Question

A current 10 a is passed for 80 minutes and 27 seconds through a cell containing dilute sulphuric acid. How many moles of hydrogen gas will be librated at the cathode?

Answer 1

Explanation:

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 48270×324×96500

= 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 48270×652×96500

= 16.25 g