In the electrochemical cell:
`Zn|ZnSO_(4)(0.01 M)||CuSO_(4)(1.0M)|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_(1) and E_(2)" "("Given, "(RT)/(F)=0.059)`-

Options

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

Answer

B

Explanation

Cell reaction, `Zn+Cu^(2+)rarrZn^(2+)+Cu`
`E_(1)=E^(@)-(0.059)/(2)log.([Zn^(2+)])/([Cu^(2+)])=E^(0)-(0.059)/(2)log.(0.01)/(1)" …[1]"`
when the cell is changed to,
`Zn|ZnSO_(4)(1M)|CuSO_(4)(0.01M)|Cu`
`E_(2)=E^(@)-(0.059)/(2)log.(1)/(0.01)" ...[2]"`
`therefore E_(1)gt E_(2)`