In the electrochemical cell:
`Zn|ZnSO_(4)(0.01 M)||CuSO_(4)(1.0 M)|Cu`, the emf of this Daniel cell is `E_(1)`. When the concentration of `ZnSO_(4)` is changed to 1.0 M and that of `CuSO_(4)` changed to 0.01 M, the emf changes to `E_(2)`. From the following, which one is the relationship between `E_(2)` and `E_(3)` ? (Given `(RT)/F= 0.059`)

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)`