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Grade: 12th pass
1.The emf of a daniel cell at 298K is E1.
Zn/ZnSO4(0.001M)||CuSO4(1.0M)/Cu.when the conc of ZnSO4 is 1M and that of CuSO4 is 0.01Mthe emf changed to E2.What is the relationship between E1 and E2
2.Given E(Fe3+/Fe)= -0.036V ,
E(Fe2+/Fe)= -0.439V
the value of standard electrode potential for the change Fe3+ + e- giving Fe2+ will be 
3 years ago

Answers : (1)

Vikas TU
11678 Points
  1. For E1 = Eocell – (0.0529/2)log(Zn2+/Cu2+) => Eocell – (1.5)(0.0529)........(1)
For E2 = Eocell – (0.059)..............(2)
Substract the eqns. (1) and (2) we get,
E1 – E2 = 0.059*2.5 = > 0.13225 VOLT
  1. Fe3+ + 3e- ==>    Fe
Fe2+ + 2e- ===> Fe
Now reversing the second reaction and addiing with the first, we get the relation as:
E1 – E2 = Ecell for  Fe3+ + e- giving Fe2+.
3 years ago
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