EMF of hydrogen electrode in terms 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))=(2.303 RT)/(F)pH`
D. `E_(H_(2))=-0.591 pH`

Answer

D

Explanation

`2H^(+)+2e^(-)rarrH_(2)`
According to nernst equation
`E=E^(@)+(2.303 RT)/(nF) log (1)/(H^(+)]^(2)`
`=-(2.303 RT)/(2xxF)-log (H^(+))^(2)=(2.303RT)/(F)pH`