Question

During electrolysis of aqueous naoh,4g of o2 gas is liberated at ntp at anode, h2 gas liberated at cathode is

Answer 1

Dear Student,

◆ Answer -

W(H2) = 0.25 g

● Explanation -

Reaction for electrolysis of aq. NaOH is given as -

2NaOH --> 2Na + O2 + H2

That means for every mole of O2 liberated at anode same no of moles are liberated at cathode.

n(O2) = n(H2)

W(O2) / M(O2) = W(H2) / M(H2)

4 / 32 = W(H2) / 2

W(H2) = 2 × 4 / 32

W(H2) = 0.25 g

Hence, 0.25 g of H2 is liberated at cathode.

Thanks dear. Hope this helps you...

Answer 2

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◆ Required Answer -

W(H2) = 0.25 g

● Explanation -

Reaction for electrolysis of aq. NaOH is given as -

2NaOH --> 2Na + O2 + H2

That means for every mole of O2 liberated at anode same no of moles are liberated at cathode.

→n(O2) = n(H2)

→W(O2) / M(O2) = W(H2) / M(H2)

→4 / 32 = W(H2) / 2

→W(H2) = 2 × 4 / 32

→W(H2) = 0.25 g

Hence, 0.25 g of H2 is liberated at cathode.

Hope it helps !