Question

The hydrogen electrode is dipped in a solution of $pH=3$ at $25^{\circ }$C. The potential of the cell would be (the value of $2.303RT/F$ is $0.059V)$

• A. 0.177 V
• B. 0.087 V
• C. -0.177 V .
• D. 0.059V

Answer 1

Answer: (C)

$pH=3$
$\therefore [H^{+}]=10^{-3}$
$\therefore [H^{+}]+e^{-}\to \frac{1}{2}{H}_2$
$E=E^0-\frac{0.059}{n}log\frac{1}{[H^{+}]}$
$=0-\frac{0.059}{1}log10^3$
$E=-0.177V$