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EMF of hydrogen electrode in term of pH ...

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

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`

Text Solution

Verified by Experts

The correct Answer is:
D
`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`
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Knowledge Check

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

    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`

  • Emf of hydrogen electrode in tern of pH is (at 1 atm pressure):

    A
    `E_(H_(2)) = (RT)/(F) xx pH`
    B
    `E_(H_(2)) = (RT)/(F)(1)/(pH)`
    C
    `E_(H_(2)) = (2.303RT)/(F) pH`
    D
    `E_(H_(2)) = -0.0591 pH`

  • Reduction potential of hydrogen electrode in terms of pH is : (at 1 atm pressure) (T = 298 kelvin )

    A
    `E=((RT)/(F))xx(PH)`
    B
    `E= (RT)/(F). (1)/(PH)`
    C
    `E= (2.303RT)/(F).(PH)`
    D
    `E=-0.0591PH`

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