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What will be the emf of given cell - `Pt |H_(2) (P_(1)) | H_((aq))^(+) | | H_(2)(P_(2)) |Pt` ?A. `(RT)/(2F) "log"_(e) (P_(1))/(P_(2))`B. `(RT)/(F) "log"_(e) (P_(1))/(P_(2))`C. `(RT)/(F) "log"_(e) (P_(2))/(P_(1))`D. `(RT)/(2F) "log"_(e)(P_(2))/(P_(1))` |
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Answer» Correct Answer - A At anode , `H_(2)(P_(1)) to 2 H^(+) + 2e^(-)` At cathode ` 2 H^(+) + 2e to H_(2)(P_(2))` `E_("cathode") = (-RT)/(2F) "ln" (P_(2))/([H^(+)]^(2))` and `E_("anode") = (-RT)/(2F) "ln" ([H^(+)]^(2))/(P_(1))` |
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