Compound A is prepared by the electrolysis of | vedclass

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(A) In the Castner-Kellner cell,the electrolysis of brine ($NaCl$ solution) is performed. At the anode (carbon),chloride ions are oxidized: $Cl^{-} \l

Vedclass · Accepted Answer

$NaOH$,$NaCl$

Vedclass · Answer

(A) In the Castner-Kellner cell,the electrolysis of brine ($NaCl$ solution) is performed. At the anode (carbon),chloride ions are oxidized: $Cl^{-} \longrightarrow \frac{1}{2} Cl_2 + e^{-}$. At the cathode (mercury),sodium ions are reduced to form a sodium amalgam: $Na^{+} + e^{-} + Hg \longrightarrow Na-Hg$. The sodium amalgam then reacts with water to produce $NaOH$: $2Na-Hg + 2H_2O \longrightarrow 2NaOH + 2Hg + H_2 \uparrow$. Thus,$A$ is $NaOH$ and $B$ is $NaCl$. Therefore,option $(A)$ is correct.

Compound $A$ is prepared by the electrolysis of an aqueous solution of $B$ using a Castner-Kellner cell with mercury as the cathode and carbon as the anode. $A$ and $B$ respectively are:

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