Emf of given Daniell cell at 298 K is `E_(1)`. `Zn|ZnSO_(4)(0.01M)||CuSO_(4(1.0M))|Cu`. The emf changed to `E_(2)` when concentration of `ZnSO_(4)` solution is 1.0 M and `CuSO_(4)` solution is 0.01 M, then what is the relation between `E_(1) and E_(2)` ?

Options

A. `E_(1) gt E_(2)`
B. `E_(1) lt E_(2)`
C. `E_(1)=E_(2)`
D. `E_(2)=0 ne E_(1)`

Answer

A

Explanation

`Zn+Cu^(2+) to Zn^(2+)+Cu`
`E=E^(@)-(0.059)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
`E_(1)=E^(@)-(0.059)/(2)"log"(0.01)/(1.0) therefore E_(1)=E^(@)+0.059`
`E_(2)=E^(@)-(0.059)/(2)"log"(1.0)/(0.01)therefore E_(2)=E^(@)-0.059`
`therefore E_(1) gt E_(2)`.