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

Answer

B

Explanation

The cell reaction is
`Zn(s) + Cu^(2+) (aq) hArr Zn^(2+) (aq) + Cu (s)`
Writing the Nernst eqn
`E_(cell) = E_(cell)^(@) - (0.059)/(2) log ((C_(Zn^(2+)))/(C_(Cu^(2+))))`
On increasing the conc. Of `Zn^(2+) (0.01 M rarr 1.0 M)` and decreasing the conc. of `Cu^(2+) (1.0 M rarr 0.01M)`,`log C_(Zn^(2+)//Cu^(2+))`
increases. This leads to decrease of `E_(cell)`. Thus `E_(2) lt E_(1)`.