Question

Complete the following ionic equations in reference to electrolysis and electroplating

$$\begin{gathered} 1.{\mathrm{CuSO}}_4\rightleftharpoons \mathrm{\_\_\_\_\_}+\mathrm{\_\_\_\_\_\_} \\ 2.{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{\_\_\_\_\_\_}+\mathrm{\_\_\_\_\_\_\_} \\ 3.{\mathrm{Cu}}^{+2}+\mathrm{\_\_\_\_}\to \mathrm{\_\_\_\_\_} \\ 4.\mathrm{Cu}-\mathrm{\_\_\_\_}\to \mathrm{\_\_\_\_\_\_} \end{gathered}$$

Answer 1

Electrolysis of ${\mathrm{CuSO}}_4$ solution

1. Dissociation of ${\mathrm{CuSO}}_4$

• Copper sulfate is dissociating into copper ions as positive ions and sulfate ions as negative ions

${\mathrm{CuSO}}_4\rightleftharpoons \mathbf{}{\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}+\mathbf{}{{\mathbf{SO}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$

2. Dissociation of ${\mathrm{H}}_2\mathrm{O}$

• The water is dissociated into hydrogen ions as positive ions and hydroxyl ions as negative ions

${\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathbf{H}}^{\mathbf{+}}\mathbf{}+\mathbf{}{\mathbf{OH}}^{\mathbf{-}}$

3. Electrode reaction

• The reduction half-reaction occurring at the cathode is given in this reaction
• Both ${\mathrm{Cu}}^{+2}$ and ${\mathrm{H}}^{+}$ ions migrate to the cathode and ${\mathrm{Cu}}^{+2}$ ions are discharged at the cathode as neutral atoms by accepting electrons. product at the cathode is a reddish-brown copper deposit

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

4. Electrode reaction

• The oxidation half-reaction occurring at the anode is given in this reaction
• Copper loses electrons more easily than ${{\mathrm{SO}}_4}^{-2}$ and ${\mathrm{OH}}^{-}$ because copper anode itself ionizes to give ${\mathrm{Cu}}^{+2}$ions

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