An electrolyte cell contains a solution of A g 2 S O 4 and have platium electrodes. A current is passed until 1.6 g of O 2 has been liberated at anode. The amount of silver deposited at cathode would be

Step-by-step solution

O 2 - - 2 e → 1 2 O 2 Equivalent weight of oxygen = 16 2 = 8 8 g O 2 = 1 g equivalent 1.6 g O 2 = 16 8 = 0.2 g equivalent O 2 1 g equivalent of Ag = 108 g Ag 0.2 g equivalent of Ag = 0.2 × 108 g Ag = 21.60 g Ag