InterviewSolution
Saved Bookmarks
| 1. |
The reduction potential of a hydrogen electrode at `pH 10` at `298K` is : `(p =1` atm)A. `0.51` voltB. 0 voltC. `-0.591` voltD. `0.059` volt |
|
Answer» Correct Answer - C `2H^(+)+2e^(-)rarr H_(2) pH = 10, [H^(+)]= 10^(-10)` `E^(@)=0` `E_("red") = (0.0591)/(2) "log" ([H^(+)]^(2))/(P_(H_(2))(g)) = (0.0591)/(2)xx2log[10^(-10)]` `= (0.0591)/(2)xx2xx[-10]= -0.591 V` |
|