1.

Emf of given Daniell cell at 298 K is E_(1). Zn|ZnSO_(4)(0.01M)||CuSO_(4(1.0M))|Cu. The emf changed to E_(2) when concentration of ZnSO_(4) solution is 1.0 M and CuSO_(4) solution is 0.01 M, then what is the relation between E_(1) and E_(2) ?

Answer»

`E_(1) GT E_(2)`
`E_(1) LT E_(2)`
`E_(1)=E_(2)`
`E_(2)=0 ne E_(1)`

Solution :`Zn+CU^(2+) to Zn^(2+)+Cu`
`E=E^(@)-(0.059)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
`E_(1)=E^(@)-(0.059)/(2)"log"(0.01)/(1.0) THEREFORE E_(1)=E^(@)+0.059`
`E_(2)=E^(@)-(0.059)/(2)"log"(1.0)/(0.01)therefore E_(2)=E^(@)-0.059`
`therefore E_(1) gt E_(2)`.


Discussion

No Comment Found

Related InterviewSolutions