Question

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ?

• A. 1.81 V
• B. 0.059 V
• C. 0.59 V
• D. 0.118 V

Answer 1

Answer: (C)

For hydrogen electrode, oxidation half
reaction is
$\underset{\text{1 atm}}{H_2}\to \underset{\text{(}AtPH10)}{2H^{+}}+2e^{-}$
If pH= 10
$H^{+}=1\times 10^{-pH}=1\times 10^{-10}$
From Nernst equation,
$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{[H^{+}{]}^2}{p_{H_2}}$
For hydrogen electrode, $E_{cell}^0=0$
$E_{cell}=-\frac{0.0591}{2}log\frac{(10^{-10}{)}^2}{1}$
$=+\frac{0.0591\times 2}{2}log\frac{1}{10^{-10}}$
$=0.0591log10^{10}$
$=0.0591\times 10=0.591V$