Question

In the electrochemical cell:
`Z|ZnSO_(4)(0.01M)||CuSO_(4)(1.0M)|Cu,`
the e.m.f. this daniel cell is `E_(1)`. When the concentration of `ZnSO_(4)` is changed to 1.0 M and that of `CuSO_(4)` is changed to 0.01 M, the e.m.f. changes to `E_(2)`. from the following, which one is the relationship between `E_(1) and E_(2)`? (Given, `(RT)/(F)=0.059`)
A. `E_(1)=E_(2)`
B. `E_(1) gt E_(2)`
C. `E_(1) gt E_(2)`
D. `E_(2)=0neE_(2)`

Answer 1

Correct Answer - C
The cell reaction is
`Zn+CuSO_(4)toZnSO_(4)+Cu,n=2`
`E=E_(cell)^(@)-(2.303RT)/(2F)"log"([ZnSO_(4)])/([CuSO_(4)])`
In 1 st case, `E_(1)=E_(cell)^(@)-(2.303RT)/(2F)"log"(0.01)/(1)`
In 2nd case, `E_(2)=E_(cell)^(@)-(2.303RT)/(2F)"log"(1)/(0.01)`
Evidently, `E_(1)gtE_(2)`.