Question

when same quantity of electricity is passed through cuso4 solution and silver nitrate solution 6.36g of copper got deposited calculate the mass of silver deposited

Answer 1

Answer:

At the cathode

Cu2+ (aq) + 2e-  ---> Cu(s)

At the anode4OH-(aq) ----> 2H2O (l) + O2(g) + 4e-

Faraday’s constant = 96500C/mol

To deposit 1 mole of copper, we need 2x96500CSo, 96500C will deposit 0.5 moles of copper = 0.5x63.5 = 31.75g

To liberate one mole of oxygen molecules,we need  4x96500CSo, 96500C will liberate = ¼  = 0.25 moles = 0.25x32 = 8g