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
$\underset{\text{1 atm}}{H_2} \rightarrow \underset{\text({At PH 10})}{2H^+} + 2e^-$
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^0_{cell}=0 $
$E_{cell}=- \frac {0.0591}{2}log \frac {(10^{-10})^2}{1} $
$ =+ \frac {0.0591 \times 2}{2}log \frac {1}{10^{-10}} $
$ =0.0591 \, log \, 10^{10} $
$ =0.0591\times 10 =0.591 \, V $