During the electrolysis of conc. H₂SO₄, it was found that H₂S₂O₈ and O₂ were liberated in a molar ratio of 3 : 1. How many moles of H₂ were found in terms of moles of H₂S₂O₈? (Express your answer as: 3 x moles of H₂; integer answer is between 0 and 9).

Let x mole of O₂ is liberated and 3x mole of H₂S₂O₈ is formed. Reactions at cathode (reduction):
$2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2\stackrel{\ominus }{\mathrm{OH}}$
Reactions at anode (oxidation):
 

Total Faradays at anode = (4x + 6x) F = 10x F.
Total Faradays at cathode = 2F = 1 mole H₂,.
10x F = Total Faradays at cathode = Total Faradays at anode
$\therefore$ 2 F at cathode = 1 mole of H₂.
$10\mathrm{xF}\text{ at cathode }\equiv \frac{1}{2\mathrm{F}}\times 10\mathrm{xF}=5\mathrm{x}\text{ mole of }{\mathrm{H}}_2\text{. }$
$\text{ Ratio }=\frac{\text{ Moles of }{\mathrm{H}}_2\text{ at cathode }}{\text{ Moles of }{\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8\text{ at anode }}=\frac{5\mathrm{x}}{3\mathrm{x}}=\frac{5}{3}$

Number of moles of H₂ =3 x$\frac{5}{3}$ = 5
Alternatively

$\text{ Equivalent ratio }=3\times 2:1\times 4=6:4$
$\text{ Total equivalent of }{\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8\text{ and }{\mathrm{O}}_2\text{ at anode }=6+4$ = 10 Eq.
So total equivalent of H₂ at cathode = 10
$\therefore \text{ moles of }{\mathrm{H}}_2(\mathrm{n}\text{ factor }=2)=\frac{10}{2}=5\mathrm{moles}$