1.

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

Answer»

`E_(H_(2)=(RT)/(F)xxpH`
`E_(H_(2))=(RT)/(F)=(1)/(pH)`
`E_(H_(2))=(2.303 RT)/(F)pH`
`E_(H_(2))=-0.591 pH`

Solution :`2H^(+)+2e^(-)rarrH_(2)`
Accordingto nernst equation
`E=E^(@)+(2.303 RT)/(NF) log (1)/(H^(+)]^(2)`
`=-(2.303 RT)/(2xxF)-log (H^(+))^(2)=(2.303RT)/(F)pH`


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