Q.

The e.m.f. of the given Daniel cell at 298 K is E1 Zn /ZnSO4(0.01M)//CuSO4 (1.0M)/Cu When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01M, the e.m.f. changed to E2. What is the relationship between E1 and E2 ?

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E1 > E2

E1 < E2

E1 = E2

answer is A.

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Detailed Solution

For metal electrodeElectrolyte concentration in anode compartment decreases SOP increases. Electrolyte concentration in cathode compartment decreases SRP decreases.E1 > E2

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