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

Answer

D

Explanation

`Zn | ZnSO_(4) (0.01 M) || CuSO_(4) (1.0 M)| Cu`
`therefore E = E_("cell")^(@) - (2.303 RT)/(2 xx F) xx log (0.01)/1`
When concentrations are charged
`therefore E_(1) = E_("cell")^(@) -(2.303 RT)/(2F) xx log 1/(0.01)`
i.e. `E_(1) gt E_(2)`