Question

A compound that liberates reddish-brown gas around the anode during electrolysis in its molten state

• A. Sodium chloride
• B. Copper(II) oxide
• C. Copper(II) sulphate
• D. Lead(II) bromide

Answer 1

The correct option is D

Lead(II) bromide

The explanation for the correct option:

Option (D). Lead(II) bromide

Redox reaction:

$\underset{\mathrm{Lead}\left(\mathrm{II}\right)\mathrm{bromide}}{{\mathrm{PbBr}}_2\left(\mathrm{s}\right)}\underset{\mathrm{Acidic}\mathrm{condition}}{\overset{\mathrm{Electrolysis}}{\to }}\underset{\underset{(\mathrm{Silver}-\mathrm{grey}\mathrm{metal})}{\mathrm{Lead}}}{\mathrm{Pb}\left(\mathrm{s}\right)}+\underset{\underset{(\mathrm{Reddish}-\mathrm{brown})}{\mathrm{Bromine}\mathrm{gas}}}{{\mathrm{Br}}_2\left(\mathrm{g}\right)}$

The reaction occurring at the anode during electrolysis of molten lead(II) bromide

• The Oxidation, the half-reaction is taking place at the anode.
• Bromide ions give up electrons to the anode and then turn into neutral bromine atoms hence the neutral bromine atoms form covalent bromine molecules.
• Product at the anode: Bromine. Reddish-brown vapors are seen at the anode and then bromine vapors are evolved from the molten lead bromide.

$$\begin{gathered} \underset{\mathrm{Bromide}\mathrm{ion}}{{\mathrm{Br}}^{-}}-\underset{\mathrm{electron}}{{\mathrm{e}}^{-}}\to \underset{\mathrm{Bromine}\mathrm{atom}}{\mathrm{Br}} \\ \underset{\mathrm{Bromine}\mathrm{atom}}{\mathrm{Br}}+\underset{\mathrm{Bromine}\mathrm{atom}}{\mathrm{Br}}\to \underset{\mathrm{Bromine}\mathrm{gas}}{{\mathrm{Br}}_2\left(\mathrm{g}\right)} \end{gathered}$$

The reaction occurring at the cathode during electrolysis of molten lead(II) bromide

• The Reduction, the half-reaction is taking place at the anode.
• The cathode is an electron donor hence the lead ions gain electrons from the cathode to metallic lead ions.
• The product at the cathode is lead, it is deposited as a silver-grey metal.

$\underset{\mathrm{Lead}\mathrm{ion}}{{\mathrm{Pb}}^{+2}}+\underset{\mathrm{electrons}}{2{\mathrm{e}}^{-}}\to \underset{\mathrm{Lead}}{\mathrm{Pb}\left(\mathrm{s}\right)}$

Thus, the correct option is (D) i.e, Lead(II) bromide is the compound that liberates reddish-brown gas around the anode during electrolysis in its molten state.