The EMFof a daniell cell at 298 K is E1ZN|ZnSo4|| CuSo4|Cu when the concentration of CuSo4 is 0.01M, the EMF changed to E2what is the relationship between E1 and E2 Share with your friends Share 2 Vartika Jain answered this Dear Student, Your questions seems incomplete as the initial concentrations and the final concentrations are missing. Only the concentration of CuSO4 is provided for E1 and rest other are missing. Hoever, here I am providing you the method to solve the question.Zn + Cu2+ → Zn2+ + Cu2+Ecell=E°cell-RTnFln[Zn2+][Cu2+]Greater the factor [Zn2+Cu2+], less is the emf. Suppose, this factor is less in case of E1 and more in case of E2, then clearly E1>E2 1 View Full Answer Hardee Shah answered this ZnSO4 0.01M, CuSO4 1.0M 3
