Correct Answer - B
(b) For hydrogent electrode, oxidation half reaction is
`underset((1 atm))(H_(2)) to underset(" (At pH 10)")(2H^(+))+2e^(-)`
If pH=10
`H^(+)=1xx10^(-pH)=1xx10^(-10)`
From Nernst equation,
`E_(cell) =E_(cell)^(@)-(0.0591)/(2)log""([H^(+)]^(2))/(p_(H_(2)))`
For hydrogen electrode, `E_(cell)^(@)=0`
`E_(cell)-(0.0591)/(2) log""((10^(-10))^(2))/(1)`
`=+(0.0591xx1)/(2) log""(1)/(10^(-10))`
`=0.0591 log 10^(10)=0.0591xx10=0.591 V`