Why Aluminium Appears Unreactive: Oxide Layers and Reactivity

An aluminium strip is placed in blue copper(II) sulfate solution. You wait a minute. Nothing much happens. Yet aluminium sits above copper in the metal activity series, so aluminium should be able to displace copper from the solution.

What is going wrong?

The answer is on the surface. Aluminium rapidly develops a thin, protective layer of aluminium oxide, \(\ce{Al2O3}\), when exposed to oxygen. This layer prevents the solution from making effective contact with the metal underneath.

The aluminium has not become less reactive. Its surface has become protected.

This is called passivation. It explains why a metal can be thermodynamically capable of reacting but show little or no visible reaction during a practical.

Before going further, make a prediction: if you disturb the oxide layer while the aluminium is immersed in copper(II) sulfate, will the reaction change direction, or will the same reaction simply become easier to observe?

It is the second. Aluminium will still be oxidised and copper(II) ions will still be reduced. Changing the surface condition changes how readily the reaction proceeds, not the underlying activity order.

01The experiment: why does aluminium appear inactive?

Start with a straightforward displacement experiment.

A student places an aluminium strip into a test tube containing copper(II) sulfate solution, \(\ce{CuSO4(aq)}\).

The solution is blue because it contains hydrated copper(II) ions, \(\mathrm{Cu}^{2+}\).

The student expects a reddish-brown copper deposit to form on the aluminium. Instead, the strip appears almost unchanged.

There are two possible explanations:

• The reaction is not thermodynamically favourable.
• The reaction is favourable, but something is preventing it from proceeding at an observable rate.

The original experiment cannot distinguish between these explanations.

Improving the investigation

A better experiment compares different surface conditions while keeping the underlying metal and reaction conditions the same.

Consider this revised investigation.

1. Place 20.0 mL of 0.100 mol L\(^{-1}\) copper(II) sulfate solution into a clear test tube.
2. Immerse an aluminium strip with its normal surface film intact.
3. Observe the strip for 60 seconds.
4. Keeping the strip submerged, use an appropriate glass instrument to disturb a small patch of its surface.
5. Continue observing the same strip and solution.

The important feature is that the metal, solution, temperature, and observation conditions remain unchanged. Only a small region of the aluminium surface has been disturbed.

What should the revised experiment show?

If the oxide layer is responsible for the lack of visible reaction, the disturbed region should become a site where copper deposition can begin.

Illustrative observations might be:

The important evidence is the localised reaction at the disturbed surface.

The original experiment suggested that aluminium did not react. The revised experiment shows that the surface condition was restricting a reaction that aluminium could undergo.

There is one experimental limitation worth recognising. Scratching also increases the local exposed surface area. That can increase the reaction rate independently of oxide removal. Repeated trials using controlled surface preparation would strengthen the conclusion.

A related school demonstration uses sodium chloride to disrupt the protective oxide film. Chloride ions can promote local breakdown of the film, allowing the reaction to become visible. The Royal Society of Chemistry describes this approach in its aluminium and copper(II) sulfate practical investigation.

02The aluminium oxide layer: small but important

Aluminium reacts readily with oxygen in the surrounding air.

The simplified equation for oxide formation is:

\[ \ce{4Al(s) + 3O2(g) -> 2Al2O3(s)} \]

The resulting aluminium oxide forms a compact, strongly attached surface layer.

At HSC level, it is useful to model this protective film as \(\ce{Al2O3}\). Real native surface films can contain more complex oxide and hydroxide structures.

Three properties explain why this film is so effective:

• It adheres strongly to the aluminium underneath.
• It is relatively chemically stable under many ordinary environmental conditions.
• It restricts the movement of reacting species and electron transfer at the metal-solution interface.

This process is called passivation.

Think of the oxide as a protective clear coat. The material underneath retains its original properties, but the coating prevents the surrounding environment from interacting with it easily.

The analogy has a limit: aluminium oxide is not paint applied to the surface. It forms through a chemical reaction involving the aluminium itself and bonds closely to the underlying metal.

How thin is the protective layer?

The surprising part is how little material is needed.

Natural aluminium oxide films are generally only a few nanometres thick. The Australian Aluminium Council’s Rolled Products – Application Data describes natural films approximately 5-10 nm thick.

Let’s estimate the relative thickness.

Suppose:

• The oxide film is 5 nm thick.
• The aluminium foil is 20 μm thick.

Convert the foil thickness into nanometres:

\[ 20\ \mu\mathrm{m}=20\,000\ \mathrm{nm} \]

The ratio of oxide thickness to aluminium thickness is:

\[ \begin{aligned} \text{Ratio} &=\frac{5}{20\,000}\\ &=0.00025 \end{aligned} \]

As a percentage:

\[ 0.00025\times100=0.025\% \]

The oxide layer is only about \(1/4000\) of the foil’s thickness.

This is an illustrative scale estimate, not a measurement of a particular piece of foil.

The important point is that a surface reaction depends on what happens at the interface. A film can represent a negligible fraction of the metal’s total thickness while controlling how the entire exposed surface behaves.

03What reaction is actually being prevented?

Now consider the reaction that should occur between aluminium and copper(II) ions.

Aluminium has a greater tendency to lose electrons than copper. Therefore, aluminium atoms can transfer electrons to copper(II) ions.

The balanced net ionic equation is:

\[ \ce{2Al(s) + 3Cu^2+(aq) -> 2Al^3+(aq) + 3Cu(s)} \]

There are two simultaneous processes.

Oxidation: aluminium loses electrons

\[ \ce{Al(s) -> Al^3+(aq) + 3e^-} \]

Aluminium atoms become aluminium ions and enter the solution.

Reduction: copper(II) ions gain electrons

\[ \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \]

Copper(II) ions gain electrons and form solid copper.

The electrons released by aluminium are accepted by copper(II) ions.

To balance the electron transfer, multiply the aluminium half-equation by two and the copper half-equation by three. Both then involve six electrons.

Why the oxide film stops the visible reaction

With an intact oxide film, copper(II) ions cannot easily interact with the underlying aluminium metal.

The film restricts the interfacial electron transfer needed for the displacement reaction.

Once the protective layer is disrupted, copper(II) ions can interact with exposed aluminium, and the redox reaction can proceed.

Notice what has not changed:

• Aluminium is still the metal being oxidised.
• Copper(II) is still the species being reduced.
• The balanced chemical equation is unchanged.
• The relative positions of aluminium and copper in the activity series are unchanged.

Only the condition of the reacting surface has changed.

04Thermodynamic prediction versus reaction rate

This experiment demonstrates an important distinction.

A reaction can be thermodynamically favourable without proceeding quickly enough to produce an obvious observation.

Thermodynamics and kinetics describe different aspects of a reaction.

For aluminium and copper(II) ions, the thermodynamic prediction favours the displacement reaction.

The oxide layer creates a kinetic limitation by restricting the reacting interface.

05A useful contrast: copper in dilute hydrochloric acid

Not every experiment with no visible reaction is explained by passivation.

Consider two separate test tubes containing dilute hydrochloric acid.

Place an aluminium strip in the first and a copper strip in the second.

The aluminium may initially show little activity before bubbles of hydrogen appear. The copper generally shows no hydrogen production under these conditions.

The two observations have different explanations.

What happens to aluminium?

The acid can first dissolve the protective oxide layer.

The simplified equation is:

\[ \ce{Al2O3(s) + 6H+(aq) -> 2Al^3+(aq) + 3H2O(l)} \]

Notice that this reaction produces water, not hydrogen gas.

Once the protective film has been sufficiently removed, the underlying aluminium can react with hydrogen ions.

\[ \ce{2Al(s) + 6H+(aq) -> 2Al^3+(aq) + 3H2(g)} \]

Hydrogen bubbles then become visible.

The initial delay is consistent with the acid removing the protective surface layer before substantial hydrogen production begins.

Why doesn’t copper produce hydrogen?

For copper to produce hydrogen from dilute hydrochloric acid, the proposed reaction would be:

\[ \ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)} \]

However, this reaction is not thermodynamically favourable under standard conditions.

Using the standard reduction potentials:

• \(\ce{Cu^2+/Cu}\): \(+0.34\ \mathrm{V}\)
• \(\ce{H+/H2}\): \(0.00\ \mathrm{V}\)

The standard cell potential for the proposed reaction is:

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=0.00-(+0.34)\\ &=-0.34\ \mathrm{V} \end{aligned} \]

The negative result indicates that the forward reaction is not thermodynamically favourable under standard conditions.

Therefore, copper’s lack of hydrogen production is not simply another example of aluminium-style passivation.

This comparison is useful:

The acid comparison assumes ordinary dilute, non-oxidising hydrochloric acid. Oxidising acids, such as nitric acid, behave differently.

06How to diagnose an unexpected aluminium result

When a practical observation seems to contradict the activity series, don’t immediately change the predicted reactivity order.

Work through the evidence.

A useful exam response separates three things:

1. The experimental observation: what actually happened?
2. The chemical prediction: what should happen according to the activity series or standard reduction potentials?
3. The explanation: what surface or kinetic limitation could account for the difference?

For a broader explanation of how displacement observations establish relative metal positions, see How to Build a Metal Activity Series from Displacement Results.

For further practice distinguishing a metal’s reactivity from its reaction rate, see How to Compare Metal Reactivity Without Confusing Reaction Rate.

07Questions and solutions

Question 1

An untreated aluminium strip is placed into a solution containing copper(II) ions.

After 60 seconds, there is no obvious visible reaction.

A student concludes that copper is more reactive than aluminium because aluminium has failed to displace copper.

(a) Explain whether this conclusion is justified.

(b) Write the balanced net ionic equation for the expected displacement reaction.

Solution 1

The student’s conclusion is not justified. Aluminium is more reactive than copper, but its protective oxide film can prevent the expected displacement reaction from becoming visible.

(a) Aluminium rapidly forms a protective aluminium oxide layer when exposed to oxygen.

This layer restricts contact between the aluminium metal and copper(II) ions in the solution.

The absence of visible change therefore does not establish that the reaction is thermodynamically unfavourable.

(b) The expected balanced net ionic equation is:

\[ \ce{2Al(s) + 3Cu^2+(aq) -> 2Al^3+(aq) + 3Cu(s)} \]

Aluminium loses electrons and is oxidised, while copper(II) ions gain electrons and are reduced.

The formation of solid copper is the expected evidence of displacement.

The key distinction is that the activity series predicts the underlying chemical tendency. An untreated surface may prevent that tendency from producing an observable reaction.

Question 2

A student investigates how surface condition affects the reaction between aluminium and copper(II) sulfate.

Three identical test tubes each contain 20.0 mL of 0.100 mol L\(^{-1}\) copper(II) sulfate solution at 25°C.

The student performs the following tests:

• Test A: An aluminium strip with its normal surface oxide is immersed in the solution.
• Test B: An aluminium strip is immersed, and a small patch is scratched while submerged.
• Test C: An untreated aluminium strip is immersed in a fresh solution, followed by the addition of a small amount of sodium chloride.

The following illustrative observations are recorded after 120 seconds.

(a) Explain the observations in Tests A and B.

(b) Explain how chloride ions contribute to the observation in Test C.

(c) A student claims that chloride ions are the reducing agent responsible for producing copper metal. Evaluate this claim.

(d) Identify one limitation of this investigation when comparing the effects of scratching and adding sodium chloride.

Solution 2

The observations support the conclusion that aluminium oxide passivates the metal surface. Disturbing the film allows the displacement reaction to proceed, while chloride ions can promote local breakdown of the protective layer.

(a) In Test A, the aluminium’s native oxide layer restricts the interaction between the underlying metal and copper(II) ions.

Consequently, little visible reaction occurs during the observation period.

In Test B, scratching disrupts the protective layer at a small region of the surface.

Aluminium can now be oxidised:

\[ \ce{Al(s) -> Al^3+(aq) + 3e^-} \]

Copper(II) ions accept these electrons:

\[ \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \]

The resulting copper metal produces the reddish-brown deposit.

(b) Chloride ions can destabilise the protective oxide layer and promote localised breakdown of passivity.

This exposes aluminium to the surrounding solution and allows the displacement reaction to occur at multiple locations.

The pits are consistent with localised attack of the aluminium surface.

(c) The claim is incorrect.

Chloride ions are not the reducing agent in the displacement reaction.

Aluminium supplies the electrons, so aluminium is the reducing agent.

Copper(II) ions accept those electrons and are reduced to copper metal.

Chloride ions influence the condition of the protective surface film. They do not supply the electrons in the net redox reaction.

(d) Adding sodium chloride changes the composition of the solution, whereas scratching mainly changes the local surface condition.

Therefore, the two treatments do not isolate exactly the same variable.

Scratching also changes the local exposed surface area.

A stronger investigation would use repeated trials, matched aluminium samples, controlled exposed areas, and consistent solution conditions.

The important conclusion is that surface disruption allows a reaction that was previously restricted. It does not make aluminium a different metal or change its position in the activity series.

Question 3

The following standard reduction potentials are provided:

(a) Calculate the standard cell potential for aluminium displacing copper from copper(II) ions.

(b) Calculate the standard cell potential for copper reacting with hydrogen ions to produce hydrogen gas.

(c) Explain why the two reactions have different experimental predictions, even though both could involve a metal appearing unchanged.

Solution 3

The aluminium-copper reaction has a standard cell potential of \(+2.00\ \mathrm{V}\), whereas the proposed copper-hydrogen reaction has a standard cell potential of \(-0.34\ \mathrm{V}\).

The first reaction is thermodynamically favourable under standard conditions, while the second is not.

(a) Aluminium is oxidised, and copper(II) ions are reduced.

Use:

\[ E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}} \]

Substitute:

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=0.34-(-1.66)\\ &=+2.00\ \mathrm{V} \end{aligned} \]

The positive result predicts a thermodynamically favourable forward reaction under standard conditions.

(b) For the proposed copper-acid reaction, copper is oxidised and hydrogen ions are reduced.

Therefore:

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=0.00-(+0.34)\\ &=-0.34\ \mathrm{V} \end{aligned} \]

The negative result predicts that the proposed forward reaction is not thermodynamically favourable under standard conditions.

(c) The two observations require different explanations.

For aluminium in copper(II) sulfate, the reaction is thermodynamically favourable, but the oxide film restricts the reaction rate.

For copper in dilute hydrochloric acid, the proposed displacement of hydrogen is thermodynamically unfavourable under standard conditions.

The distinction matters because a protective surface layer cannot explain every absence of visible reaction.

A negative experimental observation must be interpreted using both the chemical prediction and the condition of the reacting surface.

Question 4

A student places an aluminium strip and a copper strip into separate test tubes containing dilute hydrochloric acid.

Both experiments are performed at 25°C using the same acid concentration.

The following illustrative observations are recorded:

• The aluminium initially shows little visible activity. After approximately 20 seconds, bubbles begin forming.
• The copper shows no visible hydrogen production after 10 minutes.

The student concludes that both metals must be protected by oxide layers.

Evaluate this explanation.

Your response should:

(a) Explain the initial delay in the aluminium experiment.

(b) Write the relevant equations for the reactions involving the aluminium oxide and the underlying aluminium.

(c) Explain why copper does not produce hydrogen under these conditions.

Assume the acid is dilute, non-oxidising hydrochloric acid.

Solution 4

The aluminium delay is consistent with the acid dissolving the protective oxide film before the metal reacts. Copper’s lack of hydrogen production has a different explanation: the proposed reaction is thermodynamically unfavourable under the specified conditions.

(a) Aluminium has a protective surface layer of aluminium oxide.

When the strip is placed in hydrochloric acid, hydrogen ions react with this oxide layer first.

This delays the reaction between the acid and the underlying aluminium metal.

Once the protective film has been sufficiently removed, aluminium can react with hydrogen ions and release hydrogen gas.

(b) The simplified equation for oxide dissolution is:

\[ \ce{Al2O3(s) + 6H+(aq) -> 2Al^3+(aq) + 3H2O(l)} \]

The equation for the subsequent reaction of aluminium metal is:

\[ \ce{2Al(s) + 6H+(aq) -> 2Al^3+(aq) + 3H2(g)} \]

The second reaction produces hydrogen gas, which appears as bubbles.

(c) Copper does not normally displace hydrogen from dilute, non-oxidising hydrochloric acid.

The proposed reaction is:

\[ \ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)} \]

Its standard cell potential is negative:

\[ E^\circ_{\mathrm{cell}}=-0.34\ \mathrm{V} \]

Therefore, the reaction is not thermodynamically favourable under standard conditions.

The absence of bubbles is not evidence that copper has an aluminium-like passivating layer.

The tempting mistake is to use one explanation for both observations. Instead, distinguish the kinetic delay caused by aluminium oxide from the thermodynamic limitation of the proposed copper-acid reaction.

Question 5

An aluminium strip is first placed into excess copper(II) sulfate solution.

A copper coating forms on the aluminium. The strip is removed, rinsed, and transferred into a fresh solution containing silver ions, \(\mathrm{Ag}^{+}\).

For the second experiment, assume:

• The copper coating is continuous and electrically connected to the aluminium underneath.
• A small region of active aluminium remains exposed.
• The solution contains excess silver ions.
• No other chemical reactions occur.
• All deposited silver is recovered.

The following illustrative results are obtained during the second experiment:

The silver deposit forms mainly on the copper-coated surface.

A separate experiment using a copper-only strip in fresh silver nitrate produces silver metal and copper(II) ions.

Use:

• \(M(\ce{Al})=26.98\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{Ag})=107.87\ \mathrm{g\,mol^{-1}}\)

(a) Calculate the amount of aluminium oxidised and the amount of electrons released.

(b) Calculate the amount of silver deposited and the amount of electrons accepted.

(c) Determine whether the aluminium oxidation can account for the formation of the observed silver deposit.

(d) Explain why silver forms mainly on the copper coating, even though aluminium can supply the electrons.

(e) Determine the relative activity order of aluminium, copper, and silver. Explain why the absence of detectable copper(II) ions in the final solution does not prove that copper was never oxidised during the experiment.

Solution 5

The supported activity order is:

\[ \ce{Al > Cu > Ag} \]

The aluminium oxidation supplies enough electrons to account for the silver deposit. Silver forms on the copper coating because the coating provides a conductive surface for reduction; this does not mean copper atoms must supply the electrons.

(a) Calculate the amount of aluminium oxidised.

Use:

\[ n=\frac{m}{M} \]

Substitute:

\[ \begin{aligned} n(\ce{Al}) &=\frac{0.00900}{26.98}\\ &=3.34\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Each aluminium atom loses three electrons:

\[ \ce{Al -> Al^3+ + 3e^-} \]

Therefore:

\[ \begin{aligned} n(e^-) &=3(3.34\times10^{-4})\\ &=1.00\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The aluminium oxidation releases approximately \(1.00\times10^{-3}\) mol of electrons.

(b) Calculate the amount of silver deposited.

Use:

\[ n=\frac{m}{M} \]

Substitute:

\[ \begin{aligned} n(\ce{Ag}) &=\frac{0.108}{107.87}\\ &=1.00\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Silver ions are reduced according to:

\[ \ce{Ag+ + e^- -> Ag(s)} \]

Each silver ion accepts one electron.

Therefore:

\[ n(e^-)=1.00\times10^{-3}\ \mathrm{mol} \]

(c) Compare the electron amounts.

The aluminium releases:

\[ 1.00\times10^{-3}\ \mathrm{mol\ e^-} \]

The silver ions require:

\[ 1.00\times10^{-3}\ \mathrm{mol\ e^-} \]

These quantities agree to three significant figures.

Therefore, the measured aluminium oxidation is sufficient to account for the formation of all the silver.

The net ionic reaction can be written as:

\[ \ce{Al(s) + 3Ag+(aq) -> Al^3+(aq) + 3Ag(s)} \]

No net consumption of copper is required.

(d) Explain where the reaction occurs.

This is the most important conceptual part of the problem.

The aluminium and copper coating are electrically connected. Electrons released by oxidation of aluminium can travel through the metal and reach the copper-coated surface.

At that surface, silver ions gain electrons and form solid silver.

The two regions perform different roles:

• The exposed aluminium is the site of oxidation, acting as the anode.
• The copper-coated surface is the site of reduction, acting as the cathode.

The copper coating provides a conductive surface on which reduction can occur.

The location where silver forms does not necessarily identify which metal supplies the electrons.

This is similar to the principle behind galvanic corrosion, where two electrically connected metals can participate in different parts of an electrochemical process.

(e) Determine the activity order and interpret the final solution.

The first experiment established that aluminium can displace copper(II) ions:

\[ \ce{Al > Cu} \]

The separate copper-silver experiment established that copper can displace silver ions:

\[ \ce{Cu > Ag} \]

Combining the two comparisons gives:

\[ \ce{Al > Cu > Ag} \]

The absence of detectable copper(II) ions in the final solution does not prove that copper was never oxidised.

There are two possible explanations:

1. The copper coating remained largely unoxidised while aluminium supplied electrons through the connected metal structure.
2. Some copper was temporarily oxidised to \(\mathrm{Cu}^{2+}\), but those ions were subsequently reduced back to copper.

For example:

\[ \ce{Cu(s) -> Cu^2+(aq) + 2e^-} \]

followed by:

\[ \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \]

The copper cancels when these processes are combined.

This illustrates an important limitation of analysing only the final solution: a species can form during a reaction and then be consumed before the experiment ends.

The final solution tells you which species remain detectable, not necessarily every intermediate species that existed during the reaction.

The mass calculations establish that aluminium oxidation is sufficient to account for the net silver production. They do not, by themselves, identify every microscopic electron-transfer step.

This is the conceptual link to galvanic cells and galvanic corrosion: the site where oxidation occurs, the site where reduction occurs, and the metal that appears in the final deposit do not have to be the same location or substance.

Aluminium can therefore be highly reactive according to the activity series while remaining remarkably resistant to visible reaction in everyday conditions. Its protective oxide film changes what you observe, not the underlying chemistry of the metal.