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The weight in grams of O₂ formed at Pt anode during the electrolysis of aq. $K_{2} {SO}_{4}$ solution during the passage of one coulomb of electricity is

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$\frac{16}{96800}$

$\frac{8}{96500}$

$\frac{32}{96500}$

$\frac{64}{96500}$

answer is B.

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Detailed Solution

When K₂SO_{4 (aq)}is electeolysed the product formed at anode is O₂ and at cathode is H₂ .

According to Faraday's law of electrolyis ,

w= zQ

w= mass of substance .

Q= charge .

z= electrochemical equivalent .

Electrochemical equivalent (z)is defined as the mass of substnace deposited when one coulomb of electricity is passed through the electrolyte.

$z = \frac{E q u i v a l e n t w e i g h t}{96500}$

Hence the mass of oxygen O formed when one coulomb of elceticity passed = $\frac{E q u i v a l e n t w e i g h t o f O_{2}}{96500} = \frac{8}{96500}$ g

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