Question

Explain electrolysis of lead bromide.

Answer 1

The positive terminal of the battery is connected to a graphite rod (which is made the anode) and the negative terminal of the battery is connected to a steel rod (which is the cathode). The silica crucible (electrolytic cell) is filled with solid lead bromide. There would be no deflection observed in the ammeter and thus no electricity is passed through solid lead bromide. When the crucible is heated to melt the solid lead bromide, deflection in ammeter can be observed. This confirms that electricity flows through the molten lead bromide.

The ions present in molten lead bromide are lead and bromide ions:
${\mathrm{PbBr}}_2\to {\mathrm{Pb}}^{2+}+2{\mathrm{Br}}^{-}$

Reaction at cathode: Positively charged lead ions get deposited at negative electrode (cathode). Thus, the cathode is coated with a thin grey coat of lead metal. The equation is:

​${\mathrm{Pb}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{P}\mathrm{b}\left[\mathrm{Reduction}\right]$

Reaction at anode: Negatively charged bromide ions get deposited at positive electrode (anode). Bromide ions oxidise to bromine atoms, that combine together to form red vapours of bromine molecule at anode.

$$\begin{gathered} {\mathrm{Br}}^{-}\to \mathrm{Br}+{\mathrm{e}}^{-}\left[\mathrm{Oxidation}\right] \\ \mathrm{Br}+\mathrm{Br}\to {\mathrm{Br}}_2 \end{gathered}$$