Question

Write equation for the reactions taking place at the two electrodes (mentioning clearly the name of the electrode) during the electrolysis of

molten Lead bromide with inert electrodes.

Answer 1

Part 1: Reaction at the anode:-

At anode : $$\begin{gathered} \underset{\mathrm{Bromine}\mathrm{ion}}{{\mathrm{Br}}^{-}}\to \underset{}{\underset{\mathrm{Bromine}}{\mathrm{Br}\left(\mathrm{aq}\right)}+\underset{\mathrm{electron}}{{\mathrm{e}}^{-}}} \\ \underset{\mathrm{Bromine}}{\mathrm{Br}}+\underset{\mathrm{Bromine}}{\mathrm{Br}}\to \underset{\mathrm{Bromine}\mathrm{molecule}}{{\mathrm{Br}}_2} \end{gathered}$$

Part 2: Reaction at the cathode:-

1. At cathode : $\underset{\mathrm{Lead}\mathrm{ions}}{{\mathrm{Pb}}^{2+}}+\underset{\mathrm{electron}}{2{\mathrm{e}}^{-}}\to \underset{\mathrm{Lead}}{\mathrm{Pb}\left(\mathrm{s}\right)}$
2. Both the anode and cathode here are unaffected by the bromine vapors because it is made of graphite.

Therefore, at the anode : $$\begin{gathered} \underset{\mathrm{Bromine}\mathrm{ion}}{{\mathrm{Br}}^{-}}\to \underset{}{\underset{\mathrm{Bromine}}{\mathrm{Br}\left(\mathrm{aq}\right)}+\underset{\mathrm{electron}}{{\mathrm{e}}^{-}}} \\ \underset{\mathrm{Bromine}}{\mathrm{Br}}+\underset{\mathrm{Bromine}}{\mathrm{Br}}\to \underset{\mathrm{Bromine}\mathrm{molecule}}{{\mathrm{Br}}_2} \end{gathered}$$, at the cathode : $\underset{\mathrm{Lead}\mathrm{ions}}{{\mathrm{Pb}}^{2+}}+\underset{\mathrm{electron}}{2{\mathrm{e}}^{-}}\to \underset{\mathrm{Lead}}{\mathrm{Pb}\left(\mathrm{s}\right)}$