Only reactive metals such as Fe and Cu , oxodosed at anode.
Fe → Fe 2+ + 2e-
Cu → Cu 2+ + 2e-
Ag and Au , being non reactive will appear below anode as anode mud.
Inc in mass at cathode is due to Cu depositied. = 22.011 g = 0.3466 mol
Moles of Fe and Cu dissolving frm anode = itn X 96500 (where i = current and t = time)
= 140 X 482.52 X 96500 = 0.3500 mol
Moles of Fe dissolved = 0.3500 - 0.3466 = 0.0034 mol = 0.1904 g
Now % of Fe = 0.190422.260 X 100 = 0.86 5
and that of Cu = 22.011 X 10022.260 = 98.88 %