Two electrolytic cells containing silver nitrate solution and dilute sulphuric acid were connected in a series. A steady current of 2.5 A was passed through them till 1.078 g of silver was deposited ( Ag = 107.8 g `//" mol"` )
(a) How much electricity was consumed?
(b) What is the weight of oxygen gas liberated during the reaction?

Three electrolytic cells A,B and C containing solutions of zinc sulphate, silver nitrate and copper sulphate, respectively are connected in series. A steady current of 1.5 A was passed through them until 1.45g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited in the concerned cells? ( Atomic mass of Ag= 108, Zn = 65.4, Cu = 63.5)

An electrolytec cell contains a solution of AgNO_3 and have platinum electrodes. A current is passed untill 1.6 g of O_2 has been liberated at anode. The amount of silver peposited at cathode would be :

The electrochemical equivalent of silver is 0.0011180g . When an electric current of 0.5 ampere is passed through an aqueous silver nitrate solution for 200 sec, the amount of silver deposited is :

A current of 2 ampere was passed through solution of CuSO_4 and AgNO_3 in series. 0.635 g of copper was deposited .Then the weight of silver deposited will be:

Two different electrolytic cells filled with molten Cu(NO_(3))_(2) and molten Al ( NO_(3))_(2) , respectively are connected in series. When electricity is passed 2.7g Al is deposited on electrode. Calculate the weight of Cu deposited on cathode. [ Cu = 63.5, Al = 27.0 g "mol"^(-1) ]

The same amount of electricity was passed through two cells containing molten Al_2O_3 and molten NaCI. If 1.8g of Al were liberated in one cell, the amonut of Na liberated in the other cell is :

How many grams of silver could be plated out on a shield by electrolysis of a solution containing Ag^(+) ions for a period of 4 hours at a current strength of 8.5 amperes? [Molar mass of Ag =107.8g]