Question

Complete the following ionic equations in reference to electrolysis and electroplating

$$\begin{gathered} 1.{\mathrm{PbBr}}_2\rightleftharpoons \mathrm{\_\_\_\_\_}+\mathrm{\_\_\_\_\_\_} \\ 2.{\mathrm{Pb}}^{+2}+\mathrm{\_\_\_\_\_}\to \mathrm{\_\_\_\_\_\_} \\ 3.{\mathrm{Br}}^{-1}-\mathrm{\_\_\_\_\_\_}\to \mathrm{\_\_\_\_\_} \\ 4.\mathrm{Br}+\mathrm{Br}\to \mathrm{\_\_\_\_\_\_\_\_\_\_} \end{gathered}$$

Answer 1

1. Dissociation of ${\mathrm{PbBr}}_2$

• Lead bromide is dissociating into lead ions as positive ions and bromide ions as negative ions

${\mathrm{PbBr}}_2\rightleftharpoons \mathbf{}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}+\mathbf{}\mathbf{2}{\mathbf{Br}}^{\mathbf{-}}$

2. Electrode reaction

• The reduction of half-reaction occurring at the cathode is given in this reaction
• The cathode is an electron donor hence the lead ions gain electrons from the cathode to form metallic lead ions.

${\mathrm{Pb}}^{+2}+\mathbf{}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\to \mathbf{Pb}$

3. Electrode reaction

• The oxidation half-reaction occurring at the anode is given in this reaction
• Bromide ions give up electrons to the anode and then turn into neutral bromine atom

${\mathrm{Br}}^{-1}-\mathbf{}{\mathbf{e}}^{\mathbf{-}}\to \mathbf{}\mathbf{Br}$

4. Electrode reaction

• Hence the neutral bromine atoms combine together through the covalent bond to form covalent bromine molecules

$\mathrm{Br}+\mathrm{Br}\to {\mathbf{Br}}_{\mathbf{2}}$