Question

The potential of a hydrogen electrode at $pH=10$ is

• A. $-0.59V$
• B. $0.00V$
• C. $0.59V$
• D. $-0.059$

Answer 1

Answer: (A)

$H^{+}(pH=10)\mid H_2(1$ atm $)\mid Pt(s)$
Reaction : $2H^{+}\left(p^H=10\right)+2e^{-}\to H_2(1$ atm $)$
$E=E^0-\frac{0.0591}{2}\log \left(\frac{p_{H_2}}{{\left[H^{+}\right]}^2}\right)$
$=0-\frac{0.0591}{2}\log \frac{1}{{\left(10^{-10}\right)}^2}$
$=-\frac{0.0591}{2}\times 2\log \frac{1}{10^{-10}}$
$=-0.0591\times 10=-0.591$
i.e. $E=-0.591V$