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Question

In the electrochemical cell:
Zn|SO4(0.01 M)||CuSO4(1.0 M)|Cu, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 is changed to 0.01 M, the emf changes to E2. From the followings, which one is the relationship between E1 and E2?
(Given,RTF,0.059)

A
E1<E2
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B
E1>E2
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C
E2=0E1
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D
E1=E2
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Solution

The correct option is B E1>E2
Cell reaction:
Zn+Cu2+Zn2++Cu
E1=E+0.05912log10[Cu+2][Zn+2]
E1=E+0.03 log10[10.01]
E1=E+0.06

E2=E+0.05912log10[Zn+2][Cu+2]
E2=E+0.03 log10[0.011]
E2=E0.06
From above, E1>E2

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