Solveeit Logo
Login

Question

Chemistry Question on Electrochemistry

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

0.59 V

Explanation

Solution

For hydrogen electrode, oxidation half
reaction is
 (1atm)    H2→ (At pH10)    2H++2e−\, _{(1 atm)}^{\, \, \, \, H_2} \rightarrow \, _{(At \, pH 10)}^{\, \, \, \, 2H^+}+2e^-
If pH= 10
H+=1×10−pH=1×10−10H^+=1 \times 10^{-pH}=1 \times 10^{-10}
From Nernst equation,
Ecell=Ecell0−0.05912log[H+]2pH2E_{cell}=E_{cell}^0- \frac {0.0591}{2}log \frac {[H^+]^2}{p_{H_2}}
For hydrogen electrode, Ecell0=0E^0_{cell}=0
Ecell=−0.05912log(10−10)21E_{cell}=- \frac {0.0591}{2}log \frac {(10^{-10})^2}{1}
\hspace5mm =+ \frac {0.0591 \times 2}{2}log \frac {1}{10^{-10}}
\hspace5mm =0.0591 \, log \, 10^{10}
\hspace5mm =0.0591\times 10 =0.591 \, V