Question

Pt, $\frac{H_{2\left(g\right)}}{H^{+}}$ Hydrogen electrode $E_{H_2/H^{+}}^o$ = 0.0 Volt
H_2(g) $\to$ 2H⁺_(aq) + 2e^-
[H⁺] = 10^-7 M at 25^oC (for Pure water)
E = $\frac{-0.0591 }{2}\log \left(\frac{[H^{+}{]}^2}{P_{H_2}}\right)$
E = 0 = log $\frac{\left[H^{+}\right]}{P_{H_2}}=0$
$\therefore [H^{+}{]}^2=P_{H_2}$
$\therefore P_{H_2}={10}^{-14} atm$