# A current of 10 ampere is passed for 80 minutes 27 seconds through a cell containing dilute H2SO4 : How many moles of oxygen gas is liberated at anode? 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 $×$60) + 27
= 4827 seconds
Charge, Q = I t
Q = 10$×$ 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​$×$96500 C will liberate 32 g of Oxygen gas

48270 C will liberate $48270×\frac{32}{4×96500}$
= 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$×$96500 C liberate 65 g of Zinc

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

• 16
What are you looking for?