Question

Describe the electrolysis of molten solution of Lead bromide.

Answer 1

Part 1: Things used for the electrolysis process:-

1. The electrolytic cell used:- Crucible (i.e., made of silica).
2. Electrolyte molten used:- Lead bromide $\left({\mathrm{PbBr}}_2\right)$.
3. Electrodes used:- Cathode and anode both are made up of graphite plate (inert electrode).

Part 2: Electrode reactions involved:-

1. At the cathode, ${\mathrm{Pb}}^{2+}$ions migrate to the cathode. They take two electrons each and become neutral lead atoms.
2. Reaction: $\underset{\mathrm{Lead}\mathrm{ions}}{{\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)}+\underset{\mathit{E}\mathit{l}\mathit{e}\mathit{c}\mathit{t}\mathit{r}\mathit{o}\mathit{n}\mathit{s}}{\mathit{2}{\mathit{e}}^{\mathit{-}}}⟶\underset{\mathrm{Lead}}{\mathrm{Pb}\left(\mathrm{s}\right)}$.
3. Greyish white metal lead is formed at the cathode.
4. At the anode, ${\mathrm{Br}}^{-}$ions migrate to the anode. It gives an electron and becomes a bromine atom.
5. Two bromine atoms combine to form bromine molecules.
6. Reactions: $$\begin{gathered} \underset{\mathrm{Bromine}\mathrm{ion}}{{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)}⟶\underset{\mathrm{Bromine}}{\mathrm{Br}}+\underset{\mathit{E}\mathit{l}\mathit{e}\mathit{c}\mathit{t}\mathit{r}\mathit{o}\mathit{n}\mathit{s}}{{\mathit{e}}^{\mathit{-}}} \\ \underset{\mathrm{Bromine}}{\mathrm{Br}}+\underset{\mathrm{Bromine}}{\mathrm{Br}}⟶\underset{\mathrm{Bromine}\mathrm{gas}}{{\mathrm{Br}}_2\left(\mathrm{g}\right)} \end{gathered}$$
7. Dark reddish-brown fumes of bromine evolve at the anode.

Part 3: Overall reaction:-

1. $\underset{\mathrm{Lead}\mathrm{bromide}}{{\mathrm{PbBr}}_2\left(\mathrm{l}\right)}\stackrel{380°\mathrm{C}}{⟶}\underset{\mathrm{Lead}}{\mathrm{Pb}\left(s\right)}+\underset{\mathrm{Bromine}\mathrm{gas}}{{\mathrm{Br}}_2(\mathrm{g})}$

Therefore, During the electrolysis of molten Lead bromide, the greyish-white metal of lead is deposited at the cathode due to the migration of ${\mathrm{Pb}}^{2+}$ions to the cathode. Dark reddish-brown fumes of bromine evolve at the anode due to the migration of ${\mathrm{Br}}^{-}$ions to the anode. The overall reaction of this electrolysis of molten Lead bromide is $\underset{\mathrm{Lead}\mathrm{bromide}}{{\mathrm{PbBr}}_2\left(\mathrm{l}\right)}\stackrel{380°\mathrm{C}}{⟶}\underset{\mathrm{Lead}}{\mathrm{Pb}\left(s\right)}+\underset{\mathrm{Bromine}\mathrm{gas}}{{\mathrm{Br}}_2(\mathrm{g})}$.