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The cell `Pt|H_(2)(g) (1atm)|H^(+), pH = x ||` Normal calomal electrode has EMF of `0.64` volt at `25^(@)C`. The standard reduction potential of normal calomal electrode is `0.28V`. What is the `pH` of solution in anodic compartment. Take `(2.303RT)/(F) = 0.06` at `298K`.

Answer» `E_(H^(+)//H_(2))^(@) =- 0.06 log.(1)/([H^(+)]) = - 0.06pH`
`rArr 0.64 = E_("cathode") - E_("Anode") = 0.28 -(-0.06pH)`
`rArr pH = (0.64 -0.28)/(0.06) =(0.36)/(0.06) =6`


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