Electrolysis of Copper II CuSO4 solution, using: 1. Graphite or platinum, 2. Copper

Here is an explanation of the electrolysis of copper(II) sulfate ($\text{CuSO}_4$) solution for the two specified electrode types.

The ions present in an aqueous solution of copper(II) sulfate are $\text{Cu}^{2+}(aq)$, $\text{SO}_4^{2-}(aq)$, and from the dissociation of water, $\text{H}^+(aq)$ and $\text{OH}^-(aq)$.

1. Using Graphite or Platinum Electrodes (Inert Electrodes)

• At the Cathode (Negative Electrode): Cations present are $\text{Cu}^{2+}$ and $\text{H}^+$. Copper(II) ions are more easily reduced than hydrogen ions.

  $$\text{Cu}^{2+}(aq) + 2e^- \to \text{Cu}(s)$$

  Observation: A reddish-brown deposit of copper metal forms on the cathode.

• At the Anode (Positive Electrode): Anions present are $\text{SO}_4^{2-}$ and $\text{OH}^-$ (from water). Hydroxide ions (or water) are more easily oxidized than sulfate ions.

  $$2\text{H}_2\text{O}(l) \to \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-$$

  Observation: Colorless gas (oxygen) is produced at the anode. The solution becomes more acidic due to the formation of $\text{H}^+$ ions.

• Overall Reaction:

  $$2\text{CuSO}_4(aq) + 2\text{H}_2\text{O}(l) \to 2\text{Cu}(s) + \text{O}_2(g) + 2\text{H}_2\text{SO}_4(aq)$$

  Observation: The blue color of the $\text{CuSO}_4$ solution fades as $\text{Cu}^{2+}$ ions are removed, and the solution becomes acidic.

2. Using Copper Electrodes (Active Electrodes)

• At the Cathode (Negative Electrode): Cations present are $\text{Cu}^{2+}$ and $\text{H}^+$. Copper(II) ions are more easily reduced than hydrogen ions.

  $$\text{Cu}^{2+}(aq) + 2e^- \to \text{Cu}(s)$$

  Observation: A reddish-brown deposit of copper metal forms on the cathode.

• At the Anode (Positive Electrode): The anode itself is made of copper, which is an active electrode. Copper metal is more easily oxidized than water or sulfate ions.

  $$\text{Cu}(s) \to \text{Cu}^{2+}(aq) + 2e^-$$

  Observation: The copper anode gradually dissolves, and its mass decreases.

• Overall Reaction:

  $$\text{Cu}(s) \text{ (anode)} \to \text{Cu}(s) \text{ (cathode)}$$

  The net effect is the transfer of copper from the anode to the cathode. The concentration of $\text{CuSO}_4$ in the solution remains relatively constant because $\text{Cu}^{2+}$ ions are consumed at the cathode and produced at the anode at the same rate. This process is used for refining copper.