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)`

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`)A. `E_(1)=E_(2)`B. `E_(1) gt E_(2)`C. `E_(1) gt E_(2)`D. `E_(2)=0neE_(2)`
Answer» 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)`.

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