Question

The reduction potential of a hydrogen electrode containing a solution of $pH=4$ is:

• A. 0.336 V
• B. 4.059 V
• C. -0.236 V
• D. 3.941 V

Answer 1

Answer: (C)

-0.236 V

Reaction:

$H^{+}+e^{-}\to \frac{1}{2}{H}_2$

Now,

$E_{cell}=E_{cell}^{\circ }+0.059log\left[H^{+}\right]$

$E_{cell}=E_{cell}^{\circ }-0.059\times pH$

$E_{cell}=0-0.059\times 4$

$E_{cell}=-0.236V$

Option C is correct.