Correct Answer - C
( 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)`