Metal displacement reactions

A displacement reaction happens when a more reactive metal is put into a solution (or compound) of a less reactive metal's salt. The more reactive metal takes the place of the less reactive one — it "kicks it out" of the compound.

To predict whether a reaction happens, you compare the two metals on the reactivity series:

K > Na > Ca > Mg > Al > Zn > Fe > Cu > Ag > Au

(potassium most reactive at the top, gold least reactive at the bottom).

The rule: the metal you add must be higher in the series than the metal in the salt. If it is, a reaction happens and the less reactive metal is deposited as a solid. If it is lower (or the same), nothing happens — no reaction.

Classic example — zinc added to copper(II) sulfate.

Word equation:

zinc + copper(II) sulfate → zinc sulfate + copper

Balanced equation:

$\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}$

Zinc is above copper in the series, so zinc displaces copper.

What you actually see (the marking observations):

• the blue colour of the copper(II) sulfate solution fades (the blue Cu²⁺ ions are being used up);
• a brown / pink-brown solid (copper) is deposited on the zinc;
• the zinc gradually dissolves into the solution as colourless Zn²⁺ ions form;
• the mixture warms up slightly (displacement is exothermic).

The reverse does NOT work. Copper added to zinc sulfate gives no reaction, because copper is below zinc:

$\text{Cu(s)} + \text{ZnSO}_4\text{(aq)} \rightarrow \text{no reaction}$

More examples to learn the pattern:

Mixture	Reaction?	Why
Mg + ZnSO₄ → MgSO₄ + Zn	Yes	Mg is above Zn
Zn + MgSO₄	No reaction	Zn is below Mg
Fe + CuSO₄ → FeSO₄ + Cu	Yes	Fe is above Cu
Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag	Yes	Cu is above Ag (silvery deposit)
Ag + CuSO₄	No reaction	Ag is below Cu

Edexcel expects you to know both ways of defining oxidation and reduction.

Process	Oxygen definition	Electron definition
Oxidation	GAIN of oxygen	LOSS of electrons
Reduction	LOSS of oxygen	GAIN of electrons

The mnemonic ties it together:

OIL RIG — Oxidation Is Loss (of electrons); Reduction Is Gain (of electrons).

Why two definitions?

• The oxygen definition is the historical one — early chemists watched metals "gain oxygen" to form oxides.
• The electron definition is more powerful because it works even when no oxygen is involved — for example, displacement reactions and metals reacting with chlorine.

Oxygen-definition example. Magnesium burning in oxygen:

$2\text{Mg(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{MgO(s)}$

• Mg gains oxygen → Mg is oxidised.

Electron-definition example. The same reaction written with electrons:

• Mg → Mg²⁺ + 2e⁻ (magnesium loses electrons → oxidised);
• O₂ + 4e⁻ → 2O²⁻ (oxygen gains electrons → reduced).

Both definitions agree: magnesium is oxidised; oxygen is reduced. Oxidation and reduction always happen together — electrons lost by one species are gained by another. That is why we call them redox (REDuction + OXidation) reactions.

Every displacement reaction is a redox reaction, because electrons are transferred from the more reactive metal to the less reactive metal's ions.

Worked through — zinc displacing copper.

Full equation:

$\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}$

The SO₄²⁻ ion appears unchanged on both sides — it is a spectator ion. Removing it gives the ionic equation:

$\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$

Now split it into two half equations that show the electrons moving:

• Oxidation: Zn → Zn²⁺ + 2e⁻ (zinc loses 2 electrons → oxidised);
• Reduction: Cu²⁺ + 2e⁻ → Cu (copper ion gains 2 electrons → reduced).

Two electrons are transferred from each zinc atom to each copper ion.

The rule to memorise for displacement:

• the metal that goes in (more reactive) is always OXIDISED — it loses electrons and forms ions that go into solution;
• the ion already in solution (less reactive) is always REDUCED — it gains electrons and forms solid metal.

Check it on iron displacing copper:

$\text{Fe(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{Cu(s)}$

• Fe → Fe²⁺ + 2e⁻ (iron oxidised);
• Cu²⁺ + 2e⁻ → Cu (copper ion reduced).

It is easy to mix these up, so learn the key idea:

An "agent" does the OPPOSITE of its name to ITSELF.

• The oxidising agent brings about oxidation of something else. It does this by accepting electrons from the other species — so the oxidising agent is itself reduced.
• The reducing agent brings about reduction of something else. It does this by donating electrons to the other species — so the reducing agent is itself oxidised.

Apply it to displacement — Zn + Cu²⁺ → Zn²⁺ + Cu:

Species	What happens to it	Its role
Zn	loses electrons → oxidised	reducing agent (it reduced the Cu²⁺)
Cu²⁺	gains electrons → reduced	oxidising agent (it oxidised the Zn)

So in any displacement reaction:

• the more reactive metal is the reducing agent (it gives away electrons);
• the less reactive metal ion is the oxidising agent (it takes electrons).

A quick way to identify them in an exam:

1. Find the species that is oxidised (loses e⁻ / gains O). That species is the reducing agent.
2. Find the species that is reduced (gains e⁻ / loses O). That species is the oxidising agent.

Common oxidising agents: O₂, Cl₂, Cu²⁺ and other metal ions in displacement. Common reducing agents: reactive metals (Mg, Zn, Fe), carbon, hydrogen.