In the electrochemical cell : Zn|ZnSO_4 (0.01M)|CuSO_4 (1.0 M)|Cu, the emf of this Daniel cell is E_1. When the concentration of ZnSO_4, is changed to 1.0 M and that CuSO_4 changed to 0.01M, the emf changes to E_2. From the following , which one is the relationship between E_1 and E_2 ?

In the electrochemical cell : Zn|ZnSO_4 (0.01M)|CuSO_4 (1.0 M)|Cu, the

1.	In the electrochemical cell : Zn\|ZnSO_4 (0.01M)\|CuSO_4 (1.0 M)\|Cu, the emf of this Daniel cell is E_1. When the concentration of ZnSO_4, is changed to 1.0 M and that CuSO_4 changed to 0.01M, the emf changes to E_2. From the following , which one is the relationship between E_1 and E_2 ?
Answer» `E_1 LT E_2` `E_1 GT E_2` `E_2 = 0 UARR E_1` `E_1 = E_2` Solution :`E_("cell") = E_("cell")^(@) - (0.0591)/(2) "LOG" ([zn^(2+)])/([Cu^(2+)])` `E_1 = E_("cell")^(@) - (0.0591)/(2) "log" (10^(-2))/(1) "" Zn (S) to Zn^(2+)(aq) + 2e^(-)` `E_1 = E_("cell")^(@) + 0.0591 ………..(1) "" Cu^(2+)(aq) + 2e^(-) to Cu(s)` `E_2 = E_("cell")^(@) - (0.0591)/(2) log 1/(10^(-2)) "" Zn(s) + Cu^(2+)(aq) to Zn^(2+)(aq) +Cu(s)` `E_2 = E_("cell")^(@) - 0.0591 .............(2)` `:. E_1 > E_2`.