InterviewSolution
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InterviewSolution
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In the electrochemical cell `Zn|ZnSO_(4)(0.01M)||CuSO_(4)(1.0M)|Cu`, the emf of this Daniel cell is `E_(1)`. When the concentration of `ZnSO_(4)` is changed to 1.0M and that of `CuSO_(4)` changed to 0.01M, the emf changes to `E_(2)`. From the followings, which one is the relationship between `E_(1) and E_(2)` (given,`(RT)/(F)=0.059`)A. `E_(1) lt E_(2)`B. `E_(1)gtE_(2)`C. `E_(2)=0neE_(1)`D. `E_(1)=E_(2)` |
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Answer» Correct Answer - B For cell `Zn|ZnSO_(4)(0.01M)||CuSO_(4)(1M)|Cu` Cell reaction `toZn+Cu^(+2)toZn^(+2)+Cu` `E_(1)=E^(o)-(0.059)/(2)"log"(Zn^(+2))/(Cu^(+2))` `E_(1)=E^(o)-(0.059)/(2)"log"(0.01)/(1)` `=E^(o)-(0.059)/(2)"log"(1)/(100)` . . . .(i) Fore cell `Zn|ZnSO_(4)(1M)||CuSO_($)(0.01M)|Cu` `E_(2)=E^(o)-(0.059)/(2)"log"(1)/(0.01)` `=E^(o)-(0.059)/(2)log100` . . . . . ltBrgt `E_(1)gt E_(2)` |
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