During an electrolysis of conc H2SO4 , peroxydisulphuric acid (H2 S2 O8) and O2 form in an equimolar amount. The moles of H2 that will be formed simultaneously will be

Question

During an electrolysis of conc H2SO4 , peroxydisulphuric acid (H2 S2 O8) and O2 form in an equimolar amount. The moles of H2 that will be formed simultaneously will be (A) Thrice that of O2 (B) Twice that of O2 (C) Equal to that of O2 (D) Half of the of O2

Tardigrade Admin · Accepted Answer

2H2O arrow O2+4H++4e- 2SO42 - arrow S2O82 -+2e- For 1 mole each of O2 and S2O82 - , no of mole of e- = 6 mole e- So 6 mole e- 6H++6e- arrow 3H2 Hence it produces 3 moles textH text2

Q. 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

Thrice that of $O_{2}$

Twice that of $O_{2}$

Equal to that of $O_{2}$

Half of the of $O_{2}$

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

$2 S O_{4}^{2 -} \to S_{2} O_{8}^{2 -} + 2 e^{-}$

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 $H_{2}$

Q. During an electrolysis of conc H2​SO4​ , peroxydisulphuric acid (H2​S2​O8​) and O2​ form in an equimolar amount. The moles of H2​ that will be formed simultaneously will be

Thrice that of O2​

Twice that of O2​

Equal to that of O2​

Half of the of O2​