduction is $E_{H_2}=E_0^{H_2}-\frac{0.0592}{n}log\frac{P_{H_2}}{{\lfloor H^{+}\rfloor }^2}$ But $E_0^{H_2}=0V$.
Substitute values in the above equation.
$0=0-\frac{0.0592}{2}log\frac{P_{H_2}}{{\lfloor 1\times {10}^{-7}\rfloor }^2}{P}_{H_2}={10}^{-14}atm.$