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

Answer

C

Explanation

( c) Thinking process Calculate the value of `E_(cell)` i.e. `E_(1) and E_(2)` by substitutingk the respective given values in the Nernst equation.
`E_(cell)=E^(@)-(0.059)/(n) log""(|Zn^(2+)|)/(|Cu^(2+)|)`
Compare the calculated values of `E_(1) and E_(2)` and find the correct relation.
For the electrochemical cells ,
`Zn|ZnSO_(4)(0.01M)||CuSO_(4)(1M)|Cu`
Cell reaction :
`Zn+Cu^(2+) to Zn^(2+) +Cu,n=2`
`E_(1)=E^(@)-(0.059)/(2) log""(Zn^(2+))/(Cu^(2+))=E^(@)-(0.059)/(2)log""(0.01)/(1)`
`E_(1)=E^(@)-(0.059)/(2)log""(1)/(100)=(E^(@)+0.059)`
For cell,
`Zn|ZnSO_(4)(0.01M)||CuSO_(4)(1M)|Cu`
`E_(2)=E^(@)-(0.059)/(2)log""(1)/(0.01)`
`E_(2)=E^(@)-(0.059)/(2)log100`
`:. " "=(E^(@)-0.059)impliesE_(1) gt E_(2)`