Question

The hydrogen electrode is dipped in a solution of pH = 3 at ${25}^{o}C$. The reduction potential of the cell would be:

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

Answer 1

Answer: (B)

- 0.177 V

For Hydrogen electrode,

$2H_{\left(aq\right)}^{+}+2e^{-}⟶H_2\left(g\right)1atm$ $E_{cell}^o=0$

We know that,

$E_{cell}=E_{cell}^o-\frac{0.059}{2}\log \frac{P{H}_2}{[H^{+}{]}^2}$

$\because pH=3$

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

$\Rightarrow E_{cell}=\frac{-0.059}{2}\log \frac{1}{[{10}^{-3}{]}^2}$

$\Rightarrow E_{cell}=\frac{-0.059}{2}\log \left({10}^6\right)$

$\Rightarrow E_{cell}=\frac{-0.059}{2}\times 6=-0.177$ volts.

Hence, option B is correct.