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_(2)=0cancel=E_(1)`
B. `E_(1)gtE_(2)`
C. `E_(1)gtE_(2)`
D. `E_(1)=E_(2)`.

Answer

B

Explanation

The cell reaction is
`Zn+Cu^(2+)rarrZn^(2+)+Cu`
`therefore` Nernst equation is
`E_(1)=E^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
when `[Zn^(2+)]` is increased and `[Cu^(2+)]` is decreased the e.m.f. `E_(2)` will b e less than `E_(1)` or `E_(1)gtE_(2)`.