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)`
B. `E_(1) gt E_(2)`
C. `E_(1) gt E_(2)`
D. `E_(2)=0neE_(2)`

Answer

C

Explanation

The cell reaction is
`Zn+CuSO_(4)toZnSO_(4)+Cu,n=2`
`E=E_(cell)^(@)-(2.303RT)/(2F)"log"([ZnSO_(4)])/([CuSO_(4)])`
In 1 st case, `E_(1)=E_(cell)^(@)-(2.303RT)/(2F)"log"(0.01)/(1)`
In 2nd case, `E_(2)=E_(cell)^(@)-(2.303RT)/(2F)"log"(1)/(0.01)`
Evidently, `E_(1)gtE_(2)`.