If 3F of electricity is passed through the solutions of AgNO3,CuSO4 and AuCl3, the molar ratio of the cations deposited at the cathode is

Since Ag++e−→Ag,Cu2++2e−→Cu,Au3++3e−→Au,3 F of electricity will deposit 3 moles of Ag, 1.5 moles of copper, and 1 mole of gold. Therefore, the molar ratio is 3 : 1.5 : 1 or 6 : 3 : 2.