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Q.

The EMF. of a Daniel cell, ZnZnSO4(0.01M)‖CuSO4(1.0M)Cu, at 298 K is E1. when the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the EMF changed to E2. What is the relationship between E1 and E2

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a

E1

b

E1=E2

c

E2=0≠E1

d

E1>E2

answer is D.

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Detailed Solution

Cell reaction Zn+Cu++⟶Zn+++CuE1=Ecell°-0.0592log0.011.0∴E1=Ecell°+0.059VE2=Ecell°-0.0592log1.00.01∴E2=Ecell°-0.059V· Thus, E1>E2
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