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) gt E_(2)`
C. `E_2 = 0 ne E_(2)`
D. `E_1 = E_2`

Answer

B

Explanation

`Zn |ZnSO_(4) (0.01 M)||CuSO_(4) (1.0M)|Cu`
`E_(1) = E_(0) - (0.059)/(2) "log" (0.01)/(1)`
`= E_(0) - (0.059)/(2) "log" (1)/(100)`
=` E_(0) + 0.059`
When concentrations are changed
`E_(2) = E_(0) - (0.059)/(2) "log" (1)/(0.01)`
`= E_(0) - (0.059)/(2) "log" 100`
`= E_(0) - 0.059`
`therefore E_(1) gt E_(2)`