The number of electrons involved in the manufacture of 1 mole of $H_{2} O_{2}$ from 50% $H_{2} S O_{4}$ :

2 moles

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3 moles

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1 mole

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4 moles

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Solution

The correct option is A 2 moles
The equations are:

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

$2 H S O_{4}^{-} \to H_{2} S_{2} O_{8} + 2 e^{-}$ At anode

$2 H^{+} + 2 e^{-} \to H_{2}$ At cathode

$H_{2} S_{2} O_{8} + 2 H_{2} O \to 2 H_{2} S O_{4} + H_{2} O_{2}$

Hence, two moles of electrons are transferred in the formation of 1 mole of $H_{2} O_{2}$ .

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The number of electrons involved in the manufacture of 1 mole of H2O2 from 50% H2SO4:

The correct option is A 2 molesThe equations are:2H2SO4→2HSO−4+2H+2HSO−4→H2S2O8+2e− At anode2H++2e−→H2 At cathodeH2S2O8+2H2O→2H2SO4+H2O2Hence, two moles of electrons are transferred in the formation of 1 mole of H2O2.

The number of electrons involved in the manufacture of 1 mole of $H_{2} O_{2}$ from 50% $H_{2} S O_{4}$ :