The oxidation potential of a hydrogne electrode at `pH=10` and `p_(H_(2))=1atm` is

Options

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

Answer

3

Explanation

`E_(OP)=E_(OP)^(0)-(0.059)/(1)"log"([H^(+)])/(P_(H_(2)))`
`therefore[H^(+)]=10^(-10)`
`P_(H_(2))=1 atm, E_(OP)=0.59V`