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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` |
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Answer» Correct Answer - 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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