The emf of a Daniel cell at 298K is E1 Zn | ZnSO4(0.01M) || CuSO4(1.0M) | Cu when the concentration of ZnSO4 is 0.1 M and that CuSO4 is 0.01 M, the emf changed to E2. What is the relationship between E1 and E2
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a
E2=0≠E1
b
E1 > E2
c
E1 < E2
d
E1 = E2
answer is B.
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Detailed Solution
As electrolyte concentration in anode compartment is decreased SOP of Zn electrode increases or SRP of Zn electrode decreases,and EMF of cell increases.