A current of 10 ampere is passed for 80 minutes 27 seconds through a cell containing dilute H₂SO₄ :

1. How many moles of oxygen gas is liberated at anode?
2. Calculate the amount of zinc deposited at the cathode when another cell containing ZnSO₄ solution is connected in series ?

Current, I = 10 A
Time, t = 80 min and 27 second
= (80 $\times$60) + 27
= 4827 seconds
Charge, Q = I t
Q = 10$\times$ 4827
= 48270 F
Now in electrolysis of H₂SO₄, H₂ and O₂ liberated as follows:
At cathode: 
2H⁺ +  2e^-  →  H₂
At anode:
2H₂O ​→ O₂  +  4H⁺  +  4e^-
​4F liberate 1 mole of Oxygen gas or 32 g of Oxygen. 
4F or 4​$\times$96500 C will liberate 32 g of Oxygen gas

48270 C will liberate $48270\times \frac{32}{4\times 96500}$
= 4 g of Oxygen gas

(ii) Reaction of Zic is as foloows:
Zn²⁺ +  2e^-  →  Zn
2F of electricity willl liberate 1 mole of Zinc or 65 g of Zinc. 
2$\times$96500 C liberate 65 g of Zinc

48270 C will liberate $48270\times \frac{65}{2\times 96500}$
= 16.25 g