How much will the reduction potential of a hydrogen electrode change when its solution initially at pH = 0 is neutralised to pH = 7

Options

A. increase by `0.059V`
B. decrease by `0.059V`
C. increase by 0.41 V
D. decrease by 0.41 V

Answer

D

Explanation

`H^(+)+e^(-)to(1)/(2)H_(2),E=0-(0.0591)/(1)log_(10)((1)/([H^(+)]))=+0.591log_(10)[H^(+)]`
`E_(1)=0{pH=0}`
`E_(2)=+0.0591log_(10)[10^(-7)]=-0.0591xx7{atpH=7}=-0.41V`