EMF of hydrogen electrode in term of pH is (at 1 atm pressure).

Options

A. `E_(H_(2))=(RT)/(F)xxpH`
B. `E_(H_(2))=(RT)/(F).(1)/(pH)`
C. `E_(H_(2))=(2.303RT)/(F)pH`
D. `E_(H_(2))=-0.591pH`

Answer

D

Explanation

`2H^(+)+2e^(-) to H_(2)`
According to Nernst equation
`E=E^(@)+(0.0591)/(2)log""(1)/([H^(+)]^(2))`
`E=0-(0.0591)/(2)log""[H^(+)]^(2)`
`=0.0591pH`