Question

Complete the following ionic equations in reference to electrolysis and electroplating

$$\begin{gathered} 1.{\mathrm{Al}}_2{\mathrm{O}}_3\rightleftharpoons \mathrm{\_\_\_\_\_}+\mathrm{\_\_\_\_\_\_} \\ 2.{\mathrm{Al}}^{+3}+\mathrm{\_\_\_\_\_}\to \mathrm{\_\_\_\_\_\_} \\ 3.{\mathrm{O}}^{-2}-\mathrm{\_\_\_\_\_\_}\to \mathrm{\_\_\_\_\_} \end{gathered}$$

Answer 1

1. Dissociation of molten ${\mathrm{Al}}_2{\mathrm{O}}_3$

• ${\mathrm{Al}}_2{\mathrm{O}}_3$ is dissociating into Aluminum ions as positive ions and oxide ions as negative ions

${\mathrm{Al}}_2{\mathrm{O}}_3\rightleftharpoons \mathbf{2}{\mathbf{Al}}^{\mathbf{3}\mathbf{+}}+\mathbf{}\mathbf{3}{\mathbf{O}}^{\mathbf{2}\mathbf{-}}$

2. Electrode reaction

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

${\mathrm{Al}}^{+3}+\mathbf{}\mathbf{3}{\mathbf{e}}^{\mathbf{-}}\to \mathbf{Al}$

3. Electrode reaction

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

${\mathrm{O}}^{-2}-\mathbf{}{\mathbf{e}}^{\mathbf{-}}\to \mathbf{}\mathbf{O}$