Displacement Reactions and Redox

If you drop a clean iron nail into blue copper(II) sulfate solution and leave it, two things happen: the blue colour slowly fades, and the nail becomes coated in a layer of pink-brown copper. The iron has "pushed out" — displaced — the copper. This is a displacement reaction, and it follows one simple rule rooted in the reactivity series you met in the last lesson: a more reactive metal displaces a less reactive metal from its compound. Displacement reactions are also a clear example of redox, the oxidation-and-reduction idea first met in C3. This lesson, part of Topic C4 of OCR Gateway Science A, shows how to predict displacement reactions and how to analyse them as electron transfer.

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, and (Higher tier) explain a displacement reaction as redox using half-equations.

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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. The more reactive metal effectively "wants" the non-metal part of the compound more strongly, so it ends up combined with it, leaving 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 is above copper in the reactivity series, so iron takes the sulfate and the copper is set free. You can actually watch it happen: the blue colour of the copper sulfate solution fades (as blue copper ions are removed) and a deposit of orange-brown copper appears on the iron.

The reverse does not happen. If you put a piece of copper into iron(II) sulfate solution, nothing occurs, because copper is less reactive than iron and cannot displace it. This one-way behaviour is the key to predicting displacement.

Exam Tip: The rule is always "more reactive displaces less reactive." Before predicting, find both metals in the reactivity series: if the added metal 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 happen, compare the positions of the two metals:

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

If the added metal 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: 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	✗	✗	✗	—

A useful real example is the thermite reaction, in which very reactive aluminium displaces iron from iron(III) oxide, releasing so much heat that molten iron is produced: $2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow 2\text{Fe} + \text{Al}_2\text{O}_3$2Al+Fe2​O3​→2Fe+Al2​O3​. Aluminium is above iron, so the displacement goes ahead.

The same predicting principle works for the halogens of Group 7 (which you met in C2): a more reactive halogen displaces a less reactive halide from solution — for example, chlorine displaces bromine from sodium bromide, $\text{Cl}_2 + 2\text{NaBr} \rightarrow 2\text{NaCl} + \text{Br}_2$Cl2​+2NaBr→2NaCl+Br2​. The idea is identical: the more reactive element displaces the less reactive one.

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Displacement Is Redox

Displacement reactions are redox reactions. Recall from C3 that oxidation is loss of electrons and reduction is gain of electrons (the mnemonic OIL RIG — Oxidation Is Loss, Reduction Is Gain). In a displacement reaction, electrons are transferred from the more reactive metal to the ions of the less reactive metal.

Take iron displacing copper again, written as an ionic equation (the sulfate ions are spectators and cancel out):

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

Following the electrons:

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

The electrons lost by the iron are exactly the electrons gained by the copper ions, which is why oxidation and reduction always happen together.

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 it to OIL RIG and you can never get the direction wrong.

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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 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 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.

For a metal that forms a $+1$+1 or $+3$+3 ion the numbers change. For example, magnesium (forms $\text{Mg}^{2+}$Mg2+) displacing silver (forms $\text{Ag}^+$Ag+) needs two silver ions to balance the 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 atoms. Put the electrons on the right for the oxidation (loss) and on the left for the reduction (gain), and make the number of electrons match before you combine them.

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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 greater the difference in reactivity, the more energy is released, so the bigger the temperature rise. This gives a neat experimental way to order metals: add equal amounts of each powdered metal to the same volume of copper(II) sulfate solution and measure the temperature change.

For example, adding magnesium (high in the series) to copper sulfate gives a large temperature rise, because magnesium is much more reactive than copper; adding zinc gives a smaller rise; and adding 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 produce reproduces their order in the reactivity series — an experiment often set as a required-practical-style task.