O-Level Chemistry · Secondary 3 and 4
Electrolysis
Electrolysis is the decomposition of an ionic compound, molten or in solution, by passing an electric current through it. It is how reactive metals such as aluminium and sodium are extracted and how copper is purified. In O-Level Chemistry you learn to predict the product at each electrode, write half-equations and explain changes in the electrolyte. The topic builds directly on ionic bonding and redox, since every electrode reaction is either an oxidation or a reduction.
What the syllabus expects
- Describe electrolysis as the decomposition of a molten or aqueous ionic compound by electricity
- Predict the products of the electrolysis of molten binary compounds using inert electrodes
- Predict the products of the electrolysis of aqueous solutions, including dilute sulfuric acid and concentrated aqueous sodium chloride, using inert electrodes
- Describe the electrolysis of aqueous copper(II) sulfate with inert electrodes and with copper electrodes
- Write half-equations for the reactions at the anode and the cathode
- Describe a simple cell made from two different metals in an electrolyte
How electrolysis works
An electrolyte is a molten or aqueous ionic compound that conducts electricity and is decomposed by it. The ions must be free to move, which is why a solid ionic compound cannot be electrolysed. Two electrodes, often inert graphite or platinum, dip into the electrolyte. The anode is the positive electrode and the cathode is the negative electrode. Positive ions (cations) move to the cathode, where they gain electrons and are reduced. Negative ions (anions) move to the anode, where they lose electrons and are oxidised. Electrons flow through the external wires, while ions carry the charge through the electrolyte. A useful reminder: oxidation happens at the anode and reduction at the cathode.
Molten binary compounds
A molten binary compound contains only two kinds of ions, so the products are easy to predict: the metal forms at the cathode and the non-metal at the anode. For molten lead(II) bromide, at the cathode Pb²⁺ + 2e⁻ gives Pb, and molten lead collects. At the anode 2Br⁻ gives Br₂ + 2e⁻, and reddish-brown bromine vapour is seen. For molten sodium chloride, sodium forms at the cathode, Na⁺ + e⁻ gives Na, and chlorine at the anode, 2Cl⁻ gives Cl₂ + 2e⁻. Check that each half-equation balances in both atoms and charge, and that the number of electrons matches the charge on the ion.
Aqueous solutions and selective discharge
In an aqueous solution, water provides a small number of H⁺ and OH⁻ ions as well as the ions of the dissolved compound, so there is a choice at each electrode. At the cathode, if the metal is less reactive than hydrogen, such as copper or silver, the metal is deposited. If the metal is more reactive than hydrogen, such as sodium, potassium or magnesium, hydrogen forms instead: 2H⁺ + 2e⁻ gives H₂. At the anode, sulfate and nitrate ions are not discharged, so OH⁻ ions are discharged to give oxygen: 4OH⁻ gives O₂ + 2H₂O + 4e⁻. If a halide such as chloride is present at high concentration, the halogen forms instead. For dilute sulfuric acid, hydrogen forms at the cathode and oxygen at the anode in a 2 : 1 volume ratio, and the acid becomes more concentrated because water is removed. For concentrated aqueous sodium chloride, hydrogen forms at the cathode and chlorine at the anode, and the solution left behind becomes sodium hydroxide.
Copper(II) sulfate with inert and copper electrodes
With inert electrodes, copper is deposited at the cathode as a pink-brown coating, Cu²⁺ + 2e⁻ gives Cu, and oxygen is given off at the anode from hydroxide ions. Cu²⁺ ions are removed, so the blue colour of the solution fades, and the solution becomes acidic because H⁺ and SO₄²⁻ ions remain. With copper electrodes, the cathode reaction is the same, but the copper anode dissolves instead of releasing oxygen: Cu gives Cu²⁺ + 2e⁻. The anode loses mass and the cathode gains the same mass, so copper is effectively transferred from anode to cathode. The concentration of Cu²⁺ ions stays the same, so the blue colour does not change. This is the basis of purifying copper, with an impure copper anode and a pure copper cathode.
Simple cells
A simple cell turns chemical energy into electrical energy. It has two different metals dipping into an electrolyte and connected by a wire. The more reactive metal loses electrons more readily, so it becomes the negative terminal, and electrons flow through the wire from it to the less reactive metal, which is the positive terminal. In a zinc and copper cell, zinc is oxidised, Zn gives Zn²⁺ + 2e⁻. The further apart the two metals are in the reactivity series, the larger the voltage. Notice the difference from electrolysis: in a simple cell a chemical reaction produces electricity, while in electrolysis electricity drives a chemical change.
Worked example
Concentrated aqueous sodium chloride is electrolysed using inert electrodes. Name the ions present, state the product at each electrode with a half-equation and a test, and explain why the solution becomes alkaline.
- Ions present: Na⁺ and Cl⁻ from the salt, and H⁺ and OH⁻ from water.
- Cathode: sodium is more reactive than hydrogen, so H⁺ ions are discharged, 2H⁺ + 2e⁻ gives H₂.
- Test for hydrogen: it extinguishes a lighted splint with a 'pop'.
- Anode: chloride ions are at high concentration, so they are discharged in preference to OH⁻, 2Cl⁻ gives Cl₂ + 2e⁻.
- Test for chlorine: it bleaches damp litmus paper.
- H⁺ and Cl⁻ ions are removed, leaving Na⁺ and OH⁻ ions in solution, which is aqueous sodium hydroxide.
Answer: Hydrogen forms at the cathode (2H⁺ + 2e⁻ gives H₂) and chlorine at the anode (2Cl⁻ gives Cl₂ + 2e⁻). The remaining Na⁺ and OH⁻ ions make the solution sodium hydroxide, so it becomes alkaline.
Common mistakes
- Saying electrons flow through the electrolyte. Electrons flow in the wires, while ions carry the current in the electrolyte.
- Predicting sodium metal at the cathode in aqueous sodium chloride. Sodium is more reactive than hydrogen, so hydrogen forms.
- Saying sulfate ions are discharged at the anode to give sulfur or sulfur dioxide. Sulfate is not discharged, so oxygen forms from OH⁻ ions.
- Predicting oxygen at a copper anode in copper(II) sulfate. A copper anode dissolves to form Cu²⁺ ions instead.
- Writing unbalanced half-equations such as Cl⁻ gives Cl₂ + e⁻. Two chloride ions lose two electrons: 2Cl⁻ gives Cl₂ + 2e⁻.
- Mixing up the electrodes. In electrolysis the anode is positive and oxidation happens there, the cathode is negative and reduction happens there.
Exam tips
- For every electrolysis question, first list all the ions present, including H⁺ and OH⁻ from water for aqueous solutions. Then decide at each electrode in turn.
- Check whether the electrodes are inert or copper before predicting the anode product. That single word changes the answer.
- When asked about the electrolyte afterwards, look at which ions are left behind. That tells you whether it becomes acidic, alkaline or stays the same.
Practice questions
1. Easy: Name the products at each electrode when molten zinc chloride is electrolysed with inert electrodes, and write the half-equations.
Show answer
Zinc at the cathode, Zn²⁺ + 2e⁻ gives Zn. Chlorine at the anode, 2Cl⁻ gives Cl₂ + 2e⁻. In a molten binary compound the metal forms at the cathode and the non-metal at the anode.
2. Medium: Dilute sulfuric acid is electrolysed with inert electrodes. Name the gas at each electrode, give the ratio of their volumes and explain why the acid becomes more concentrated.
Show answer
Hydrogen at the cathode and oxygen at the anode, in a 2 : 1 volume ratio. 4H⁺ + 4e⁻ gives 2H₂ while 4OH⁻ gives O₂ + 2H₂O + 4e⁻, so the same 4 electrons give 2 molecules of hydrogen for each molecule of oxygen. Overall, water is decomposed, so the same amount of acid is left in less water.
3. Harder: Aqueous copper(II) sulfate is electrolysed using copper electrodes. Describe what happens to the mass of each electrode and to the colour of the solution, with half-equations.
Show answer
The cathode gains mass as copper is deposited, Cu²⁺ + 2e⁻ gives Cu. The anode loses the same mass as it dissolves, Cu gives Cu²⁺ + 2e⁻. Copper ions enter the solution at the same rate as they are removed, so the concentration of Cu²⁺ is unchanged and the blue colour stays the same.
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