Question

With respect to the electrolysis of lead bromide, answer the following questions:
(i) Why is the lead bromide maintained in a molten state?
(ii) Why is the electrolytic cell made of silica?
(iii) Name the ions present in the electrolyte.
(iv) State what you observe at the (a) cathode (b) anode.
(v) Summarize the electrode reactions.

Answer 1

(i)

• In molten state lead bromide decomposes in to${\mathrm{Pb}}^{2+}\mathrm{and}{\mathrm{Br}}^{-}$ions.
• Due to ability of these ions to freely move in the molten form, it is able to conduct electricity and the process of electrolysis can be carried out.
• In solid state, due to absence of these ions in free state, it is a non conductor of electricity.

(ii)

• The lead bromide is maintained in a molten state because Ions can freely migrate in the molten state.
• The electrolytic cell is made of silica. This is due to the fact that silica is both heat resistant and a poor conductor of electricity.

(iii)

• The ions which are present in the electrolyte are Lead ion $\left({\mathrm{Pb}}^{2+}\right)$and Bromide ion $\left({\mathrm{Br}}^{-}\right).$

(iv)

• At the cathode, $Pb$(Lead) is deposited, whereas at the anode, $B{r}_2$ (Bromine) is produced.

(v)

• Reactions taking place are :
  At cathode (reduction): ${\mathrm{Pb}}^{2+}\left(l\right)+2{\mathrm{e}}^{-}⟶\mathrm{Pb}\left(s\right)$
  At anode (oxidation) : $2{\mathrm{Br}}^{-}\left(l\right)⟶{\mathrm{Br}}_2\left(l\right)+2e^{-}$
  The overall reaction is that the Lead (II) Bromide undergoes electrolysis to form Lead and Bromine is:
  ${\mathrm{PbBr}}_2\left(l\right)\stackrel{\text{Electrolysis}}{⟶}\mathrm{Pb}\left(s\right)+{\mathrm{Br}}_2\left(l\right)$