Question

Write the equations for the reaction which takes place at the electrodes during the electrolysis of Lead bromide.

Answer 1

Electrolysis of Molten Lead Bromide $\left({\mathrm{PbBr}}_2\right)$:

• Electrolysis of Lead bromide is done by heating solid Lead bromide in a silica crucible.
• Lead bromide will undergo electrolysis when a current of $3$ amperes at a voltage of $12$ volts is applied from a storage battery across the electrodes into molten lead bromide.

Reaction at cathode:

• The positively charged ${\mathrm{Pb}}^{2+}$ ions are discharged at the cathode where these ions are converted into a metallic lead by accepting electrons from the cathode.
• Grey deposition of Lead is obtained at the cathode.

${\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Pb}\left(\mathrm{s}\right)$

Reaction at anode:

• The negatively charged ${\mathrm{Br}}^{-}$ ions are discharged at the anode where these ions lose electrons to form neutral bromine atoms.
• These atoms then combine to form bromine molecules by forming a covalent bond between them.
• The reddish-brown vapor of ${\mathrm{Br}}_2$ is liberated at anode.

$$\begin{gathered} [{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)-{\mathrm{e}}^{-}\to \mathrm{Br}]\times 2\left(\mathrm{atom}\right) \\ \mathrm{Br}+\mathrm{Br}\to {\mathrm{Br}}_2\uparrow \left(\mathrm{molecule}\right) \end{gathered}$$