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)` ?

Options

A. `E_(1) lt E_(2)`
B. `E_(1)gtE_(2)`
C. `E_(2)=0neE_(1)`
D. `E_(1)=E_(2)`

Answer

B

Explanation

For cell
`Zn|ZnSO_(4)(0.01M)||CuSO_(4)(1M)|Cu`
Cell reaction `toZn+Cu^(+2)toZn^(+2)+Cu`
`E_(1)=E^(o)-(0.059)/(2)"log"(Zn^(+2))/(Cu^(+2))`
`E_(1)=E^(o)-(0.059)/(2)"log"(0.01)/(1)`
`=E^(o)-(0.059)/(2)"log"(1)/(100)` . . . .(i)
Fore cell
`Zn|ZnSO_(4)(1M)||CuSO_($)(0.01M)|Cu`
`E_(2)=E^(o)-(0.059)/(2)"log"(1)/(0.01)`
`=E^(o)-(0.059)/(2)log100` . . . . . ltBrgt `E_(1)gt E_(2)`