In electrolysis of aqueous NaOH at NTP,2.8 L | vedclass

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Question

(B) The electrolysis of aqueous $NaOH$ involves the following reactions:Anode: $4 OH^{-} \rightarrow O_{2(g)} + 2 H_{2}O + 4 e^{-}$Cathode: $4 H_{2}O

Vedclass · Accepted Answer

$5.6$

Vedclass · Answer

(B) The electrolysis of aqueous $NaOH$ involves the following reactions:Anode: $4 OH^{-} ightarrow O_{2(g)} + 2 H_{2}O + 4 e^{-}$Cathode: $4 H_{2}O + 4 e^{-} ightarrow 2 H_{2(g)} + 4 OH^{-}$The overall reaction is $2 H_{2}O ightarrow 2 H_{2(g)} + O_{2(g)}$.From the stoichiometry,the molar ratio of $H_{2}$ to $O_{2}$ is $2 : 1$.Since the volume of gas is directly proportional to the number of moles at $NTP$,the volume of $H_{2}$ liberated at the cathode will be twice the volume of $O_{2}$ liberated at the anode.Given volume of $O_{2} = 2.8 \ L$.Volume of $H_{2} = 2 	imes 2.8 \ L = 5.6 \ L$.Therefore,the correct option is $B$.

In electrolysis of aqueous $NaOH$ at $NTP$,$2.8 \ L$ of oxygen gas is liberated at the anode. The amount of gas liberated at the cathode will be ............ $L$.

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