Redox in displacement reactions

This page covers one Form 5 content standard from Redox Equilibrium: redox in displacement reactions. You met simple displacement in Form 4; here you must explain it as electron transfer, writing half-equations, tracking oxidation numbers, and even setting up a displacement so that the two halves happen in separate beakers linked by a wire.

Metal displacement as redox

A more reactive metal displaces a less reactive metal from a solution of its salt. For example, adding zinc to blue copper(II) sulfate solution deposits brown copper and fades the blue colour:

Zn + CuSO₄ → ZnSO₄ + Cu

Stripping out the spectator sulfate ion gives the ionic equation Zn + Cu²⁺ → Zn²⁺ + Cu, and this is clearly redox:

• Oxidation: Zn → Zn²⁺ + 2e-, zinc loses electrons; its oxidation number rises from 0 to +2.
• Reduction: Cu²⁺ + 2e- → Cu, copper(II) gains electrons; its oxidation number falls from +2 to 0.

Zinc is the reducing agent (it is oxidised); the copper(II) ion is the oxidising agent (it is reduced). The rule to remember is that the more reactive metal is oxidised, it is the one that gives its electrons away.

Halogen displacement as redox

A more reactive halogen displaces a less reactive halogen from a solution of its halide. Reactivity of the halogens decreases down Group 17, so chlorine displaces bromine and iodine, and bromine displaces iodine. For example:

Cl₂ + 2KBr → 2KCl + Br₂

Ionically, Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂:

• Reduction: Cl₂ + 2e- → 2Cl⁻, chlorine gains electrons; oxidation number falls from 0 to -1.
• Oxidation: 2Br⁻ → Br₂ + 2e-, bromide loses electrons; oxidation number rises from -1 to 0.

Here the more reactive halogen (chlorine) is reduced, it is the one that takes electrons. Notice the mirror image with metals: the more reactive metal is oxidised, but the more reactive halogen is reduced. Both patterns simply say that the more reactive element wins the tug-of-war for electrons.

Redox reactions at a distance

A powerful demonstration separates the oxidation and reduction into two beakers joined by a carbon electrode in each and a connecting wire, with the solutions linked by a filter paper soaked in electrolyte (a salt bridge). Put a reducing agent (for example acidified iron(II) sulfate) around one electrode and an oxidising agent (for example acidified potassium manganate(VII), or chlorine/bromine water) around the other. No reactants touch, yet a reaction still occurs: electrons flow through the wire from the species being oxidised to the species being reduced. The colour changes at each electrode confirm the redox, and the direction of electron flow tells you which side is oxidation. This shows that a redox reaction is fundamentally a transfer of electrons, which do not need the two reactants to be mixed.

Worked example

Question. A clean iron nail is placed in copper(II) sulfate solution. A brown solid coats the nail and the blue solution turns pale green. Write the ionic equation and the two half-equations, and identify the oxidising and reducing agents, using oxidation numbers to justify your answer.

Step 1, write the ionic equation. Iron is more reactive than copper, so iron displaces copper: Fe + Cu²⁺ → Fe²⁺ + Cu. (The pale green colour is the Fe²⁺ ion; the brown solid is copper metal.)

Step 2, write the half-equations.

• Oxidation: Fe → Fe²⁺ + 2e-
• Reduction: Cu²⁺ + 2e- → Cu

Step 3, check with oxidation numbers. Iron goes from 0 to +2 (an increase, so oxidation). Copper goes from +2 to 0 (a decrease, so reduction). The increase (2) equals the decrease (2), so electrons balance.

Step 4, name the agents. Iron is oxidised, so iron is the reducing agent. The copper(II) ion is reduced, so Cu²⁺ is the oxidising agent.

Answer. Fe + Cu²⁺ → Fe²⁺ + Cu; iron is oxidised (0 to +2, reducing agent) and copper(II) is reduced (+2 to 0, oxidising agent).

Practice question

Chlorine water is added to colourless potassium bromide solution and the mixture turns orange-brown. Explain, using half-equations, why this is a redox reaction and name the oxidising agent.

Answer. Chlorine is more reactive than bromine, so it displaces bromine: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂ (the orange-brown colour is bromine). Reduction: Cl₂ + 2e- → 2Cl⁻ (chlorine gains electrons, 0 to -1). Oxidation: 2Br⁻ → Br₂ + 2e- (bromide loses electrons, -1 to 0). Both oxidation and reduction occur, so it is redox. Chlorine is reduced, so chlorine is the oxidising agent (and bromide is the reducing agent).

Exam tip

State the reactivity comparison first (“chlorine is more reactive than bromine, so it displaces it”), because that one sentence earns the reasoning mark and tells you the direction of the reaction. Then give half-equations with balanced electrons and back them with oxidation-number changes. Watch the flip you must not confuse: the more reactive metal is oxidised, but the more reactive halogen is reduced. For the “reaction at a distance” question, be ready to state that electrons travel through the wire and to name which electrode is the site of oxidation. These points recur across SPM Chemistry Paper 2.

Where this fits in the chapter

Displacement is the bridge between plain redox and the electrochemistry to come. The “reaction at a distance” is really a voltaic cell in disguise, so this standard leads directly into cells and the reactivity series.

• Up to the chapter hub: Redox Equilibrium.
• Sideways for practice: the chapter’s worked examples and practice questions.

In our online 1-to-1 SPM Chemistry lessons, English medium, from RM50 per hour, we make sure you can always name the oxidising and reducing agent correctly in a displacement, because the “more reactive metal is oxidised / more reactive halogen is reduced” flip is a favourite trap in the exam.