How to explain the electrolysis of copper(II) sulfate

Copper(II) sulfate solution is one of the most-tested electrolysis examples in SPM, and it is a favourite for a reason: the same solution gives completely different results depending on whether you use carbon or copper electrodes. Understand why, and you understand the whole logic of selective discharge in redox equilibrium. This guide walks through both cases with correct half-equations, the observations you must be able to state, and the reasoning that earns the marks.

First, list the ions present

Before touching either electrode, always start by listing every ion in the solution. Copper(II) sulfate solution contains ions from two sources:

• From the salt CuSO₄: copper(II) ions, Cu²⁺, and sulfate ions, SO₄²⁻.
• From the water itself: hydrogen ions, H⁺, and hydroxide ions, OH⁻.

So four ions are moving: two positive (Cu²⁺ and H⁺) drift to the cathode, and two negative (SO₄²⁻ and OH⁻) drift to the anode. At each electrode, only one ion is discharged, and working out which is the whole skill.

Case 1: carbon (inert) electrodes

Carbon electrodes take no part in the reaction, they are inert, so the ions themselves are discharged.

At the cathode (negative electrode): Cu²⁺ and H⁺ both arrive. Selective discharge favours the ion of the metal lower in the electrochemical series, copper is below hydrogen, so Cu²⁺ is discharged in preference to H⁺. Copper ions gain electrons and are reduced to copper metal, which deposits as a pink/brown solid on the electrode:

Cu²⁺ + 2e⁻ → Cu

At the anode (positive electrode): SO₄²⁻ and OH⁻ both arrive. Sulfate is not discharged; the hydroxide ion is discharged instead, giving off oxygen gas (colourless gas, relights a glowing splinter):

4OH⁻ → O₂ + 2H₂O + 4e⁻

What happens to the solution: copper ions are removed at the cathode while hydroxide ions are removed at the anode. The Cu²⁺ concentration falls, so the blue colour of the solution fades. Meanwhile H⁺ and SO₄²⁻ are left behind, so the solution effectively becomes dilute sulfuric acid and turns more acidic. These are exactly the observations recorded in the electrolysis of copper(II) sulfate experiment.

Case 2: copper electrodes

Now repeat the electrolysis using two copper electrodes. The cathode behaves the same way, Cu²⁺ is discharged and copper deposits, but the anode behaves completely differently, and that difference is the point of the question.

At the anode (positive electrode): a copper anode is not inert. Instead of discharging any ion from the solution, the copper metal of the anode itself dissolves, releasing Cu²⁺ ions into the solution and giving up electrons:

Cu → Cu²⁺ + 2e⁻

At the cathode (negative electrode): the same reduction as before:

Cu²⁺ + 2e⁻ → Cu

What happens overall: copper dissolves from the anode and deposits on the cathode, so the anode loses mass and the cathode gains mass by the same amount. Copper is effectively transferred from one electrode to the other. Because every Cu²⁺ removed at the cathode is replaced by one released at the anode, the concentration of Cu²⁺ stays constant and the blue colour does not change. This is the principle behind purifying copper and electroplating an object with copper.

The one comparison examiners want

The whole topic comes down to a single contrast you should be ready to state in a sentence:

• With carbon (inert) electrodes, the anode releases oxygen, the blue colour fades, and the solution becomes acidic.
• With copper (active) electrodes, the anode dissolves, the blue colour stays the same, and copper is transferred from anode to cathode.

The cathode reaction is identical in both; only the anode differs, because an active copper anode offers an easier reaction, its own atoms dissolving, than discharging hydroxide.

The factors behind selective discharge

If a question changes the conditions, three factors decide which ion is discharged:

1. Position in the electrochemical series, the main factor. At the cathode, the less electropositive metal ion (lower in the series) is discharged; at an inert anode, the ion lower in the discharge order for anions (halides and hydroxide are discharged in preference to sulfate).
2. Concentration, a very high concentration of an ion can lead to its discharge even if the series would not normally favour it (as with concentrated chloride solutions).
3. Type of electrode, an active electrode like copper can dissolve instead of an ion being discharged, which is exactly what changes at the anode here.

Common mistakes to avoid

• Writing the wrong sign of electrode. Reduction (gaining electrons) happens at the cathode, the negative electrode; oxidation at the anode.
• Discharging sulfate. SO₄²⁻ is not discharged in these conditions, it stays in solution.
• Saying the blue colour fades with copper electrodes. It does not, the Cu²⁺ is replenished from the anode.
• Unbalanced half-equations. Electrons and charges must balance: 4OH⁻ → O₂ + 2H₂O + 4e⁻, not a rougher version.

Pulling it together

Copper(II) sulfate rewards a clear routine: list the ions, decide the discharge at each electrode, write the balanced half-equation, then state the observation. Do that twice, once for carbon and once for copper, and you have the full answer, including why one blue solution fades and the other does not. You can compare it with other cells in electrolysis of aqueous solutions. If half-equations or the inert-versus-active distinction keep tangling you up, a teacher can straighten it out quickly. Our online one-to-one lessons run in English from RM50 an hour, with a paid one-hour trial.