Question

The reduction potential of hydrogen electrode when placed in a buffer solution is found to be - 0.413V. The pH of the buffer is:

• A. 10
• B. 4
• C. 7
• D. 12

Answer 1

Answer: (C)

7

Cell reaction:

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

Using Nerst equation and SHE $=0$ we get

$E_{cell}=-\frac{0.059}{n}\log K_{eq}$

$K_{eq}=\frac{1}{\left[H^{+}\right]}$

$E_{cell}=\frac{0.059}{1}\log \left(\right[H^{+}\left]\right)$

$\log \left(\right[H^{+}\left]\right)=-\frac{0.413}{0.059}=-7$

$-\log \left(\right[H^{+}\left]\right)=7=pH$

So, C is the correct option.