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) gt E_(2)`
D. `E_(2)=0 !=E_(1)`

Answer

A

Explanation

For the Deniell cell, the cell reaction is
`Zn(s)+Cu^(2+)(aq.)hArr Cu(s)+Zn^(2+)(aq.)`
Applying Nernst equation, we have (at `2981^(@)C`)
`E_("cell") = E_("cell")^(@) - (0.0591 V)/(n) "10g" (C_(Zn^(2+)))/(C_(Zn^(2+)`
Since `E_("cell")^(@)` is positive, `E_("cell")` will decrease if we increase the magnitude of concentration ratio, i.e., `C_(Zn^(2+))//C_(Cu^(2+))`. Since
`(0.01 M)/(1.0 M) lt (1.0 M)/(0.01 M)` , we have `E_(1) gt E_(2)`.