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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) |
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Answer» `E_(1) lt E_(2)` `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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