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Question: Answered & Verified by Expert
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 1.0M and that of CuSO4 changed to 0.01M, the emf changes to E2 . From the followings, which one is the relationship between E1 and E2 ?
(Given, RTF=0.059 )
ChemistryElectrochemistryNEET
Options:
  • A E1<E2
  • B E1>E2
  • C E2=0E1
  • D E1=E2
Solution:
2087 Upvotes Verified Answer
The correct answer is: E1>E2
E =E o 0.059 2 log 10 Zn 2+ Cu 2+
E 1 =E° 0.059 2 log 10 0.01 (1) 2
E 2 =E° 0.059 2 log 10 1 (0.01) 2
Hence E1>E2

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