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The reduction potential of a hydrogen electrode at `pH 10` at `298K` is : `(p =1` atm)A. 0.51 VB. 0.000 VC. 0.59 VD. 0.059 V |
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Answer» Correct Answer - C `E = E^(@) - (0.059)/(n) " log " _(10) [H^(+)]` Given `pH = 10 , n =1 therefore [H^(+)] = 10^(-10)` , `E_(SHE)^(@) = 0 , 10 = - "log"_(10) [ H^(+)]` `E = E^(@) - 0.059 xx log 10^(-10) = 0-0.059 xx (-10)` `= 0 + 0.59 = 0.59` V |
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