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

Answer

C

Explanation

Nernst equation for the given cell representation is given as `E= E^(0)= - (0.0591)/(2) "log" ([Zn^(2+))/(Cu^(2+)])`
Value of EMF will be greater if `[Cu^(2+)] gt [Zn^(2+)]`