How to Build a Metal Activity Series from Displacement Results

A zinc strip placed in copper(II) sulfate solution develops a reddish-brown coating of copper. Put copper into zinc sulfate solution, and nothing obvious happens. Now try an unpolished aluminium strip in copper(II) sulfate: it might also appear unchanged, even though aluminium is more active than copper.

These observations create a problem. If you simply rank metals according to what you see, you could produce the wrong activity series.

The solution is to treat each displacement reaction as a comparison between two metals, connect the comparisons, and investigate negative results before accepting them. You also need to distinguish a reaction that is thermodynamically favourable from one that happens quickly enough to observe.

01Turn each displacement result into a comparison

Suppose you place a piece of metal M into a solution containing ions of another metal, N.

If M displaces N from solution, the reaction is:

\[
\ce{M(s) + N^2+(aq) -> M^2+(aq) + N(s)}
\]

Two things happen:

• Metal M loses electrons and is oxidised.
• Metal ions N²⁺ gain electrons and are reduced to solid metal N.

The half-equations are:

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

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

This gives you the most important rule for constructing an activity series:

If metal M displaces the ions of metal N, then M is more active than N.

Write this as:

\[
\boxed{\ce{M > N}}
\]

The greater-than sign means more active than. It does not mean that M reacts faster in every experiment.

For example:

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

Zinc displaces copper ions, so:

\[
\ce{Zn > Cu}
\]

This comparison comes from the chemical change, not simply the appearance of the solution.

What counts as evidence of displacement?

Look for evidence that the original metal is being oxidised and the metal ions in solution are being reduced.

Ideally, use more than one observation to confirm a reaction.

A colour change alone is not enough. Neither is a disappearing metal strip. You need to establish which species gained electrons and what product formed.

What if the metals form ions with different charges?

The comparison still works. You simply need to balance the electrons correctly.

For example, aluminium displaces copper from copper(II) ions:

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

Aluminium loses three electrons per atom, while each copper ion gains two.

The balanced equation confirms that aluminium is oxidised and copper ions are reduced.

Therefore:

\[
\ce{Al > Cu}
\]

The coefficients in the balanced equation tell you the reacting mole ratios. They do not determine the relative positions in the activity series.

02Connect the results to build an activity series

You rarely need to test every possible combination of metals. A sequence of reliable displacement reactions can establish a much larger order.

Consider the following illustrative experimental results.

Illustrative data:

Each positive result gives you one relationship.

Connect them:

\[
\ce{Mg > Ni > Cu > Ag}
\]

This is your experimentally supported activity series for the four metals.

Use transitivity to fill the gaps

You can use an important mathematical property called transitivity.

If:

\[
\ce{A > B}
\]

and:

\[
\ce{B > C}
\]

then:

\[
\ce{A > C}
\]

This means that if magnesium displaces nickel ions, and nickel displaces copper ions, magnesium must also be able to displace copper ions under comparable conditions.

You do not need a separate experiment for every possible pair.

However, there is a limit.

Suppose aluminium also displaces copper ions. You now know:

\[
\ce{Al > Cu}
\]

But can you place aluminium above or below magnesium and nickel?

Not from these results alone.

You have established:

\[
\ce{Mg > Ni > Cu}
\]

and:

\[
\ce{Al > Cu}
\]

The relative positions of aluminium and magnesium, and aluminium and nickel, remain undetermined.

This is an important distinction: experimental results may establish a partial order rather than a complete activity series.

Do not force every metal into one continuous sequence when the evidence does not support it.

For the broader activity-series foundation and its practical applications, see The Metal Activity Series: Practical HSC Chemistry Guide.

03Why a metal can appear less active than it really is

Now consider aluminium.

Aluminium is more active than copper, so you would predict that aluminium should displace copper ions:

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

But an unpolished aluminium strip may show little or no visible change when placed in copper(II) sulfate solution.

Does that mean copper is more active than aluminium?

No. The problem may be the aluminium surface.

Aluminium oxide and passivation

When aluminium is exposed to oxygen, a thin layer of aluminium oxide forms on its surface.

\[
\ce{Al2O3}
\]

This layer is strongly attached to the aluminium and acts as a protective barrier.

It prevents the solution from making direct contact with much of the aluminium underneath.

The process is called passivation.

This introduces two different questions.

Thermodynamics asks: Is the overall reaction favourable under the stated conditions?

Kinetics asks: How quickly does the reaction proceed?

For aluminium and copper ions, the overall displacement reaction is thermodynamically favourable under ordinary aqueous conditions.

However, the oxide layer can prevent the reaction from proceeding at an observable rate.

Think of a ball sitting on a downhill slope behind a closed gate. The slope indicates the direction in which the ball would move, but the gate prevents it from getting started.

The downhill direction represents the thermodynamic driving force. The gate represents a kinetic barrier.

The analogy has limits: chemical reactions involve electron transfer and interactions between particles, rather than a physical ball rolling down a slope. However, it captures the difference between a favourable reaction and a reaction that actually proceeds quickly.

What happens when the oxide layer is removed?

If the aluminium surface is freshly abraded, the exposed metal can react with copper ions.

You may observe:

• A reddish-brown copper deposit forming.
• The aluminium surface becoming rougher as aluminium atoms are oxidised.
• The blue solution becoming paler as copper(II) ions are consumed.

The earlier absence of visible change was therefore not reliable evidence that aluminium was less active than copper.

It was evidence that the reaction was restricted under the original surface conditions.

Compare thermodynamic and kinetic explanations

Surface condition is not the only factor affecting reaction rate. Temperature, concentration, surface area, and mixing can also influence how quickly a displacement reaction becomes visible.

This is why you should never rank metals simply by timing how quickly a coating appears.

For the underlying periodic explanation of why metals differ in their tendency to lose electrons, see Periodic Patterns in Metal Activity: HSC Chemistry Explained.

04Check which species is actually being reduced

There is another mistake worth avoiding.

A metal strip can dissolve in a solution without displacing the metal associated with that solution.

Consider copper placed in a solution containing iron(III) ions.

The copper can dissolve because iron(III) ions accept electrons and become iron(II) ions.

The reaction is:

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

Copper is oxidised:

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

Iron(III) ions are reduced:

\[
\ce{2Fe^3+(aq) + 2e^- ->
2Fe^2+(aq)}
\]

Notice what has not happened.

No solid iron has formed.

Therefore, this reaction does not establish that copper displaces iron from its ions. It is a redox reaction, but it is not a metal-displacement reaction producing iron metal.

This distinction matters because iron(III) ions and iron(II) ions are different chemical species with different oxidising abilities.

To compare iron and copper using a conventional metal displacement experiment, you would investigate a suitable comparison involving solid iron and copper ions, or solid copper and iron(II) ions.

Before adding an arrow to your activity series, ask:

1. Which metal was originally present as a solid?
2. Which ions were present in solution?
3. Which species gained electrons?
4. Did those ions actually become solid metal?

If the product is not the displaced metal, you cannot automatically use that experiment to rank the two metals.

05A reliable method for interpreting experimental results

When given a table of displacement observations, work through the following process.

Step 1: Identify the reactants.

Record the solid metal and the metal ions in solution. Pay attention to the ionic charge.

Step 2: Confirm the displacement.

Look for evidence that the metal ions have been reduced to solid metal.

Step 3: Write the reaction.

Balance the net ionic equation, including electron transfer.

Step 4: Establish the comparison.

If the solid metal displaces the ions, place it above the metal that formed.

Step 5: Investigate negative results.

Before concluding that no displacement is thermodynamically favourable, consider:

• Was the metal surface clean?
• Could an oxide coating have blocked the reaction?
• Was the observation period long enough?
• Were the solution concentrations and temperatures suitable?
• Was the expected metal deposit difficult to identify?

Step 6: Connect the comparisons.

Use transitivity to establish relationships between metals that were not directly tested.

Step 7: Check for contradictions.

Suppose your results suggest:

\[
A>B,\qquad B>C,\qquad C>A
\]

This creates a cycle.

A consistent activity series cannot contain these three relationships under the same conditions.

At least one comparison needs to be investigated. Possible explanations include an incorrect identification of the deposited substance, different surface conditions, contamination, or a change in the chemical species present.

Do not simply choose whichever result looks most convincing. Identify the experiment that needs repeating and explain why.

The strongest activity series is not necessarily the longest one. It is the order that your experimental evidence can actually support.

06Questions and solutions

Question 1

Three metals, X, Y, and Z, each form 2+ ions.

The following experiments are performed:

• X displaces Y²⁺ from solution.
• Y displaces Z²⁺ from solution.

Construct the activity series and predict whether X will displace Z²⁺.

Solution 1

The activity order is:

\[
\boxed{X>Y>Z}
\]

Therefore, X is predicted to displace Z²⁺.

X displaces Y²⁺, establishing X > Y. Y displaces Z²⁺, establishing Y > Z.

Using transitivity, X must be more active than Z.

The predicted reaction is:

\[
\ce{X(s) + Z^2+(aq) ->
X^2+(aq) + Z(s)}
\]

This assumes the metals form 2+ ions and the reaction occurs under comparable conditions.

Question 2

Four metals, A, B, C, and D, form 2+ ions.

The following results are obtained under controlled conditions:

Assume the negative result is reliable and there are no competing reactions.

Construct the strongest activity order supported by the data. Can A and D be ranked relative to each other?

Solution 2

The results establish:

\[
\boxed{D>B>C}
\]

and:

\[
\boxed{A>B}
\]

However, the relative positions of A and D cannot be determined.

The reasoning is as follows.

Step 1: Translate the positive displacement results.

A displaces B²⁺:

\[
A>B
\]

B displaces C²⁺:

\[
B>C
\]

D displaces C²⁺:

\[
D>C
\]

Step 2: Interpret the negative result.

B does not displace D²⁺ under the stipulated reliable experimental conditions.

This supports D being more active than B:

\[
D>B
\]

Step 3: Connect the relationships.

We now have:

\[
D>B>C
\]

and:

\[
A>B>C
\]

Both A and D are above B, but no experiment or chain of comparisons establishes which is more active.

The correct conclusion is that the data support a partial order, not a complete activity series.

This is a common place to overinterpret experimental results. Two metals being more active than the same third metal does not establish their relative positions.

Question 3

A freshly abraded aluminium strip reacts with excess copper(II) ions.

Experimental analysis shows that:

• 0.0100 mol of aluminium atoms are oxidised.
• 0.0150 mol of copper(II) ions are reduced.

Assume aluminium forms Al³⁺ ions and copper(II) ions form solid copper.

Calculate the number of moles of electrons transferred, write the balanced net ionic equation, and state what the experiment establishes about the activity series.

Solution 3

The experiment establishes that aluminium is more active than copper, with 0.0300 mol of electrons transferred.

Step 1: Calculate the electrons released by aluminium.

Each aluminium atom loses three electrons:

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

Therefore:

\[
n(e^-)=0.0100\times3
\]

\[
n(e^-)=0.0300\text{ mol}
\]

Step 2: Calculate the electrons accepted by copper ions.

Each copper(II) ion gains two electrons:

\[
\ce{Cu^2+ + 2e^- -> Cu}
\]

Therefore:

\[
n(e^-)=0.0150\times2
\]

\[
n(e^-)=0.0300\text{ mol}
\]

Both half-reactions involve the same amount of electrons, confirming the electron balance.

Step 3: Write the balanced overall reaction.

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

Step 4: Establish the activity relationship.

Aluminium is oxidised while copper ions are reduced.

Therefore:

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

The experiment establishes the relative activity of aluminium and copper. The amount of electron transfer does not tell us how many times more active one metal is than another.

Question 4

Two equal aluminium strips are placed into separate beakers containing identical 0.10 mol L⁻¹ copper(II) sulfate solutions at 25 °C.

• Strip 1 is used without surface preparation. No visible reaction occurs after five minutes.
• Strip 2 is freshly abraded immediately before immersion. A reddish-brown deposit appears after 90 seconds, and the blue solution becomes paler.

A student concludes that strip 2 is more active than strip 1 because it reacts faster.

Evaluate this conclusion. Explain the different observations and suggest a suitable follow-up experiment.

Solution 4

The student’s conclusion is incorrect. Both strips are aluminium, so they have the same underlying chemical activity.

The difference is their surface condition.

Step 1: Explain the first observation.

The untreated aluminium strip has a protective aluminium oxide coating.

This coating restricts contact between the aluminium and copper(II) ions. As a result, the displacement reaction may be too slow to produce visible changes during the observation period.

Step 2: Explain the second observation.

Abrading the second strip exposes fresh aluminium.

The aluminium can now lose electrons to copper(II) ions:

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

Copper ions gain those electrons:

\[
\ce{Cu^2+ + 2e^- -> Cu}
\]

The reddish-brown deposit is evidence of copper formation.

The blue solution becoming paler is consistent with a reduction in the concentration of copper(II) ions.

Step 3: Distinguish activity from reaction rate.

The faster visible reaction does not mean that the second strip is more active.

Removing the oxide layer improves access to the underlying metal. Abrasion may also increase the effective surface area, which can increase reaction rate.

The experiment supports:

\[
\ce{Al > Cu}
\]

It does not establish two different activity levels for the two aluminium strips.

Step 4: Propose a fair follow-up.

Prepare identical aluminium strips using the same polishing procedure, ensuring that they have comparable exposed surface areas and roughness.

Transfer one strip immediately into the solution and allow the other to develop a surface oxide layer for a controlled period before immersion.

Keep the following conditions consistent:

• Solution concentration and volume.
• Temperature.
• Exposed metal surface area.
• Reaction time.
• Mixing conditions.

Repeat the tests and compare the observations.

This would help investigate how the oxide layer affects the reaction rate without incorrectly treating surface condition as a change in the metal’s activity.

Question 5

A student investigates the relative activity of zinc, aluminium, copper, silver, and iron.

All aqueous solutions are prepared at the same temperature and concentration.

The following observations are recorded:

Further analysis of the final experiment shows that 2.00 mmol of iron(III) ions are consumed for every 1.00 mmol of copper dissolved.

A student proposes the following conclusions:

• Zinc is more active than copper.
• Copper is more active than silver.
• Aluminium is less active than copper because it initially showed no visible reaction.
• Copper is more active than iron because it dissolves in an iron-containing solution.

Evaluate all four conclusions.

Construct the strongest activity relationships supported by the evidence. Explain whether a complete activity series containing all five metals can be established.

Solution 5

The evidence supports:

\[
\boxed{\ce{Zn > Cu > Ag}}
\]

and:

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

However, the relative positions of zinc and aluminium are not established, and the experiment involving iron(III) ions does not establish iron’s position in the activity series.

The student’s first two conclusions are correct. The final two are not.

Step 1: Analyse zinc in copper(II) solution.

Zinc dissolves, and copper metal forms.

The reaction is:

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

Zinc is oxidised, and copper ions are reduced.

Therefore:

\[
\ce{Zn > Cu}
\]

The first conclusion is correct.

Step 2: Analyse copper in silver ion solution.

Copper displaces silver ions:

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

Therefore:

\[
\ce{Cu > Ag}
\]

The second conclusion is also correct.

Combining the first two results:

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

Step 3: Reconsider the aluminium experiment.

The untreated aluminium strip showed no visible reaction after 60 seconds.

However, this is not sufficient evidence that copper is more active than aluminium.

The aluminium surface may have been protected by aluminium oxide.

When the surface was freshly abraded, copper deposited on the aluminium.

The confirmed displacement reaction is:

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

Therefore:

\[
\ce{Al > Cu}
\]

The third conclusion is incorrect.

Notice that both zinc and aluminium are established as more active than copper:

\[
\ce{Zn > Cu}
\]

\[
\ce{Al > Cu}
\]

But these two relationships do not establish whether zinc or aluminium is more active.

You cannot simply assume their relative positions.

Step 4: Analyse the iron(III) experiment.

Copper dissolves in the iron(III) solution, but this does not mean that copper displaces iron.

The iron(III) ions gain electrons and become iron(II) ions.

The balanced reaction is:

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

Check the electron transfer.

One mole of copper atoms loses two moles of electrons:

\[
1.00\text{ mmol Cu}
\rightarrow
2.00\text{ mmol }e^-
\]

Two moles of iron(III) ions each gain one mole of electrons:

\[
2.00\text{ mmol Fe}^{3+}
\rightarrow
2.00\text{ mmol Fe}^{2+}
\]

The measured 2:1 mole ratio is consistent with this reaction.

Crucially, no solid iron forms.

The iron(III) ions are reduced to iron(II) ions, not to metallic iron.

Therefore, this experiment cannot establish whether copper is above or below iron in the metal activity series.

The fourth conclusion is incorrect.

Step 5: State the strongest defensible conclusion.

The experimental evidence establishes:

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

and:

\[
\ce{Al > Cu}
\]

By transitivity, both zinc and aluminium are also more active than silver.

However:

• Zinc cannot be ranked relative to aluminium from these results.
• Iron cannot be positioned because no suitable iron-metal displacement comparison was performed.

A complete five-metal activity series is therefore not justified.

A further valid comparison between aluminium and zinc would be needed to establish their relative positions. Iron would also need to be connected to the existing order using an appropriate metal-displacement experiment or independent redox evidence.

The central skill in this question is deciding what the evidence actually establishes, rather than forcing every observation into a familiar activity series.

07What to remember when building an activity series

The method comes down to three decisions:

1. Confirmed displacement: The solid metal that displaces another metal’s ions is more active.
2. No visible reaction: Investigate surface condition and reaction kinetics before treating the result as evidence of relative activity.
3. Incomplete or conflicting evidence: Use transitivity where justified, leave untested relationships unresolved, and investigate any contradictory comparisons.

Most importantly, keep thermodynamic prediction separate from reaction speed. A metal can be highly active but appear unreactive because its surface prevents the reaction from getting started.

Once you can distinguish these situations, you can interpret experimental results much more reliably. The next useful step is connecting the activity series to galvanic cells, where the same tendency for metals to lose electrons helps explain how electrical energy can be produced by a spontaneous redox reaction.