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The reduction potential of a hydrogen electrode at `pH 10` at `298K` is : `(p =1` atm)A. `-0.0592` VB. `+0.0592` VC. `-0.592` VD. `0.592` V |
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Answer» Correct Answer - C `E_(OP) = E_(OP)^(@) - (0.0592)/(1) "log"_(10) ([H^(+)])/((P_(H_(2)))^(1//2))` `(1)/(2) H_(2) to H^(+) + e^(-)` `because [H^(+)] = 10^(-10) , P_(H_(2)) = 1 ` atm and `E_(OP (H_(2)))^(@) = 0` `E_(OP) = 0.592 V` `therefore E_(Red) = -0.592` V |
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