Saved Bookmarks
| 1. |
Two students use same stoch solution of ZnSO_(4) and a solution of CuSO_(4). The e.m.f. of one cell is 0.03 V higher than the other. The concentration of CuSO_(4) in the cell with higher e.m.f. value is 0.5 M. find out the concentration of CuSO_(4) in the other cell (2.303RT/F=0.06) |
|
Answer» Solution :The TWO cells may represented as `Zn|Zn^(2+)(conc=c)||Cu^(2+)(c=?)|Cu,EMF=E_(1)` `Zn|Zn^(2+)(conc.=c)||Cu^(2+)(0.5M)|Cu,EMF=E_(2)` The CELL reaction is: `Zn+Cu^(2+)toZn^(2+)+Cu` `thereforeE_(1)=E^(@)-(2.303RT)/(2F)"log"(c)/([Cu^(2+)])` `E_(2)=E^(@)-(2.303RT)/(2F)"log"(c)/(0.5)` `thereforeE_(2)-E_(1)=(2.303RT)/(2F)("log"(c)/([Cu^(2)])-"log"(c)/(0.5))=0.03V` (given) `THEREFORE(0.06)/(2)"log"(0.5)/([Cu^(2+)])=0.03` or `"log"(0.5)/([Cu^(2+)])=1` or `(0.5)/([Cu^(2+)])=10` or `[Cu^(2+)]=(0.5)/(10)=0.05M` |
|