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ChemistryElectrochemistryMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

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

Options:

Answer:

Solution:

$2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$
$2 S O_{4}^{2 -} \to S_{2} O_{8}^{2 -} + 2 e^{-}$
For 1 mole each of $O_{2}$ and $S_{2} O_{8}^{2 -}$ , no of mole of $e^{-}$ = 6 mole $e^{-}$
So 6 mole $e^{-}$
$6 H^{+} + 6 e^{-} \to 3 H_{2}$
Hence it produces 3 moles $H_{2}$

Stream:NTA_ABHYASSubject:ChemistryTopic:ElectrochemistrySubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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