Metal Reactivity and Displacement Reactions for the ESAT

Updated July 2026

Mastering the reactivity series is essential for predicting the outcome of chemical reactions in the ESAT. This topic covers how displacement reactions occur when a more reactive metal displaces a less reactive metal ion from its compound. You will learn to use experimental observations to establish a hierarchy of metal reactivity.

Core concept

A displacement reaction is a redox process where a more reactive metal, which has a higher tendency to lose electrons, displaces a less reactive metal from its salt solution or oxide. This occurs because the more reactive metal is a stronger reducing agent than the metal it replaces.

The Reactivity Series of Metals

The reactivity of a metal is defined by its tendency to form positive ions by losing electrons. Metals that lose their outer shell electrons most easily are the most reactive and are positioned at the top of the reactivity series. For the ESAT, you must be familiar with the following order, from most to least reactive: Potassium, Sodium, Lithium, Calcium, Magnesium, Aluminium, [Carbon], Zinc, Iron, [Hydrogen], Copper, Silver, and Gold. While carbon and hydrogen are non-metals, they are often included in the series as a reference point for extraction processes and reactions with acids.

Understanding Displacement Reactions

A displacement reaction occurs when a more reactive metal is added to a solution of a less reactive metal salt. The more reactive metal 'displaces' the less reactive metal from the compound, resulting in the more reactive metal forming a salt solution and the less reactive metal appearing as a solid precipitate.

If the metal added is less reactive than the metal already in the salt solution, no reaction will take place. This clear 'yes or no' outcome allows chemists to determine the relative positions of metals in the reactivity series through simple experimentation.

Observations in Aqueous Solutions

When conducting displacement reactions in the laboratory, specific physical changes provide evidence that a reaction has occurred. These typically include:

A change in the color of the solution. This happens if the transition metal ions in the solution are replaced by different ions. For example, blue copper(II) sulfate solutions become colorless as copper ions are removed and replaced by colorless magnesium or zinc ions.
The formation of a solid. The displaced, less reactive metal will precipitate out of the solution. This often appears as a dark grey or brown coating on the surface of the more reactive metal.
A temperature change. Displacement reactions are typically exothermic, meaning they release heat energy to the surroundings.

Worked Example: Magnesium and Copper(II) Sulfate

Consider the reaction between magnesium ribbon and an aqueous solution of copper(II) sulfate. Because magnesium is higher in the reactivity series than copper, it will displace the copper ions.

Full Chemical Equation: $Mg(s) + CuSO_{4}(aq) \rightarrow MgSO_{4}(aq) + Cu(s)$

Observations: The blue color of the copper(II) sulfate solution gradually fades to colorless as magnesium sulfate forms. The magnesium ribbon becomes coated in a brown-orange solid, which is the displaced copper metal.

Ionic Equations and Redox in Displacement

Displacement reactions are redox reactions because they involve the transfer of electrons. We can represent this by writing an ionic equation that omits the spectator ions (ions that do not change during the reaction).

In the magnesium and copper(II) sulfate example, the sulfate ion ( $SO_{4}^{2-}$ ) remains unchanged in the aqueous state. The net ionic equation is: $Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)$

This equation can be broken down into two half-equations:

Oxidation (loss of electrons): $Mg(s) \rightarrow Mg^{2+}(aq) + 2e^{-}$

Reduction (gain of electrons): $Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)$

Magnesium acts as the reducing agent because it provides electrons to the copper ions. The copper(II) ions act as the oxidising agent because they accept electrons from the magnesium.

Establishing the Order of Reactivity

You can deduce the order of reactivity for unknown metals by reacting them with the salt solutions of the other metals.

Experimental Method:

Place a small sample of a metal (e.g., Metal X) into a test tube containing a solution of another metal's salt (e.g., Metal Y Nitrate).
Observe if a reaction occurs (color change or solid formation).
If a reaction occurs, Metal X is more reactive than Metal Y.
If no reaction occurs, Metal Y is more reactive than Metal X.

By repeating this process with multiple metals and solutions, a complete reactivity series can be constructed. For instance, if Metal A displaces Metal B, and Metal B displaces Metal C, the order of reactivity must be A > B > C.

Key takeaways

A more reactive metal will always displace a less reactive metal from its salt solution.
The reactivity of a metal is its tendency to lose electrons and form positive ions.
Displacement reactions are redox processes where the more reactive metal is oxidised and the less reactive metal ion is reduced.
The reactivity series order can be deduced experimentally by observing which metals react with specific salt solutions.
If no visible change occurs when a metal is added to a solution, the added metal is less reactive than the metal in the solution.

Tips

When answering ESAT questions, always look for the state symbols. If a metal is listed as $(s)$ and is lower in the reactivity series than the metal in the $(aq)$ solution, the answer is 'no reaction'. Always check for the presence of an oxide layer on metals like aluminium, which can sometimes hinder the start of a displacement reaction.

Cautions

A common mistake is forgetting to balance the charges in the ionic equation. For example, when Magnesium ( $2+$ ) reacts with Silver ions ( $1+$ ), you need two silver ions for every one magnesium atom to balance the electron transfer: $Mg + 2Ag^{+} \rightarrow Mg^{2+} + 2Ag$ .

Insight

The reactivity series is effectively a simplified version of the Electrochemical Series used in advanced chemistry. The position of a metal in the series is determined by its standard electrode potential ( $E^{\theta}$ ); the more negative the potential, the more reactive the metal and the stronger its ability to act as a reducing agent.

Frequently asked questions

What is a spectator ion in a displacement reaction?

A spectator ion is an ion that exists in the same form on both the reactant and product sides of a chemical equation. It does not participate in the transfer of electrons. In the reaction between $Zn$ and $CuSO_{4}$ , the sulfate ion ( $SO_{4}^{2-}$ ) is the spectator ion.

Can carbon displace aluminium from its oxide?

No, carbon cannot displace aluminium because aluminium is more reactive than carbon. Metals above carbon in the reactivity series must be extracted from their ores using electrolysis rather than reduction with carbon.

Why is gold at the bottom of the reactivity series?

Gold is extremely unreactive because it does not easily lose its outer electrons to form ions. This stable electronic structure is why gold is found in its native (elemental) state in the Earth's crust.

Will a reaction occur if I put a copper wire into a silver nitrate solution?

Yes, a reaction will occur because copper is more reactive than silver. The copper will displace the silver, forming blue copper(II) nitrate solution and solid silver crystals. The ionic equation is $Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$ .

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