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Question

In the electrochemical cell:
Zn|ZnSO4(0.01M)| |CuSO4(1.0M)|Cu, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 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
E2=0E1
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B
E1=E2
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C
E1<E2
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D
E1>E2
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Solution

The correct option is B E1>E2
Ecell=E0cell0.059nlog[Zn2+][Cu2+]
E1=1.10.0592log0.011.0
E1=1.10.0592×(2)
E1=1.1(0.0592)
E1=1.16V
Again,

E2=1.10.0592log1.00.01
E1=1.10.0592×(2)
E1=1.1(0.059)
E2=1.041 V

Hence, E1>E2

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