Worked examples: Redox Equilibrium

Work through each example by covering the answer, attempting it, then checking your method. Every example flags the slip that most often costs marks. These are original questions in SPM style, not past-year papers.

Example 1, Assigning an oxidation number

Question. Determine the oxidation number of manganese in the permanganate ion, MnO₄⁻.

Solution. Oxygen is −2, and there are four oxygen atoms, giving 4 × (−2) = −8. The whole ion carries a charge of −1, so the oxidation numbers must sum to −1. Let manganese be x: x + (−8) = −1, so x = +7. Manganese has an oxidation number of +7 in MnO₄⁻.

Common slip: setting the sum to 0 instead of to the ion charge. In a polyatomic ion the total equals the ion’s charge, not zero.

Example 2, Identifying oxidising and reducing agents

Question. In the reaction between magnesium and dilute hydrochloric acid, Mg + 2HCl → MgCl₂ + H₂, identify the oxidising agent and the reducing agent.

Solution. Magnesium goes from 0 to +2, it is oxidised, so it is the reducing agent. Hydrogen goes from +1 (in HCl) to 0 (in H₂), it is reduced, so the hydrogen ion (the acid) is the oxidising agent.

Common slip: naming the agent by what happens to it. The species that is oxidised is the reducing agent; the species that is reduced is the oxidising agent.

Example 3, Writing half-equations for displacement

Question. A zinc rod is placed in copper(II) sulfate solution. Write the half-equation for each electrode change and the overall ionic equation.

Solution. Oxidation: Zn → Zn²⁺ + 2e⁻. Reduction: Cu²⁺ + 2e⁻ → Cu. The electrons balance (two each), so the overall ionic equation is Zn + Cu²⁺ → Zn²⁺ + Cu.

Common slip: leaving electrons unbalanced when combining halves. Multiply so that electrons lost equal electrons gained before adding.

Example 4, Electrolysis of a molten compound

Question. Molten lead(II) bromide is electrolysed using carbon electrodes. Name the product at each electrode and write the half-equations.

Solution. Only Pb²⁺ and Br⁻ are present. At the cathode: Pb²⁺ + 2e⁻ → Pb (grey lead forms). At the anode: 2Br⁻ → Br₂ + 2e⁻ (brown bromine vapour forms).

Common slip: writing water half-equations for a molten compound. In a molten electrolyte there is no water, so only the compound’s own ions are discharged.

Example 5, Selective discharge in aqueous solution

Question. Dilute sodium chloride solution is electrolysed with carbon electrodes. Predict and explain the product at each electrode.

Solution. Ions present: Na⁺, H⁺, Cl⁻, OH⁻. At the cathode, H⁺ is discharged in preference to Na⁺ because hydrogen is lower (less reactive) in the electrochemical series, giving hydrogen gas: 2H⁺ + 2e⁻ → H₂. At the anode, because the chloride is dilute, OH⁻ is discharged in preference to Cl⁻, giving oxygen: 4OH⁻ → 2H₂O + O₂ + 4e⁻.

Common slip: forgetting that concentration matters. With concentrated chloride, chlorine would be discharged at the anode instead of oxygen.

Example 6, Effect of the electrode

Question. Copper(II) sulfate solution is electrolysed once with carbon electrodes and once with copper electrodes. State the anode product in each case.

Solution. With carbon (inert) electrodes, OH⁻ is discharged and oxygen gas is released at the anode. With copper (active) electrodes, the copper anode itself dissolves, Cu → Cu²⁺ + 2e⁻, so no gas forms and the anode loses mass. This is the principle behind electroplating and copper purification.

Common slip: giving the same anode product for both. The type of electrode changes the anode reaction.

Example 7, Voltaic cell terminals

Question. A simple cell is made from magnesium and copper in dilute sulfuric acid. State the negative terminal, the direction of electron flow, and which metal dissolves.

Solution. Magnesium is more reactive, so it is the negative terminal; it is oxidised and dissolves (Mg → Mg²⁺ + 2e⁻). Electrons flow through the external wire from magnesium to copper. Copper, the less reactive metal, is the positive terminal.

Common slip: naming the less reactive metal as negative. In a voltaic cell the more reactive metal is always the negative terminal.

Example 8, Choosing an extraction method

Question. Explain why aluminium is extracted by electrolysis while iron is extracted using carbon.

Solution. Aluminium is high in the reactivity series and holds its oxide too strongly for carbon to remove, so electrolysis of the molten compound is used. Iron is lower in the series, so carbon (a cheaper reducing agent) is reactive enough to reduce iron oxide to iron. Position in the reactivity series decides the method.

Common slip: saying carbon can extract any metal. Carbon cannot reduce the oxide of a metal more reactive than itself.

Example 9, Halogen displacement

Question. Chlorine water is added to potassium bromide solution and the mixture turns brown. Write the two half-equations and name the substance oxidised.

Solution. Reduction: Cl₂ + 2e⁻ → 2Cl⁻ (chlorine gains electrons). Oxidation: 2Br⁻ → Br₂ + 2e⁻ (bromide loses electrons, forming the brown bromine that colours the mixture). The bromide ion is oxidised, and chlorine is the oxidising agent because it is more reactive than bromine.

Common slip: writing Br → Br⁻. Bromide starts as the ion Br⁻ and is oxidised to Br₂; the electrons come off, they are not added.

Example 10, Redox from oxidation-number change

Question. In the reaction 2FeCl₂ + Cl₂ → 2FeCl₃, show that it is a redox reaction and identify what is oxidised.

Solution. Iron rises from +2 (in FeCl₂) to +3 (in FeCl₃), it is oxidised. Chlorine falls from 0 (in Cl₂) to −1 (in FeCl₃), it is reduced. Because one element rises and another falls, it is a redox reaction; iron(II) is oxidised and chlorine is the oxidising agent.

Common slip: checking only the chlorine and missing the change in iron. Always track every element that changes oxidation number.

Using these examples

Notice that every answer either applies the oxidation-number rules, balances electrons in a half-equation, or justifies a product using the three discharge factors, the habits this chapter rewards. Once the method feels automatic, move on to the practice questions and mark yourself the same way. A one-to-one teacher can check that your half-equations balance and your electrode signs are consistent, which is exactly where marks are gained or lost in the SPM Chemistry written papers.