Question

During the electrolysis of copper (II) sulphate solution using platinum as cathode and carbon as anode.
What do you observe at the cathode and at the anode?
What changes are noticed in the electrolyte?
Write the reactions at the cathode and at the anode.

Answer 1

Part 1: Observation at anode and cathode:-

1. During this electrolysis of Copper (II) sulphate solution, a red shiny metal of copper is deposited at the cathode.
2. And a colourless and odourless gas of oxygen is deposited at the anode.

Part 2: Changes in the electrolyte:-

1. Here the electrolyte is the Copper (II) sulphate solution ${\mathrm{CuSO}}_4$.
2. It slowly changes its colour from blue to colourless.

Part 3:- Reactions at cathode and anode:-

1. At the cathode, $\underset{\mathrm{Copper}\mathrm{ion}}{{\mathrm{Cu}}^{2+}\left(\mathrm{s}\right)}+\underset{\mathrm{Electron}}{2{\mathrm{e}}^{-}}\to \underset{\mathrm{Copper}}{\mathrm{Cu}\left(\mathrm{s}\right)}$
2. At the anode, $$\begin{gathered} \underset{\mathrm{Hydroxide}}{{\mathrm{OH}}^{-}\left(\mathrm{l}\right)}\to \underset{\mathrm{Hydroxide}}{\mathrm{OH}\left(\mathrm{l}\right)}+\underset{\mathrm{Electron}}{{\mathrm{e}}^{-}} \\ \underset{\mathrm{Hydroxide}}{4\mathrm{OH}\left(\mathrm{l}\right)}\to \underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2\left(\mathrm{g}\right)} \end{gathered}$$

Therefore,

1. a red shiny metal of copper is deposited at the cathode and a colourless and odourless gas of oxygen is deposited at the anode.
2. The electrolyte slowly changes its colour from blue to colourless.
3. At the cathode, $\underset{\mathrm{Copper}\mathrm{ion}}{{\mathrm{Cu}}^{2+}\left(\mathrm{s}\right)}+\underset{\mathrm{Electron}}{2{\mathrm{e}}^{-}}\to \underset{\mathrm{Copper}}{\mathrm{Cu}\left(\mathrm{s}\right)}$, at the anode $$\begin{gathered} \underset{\mathrm{Hydroxide}}{{\mathrm{OH}}^{-}\left(\mathrm{l}\right)}\to \underset{\mathrm{Hydroxide}}{\mathrm{OH}\left(\mathrm{l}\right)}+\underset{\mathrm{Electron}}{{\mathrm{e}}^{-}} \\ \underset{\mathrm{Hydroxide}}{4\mathrm{OH}\left(\mathrm{l}\right)}\to \underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2\left(\mathrm{g}\right)} \end{gathered}$$