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)-E_(2)=0`
B. `E_(1) lt E_(2)`
C. `E_(1) gt E_(2)`
D. `E_(1)=10^(2)E_(2)`

Answer

C

Explanation

`underset(n=2)(Zn)+Cu underset(Q=(Zn^(2+))/(Cu^(2+)))(SO_(4)hArrZn)SO_(4)+Cu`
`E_(1)-E_(2)=-(0.059)/(2)(log((0.01)/(1))-log((1)/(0.01))`
`=-(0.059)/(2)(log((1)/(100))-log100)`
`=(0.059)/(2)(-log100-log100)=+(0.059)/(2)xx2impliesSo,E_(1)gtE_(2s)`