Question

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

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

Answer 1

Answer: (A)

$H^{+}(p H=10) \mid H_{2}(1$ atm $) \mid P t(s)$
Reaction : $2 H^{+}\left(p^{H}=10\right)+2 e^- \rightarrow H_{2}(1$ 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. $E=-0.591\, V$