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)

In the electrochemical cell: Z|ZnSO_(4)(0.01M)||CuSO_(4)(1.0M)|Cu, t

1.	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)
Answer» `E_(1)=E_(2)` `E_(1) gt E_(2)` `E_(1) gt E_(2)` `E_(2)=0neE_(2)` Solution :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)`.