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 $2.303\frac{RT}{F}$is $0.059V$

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

Answer 1

The correct option is

A

– 0.177

V

Since the pH = 3

Hence

$\left[H^{+}\right]={10}^{-3}$

$Thusthereactionmaybegivenas$

$\left[H^{+}\right]+e^{-}⟶\frac{1}{2}{H}_2$

$SinceE=E^o-\frac{0.059}{n}log\frac{1}{\left[H^{+}\right]}$

Reduction potential of hydrogen electrode \(=E_H = \frac{-2.303~RT}{F} log \frac{1}{[H^+]}= -0.059~

$=0-\frac{0.059}{1}log{10}^3$

$E=-0.177V$