Why Cleaning a Metal Surface Changes an Experiment’s Result

Two identical magnesium ribbons are placed in separate flasks of dilute hydrochloric acid. One ribbon has just been polished with abrasive paper. The other is taken straight from its container.

The polished ribbon starts producing hydrogen almost immediately. The untreated ribbon shows very little visible activity at first, then begins bubbling. Eventually, both produce approximately the same volume of gas.

Why?

Cleaning changes the condition of the metal’s surface. It can remove oxide layers, grease, and other substances that prevent the solution from contacting the underlying metal. This changes how quickly the reaction proceeds and how much product is visible during a particular observation period.

It does not automatically change the metal’s position in the activity series or make a thermodynamically unfavourable reaction favourable.

There is also an important complication: some oxide coatings react with the solution themselves. If the solution is the limiting reagent, removing the coating can even change the final amount of product. Understanding this distinction is the key to interpreting experiments correctly.

01The experiment: two surfaces, one chemical reaction

Consider an investigation designed to test whether cleaning magnesium changes its reaction with hydrochloric acid.

Two identical conical flasks each contain:

• \(25.0\ \mathrm{mL}\) of \(0.20\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
• A magnesium ribbon containing \(0.0243\ \mathrm{g}\) of magnesium metal.
• The same temperature and reaction conditions.

One magnesium ribbon is polished immediately before the experiment. The other is left untreated.

Each flask is fitted with a bung connected to a gas syringe. The magnesium is initially held above the acid in a small dry holder. Tilting the flask brings the acid into contact with the ribbon and starts the reaction.

The initial gas syringe reading is recorded before mixing, so any gas collected can be measured as a change in volume.

Before looking at the results, make a prediction.

If both ribbons contain the same amount of magnesium and the acid is in excess, should cleaning change the final volume of hydrogen, or just how quickly the hydrogen appears?

The expected result is a difference in reaction rate, not theoretical gas yield. Both samples contain the same amount of reacting magnesium.

The reaction is:

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

Magnesium atoms lose electrons to form magnesium ions. Hydrogen ions gain those electrons to form hydrogen gas.

Illustrative experimental results

The following data are illustrative, not measurements from a particular laboratory investigation. Assume both samples contain the same amount of magnesium metal, the acid is in excess, and all other experimental conditions are controlled.

Three observations matter.

1. The cleaned magnesium produces hydrogen sooner.

Its gas syringe reading increases immediately. The untreated sample shows an initial delay.

2. The cleaned magnesium has a greater initial reaction rate.

During the first 30 seconds, it produces \(18\ \mathrm{mL}\) of hydrogen, compared with only \(5\ \mathrm{mL}\) from the untreated sample.

3. Both samples eventually produce approximately the same volume of hydrogen.

The difference is mainly in how quickly the reaction proceeds, rather than how much magnesium ultimately reacts.

This is the first important distinction: a reaction can be occurring slowly, or producing too little visible product to detect, without being thermodynamically unfavourable.

02What is actually happening at the metal surface?

A piece of metal might look perfectly clean, but its surface is not necessarily chemically bare.

When metals are exposed to air, they can react with oxygen, moisture, and other substances. This produces thin surface layers containing metal oxides, hydroxides, or other compounds.

Some layers are barely noticeable. Others are protective enough to substantially slow further reactions.

For magnesium, a simplified model is a surface coating of magnesium oxide.

The coating can interfere with the reaction in two ways.

First, it reduces the amount of underlying magnesium directly accessible to the acid. Hydrogen ions cannot react as easily with the metal when a surface layer separates them.

Second, the acid can react with the oxide itself.

The simplified reaction is:

\[ \ce{MgO(s) + 2H+(aq) -> Mg^2+(aq) + H2O(l)} \]

Notice something important: this reaction produces water, not hydrogen gas.

The acid is reacting with the oxide coating before, or while, the underlying magnesium becomes accessible.

Once the acid reaches exposed magnesium, the hydrogen-producing reaction can proceed:

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

A useful analogy is a thin raincoat covering the metal. The acid reaches the coating first rather than directly contacting the magnesium underneath.

But the analogy has a limit. A raincoat is not normally dissolved by the rain. A metal oxide is a chemical substance that can react with the solution surrounding it.

This is why cleaning can remove an initial delay: it exposes more of the underlying metal to the reacting solution.

Real magnesium surfaces can contain mixtures of oxides, hydroxides, and adsorbed contaminants, so the pure MgO equation is a simplified model rather than a complete description of every surface.

03Different ways cleaning changes an experiment

Cleaning is not a single chemical process. Different preparation methods can affect different properties of the metal.

The distinction between removing a coating and increasing surface area is particularly important.

Suppose you rub a magnesium ribbon with coarse abrasive paper. You might remove some oxide, but you could also create a rougher surface.

A rougher surface can expose more metal to the solution. The reaction may therefore become faster for two reasons:

• More of the underlying metal is accessible.
• The actual surface area available for reaction may have increased.

If you attribute the entire change to oxide removal, you may be overlooking the second effect.

Not all oxide layers behave the same way

The type of oxide matters.

Consider three common metals.

Aluminium is particularly interesting.

It is a reactive metal, but it forms a thin, adherent aluminium oxide layer when exposed to air. This process is called passivation.

Passivation occurs when a surface layer substantially protects the metal underneath from further reaction.

This creates an interesting experimental problem.

You can polish aluminium to remove much of its existing oxide layer, but a new layer begins forming when the metal is exposed to air.

Therefore, the time between cleaning and starting the experiment can influence the observation.

A freshly polished aluminium strip is not guaranteed to remain oxide-free until it reaches the reaction vessel.

04Cleaning versus thermodynamic prediction

Now connect surface preparation to the metal activity series.

The activity series predicts the relative tendency of metals to lose electrons. It helps determine whether a metal displacement reaction is expected under the usual aqueous conditions considered in HSC Chemistry.

For example, zinc can displace copper ions from solution:

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

Zinc is more reactive than copper, so zinc atoms can lose electrons while copper ions gain them.

Cleaning the zinc can make this reaction easier to observe by exposing more accessible metal.

However, cleaning does not change the underlying thermodynamic prediction.

It helps to separate three different questions:

The activity series is not a stopwatch. It tells you about the expected reaction direction, not how quickly a particular sample will produce visible evidence.

A contrasting case: cleaning copper does not make it displace zinc

Imagine polishing a copper strip until it is shiny, then placing it into zinc sulfate solution.

You might expect a very clean surface to react more readily.

But no zinc displacement is expected:

\[ \ce{Cu(s) + Zn^2+(aq) -> no\ displacement} \]

Copper is less reactive than zinc. It does not readily lose electrons to reduce zinc ions to zinc metal under these conditions.

Polishing the copper may remove surface contamination, but it cannot change the underlying chemical tendency of copper.

This gives you a useful rule:

Cleaning can reveal or accelerate a reaction that is already favourable. It cannot make an unfavourable displacement reaction favourable simply by exposing a cleaner surface.

The opposite situation occurs with aluminium and copper(II) ions.

Aluminium is more reactive than copper, so the displacement reaction is predicted:

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

However, the protective aluminium oxide layer can prevent the copper ions from accessing the underlying metal effectively.

The predicted reaction and the visible experimental result can therefore look quite different.

For a fuller explanation of the electron-transfer predictions, see Metal Displacement Reactions: Predicting HSC Chemistry Changes.

05How to design a fair surface-cleaning experiment

There is a problem with the original magnesium investigation.

The polished ribbon and untreated ribbon have different surface conditions, but polishing may have changed more than one variable.

It could have:

• Removed oxide.
• Removed grease or other contaminants.
• Increased or decreased microscopic surface roughness.
• Removed a small amount of magnesium.
• Changed the time between preparation and immersion.

If the polished ribbon reacts faster, you cannot automatically conclude that oxide removal alone caused the difference.

The original investigation tests the effect of two surface-preparation conditions. It does not necessarily isolate one particular surface property.

Improving the experimental design

A better investigation should control the variables as carefully as possible.

If you want to investigate the overall effect of cleaning, compare a standardised cleaning procedure with an untreated sample.

If you want to investigate oxide removal specifically, you need to consider other effects of the cleaning method, particularly changes in surface roughness and metal mass. Even a chemical cleaning treatment can etch the underlying metal.

Repeat the experiment using several samples rather than relying on one comparison.

Most importantly, measure the reaction over time rather than recording only whether bubbles appear.

How to interpret the results

The shape of your results can help identify what is happening.

One final detail is easy to miss.

A surface oxide does not always act only as a physical barrier. It can also be a reactant.

That means removing an oxide layer can sometimes change the final product quantity, not just the reaction rate. Whether this happens depends on the chemical reaction and which reactant is limiting.

06Questions and solutions

Question 1

A zinc strip is polished immediately before being placed into copper(II) sulfate solution. An identical untreated zinc strip is placed into a separate sample of the same solution.

The polished strip develops a reddish-brown copper coating sooner.

(a) Write the balanced net ionic equation for the reaction.

(b) Explain why polishing the zinc can change the observed reaction rate without changing the predicted reaction direction.

Solution 1

(a) The balanced net ionic equation is:

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

(b) Polishing increases the accessibility of the underlying zinc surface, allowing copper ions to interact more readily with the metal.

An untreated zinc strip may have an oxide layer or surface contamination that restricts contact between the reacting species.

Removing some of this material can increase the number of accessible reaction sites and increase the initial reaction rate.

However, zinc remains the same metal before and after cleaning. Its position in the activity series does not change.

Zinc is more reactive than copper, so the displacement reaction is predicted in both cases.

The difference is in how quickly the reaction becomes observable, not whether the underlying displacement is favourable.

Question 2

The following results are obtained when equal amounts of cleaned and untreated magnesium react with excess hydrochloric acid.

Use the illustrative data from the magnesium investigation earlier in this article.

(a) Estimate the time taken for each sample to produce \(20.0\ \mathrm{mL}\) of hydrogen. Use linear interpolation between the nearest measurements.

(b) Calculate the ratio of the time taken by the untreated sample to the time taken by the cleaned sample.

(c) Explain why this time ratio should not be interpreted as a ratio of the intrinsic reactivities of the two magnesium samples.

Solution 2

(a) The cleaned sample reaches \(20.0\ \mathrm{mL}\) between 30 and 40 seconds.

During this interval, the volume increases from \(18\ \mathrm{mL}\) to \(22\ \mathrm{mL}\).

The required increase is:

\[ 20-18=2\ \mathrm{mL} \]

The total increase over the interval is \(4\ \mathrm{mL}\), so the required increase is half the interval.

\[ t=30+\frac{2}{4}\times10 \]

\[ t=35\ \mathrm{s} \]

The untreated sample reaches \(20.0\ \mathrm{mL}\) between 50 and 60 seconds.

During this interval, the volume increases from \(17\ \mathrm{mL}\) to \(22\ \mathrm{mL}\).

The required increase is:

\[ 20-17=3\ \mathrm{mL} \]

Therefore:

\[ t=50+\frac{3}{5}\times10 \]

\[ t=56\ \mathrm{s} \]

The estimated times are:

• Cleaned magnesium: \(35\ \mathrm{s}\).
• Untreated magnesium: \(56\ \mathrm{s}\).

(b) The time ratio is:

\[ \frac{t_{\text{untreated}}} {t_{\text{cleaned}}} = \frac{56}{35} \]

\[ =1.6 \]

The untreated sample takes approximately 1.6 times as long to produce \(20.0\ \mathrm{mL}\) of hydrogen.

(c) This difference reflects the effect of surface preparation on the measured reaction rate.

The untreated magnesium initially has less accessible metal surface because of its surface coating. As the coating is removed or dissolved, more magnesium becomes available to react.

The time ratio does not represent a change in the intrinsic chemical reactivity of magnesium.

It is also an estimate based on linear interpolation. Real reaction rates are not necessarily constant over the measured intervals.

Question 3

Two zinc samples are placed separately into identical copper(II) sulfate solutions.

Sample A is a flat, polished zinc strip with an exposed surface area of \(2.0\ \mathrm{cm^2}\).

Sample B is an untreated, folded zinc foil with an exposed surface area of \(4.0\ \mathrm{cm^2}\).

Assume the supplied areas include all wetted surfaces.

After \(60.0\ \mathrm{s}\), the following masses of copper have formed:

(a) Calculate the average copper deposition rate per unit exposed surface area for each sample.

(b) Which sample has the greater area-normalised deposition rate?

(c) Can you conclude that polishing alone caused the difference? Explain.

Solution 3

(a) Sample A has an average copper deposition rate per unit area of \(1.00\times10^{-3}\ \mathrm{g\,cm^{-2}\,s^{-1}}\), while Sample B has a rate of \(7.50\times10^{-4}\ \mathrm{g\,cm^{-2}\,s^{-1}}\).

The required equation is:

\[ \text{Rate per unit area} = \frac{\text{Mass deposited}} {\text{Area}\times\text{Time}} \]

For Sample A:

\[ \text{Rate}_A = \frac{0.120}{2.0\times60.0} \]

\[ =0.00100\ \mathrm{g\,cm^{-2}\,s^{-1}} \]

For Sample B:

\[ \text{Rate}_B = \frac{0.180}{4.0\times60.0} \]

\[ =0.000750\ \mathrm{g\,cm^{-2}\,s^{-1}} \]

(b) Sample A has the greater area-normalised deposition rate.

The ratio is:

\[ \frac{0.00100}{0.000750}=1.33 \]

Its average copper deposition rate per unit exposed area is approximately 1.33 times that of Sample B.

Notice that Sample B actually produces more copper overall. This is because it has twice the exposed surface area.

(c) No. The results do not isolate the effect of polishing.

The two samples have different shapes, surface areas, and surface conditions. Folding the foil may also change how effectively different parts of its surface contact the solution.

Although dividing by surface area provides a fairer comparison than using total copper mass alone, the experiment still has multiple uncontrolled differences.

The results are consistent with Sample A having a more accessible surface, but they do not prove that polishing alone caused the difference.

A better investigation would use zinc samples with identical dimensions and surface areas, changing only the defined surface-preparation treatment.

Question 4

Two aluminium strips are cut from the same sheet and polished using the same abrasive material.

• Strip A is immersed in copper(II) sulfate solution within 10 seconds of polishing.
• Strip B is left exposed to air for 10 minutes before being immersed in an identical solution.

In an illustrative investigation, Strip A develops a reddish-brown coating within two minutes, while Strip B shows no visible change during the same period.

A student concludes that Strip A is more reactive than Strip B.

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

(b) Evaluate the student’s conclusion using your knowledge of aluminium’s surface chemistry.

(c) Describe a follow-up experiment that could investigate whether exposure to air after polishing affects the reaction.

Solution 4

(a) The balanced net ionic equation is:

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

Aluminium atoms are oxidised to aluminium ions, while copper(II) ions are reduced to copper metal.

(b) The student’s conclusion is not justified.

Both strips are aluminium, so their underlying chemical reactivity is the same.

The difference is their surface condition.

Aluminium forms a compact, protective aluminium oxide layer when exposed to air. This process occurs rapidly, meaning Strip A may already have developed a thin oxide layer before entering the solution.

Strip B has been exposed to air for considerably longer. Its surface may therefore have a different oxide condition, potentially making the reaction more difficult to observe.

The difference in copper deposition could be related to surface passivation rather than a difference in the intrinsic reactivity of the two strips.

Importantly, the experiment does not establish that Strip A is completely oxide-free. Nor does one observation prove that oxide growth is the only explanation.

(c) A suitable follow-up would investigate the effect of the time between polishing and immersion.

Prepare several identical aluminium strips from the same sheet.

Polish each strip using the same method, then immerse them after different controlled delays, such as:

• 10 seconds.
• 30 seconds.
• 2 minutes.
• 10 minutes.

Keep the following conditions constant:

• Aluminium dimensions and exposed area.
• Copper(II) sulfate concentration and volume.
• Temperature.
• Polishing method.
• Reaction observation time.

Repeat each condition and measure the mass of copper deposited after a fixed period.

If the amount of copper deposited decreases as the exposure time increases, this would support the idea that changes to the surface layer are affecting the reaction.

The investigation would provide evidence about surface reformation, rather than suggesting that aluminium itself has become less reactive.

Question 5

A chemistry student investigates whether removing an oxide coating can change the final amount of hydrogen produced in a metal-acid reaction.

Two magnesium samples each contain exactly \(0.100\ \mathrm{g}\) of underlying magnesium metal.

• Sample A has been cleaned, removing its magnesium oxide coating.
• Sample B retains a coating containing \(1.00\times10^{-5}\ \mathrm{mol}\) of magnesium oxide.

The oxide amount is deliberately specified for this illustrative model. It is not a claim about the typical mass of oxide on an ordinary magnesium ribbon.

Each sample is placed separately into \(1.00\ \mathrm{mL}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.

Assume:

• The magnesium oxide reacts completely with the acid.
• The magnesium metal also reacts wherever acid can reach it.
• The magnesium metal is in excess, so hydrochloric acid is the limiting reagent.
• No hydrogen escapes from the apparatus.
• The only relevant reactions are:

\[ \ce{MgO(s) + 2H+(aq) -> Mg^2+(aq) + H2O(l)} \]

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

Use a molar gas volume of \(24.8\ \mathrm{L\,mol^{-1}}\) at SATP.

(a) Calculate the theoretical volume of hydrogen produced by Sample A.

(b) Calculate the theoretical volume of hydrogen produced by Sample B.

(c) Explain why the two samples produce different final gas volumes, even though they contain the same amount of magnesium metal.

(d) Predict what would happen to the difference in final hydrogen volume if both samples were instead supplied with a sufficiently large excess of hydrochloric acid.

Solution 5

(a) Sample A produces \(1.24\ \mathrm{mL}\) of hydrogen.

(b) Sample B produces \(0.992\ \mathrm{mL}\) of hydrogen.

The difference arises because the magnesium oxide coating consumes hydrogen ions without producing hydrogen gas.

Step 1: Calculate the initial amount of hydrogen ions.

The concentration of hydrochloric acid is:

\[ C=0.100\ \mathrm{mol\,L^{-1}} \]

The volume is:

\[ V=1.00\times10^{-3}\ \mathrm{L} \]

Using \(n=CV\):

\[ n(\mathrm{H^+}) = 0.100\times1.00\times10^{-3} \]

\[ =1.00\times10^{-4}\ \mathrm{mol} \]

This is the total amount of hydrogen ions initially available.

Step 2: Calculate the hydrogen produced by the cleaned sample.

Sample A has no magnesium oxide coating.

All available hydrogen ions can therefore react with magnesium metal.

From the balanced equation:

\[ \ce{Mg + 2H+ -> Mg^2+ + H2} \]

Two moles of hydrogen ions produce one mole of hydrogen gas.

Therefore:

\[ n(\mathrm{H_2}) = \frac{1.00\times10^{-4}}{2} \]

\[ =5.00\times10^{-5}\ \mathrm{mol} \]

Convert this amount into a gas volume:

\[ V=nV_m \]

\[ V=5.00\times10^{-5}\times24.8 \]

\[ =0.00124\ \mathrm{L} \]

Therefore:

\[ V=1.24\ \mathrm{mL} \]

Step 3: Calculate how much acid the oxide coating consumes.

Sample B contains:

\[ n(\mathrm{MgO})=1.00\times10^{-5}\ \mathrm{mol} \]

From the oxide reaction:

\[ \ce{MgO + 2H+ -> Mg^2+ + H2O} \]

Each mole of magnesium oxide consumes two moles of hydrogen ions.

Therefore:

\[ n(\mathrm{H^+})_{\text{consumed}} = 2\times1.00\times10^{-5} \]

\[ =2.00\times10^{-5}\ \mathrm{mol} \]

Step 4: Calculate the hydrogen ions remaining for the metal reaction.

\[ n(\mathrm{H^+})_{\text{remaining}} = 1.00\times10^{-4} - 2.00\times10^{-5} \]

\[ =8.00\times10^{-5}\ \mathrm{mol} \]

These remaining hydrogen ions react with the underlying magnesium metal.

Step 5: Calculate the hydrogen produced by Sample B.

Again, two moles of hydrogen ions produce one mole of hydrogen gas.

\[ n(\mathrm{H_2}) = \frac{8.00\times10^{-5}}{2} \]

\[ =4.00\times10^{-5}\ \mathrm{mol} \]

Convert to volume:

\[ V=4.00\times10^{-5}\times24.8 \]

\[ =9.92\times10^{-4}\ \mathrm{L} \]

Therefore:

\[ V=0.992\ \mathrm{mL} \]

Step 6: Compare the final results.

The coated sample produces \(20\%\) less hydrogen.

This is not simply a difference in reaction rate.

The magnesium oxide coating consumes some of the limited hydrogen ions. Those ions form water rather than hydrogen gas.

The amount of hydrogen produced is therefore reduced, even though the two samples contain exactly the same amount of underlying magnesium metal.

(c) The final gas volumes differ because the oxide coating is an additional reactant.

It consumes acid without generating hydrogen, leaving fewer hydrogen ions available for the reaction with magnesium metal.

The key point is that surface cleaning can affect both reaction kinetics and stoichiometry, depending on the experimental conditions.

(d) If hydrochloric acid is supplied in sufficiently large excess, both samples should eventually produce the same amount of hydrogen.

The magnesium metal is then the limiting reagent rather than the acid.

Each sample contains:

\[ n(\mathrm{Mg})=\frac{0.100}{24.3} \]

\[ =0.00412\ \mathrm{mol} \]

The mole ratio between magnesium and hydrogen is \(1:1\).

Therefore, both samples produce:

\[ n(\mathrm{H_2})=0.00412\ \mathrm{mol} \]

The theoretical hydrogen volume for each is:

\[ V=0.00412\times24.8 \]

\[ =0.102\ \mathrm{L} \]

\[ =102\ \mathrm{mL} \]

The oxide coating still consumes some acid, but sufficient acid remains for all the magnesium metal to react.

This is the important distinction: surface cleaning can change the reaction rate, and it can change the final yield when the surface coating consumes a limiting reagent.

When interpreting a metal practical, examine the initial gradient of the reaction-rate graph and its final plateau separately. The gradient tells you how quickly the reaction proceeds; the plateau tells you how much product has formed. Checking the limiting reagent then helps you decide whether a difference comes from kinetics, stoichiometry, or both.

For further practical troubleshooting, see Why Metal Reactivity Practicals Give Unexpected Results.