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

Answer

A

Explanation

For the given cell , `Zn + Cu^2+ to Zn^(2+) + Cu`
`E = E^(Theta) - (0.059)/(2) "log" ([Zn^(2+)])/([Cu^(2+)])`
`therefore " " E_(1) = E^(Theta)- (0.059)/(2) "log" (0.01)/(1.0) = E^(@) + 0.059`
`E_(2) = E^(Theta) - (0.059)/(2) "log" (1.0)/(0.01) = E^(@) - 0.059`
`therefore E_1 gt E_2`