Question

The emf of a Daniel cell at 298K is `E_(1)` `Zn|underset((0.01" "M))(ZnSO_(4))||underset((1.0" M"))(CuSO_(4))|Cu` when the concentration of `ZnSO_(4)` is 1.0 M and that of `CuSO_(4)` is 0.01M the emf changed to `E_(2)` what is the relationship between `E_(1) and E_(2)`.
A. `E_(2)=0neE_(1)`
B. `E_(1)gtE_(2)`
C. `E_(1)ltE_(2)`
D. `E_(1)=E_(2)`

Answer 1

Correct Answer - B
`E_(1)=E_(o)-(0.0591)/(2)log((100)/(0.01))=E_(o)=(0.0591)/(2)xx4`
`thereforeE_(1) gt E_(2)`.