Question

In an electrolysis of an aqueous solution of sodium sulphate,2.4L of oxygen at STP was liberated at anode.The volume of hydrogen at STP liberated at cathode would be

Answer 1

Hello Dear.

We know, for every 1 mole of oxygen liberated at Anode, 2 moles of Hydrogen are liberated at Cathode.

Now, Mass of 1 mole of Oxygen = 32 g/mole.
Mass of one mole of Hydrogen = 2 g/mole

∵ For 1 mole of Oxygen liberated at Anode, 2 mole of H₂ is liberated at Cathode.
∴ For every 32 g of oxygen liberated at anode, 2 × 2 mole of H₂ is liberated at Cathode.
∴ For every 22.4 liter of Oxygen liberated at Anode, 2 × 22.4 liter of H₂ is liberated at cathode.

∴ For every 1 liter of oxygen liberated at anode, 2 liter of Hydrogen is liberated at cathode.

∴ For every 2.4 liter of oxygen liberated at anode, 2 × 2.4 = 4.8 liter of Hydrogen is liberated at cathode. 


Hence, the Volume of the Hydrogen liberated at the cathode is 4.8 liter at S.T.P.


Hope it helps.

Answer 2

Answer:

4.8 sttp because of hydrogen and oxygen h..32 o..2