Question

The reduction potential of hydrogen electrode having pH = 10 is

• A. 0.59V
• B. Zero Volt
• C. −0.59V
• D. – 0.059V

Answer 1

Answer: (C)

Explanation:

Step 1:

The electrode potential can be calculated as:

$E_{cell}=E_{cell}^o-\frac{0.0591}{1}log\frac{product}{reactant}$

where $E_{cell}$ is the potential difference between two half cells.

$E_{cell}^o$ will be zero for hydrogen electrode.

Step 2:

The reaction for hydrogen electrode can be written as:

$H^{+}+e^{-}⟶\frac{1}{2}{H}_2$

The pH is given as 10.
Thus, $\left[H^{+}\right]$ = ${10}^{-10}$ mol/L
$E=E^0-\frac{0.059}{1}log\left(\frac{1}{{10}^{-10}}\right)$
E = 0−0.059 $\times$ 10 =− 0.59V
Hence, option (B) is correct.