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(D) The electrode reactions during the electrolysis of aqueous $Na_{2}SO_{4}$ are:Anode: $2H_{2}O \rightarrow O_{2} + 4H^{+} + 4e^{-}$Cathode: $2H_{2}

Vedclass · Accepted Answer

$4.8$

Vedclass · Answer

(D) The electrode reactions during the electrolysis of aqueous $Na_{2}SO_{4}$ are:Anode: $2H_{2}O ightarrow O_{2} + 4H^{+} + 4e^{-}$Cathode: $2H_{2}O + 2e^{-} ightarrow H_{2} + 2OH^{-}$From the anode reaction,$4 \, 	ext{mol}$ of electrons produce $1 \, 	ext{mol}$ of $O_{2}$.From the cathode reaction,$2 \, 	ext{mol}$ of electrons produce $1 \, 	ext{mol}$ of $H_{2}$.Given volume of $O_{2} = 2.4 \, L$ at $STP$.Moles of $O_{2} = \frac{2.4}{22.4} \, 	ext{mol}$.Since $1 \, 	ext{mol}$ of $O_{2}$ requires $4 \, 	ext{mol}$ of electrons,$2.4/22.4 \, 	ext{mol}$ of $O_{2}$ requires $(2.4/22.4) 	imes 4 \, 	ext{mol}$ of electrons.These same electrons pass through the cathode. Since $2 \, 	ext{mol}$ of electrons produce $1 \, 	ext{mol}$ of $H_{2}$,the moles of $H_{2}$ produced = $\frac{1}{2} 	imes 	ext{moles of electrons} = \frac{1}{2} 	imes \frac{2.4}{22.4} 	imes 4 = \frac{2.4}{22.4} 	imes 2 \, 	ext{mol}$.Volume of $H_{2} = 	ext{moles} 	imes 22.4 \, L = \frac{2.4}{22.4} 	imes 2 	imes 22.4 = 4.8 \, L$.

During electrolysis of an aqueous solution of $Na_{2}SO_{4}$,$2.4 \, L$ of $O_{2}$ at $STP$ was liberated at the anode. The volume of $H_{2}$ at $STP$ liberated at the cathode is ............. $L$.

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