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

Answer

D

Explanation

Decreasing concentration of cathodic half cell and /or increasing concentration of anodic half cell decreases the e.m.f. of the cell.