The quantity of electricity needed to electrolyse completely 1 M solution of CuSO4,Bi2(SO4)3,AlCl3 and AgNO3 each will be

CuSO4+2e−→Cu+SO4−2Bi2(SO4)3+6e−→2Bi+3SO4−2AlCl3+3e−→Al+3Cl−AgNO3+e−→Ag+NO3−Since, in 1 M CuSO4 solution, 1 M Bi2(SO4)3 solution, 1 M AlCl3 solution and 1 M AgNO3 solution, 2 moles electron, 6 moles electron, 3 moles electron are needed to deposit Cu, Bi, Al and Ag at the cathode respectively. But one mole electron = 1 F electricity That's why number of faradays required to deposit 1 M of each CuSO4,Bi2(SO4)3,AlCl3 and AgNO3 solution are 2 F, 6 F, 3 F and 1 F respectively AlternativelyNumber of moles = Number of equivalents ×Valency1 equivalent is deposited by 1 Faraday