InterviewSolution
Saved Bookmarks
| 1. |
The emf `(E^(@))` of the following cels are : `Ag|Ag^(+) (1 M)||Cu^(2+) (1 M)|Cu, E^(@)=-0.46` volt `Zn|Zn^(2+) (1 M)||Cu^(2+) (1 M)|Cu, E^(@)=+1.10` volt Calculate the emf of the cell : `Zn|Zn^(2+) (1 M)||Ag^(+) (1 M)|Ag` |
|
Answer» `Zn|Zn^(2+) (1 M)||Ag^(+) (1 M)|Ag` `E_(cell)=E_(o x(Zn//Zn^(2+)))+E_(red (Ag^(+)//Ag))` With the help of the following two cells, the above equation can be obtained : `Ag|Ag^(+) (1 M)||Cu^(2+) (1M)|Cu, E^(@)=-0.46" volt"` or `Cu|Cu^(2+) (1 M)||Ag^(+) (1 M) |Ag, E^(@)" will be "+0.46" volt"` or `+0.46=E_("ox"(Cu//Cu^(2+)))+R_("red "(Ag^(+)//Ag))` ...(i) `Zn|Zn^(2+) (1 M)||Cu^(2+)|Cu, E^(@)=+1.10" volt"` `+1.10=E_(o x (Zn//Zn^(2+)))+E_(red (Cu^(2+)//Cu))` ...(ii) Adding eqs. (i) and (ii), `+1.56=E=E_("red "(Ag^(+)//Ag))+E_(o x (Zn//Zn^(2+)))+E_("red "(Cu^(2+)//Cu))` since, `E_(o x(Cu//Cu^(2+)))=-E_("red "(Cu^(2+)//Cu))` So, `+1.56=E_(o x (Zn//Zn^(2+)))+E_("red "(Ag^(+)//Ag))` Thus, the emf of the following cell is `Zn|Zn^(2+)(1 M)||Ag^(+) (1 M)|Ag` is `+ 1.56` volt |
|