InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The emf of a Daniell cell at `298 K` is `E_(1)` `Zn|ZnSO_(4)(0.01 M)||CuSO_(4)(1.0M)|Cu` When the concentration of `ZNSO_(4)` is `1.0 M` and that of `CuSO_(4)` is `0.01 M`, the `emf` changed to `E_(2)`. What is the relationship between `E_(1)` and `E(2)` ?A. `E_(2)=0cancel=E_(1)`B. `E_(1)gtE_(2)`C. `E_(1)gtE_(2)`D. `E_(1)=E_(2)`. |
|
Answer» Correct Answer - B The cell reaction is `Zn+Cu^(2+)rarrZn^(2+)+Cu` `therefore` Nernst equation is `E_(1)=E^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])` when `[Zn^(2+)]` is increased and `[Cu^(2+)]` is decreased the e.m.f. `E_(2)` will b e less than `E_(1)` or `E_(1)gtE_(2)`. |
|