Displacement Reactions and Redox

Stand a clean iron nail in a beaker of blue copper(II) sulfate solution and come back an hour later: the blue has faded and the nail is coated in a fuzz of pink-brown copper. The iron has effectively pushed the copper out of its compound — it has displaced it. This is a displacement reaction, and it obeys a single rule that follows straight on from the reactivity series you built last lesson: a more reactive metal displaces a less reactive metal from its compound. Displacement reactions are also a clear window onto redox — oxidation and reduction happening together — which lets us describe them not just in terms of oxygen but in terms of electrons. This lesson is part of Topic C4 of OCR Gateway Combined Science A.

By the end of this lesson you should be able to predict the outcome of a displacement reaction from the reactivity series, write word and balanced symbol equations for displacements, describe redox in terms of oxygen and of electrons using OIL RIG, and (Higher tier) write half-equations for the electron transfer.

This lesson builds AO1 recall of the displacement rule and the meaning of redox, AO2 application when you write balanced symbol equations and (Higher) half-equations for the electron transfer, and AO3 analysis when you predict a displacement outcome and identify what is oxidised and reduced.

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The Displacement Rule

A displacement reaction is one in which a more reactive metal takes the place of a less reactive metal in its compound. You can picture the more reactive metal as "wanting" the non-metal part of the compound more strongly, so it ends up combined with it and leaves the less reactive metal behind as the free element.

The classic example is iron displacing copper from copper(II) sulfate solution:

$\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}$Fe+CuSO4​→FeSO4​+Cu

Iron sits above copper in the reactivity series, so iron takes the sulfate and the copper is set free. You can watch it happen: the blue colour of the copper sulfate solution fades as blue copper ions leave the solution, and a coating of orange-brown copper builds up on the iron.

The reverse does not occur. Put a piece of copper into iron(II) sulfate solution and nothing happens, because copper is less reactive than iron and cannot displace it. This one-way behaviour is exactly what makes displacement so useful for predictions.

Exam Tip: The rule is always "more reactive displaces less reactive." Before predicting, find both metals in the reactivity series: if the metal you add is higher than the metal in the compound, a reaction happens; if it is lower, nothing happens.

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Predicting Displacement from the Reactivity Series

To decide whether a displacement reaction will take place, compare the positions of the two metals:

• the metal added (the pure element);
• the metal in the compound (present in solution as ions).

If the metal added is more reactive (higher in the series), it displaces the other metal and a reaction occurs. If it is less reactive (lower), there is no reaction.

The grid below shows the results of adding four metals (down the side) to four metal-sulfate solutions (across the top). A tick means a displacement happens; a cross means no reaction. Notice the clear diagonal pattern: a metal can only displace those less reactive than itself.

Metal added ↓ / Solution →	Magnesium sulfate	Zinc sulfate	Iron(II) sulfate	Copper(II) sulfate
Magnesium	—	✓ displaces Zn	✓ displaces Fe	✓ displaces Cu
Zinc	✗	—	✓ displaces Fe	✓ displaces Cu
Iron	✗	✗	—	✓ displaces Cu
Copper	✗	✗	✗	—

The very same predicting principle carries over to the halogens of Group 7, which you will meet in detail later in this topic: a more reactive halogen displaces a less reactive halide from solution — for example, chlorine displaces bromine, $\text{Cl}_2 + 2\text{NaBr} \rightarrow 2\text{NaCl} + \text{Br}_2$Cl2​+2NaBr→2NaCl+Br2​. Metal or halogen, the idea is identical: the more reactive element takes the place of the less reactive one.

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Displacement Is Redox

Displacement reactions are redox reactions. You may already have met the oxygen definitions — oxidation is gain of oxygen, reduction is loss of oxygen — but redox is defined more generally in terms of electrons: oxidation is loss of electrons and reduction is gain of electrons. The mnemonic is OIL RIG — Oxidation Is Loss, Reduction Is Gain (of electrons). In a displacement reaction, electrons are handed from the more reactive metal to the ions of the less reactive metal.

Take iron displacing copper again, this time written as an ionic equation. The sulfate ions take no part — they are spectator ions — so they cancel from both sides, leaving:

$\text{Fe} + \text{Cu}^{2+} \rightarrow \text{Fe}^{2+} + \text{Cu}$Fe+Cu2+→Fe2++Cu

Following the electrons:

• The iron atom loses two electrons to become $\text{Fe}^{2+}$Fe2+ — it is oxidised (the more reactive metal, which loses electrons).
• The copper ion gains two electrons to become copper metal — it is reduced (the less reactive metal ion, which gains electrons).

The electrons lost by the iron are precisely the electrons gained by the copper ions, which is why oxidation and reduction always happen together — you cannot have one without the other.

Exam Tip: In any displacement reaction the more reactive metal is oxidised (loses electrons) and the less reactive metal ion is reduced (gains electrons). Tie every case back to OIL RIG and you will never get the direction the wrong way round.

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Higher Tier: Half-Equations

Higher tier only. The two electron changes can each be written as a half-equation, balanced for both atoms and charge. For iron displacing copper:

• Oxidation (iron loses electrons):

$\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$Fe→Fe2++2e−

• Reduction (copper ion gains electrons):

$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$Cu2++2e−→Cu

The two electrons released by the iron are the two picked up by the copper ion, so the electrons cancel when the half-equations are added together, giving back the overall ionic equation $\text{Fe} + \text{Cu}^{2+} \rightarrow \text{Fe}^{2+} + \text{Cu}$Fe+Cu2+→Fe2++Cu.

The numbers change if a metal forms a $+1$+1 or $+3$+3 ion. Magnesium (which forms $\text{Mg}^{2+}$Mg2+) displacing silver (which forms $\text{Ag}^+$Ag+) needs two silver ions to soak up the two electrons: $\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$Mg→Mg2++2e− and $2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag}$2Ag++2e−→2Ag.

Exam Tip: A half-equation must balance for charge as well as for atoms. Put the electrons on the right for the oxidation (loss) and on the left for the reduction (gain), and make the numbers of electrons match before you combine the two halves.

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Using Temperature Rise to Order Metals

Displacement reactions are exothermic — they release energy and warm the mixture — and the size of the temperature rise depends on how far apart the two metals are in the reactivity series. The bigger the gap in reactivity, the more energy is released and the larger the temperature rise. This gives a tidy experimental route to ordering metals: add equal amounts of each powdered metal to the same volume of copper(II) sulfate solution and measure the temperature change.

Adding magnesium (high in the series) to copper sulfate gives a large temperature rise, because magnesium is far more reactive than copper; adding zinc gives a smaller rise; and a metal only just above copper gives the smallest rise of all. A metal below copper produces no temperature change because no reaction occurs. Listing the metals in order of the temperature rise they cause reproduces their order in the reactivity series — a neat, fair-test-style task.

Metal added to copper sulfate	Temperature rise	Reason
Magnesium	Large	Far above copper — big reactivity gap
Zinc	Moderate	Above copper, smaller gap
Iron	Small	Just above copper
Copper (or below)	None	No displacement occurs

Exam Tip: In a displacement experiment, a larger temperature rise means a bigger reactivity gap. Ranking metals by the temperature rise they produce in the same salt solution gives their order in the reactivity series — and no temperature change means no reaction at all.

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Worked Examples

Worked Example 1 — Predicting whether displacement happens