The reduction potential of a hydrogen electrode at `pH 10` at `298K` is : `(p =1` atm)

Options

A. `0.51` V
B. `0.00` V
C. `-0.59` V
D. `0.059` V

Answer

A

Explanation

`E_(OP)=E_(OP)^(@)-(0.0592)/(1)`
`log""([H^(+)])/([p_(H_(2))]^(1//2))1/2H_(2)toH^(+)+e^(-)`
`because[H^(+)]=10^(-10),p_(H_(2))=1`atm
`E_(OP)=0.59V`
`thereforeE_(Red)=-0.592V`