Question

During an electrolysis of conc $H_{2}SO_{4}$ , peroxydisulphuric acid $\left(H_{2} S_{2} O_{8}\right)$ and $O_{2}$ form in an equimolar amount. The moles of $H_{2}$ that will be formed simultaneously will be

• A. Thrice that of $O_{2}$
• B. Twice that of $O_{2}$
• C. Equal to that of $O_{2}$
• D. Half of the of $O_{2}$

Answer 1

Answer: (A)

$2H_{2}O \rightarrow O_{2}+4H^{+}+4e^{-}$

$2SO_{4}^{2 -} \rightarrow S_{2}O_{8}^{2 -}+2e^{-}$

For 1 mole each of $O_{2}$ and $S_{2}O_{8}^{2 -}$ , no of mole of $e^{-}$ = 6 mole $e^{-}$

So 6 mole $e^{-}$

$6H^{+}+6e^{-} \rightarrow 3H_{2}$

Hence it produces 3 moles $\text{H}_{\text{2}}$