How to Identify an Unknown Metal Using Displacement Tests

An unlabelled metal strip is placed into several different metal-ion solutions. A reddish-brown coating appears in one test tube, a silver coating forms in another, and two solutions show no obvious change.

Which metal is the strip?

The reliable way to identify it is to use the displacement results to place the unknown metal between known metals in the activity series. Positive results tell you which metals the unknown can displace. Reliable negative results tell you which metals it cannot displace. Combining these comparisons narrows down the identity.

There is one complication: a metal can appear unreactive because its surface is protected by an oxide layer, or because the reaction is too slow to observe. You need to account for this before treating every negative result as proof that a metal is less reactive.

01Understand what each displacement test tells you

Imagine placing an unknown metal, X, into a solution containing ions of another metal, Y.

If X is more reactive than Y, X can transfer electrons to the Y ions. The Y ions become solid metal, while X atoms enter the solution as positive ions.

For metals that both form 2+ ions, the general reaction is:

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

Two electron-transfer processes occur:

Oxidation: X loses electrons.

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

Reduction: Y ions gain electrons.

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

The unknown metal is oxidised, while the metal ions in solution are reduced.

The most important rule is:

\[
\boxed{\text{If X displaces Y, then X > Y}}
\]

Here, the greater-than sign means more reactive than. It does not mean that X necessarily reacts faster.

For example:

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

Zinc displaces copper from copper(II) ions, so:

\[
\ce{Zn > Cu}
\]

You can use this relationship even if the two metals form ions with different charges. You simply need to balance the electrons correctly.

What counts as a positive displacement result?

Look for evidence that the metal ions in solution have actually become solid metal.

A colour change alone is not always enough to identify the reaction. Ideally, confirm the identity of the deposited metal and check whether the original strip is dissolving.

Also, be careful about which metal is being tested.

If an unknown metal X is placed into copper(II) sulfate and copper forms, you know that X is more reactive than copper.

If a copper strip is placed into a solution containing X ions and nothing happens, that gives a different comparison. You cannot interpret both arrangements in the same direction without thinking about which metal is the solid and which is initially in solution.

02Use several tests to bracket the unknown metal

You rarely need to test every possible combination of metals.

Instead, use positive and negative results to establish an interval in the activity series.

The method is straightforward:

1. Write down the possible identities of the unknown metal.
2. Record which metal-ion solutions produce a confirmed displacement reaction.
3. Convert every positive result into an activity comparison.
4. Use reliable negative results to establish upper limits on the unknown metal’s activity.
5. Combine the comparisons and eliminate candidates that do not fit.
6. Check whether surface effects or slow reactions could have affected any negative results.

Positive and negative results

Suppose an unknown metal X is tested in solutions containing zinc ions and iron(II) ions.

Combining the results gives:

\[
\boxed{\ce{Fe > X > Zn}}
\]

The unknown metal must be more reactive than zinc but less reactive than iron.

This is called bracketing the unknown metal.

Notice what you have not established: the exact identity of X.

If the candidate list contains three metals between iron and zinc, these tests alone cannot distinguish them. You would need another suitable metal-ion solution to narrow the interval.

03Why a negative displacement test can be misleading

Here is where many students make an incorrect prediction.

Suppose an aluminium strip is placed into copper(II) sulfate solution. You might expect a reddish-brown copper coating because aluminium is more reactive than copper.

Yet an unpolished aluminium strip can show little or no visible change.

Does this mean copper is more reactive than aluminium?

No.

The problem may be the aluminium surface.

Aluminium oxide and passivation

Aluminium reacts with oxygen in the air to form a thin layer of aluminium oxide:

\[
\ce{Al2O3}
\]

This layer is strongly attached to the metal and protects the aluminium underneath.

It acts as a barrier between the aluminium and the copper(II) ions in solution.

The process is called passivation.

If the aluminium surface is freshly abraded and immediately placed into copper(II) sulfate, the reaction can become observable:

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

You may observe:

• A reddish-brown copper coating forming.
• Aluminium gradually dissolving.
• The blue solution becoming paler as copper(II) ions are consumed.

The earlier negative result did not establish that aluminium was less reactive than copper. It showed that no visible reaction occurred with the original surface under the original test conditions.

Thermodynamics versus kinetics

To interpret these results properly, you need to distinguish two ideas.

Thermodynamics tells you whether a reaction is energetically favourable.

Kinetics tells you how quickly the reaction proceeds.

Think of a ball sitting on a downhill slope behind a closed gate.

The slope represents the thermodynamic driving force. The gate represents a kinetic barrier.

The ball has a downhill direction available, but it cannot move freely until the gate is opened.

The analogy has a limit: chemical reactions involve interactions between particles and electron transfer, rather than a physical object rolling downhill. However, it captures the distinction between a favourable reaction and an observable reaction rate.

In a displacement experiment:

Other factors can also affect the observed rate:

• Surface area of the metal.
• Concentration of metal ions.
• Temperature.
• Time allowed for the reaction.
• Formation of a coating that blocks further contact.

These factors can change how quickly you observe a reaction without necessarily changing the underlying activity order.

Using standard reduction potentials to check a prediction

Sometimes a question provides a table of standard reduction potentials.

You can use these values to determine whether a displacement reaction is thermodynamically favourable under standard conditions.

The relevant equation is:

\[
E^\circ_{\text{cell}}
=
E^\circ_{\text{red,cathode}}
–
E^\circ_{\text{red,anode}}
\]

Both values are taken from a table of standard reduction potentials.

For a displacement reaction:

• The metal ions being reduced form the cathode half-reaction.
• The original metal being oxidised forms the anode half-reaction.

The sign of the calculated cell potential tells you the predicted direction:

A positive cell potential does not guarantee that a visible deposit will form immediately. It tells you the thermodynamic direction, not the reaction rate.

04Use mass changes as additional evidence

Sometimes a displacement experiment includes measurements of the metal strip’s mass before and after the reaction.

These measurements can provide another way to identify an unknown metal, but there is a common mistake to avoid.

Suppose an unknown metal M reacts with copper(II) ions:

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

For every mole of copper deposited, one mole of M dissolves.

The strip gains the mass of the deposited copper but loses the mass of the dissolved unknown metal.

Therefore:

\[
\Delta m_{\text{strip}}
=
m_{\text{Cu deposited}}
–
m_{\text{M dissolved}}
\]

For a 1:1 reaction:

\[
\Delta m_{\text{strip}}
=
n(M_{\ce{Cu}}-M_{\ce{M}})
\]

where:

• \(\Delta m_{\text{strip}}\) is the change in strip mass, in grams.
• \(n\) is the number of moles of metal deposited and dissolved, in mol.
• \(M_{\ce{Cu}}\) is the molar mass of copper, in g mol\(^{-1}\).
• \(M_{\ce{M}}\) is the molar mass of the unknown metal, in g mol\(^{-1}\).

A positive mass change does not mean that only copper has been deposited. You must account for the metal that dissolved.

Likewise, the strip can lose mass even though a new metal coating forms.

This is a useful independent check when you are narrowing down possible identities.

05A reliable decision method for exam questions

When you are given several displacement results, use this sequence.

Three common mistakes are worth avoiding.

Mistake 1: Treating no visible change as proof of no reaction.

A protective oxide layer, a slow reaction, or an insufficient observation period can hide a favourable displacement reaction.

Mistake 2: Ranking metals by how quickly they react.

A metal that reacts visibly in 10 seconds is not automatically more active than one that takes several minutes. Reaction rate and relative activity are different properties.

Mistake 3: Forcing a single identity when the evidence only gives a range.

If two candidate metals satisfy every reliable test, you cannot choose between them without further evidence.

For more practice constructing a complete activity series from multiple displacement results, see How to Build a Metal Activity Series from Displacement Results.

Also remember that displacement tests are different from precipitation tests. Displacement involves electron transfer and the formation of a solid metal. Precipitation involves dissolved ions combining to form an insoluble compound. For the second type of identification, see How to Identify Two Unknown Ions Using Precipitation Tests.

06Questions and solutions

The following questions move from straightforward identification towards situations where the experimental conditions make the answer less obvious.

Question 1

An unknown metal R is known to be one of the following:

• Zinc
• Iron
• Copper
• Silver

Three separate displacement tests produce these results:

All metal surfaces are clean, the tests are conducted under comparable conditions, and the deposits have been correctly identified.

Identify R and write a balanced net ionic equation for one of the displacement reactions.

Solution 1

R is iron (Fe).

The copper and silver deposits establish that R is more reactive than both copper and silver.

Step 1: Interpret the positive copper displacement result.

Copper is deposited from copper(II) ions, so:

\[
\ce{R > Cu}
\]

Step 2: Interpret the positive silver displacement result.

Silver is deposited, so:

\[
\ce{R > Ag}
\]

Step 3: Use the negative zinc test.

Since no displacement occurs in zinc sulfate, and the test is assumed to be reliable:

\[
\ce{Zn > R}
\]

Combining the results:

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

Step 4: Identify the candidate.

The only candidate between zinc and copper is iron.

Therefore:

\[
\boxed{R=\ce{Fe}}
\]

One balanced net ionic equation is:

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

Iron loses two electrons, while copper(II) ions gain two electrons.

The important reasoning is that the positive results establish lower limits on R’s activity, while the reliable negative result establishes an upper limit.

Question 2

Two unknown metals, A and B, are tested using separate metal-ion solutions.

The following illustrative results are recorded:

Assume the strips have been freshly cleaned, the solutions have suitable concentrations, and the negative result for A in zinc sulfate is reliable.

(a) Determine the relative activity order of zinc, A, B, copper, and silver.

(b) A student concludes that A is more active than B because A reacts faster with copper(II) ions. Explain why this conclusion is incorrect.

Solution 2

The correct activity order is:

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

B is more active than A, despite reacting more slowly with copper(II) ions.

Step 1: Interpret B’s reaction with zinc sulfate.

B displaces zinc from zinc ions:

\[
\ce{B > Zn}
\]

Step 2: Interpret A’s reaction with zinc sulfate.

A does not displace zinc. Because this is a validated negative result:

\[
\ce{Zn > A}
\]

Combining these:

\[
\ce{B > Zn > A}
\]

Step 3: Interpret the copper and silver tests.

Both metals displace copper and silver:

\[
\ce{A > Cu}
\]

\[
\ce{B > Cu}
\]

Silver is less reactive than copper, so the complete justified order is:

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

Step 4: Explain the difference in reaction times.

The fact that A deposits copper more quickly does not establish that A is more active.

Reaction rate can be affected by surface condition, surface area, concentration, and other kinetic factors.

The zinc displacement tests establish the relative activity order. The different reaction times in copper(II) sulfate describe how quickly the reactions occur, not which metal is more reactive.

This is an important distinction: the fastest reaction does not necessarily involve the most reactive metal.

Question 3

An unknown metal M is known to form 2+ ions and is one of the following:

A clean strip of M is placed into 100.0 mL of copper(II) sulfate solution.

The initial copper(II) ion concentration is 0.400 mol L\(^{-1}\). After the reaction, the concentration has fallen to 0.200 mol L\(^{-1}\).

The strip’s mass has increased by 0.154 g.

Assume:

• The solution volume remains constant.
• The reaction is \(\ce{M + Cu^2+ -> M^2+ + Cu}\).
• All deposited copper remains attached to the strip.
• There are no competing reactions or losses of material.

(a) Calculate the number of moles of copper deposited.

(b) Calculate the mass of the unknown metal that dissolved.

(c) Identify M and explain why the measured mass increase is not equal to the mass of copper deposited.

Solution 3

The unknown metal is iron (Fe), with a molar mass of approximately 55.8 g mol\(^{-1}\).

The key is to remember that copper deposits while the original metal dissolves. The measured mass change is the difference between these two masses.

Step 1: Calculate the initial amount of copper(II) ions.

Use:

\[
n=cV
\]

where \(n\) is amount in mol, \(c\) is concentration in mol L\(^{-1}\), and \(V\) is volume in L.

\[
\begin{aligned}
n_{\text{initial}}
&=0.400\times0.1000\\
&=0.0400\text{ mol}
\end{aligned}
\]

Step 2: Calculate the final amount of copper(II) ions.

\[
\begin{aligned}
n_{\text{final}}
&=0.200\times0.1000\\
&=0.0200\text{ mol}
\end{aligned}
\]

Therefore, the amount of copper(II) ions consumed is:

\[
\begin{aligned}
n_{\mathrm{Cu}^{2+}}
&=0.0400-0.0200\\
&=0.0200\text{ mol}
\end{aligned}
\]

Step 3: Calculate the mass of copper deposited.

The reaction has a 1:1 mole ratio between copper ions and copper atoms.

Using the molar mass of copper, 63.55 g mol\(^{-1}\):

\[
\begin{aligned}
m_{\ce{Cu}}
&=nM\\
&=0.0200\times63.55\\
&=1.271\text{ g}
\end{aligned}
\]

Step 4: Calculate the mass of M that dissolved.

The strip gained 0.154 g overall.

Therefore:

\[
\begin{aligned}
m_{\text{M dissolved}}
&=m_{\ce{Cu}}-\Delta m_{\text{strip}}\\
&=1.271-0.154\\
&=1.117\text{ g}
\end{aligned}
\]

Step 5: Calculate the molar mass of M.

The reaction consumes one mole of M for every mole of copper(II) ions reduced.

Therefore:

\[
n_{\text{M}}=0.0200\text{ mol}
\]

The molar mass is:

\[
\begin{aligned}
M_{\text{M}}
&=\frac{m}{n}\\
&=\frac{1.117}{0.0200}\\
&=55.85\text{ g mol}^{-1}
\end{aligned}
\]

Thus:

\[
\boxed{M_{\text{M}}\approx55.8\text{ g mol}^{-1}}
\]

This matches iron.

Step 6: Explain the mass change.

The strip gained 1.271 g of copper but lost 1.117 g of iron.

The net mass increase was:

\[
1.271-1.117=0.154\text{ g}
\]

It would be incorrect to treat the measured mass increase as the mass of copper deposited.

The dissolved metal must be included in the mass balance.

Question 4

An unknown metal R is tested in copper(II) sulfate solution.

Two identical strips are prepared:

• Strip 1 is left with its original surface.
• Strip 2 is freshly polished immediately before the experiment.

Both strips are placed into separate, identical samples of copper(II) sulfate at the same temperature and concentration.

After 30 minutes:

• Strip 1 shows no visible change.
• Strip 2 develops a reddish-brown coating.

A student argues that the results are contradictory and that the experiment cannot be used to determine the activity of R.

(a) Explain whether the results are actually contradictory.

(b) State what can be concluded about R’s relative activity compared with copper.

(c) Explain how a protective surface layer could produce the different observations.

(d) Suggest one further experimental control that would strengthen the conclusion that surface passivation is responsible.

Solution 4

The results are not contradictory. They indicate that the condition of the metal surface affects the observed reaction.

The positive displacement result establishes that R is more reactive than copper.

(a) Explain the apparent contradiction.

The two strips have different surface conditions.

The original strip may have a protective oxide layer that prevents copper(II) ions from making effective contact with the underlying metal.

Polishing can remove or disrupt this barrier, allowing electron transfer to occur.

Therefore, the two results can be explained by different reaction rates at the two surfaces.

(b) Determine the activity relationship.

The polished strip produces solid copper.

Therefore, R displaces copper from copper(II) ions:

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

The negative result from the original strip cannot be used to reverse this conclusion because the surface may have prevented an observable reaction.

(c) Explain the role of passivation.

A protective oxide layer can separate the solution from the metal underneath.

Although the overall reaction may be thermodynamically favourable, the barrier reduces the rate at which reactants can interact.

Removing or disrupting the layer exposes fresh metal and allows the reaction to proceed.

(d) Suggest a further control.

Repeat the experiment using several pairs of identical strips.

Keep their dimensions, exposed surface areas, solution concentrations, volumes, temperatures, and reaction times constant.

Prepare one strip from each pair using a controlled surface-cleaning procedure, while leaving the other untreated.

Record the amount of copper deposited on each strip, rather than relying only on visual observations.

A known reactive metal in copper(II) sulfate can also be used as a positive control to confirm that the solution and experimental conditions permit displacement.

If the cleaned strips consistently react while the untreated strips do not, this supports the explanation that a surface barrier is limiting the reaction.

The experiment supports a surface-related effect, although the observation alone does not establish the exact chemical composition of the protective layer.

Question 5

An unknown metal P is known to be either cobalt, nickel, copper, or silver.

Four separate tests are performed using freshly cleaned P strips.

A student concludes that P must be nickel because it does not displace nickel ions.

However, the standard reduction potentials provided are:

\[
E^\circ_{\ce{Co^2+/Co}}=-0.28\text{ V}
\]

\[
E^\circ_{\ce{Ni^2+/Ni}}=-0.25\text{ V}
\]

Assume the tests are conducted at approximately 25°C. The nickel-ion test relies on visual detection after 60 seconds, and no evidence has been supplied that this observation period is sufficient to detect every thermodynamically favourable reaction.

(a) Use the first three results to establish the strongest justified activity range for P.

(b) Calculate the standard cell potential for cobalt displacing nickel from nickel(II) ions.

(c) Explain why the absence of a visible nickel deposit does not necessarily prove that P is nickel.

(d) State whether the unknown can be uniquely identified from the available evidence. Suggest a further measurement that could distinguish cobalt from nickel.

Solution 5

The unknown cannot be uniquely identified from the available evidence. Both cobalt and nickel remain possible.

The key is that a thermodynamically favourable reaction does not necessarily produce a visible deposit within 60 seconds.

(a) Establish the activity range.

First, P does not displace iron from iron(II) ions.

Assuming this negative result is reliable:

\[
\ce{Fe > P}
\]

P does displace copper:

\[
\ce{P > Cu}
\]

It also displaces silver:

\[
\ce{P > Ag}
\]

The first three tests therefore establish:

\[
\boxed{\ce{Fe > P > Cu > Ag}}
\]

The relevant known activity order is:

\[
\ce{Fe > Co > Ni > Cu > Ag}
\]

Both cobalt and nickel fit the experimental results.

The first three tests cannot distinguish them.

(b) Calculate the cell potential for cobalt displacing nickel.

The proposed reaction is:

\[
\ce{Co(s) + Ni^2+(aq) -> Co^2+(aq) + Ni(s)}
\]

Cobalt is oxidised, while nickel ions are reduced.

Using standard reduction potentials:

\[
E^\circ_{\text{cell}}
=
E^\circ_{\text{red,cathode}}
–
E^\circ_{\text{red,anode}}
\]

Substitute:

\[
\begin{aligned}
E^\circ_{\text{cell}}
&=-0.25-(-0.28)\\
&=-0.25+0.28\\
&=\boxed{+0.03\text{ V}}
\end{aligned}
\]

The positive cell potential indicates that cobalt displacement of nickel is thermodynamically favourable under standard conditions.

(c) Explain why the negative nickel test is inconclusive.

Consider the two possible identities.

If P is nickel, placing it into nickel(II) ions produces no net displacement reaction. The solid metal and dissolved ions belong to the same redox couple.

If P is cobalt, the reaction with nickel(II) ions is thermodynamically favourable:

\[
\ce{Co(s) + Ni^2+(aq) -> Co^2+(aq) + Ni(s)}
\]

However, the calculated cell potential is only +0.03 V.

This tells you the predicted thermodynamic direction, not how quickly the reaction proceeds or how much nickel must deposit before you can see it.

A short visual test may fail to detect a small amount of deposited nickel.

Therefore, the absence of a visible deposit after 60 seconds does not automatically exclude cobalt.

The tempting conclusion is that no nickel deposit means P must be nickel. The problem is that the test has not established whether a cobalt strip would produce a detectable deposit within the same observation period.

(d) Determine whether the identity is unique and propose a further measurement.

The identity is not unique.

The first three tests are consistent with both cobalt and nickel, and the final test has insufficient sensitivity and observation time to eliminate cobalt.

One suitable further investigation would be to construct a carefully controlled galvanic cell using the unknown metal and a known nickel half-cell.

A high-impedance voltmeter could measure the cell potential under standardised conditions.

If P is cobalt, the potential difference relative to the nickel half-cell should be approximately 0.03 V, with cobalt acting as the anode.

If P is nickel, the two half-cells involve the same redox couple, so there should be no standard potential difference.

This provides an independent comparison that does not rely on waiting for a visible metal coating to form.

The deeper lesson is that experimental evidence has limits. A negative observation only rules out a candidate when the test was capable of detecting the reaction that candidate predicts.

When identifying an unknown metal, keep three ideas separate: what the experiment actually shows, what the activity series predicts, and how quickly the reaction can be observed. Combining all three is what turns a collection of displacement results into a defensible chemical identification.