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

Options

A. `0.59 V`
B. `0.118 V`
C. `1.18 V`
D. `0.059 V`

Answer

A

Explanation

For hydrogen gas elctrode, the Nernst equation is written as
`2H^(+)(aq.)+2e^(-)hArr H_(2)g`
or `H^(+)(aq.)+e^(-)(1)/(2)H_(2)(g)`
`E_H^(+)//H_(2) = E_(H^(+))^(ɵ)-(0.059 V)/(1)"log"(p_(H_(2))^(1//2))/(C_(H^(+)))`
Since log `1//C_(H^(+))=pH`, we can write
`E_H^(+)//H_(2) = E_(H^(+))^(ɵ)-(0.059 V) E_H^(+)//H_(2) = E_(H^(+))^(ɵ)(pH)`
Substituting the values: `E_(H^(+))^(ɵ) = 0.0` and `p_(H_(2)) = 1` atm, we get
`E_(H^(+))^(ɵ) = (0.0 V)-(0.059 V)(1)^(1//2)(10)`
`= -0.59 V`
This is the reducation potential. The Oxidation potential would be negative of this, i.e., `0.59 V`.