Why Copper Does Not Usually React with Dilute Hydrochloric Acid
Discover why copper cannot normally displace hydrogen from dilute hydrochloric acid. Explore the metal activity series, standard reduction potentials, and how surface oxides and dissolved oxygen can cause different observations.
A clean piece of copper placed in dilute hydrochloric acid usually does nothing noticeable. There are no sustained bubbles of hydrogen, and the copper metal does not gradually disappear.
Yet magnesium reacts readily with the same acid. Both metals are surrounded by hydrogen ions, so why does one release hydrogen gas while the other does not?
The answer is that copper is less reactive than hydrogen. It does not readily lose electrons to hydrogen ions, so it cannot normally displace hydrogen from dilute hydrochloric acid. More precisely, the proposed reaction is thermodynamically unfavourable under standard conditions. It is not simply a reaction that needs more time.
Before looking at the equations, make a prediction: if you grind the copper into a fine powder, will it suddenly start producing hydrogen?
It won’t, under ordinary dilute hydrochloric acid conditions. Grinding increases the exposed surface area, but it does not change the underlying thermodynamic preference for electron transfer. That distinction is important when explaining why copper behaves differently from more reactive metals.
01What needs to happen for a metal to react with hydrochloric acid?
Hydrochloric acid is a strong acid. In water, it ionises essentially completely:
\[ \ce{HCl(aq) -> H+(aq) + Cl-(aq)} \]
The hydrogen ions can accept electrons and combine to form hydrogen gas.
For a metal to produce hydrogen in this way, two processes must occur:
- The metal must lose electrons. This is oxidation.
- Hydrogen ions must gain those electrons. This is reduction.
Consider magnesium. Its atoms readily lose electrons:
\[ \ce{Mg(s) -> Mg^2+(aq) + 2e-} \]
The hydrogen ions accept those electrons:
\[ \ce{2H+(aq) + 2e- -> H2(g)} \]
Adding the half-equations gives:
\[ \ce{Mg(s) + 2H+(aq) -> Mg^2+(aq) + H2(g)} \]
This produces the familiar bubbles of hydrogen.
Copper would need to undergo the equivalent oxidation process:
\[ \ce{Cu(s) -> Cu^2+(aq) + 2e-} \]
Those electrons would then need to reduce hydrogen ions:
\[ \ce{2H+(aq) + 2e- -> H2(g)} \]
The proposed overall reaction is:
\[ \ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)} \]
The equation is balanced, but that doesn’t mean the reaction will actually occur. We still need to establish whether copper can readily undergo oxidation while hydrogen ions undergo reduction.
02Copper sits below hydrogen in the metal activity series
The metal activity series ranks metals according to their tendency to undergo oxidation.
A simplified section of the series is:
\[ \ce{Mg > Zn > Fe > H > Cu} \]
Hydrogen is included as a reference point, even though it is not a metal.
The important rule is:
A metal above hydrogen can generally displace hydrogen from a dilute, non-oxidising acid. A metal below hydrogen generally cannot.
This gives us the following comparison:
| Substance | Position relative to hydrogen | Behaviour with dilute HCl |
|---|---|---|
| Magnesium | Above hydrogen | Reacts rapidly, producing H2 |
| Zinc | Above hydrogen | Reacts, producing H2 |
| Iron | Above hydrogen | Reacts more slowly, producing H2 |
| Copper | Below hydrogen | No normal hydrogen-producing reaction |
Copper is below hydrogen because copper atoms have a lower tendency to lose electrons than hydrogen gas has to be oxidised under the relevant standard conditions.
In practical terms, copper does not readily supply the electrons needed to turn hydrogen ions into hydrogen gas.
This is why simply having hydrogen ions present is not enough to guarantee a reaction.
For a broader explanation of how these observations establish the activity series, see Metal Reactivity with Dilute Acids: HSC Chemistry Guide.
03Using standard reduction potentials to prove the prediction
The activity series gives us a useful qualitative prediction. Standard reduction potentials allow us to test that prediction quantitatively.
A standard reduction potential measures the tendency of a reduction half-reaction to occur.
For our comparison, the relevant values at 25°C are:
| Reduction half-equation | Standard reduction potential |
|---|---|
| \(\ce{2H+(aq) + 2e- -> H2(g)}\) | 0.00 V |
| \(\ce{Cu^2+(aq) + 2e- -> Cu(s)}\) | +0.34 V |
These values use the standard hydrogen electrode as the reference, which is assigned a potential of 0.00 V.
Standard conditions use unit activities for dissolved species, approximately 1 mol L\(^{-1}\) for ordinary HSC calculations, and a standard gas pressure.
Notice that copper’s reduction potential is more positive than hydrogen’s.
This means that \(\mathrm{Cu}^{2+}\) has a greater tendency to gain electrons than \(\mathrm{H}^{+}\) under standard conditions.
Consequently, copper metal is not readily oxidised by hydrogen ions.
Calculating the standard cell potential
We can calculate the potential for the proposed reaction using:
\[ E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}} \]
Both values in this equation are taken from the table of standard reduction potentials.
- The cathode is where reduction occurs.
- The anode is where oxidation occurs.
For copper reacting with hydrochloric acid:
- Cathode: hydrogen ions are reduced to hydrogen gas.
- Anode: copper is oxidised to copper(II) ions.
Therefore:
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\ce{H+/H2}} -E^\circ_{\ce{Cu^2+/Cu}}\\ &=0.00-(+0.34)\\ &=-0.34\ \mathrm{V} \end{aligned} \]
The negative cell potential tells us that the proposed reaction is not spontaneous under standard conditions.
The reverse reaction is thermodynamically favoured:
\[ \ce{Cu^2+(aq) + H2(g) -> Cu(s) + 2H+(aq)} \]
This is the quantitative reason copper does not normally displace hydrogen from dilute hydrochloric acid.
Worked example: Will copper react with dilute hydrochloric acid?
A student proposes that copper reacts with dilute hydrochloric acid according to:
\[ \ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)} \]
Given:
\[ E^\circ_{\ce{Cu^2+/Cu}}=+0.34\ \mathrm{V} \]
\[ E^\circ_{\ce{H+/H2}}=0.00\ \mathrm{V} \]
Calculate the standard cell potential and determine whether the proposed reaction is spontaneous.
Step 1
Copper loses electrons, so it is oxidised at the anode.
Hydrogen ions gain electrons, so they are reduced at the cathode.
Step 2
\[ E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}} \]
Step 3
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=0.00-(+0.34)\\ &=-0.34\ \mathrm{V} \end{aligned} \]
Step 4
The negative value indicates that the proposed reaction is thermodynamically unfavourable under standard conditions.
Therefore, copper does not normally react with dilute hydrochloric acid to produce hydrogen gas.
The important exam point is that a negative cell potential is not evidence of a particularly slow reaction. It indicates that the proposed reaction is not thermodynamically favoured as written.
04Is copper’s lack of reaction caused by its surface?
There is another complication worth understanding.
Copper exposed to air can develop a thin surface layer containing copper oxides. This means that the substance touching the hydrochloric acid may not be pure copper metal.
Could that explain why copper appears to react in some experiments?
Sometimes, yes, but we need to identify exactly which substance is reacting.
1. Copper oxide can react with hydrochloric acid
Copper(II) oxide, \(\ce{CuO}\), reacts with hydrochloric acid:
\[ \ce{CuO(s) + 2HCl(aq) -> CuCl2(aq) + H2O(l)} \]
You may observe:
- The black copper oxide gradually disappears.
- A blue-green solution forms as copper(II) ions enter the solution.
- No hydrogen gas is produced.
This is an acid-base reaction involving a metal oxide, not the oxidation of copper metal.
Notice that copper has an oxidation state of +2 in both \(\ce{CuO}\) and \(\ce{CuCl2}\). It does not lose additional electrons during this reaction.
Therefore, the disappearance of a black coating does not demonstrate that copper metal has displaced hydrogen.
Worked example: Is there enough hydrochloric acid to dissolve a copper oxide coating?
A sample contains \(0.159\ \mathrm{g}\) of pure copper(II) oxide. It is treated with \(20.0\ \mathrm{mL}\) of \(0.200\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.
Calculate whether there is enough hydrochloric acid to react completely with the copper oxide.
Assume the reaction follows the balanced equation:
\[ \ce{CuO + 2HCl -> CuCl2 + H2O} \]
Use:
\[ M(\ce{CuO})=79.55\ \mathrm{g\,mol^{-1}} \]
Step 1
\[ \begin{aligned} n(\ce{CuO}) &=\frac{m}{M}\\ &=\frac{0.159}{79.55}\\ &=0.001999\ \mathrm{mol} \end{aligned} \]
Step 2
The balanced equation gives a mole ratio of:
\[ \ce{CuO:HCl}=1:2 \]
Therefore:
\[ \begin{aligned} n(\ce{HCl})_{\mathrm{required}} &=2(0.001999)\\ &=0.003998\ \mathrm{mol} \end{aligned} \]
Step 3
Convert the volume into litres:
\[ 20.0\ \mathrm{mL}=0.0200\ \mathrm{L} \]
Use \(n=cV\):
\[ \begin{aligned} n(\ce{HCl})_{\mathrm{supplied}} &=cV\\ &=0.200\times0.0200\\ &=0.00400\ \mathrm{mol} \end{aligned} \]
Step 4
The acid supplied is \(0.00400\ \mathrm{mol}\), while the reaction requires \(0.003998\ \mathrm{mol}\).
Therefore, there is just enough hydrochloric acid to react completely with the copper oxide.
The black coating can disappear even though the underlying copper metal does not react with the acid to produce hydrogen.
This is a useful reminder: always identify the actual reactant at the metal surface before deciding what a laboratory observation means.
2. Dissolved oxygen can cause copper corrosion
There is a second way copper can undergo oxidation in an acidic solution.
Ordinary hydrochloric acid exposed to air contains dissolved oxygen. Oxygen is an oxidising agent, meaning it can accept electrons from another substance.
Copper can therefore undergo corrosion if dissolved oxygen is available to accept the electrons.
The oxidation half-equation is:
\[ \ce{Cu(s) -> Cu^2+(aq) + 2e-} \]
Oxygen is reduced in acidic conditions:
\[ \ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)} \]
Combining the half-equations gives:
\[ \ce{2Cu(s) + O2(g) + 4H+(aq) -> 2Cu^2+(aq) + 2H2O(l)} \]
Notice what has changed.
The copper is still being oxidised, but hydrogen ions are not accepting the electrons. Oxygen is.
The chloride ions can also help keep dissolved copper species in solution by forming chloride-containing complexes.
This means that copper can undergo some corrosion in oxygen-containing hydrochloric acid without producing hydrogen gas.
The distinction is important:
- Copper with hydrogen ions alone: the usual hydrogen-producing reaction is thermodynamically unfavourable.
- Copper with oxygen and acidic solution: oxygen can act as the oxidising agent and drive copper corrosion.
Oxygen-assisted corrosion of copper in hydrochloric acid solutions has also been investigated in modern corrosion research. See Scientific Reports (2025).
05Why copper reacts with nitric acid but not dilute hydrochloric acid
Now consider a useful comparison.
Copper does not normally react with dilute hydrochloric acid, but it does react with nitric acid.
Does this mean the activity series is wrong?
No. The difference is the oxidising agent.
Hydrochloric acid is a non-oxidising acid for this comparison. Its hydrogen ions are not able to oxidise copper under ordinary dilute conditions.
Nitric acid is different. It contains nitrate ions, which can act as oxidising agents and accept electrons from copper.
For concentrated nitric acid, the overall reaction is:
\[ \ce{Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)} \]
The observations include:
- The copper gradually dissolves.
- A blue solution containing copper(II) ions forms.
- Brown nitrogen dioxide gas is produced.
Copper is oxidised from an oxidation state of 0 to +2.
Nitrogen in nitrate is reduced from +5 to +4 in nitrogen dioxide.
The nitrate ions, rather than hydrogen ions, provide the relevant oxidising power.
This comparison exposes a common misconception:
A strong acid is not automatically a strong oxidising agent.
Acid strength describes how readily an acid donates protons. Oxidising strength describes how readily a species accepts electrons.
They are different chemical properties.
If you need to revisit the difference between acid strength and concentration, see Strong vs Weak, Concentrated vs Dilute Acids and Bases.
06Thermodynamic limitations versus kinetic limitations
This is where students sometimes lose marks by using the wrong explanation.
Thermodynamics and kinetics answer different questions.
| Thermodynamics | Kinetics |
|---|---|
| Predicts whether a reaction is energetically favourable in a specified direction. | Describes how quickly a reaction proceeds. |
| Can be investigated using standard cell potentials. | Can be investigated using reaction-rate measurements. |
| A positive cell potential indicates a favourable reaction as written under standard conditions. | Reaction rate depends on factors such as temperature, concentration, surface area, and activation energy. |
For example, imagine a reaction has a positive standard cell potential but proceeds very slowly.
It may have a substantial activation energy barrier, or a protective surface layer may prevent reactants from reaching the metal.
That is a kinetic limitation.
Copper reacting with hydrogen ions is different. The proposed reaction has a negative standard cell potential under standard conditions.
Simply increasing the surface area or waiting longer does not turn that unfavourable reaction into a normal hydrogen-producing reaction.
There are three different situations to keep separate:
| Observation | Correct interpretation |
|---|---|
| Copper in dilute, non-oxidising HCl produces no hydrogen. | The proposed Cu/H+ redox reaction is thermodynamically unfavourable under standard conditions. |
| Copper oxide dissolves in HCl. | The oxide reacts with acid; this is not hydrogen displacement by copper metal. |
| Copper slowly corrodes in oxygen-containing HCl. | Oxygen acts as an alternative oxidising agent. |
The important distinction is that surface conditions can affect what you observe, but they do not make the ordinary copper-hydrogen displacement reaction thermodynamically favourable.
07How would you investigate copper’s behaviour experimentally?
A useful investigation should distinguish between two possibilities:
- Copper is genuinely unable to displace hydrogen under the tested conditions.
- A reaction is occurring, but the observations or experimental design are not sensitive enough to detect it.
A suitable investigation would compare copper and magnesium in identical hydrochloric acid solutions.
The original experimental design
Prepare two identical reaction vessels.
- Add \(25.0\ \mathrm{mL}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) hydrochloric acid to each.
- Maintain both solutions at \(25^\circ\mathrm{C}\).
- Place a cleaned copper strip in the first vessel.
- Place a magnesium ribbon with a comparable exposed surface area in the second vessel.
- Seal each vessel with a stopper and connect a delivery tube to an inverted graduated cylinder filled with water.
The gas produced at each metal surface travels through the delivery tube and collects in the graduated cylinder, displacing water.
What should the experiment show?
Before conducting the investigation, predict the observations.
| Measurement | Copper vessel | Magnesium vessel |
|---|---|---|
| Visible bubbling | No sustained bubbling | Sustained bubbling |
| Hydrogen collected | No detectable hydrogen | Hydrogen collected |
| Metal dissolving | No normal observable dissolution | Magnesium gradually dissolves |
Magnesium acts as a positive control.
If magnesium produces hydrogen, we have evidence that the hydrochloric acid and gas-collection apparatus are functioning.
If copper produces no detectable hydrogen, the observation supports the activity-series prediction.
However, we must be careful with the conclusion.
No visible bubbles do not prove that absolutely no copper corrosion has occurred. A small amount of copper might dissolve through an oxygen-assisted process without producing hydrogen.
Improving the experiment
The original design does not control dissolved oxygen.
To investigate this, prepare two additional vessels containing identical hydrochloric acid solutions and identical cleaned copper strips.
- Vessel A contains deoxygenated acid and is sealed to minimise oxygen entering.
- Vessel B contains air-saturated acid and is exposed to an oxygen supply.
Keep temperature, acid concentration, volume, and copper surface area the same.
Measure the change in copper mass and the amount of hydrogen produced.
If the oxygen-containing solution shows greater copper loss without measurable hydrogen production, that supports oxygen-assisted corrosion rather than direct displacement of hydrogen.
This revised design answers a more precise question: is copper being oxidised by hydrogen ions, or is another oxidising agent responsible?
The key lesson is to distinguish the reaction you are trying to investigate from other processes that can produce similar-looking observations.
08A decision process for HSC questions
When a question asks why copper does not react with dilute hydrochloric acid, work through these steps.
| Step | What to identify | What to conclude |
|---|---|---|
| 1 | Is the acid oxidising or non-oxidising? | Dilute HCl is a non-oxidising acid. |
| 2 | Where is copper relative to hydrogen? | Copper is below hydrogen in the activity series. |
| 3 | What electron transfer is proposed? | Cu is oxidised and H+ is reduced. |
| 4 | What does the standard cell potential show? | \(E^\circ_{\mathrm{cell}}=-0.34\ \mathrm{V}\). |
| 5 | Could another process explain the observation? | Consider copper oxide or dissolved oxygen. |
A strong HSC explanation should connect all three levels:
- The activity series predicts that copper cannot normally displace hydrogen.
- The redox half-equations show the electron transfer that would be required.
- The negative standard cell potential explains why that proposed reaction is thermodynamically unfavourable.
Avoid writing only that copper is unreactive. Explain what copper would need to do and why that process is not favoured.
09Questions and solutions
The following questions move from a standard activity-series prediction to more demanding applications of redox chemistry, standard reduction potentials, and experimental evidence.
Question 1
A student places a cleaned copper strip into \(25.0\ \mathrm{mL}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) dilute hydrochloric acid at \(25^\circ\mathrm{C}\).
After 10 minutes, there are no sustained gas bubbles, and the copper strip shows no visible loss of mass.
(a) Predict whether copper reacts with the hydrochloric acid to produce hydrogen.
(b) Explain your prediction using the metal activity series.
(c) Write the balanced ionic equation for the proposed reaction and explain why it does not normally proceed.
Solution 1
Copper does not normally react with dilute hydrochloric acid to produce hydrogen gas.
(a) No sustained bubbling or visible dissolution is expected under the stated conditions.
(b) Copper is below hydrogen in the metal activity series. Therefore, copper does not readily lose electrons to hydrogen ions and cannot normally displace hydrogen from a dilute, non-oxidising acid.
(c) The proposed ionic equation is:
\[ \ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)} \]
For this reaction to occur, copper must be oxidised:
\[ \ce{Cu(s) -> Cu^2+(aq) + 2e-} \]
At the same time, hydrogen ions must be reduced:
\[ \ce{2H+(aq) + 2e- -> H2(g)} \]
However, this overall electron-transfer process is thermodynamically unfavourable under standard conditions.
Chloride ions are spectators in the proposed redox process. They do not provide the oxidising power needed to make copper dissolve and release hydrogen.
Question 2
At \(25^\circ\mathrm{C}\), a student investigates the reactions of zinc and copper with dilute hydrochloric acid.
The following standard reduction potentials are provided:
| Reduction half-equation | \(E^\circ\) |
|---|---|
| \(\ce{2H+ + 2e- -> H2}\) | 0.00 V |
| \(\ce{Cu^2+ + 2e- -> Cu}\) | +0.34 V |
| \(\ce{Zn^2+ + 2e- -> Zn}\) | -0.76 V |
(a) Calculate the standard cell potential for the proposed reaction between zinc and hydrogen ions.
(b) Calculate the standard cell potential for the proposed reaction between copper and hydrogen ions.
(c) Identify which metal is expected to react with dilute hydrochloric acid and explain why the other does not.
(d) Explain why a negative standard cell potential should not be described simply as a slow reaction.
Solution 2
Zinc is expected to react with dilute hydrochloric acid, while copper is not expected to produce hydrogen under ordinary conditions.
(a) For zinc:
\[ \ce{Zn(s) + 2H+(aq) -> Zn^2+(aq) + H2(g)} \]
Hydrogen ions are reduced at the cathode, while zinc is oxidised at the anode.
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\mathrm{cathode}} -E^\circ_{\mathrm{anode}}\\ &=0.00-(-0.76)\\ &=+0.76\ \mathrm{V} \end{aligned} \]
The positive potential indicates that the reaction is thermodynamically favourable under standard conditions.
(b) For copper:
\[ \ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)} \]
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=0.00-(+0.34)\\ &=-0.34\ \mathrm{V} \end{aligned} \]
The negative potential indicates that the proposed reaction is thermodynamically unfavourable under standard conditions.
(c) Zinc is expected to react because its oxidation is favourable when coupled with the reduction of hydrogen ions.
Copper is not expected to react because its oxidation is unfavourable when coupled with hydrogen ion reduction.
The difference is consistent with their positions in the activity series:
\[ \ce{Zn > H > Cu} \]
(d) A negative cell potential describes the thermodynamic direction of a reaction under the specified standard conditions. It does not describe its reaction rate.
A reaction with a positive cell potential may still proceed slowly because of activation energy or surface barriers. However, the copper-hydrogen reaction has an unfavourable thermodynamic direction under standard conditions, rather than simply a high activation energy.
Question 3
Two copper samples are placed into separate beakers containing identical volumes and concentrations of dilute hydrochloric acid.
- Sample A has a thin black surface coating identified as copper(II) oxide.
- Sample B has been polished to expose clean copper metal.
After five minutes, the following observations are recorded:
| Observation | Sample A | Sample B |
|---|---|---|
| Surface appearance | Black coating disappears | Copper remains unchanged |
| Solution appearance | Becomes blue-green | Remains essentially colourless |
| Gas production | No detectable gas | No detectable gas |
Explain:
(a) Which substance in Sample A reacts with hydrochloric acid?
(b) Write the balanced chemical equation.
(c) Explain why the observations do not demonstrate that copper metal reacts with dilute hydrochloric acid.
Solution 3
The copper(II) oxide coating in Sample A reacts with hydrochloric acid, not the underlying copper metal.
(a) The black coating is copper(II) oxide, \(\ce{CuO}\).
It reacts with hydrochloric acid to produce aqueous copper(II) chloride and water.
(b) The balanced equation is:
\[ \ce{CuO(s) + 2HCl(aq) -> CuCl2(aq) + H2O(l)} \]
(c) Copper is already in the +2 oxidation state in copper(II) oxide.
It remains in the +2 oxidation state in copper(II) chloride.
Therefore, copper is not being oxidised further, and hydrogen ions are not being reduced to hydrogen gas.
The blue-green solution is consistent with the presence of dissolved copper(II) species.
Sample B provides the more direct comparison. Its polished copper surface shows no observable reaction with the dilute hydrochloric acid.
The disappearance of the black coating is evidence of an acid-oxide reaction, not evidence that copper metal can displace hydrogen.
Question 4
A student places equal masses of clean copper turnings into two separate test tubes.
- Test tube A contains dilute hydrochloric acid.
- Test tube B contains concentrated nitric acid.
The student observes no noticeable reaction in test tube A. In test tube B, the copper gradually dissolves, a blue solution forms, and a brown gas is released.
Explain the different observations.
In your response:
(a) Identify the oxidising species responsible for the reaction in test tube B.
(b) Write the balanced overall equation for copper reacting with concentrated nitric acid.
(c) Explain why the reaction in test tube B does not contradict the activity series.
(d) Explain why the difference cannot be attributed simply to the strength of the acids.
Solution 4
Copper reacts with concentrated nitric acid because nitrate ions act as oxidising agents, whereas hydrogen ions in dilute hydrochloric acid cannot normally oxidise copper.
(a) The nitrate ions, \(\ce{NO3-}\), are responsible for accepting electrons in the nitric acid reaction.
Copper atoms lose electrons to form copper(II) ions:
\[ \ce{Cu(s) -> Cu^2+(aq) + 2e-} \]
Nitrogen in nitrate is reduced from an oxidation state of +5 to +4 in nitrogen dioxide.
(b) The balanced equation is:
\[ \ce{Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)} \]
(c) The activity series predicts whether a metal can displace hydrogen from a non-oxidising acid.
It does not mean that copper can never be oxidised by another oxidising agent.
In nitric acid, nitrate ions provide an alternative pathway for accepting electrons. Copper can therefore dissolve without hydrogen ions being reduced to hydrogen gas.
(d) Both hydrochloric acid and nitric acid are strong acids in water.
However, acid strength and oxidising strength describe different chemical properties.
Hydrochloric acid supplies hydrogen ions and chloride ions. Under ordinary dilute conditions, neither provides a favourable route for oxidising copper.
Nitric acid supplies nitrate ions, which can accept electrons and drive the oxidation of copper.
Therefore, the difference is explained by the identity of the oxidising species, not simply by how strongly the acids ionise.
Question 5
A research group investigates whether dissolved oxygen can explain the corrosion of copper in hydrochloric acid.
Two sealed flasks are prepared. Each contains:
- \(50.0\ \mathrm{mL}\) of \(0.200\ \mathrm{mol\,L^{-1}}\) hydrochloric acid
- A clean copper foil sample with an initial mass of \(0.127\ \mathrm{g}\)
The conditions differ as follows:
- Flask A contains deoxygenated hydrochloric acid and negligible oxygen in its sealed headspace.
- Flask B contains air in a \(250\ \mathrm{mL}\) headspace at \(25^\circ\mathrm{C}\) and \(100\ \mathrm{kPa}\).
After 48 hours, the following observations are recorded.
| Observation | Flask A | Flask B |
|---|---|---|
| Copper mass loss | Below measurement detection | 0.127 g |
| Dissolved copper ions | No measurable increase | Present |
| Hydrogen collected | No detectable hydrogen | No detectable hydrogen |
The following information is available:
| Quantity | Value |
|---|---|
| Molar mass of copper | \(63.55\ \mathrm{g\,mol^{-1}}\) |
| Molar gas volume at experimental conditions | \(24.8\ \mathrm{L\,mol^{-1}}\) |
| Oxygen fraction in air | 21% |
| \(E^\circ_{\ce{Cu^2+/Cu}}\) | +0.34 V |
| \(E^\circ_{\ce{2H+/H2}}\) | 0.00 V |
| \(E^\circ_{\ce{O2/H2O}}\) | +1.23 V |
Assume:
- All copper lost from Flask B enters solution as copper(II) ions.
- Oxygen is the only oxidising agent responsible for the copper loss.
- Oxygen is reduced to water.
- Chloride complex formation does not change the electron balance.
- All oxygen initially present in the headspace is available for the reaction.
- No other reactions occur.
Answer the following:
(a) Write the balanced net ionic equation for the oxygen-driven corrosion of copper.
(b) Calculate the initial amount of oxygen in Flask B. Then calculate the amount of oxygen required to account for the observed copper loss. Determine whether the available oxygen is sufficient.
(c) Calculate the amount of hydrogen ions initially present in the hydrochloric acid. Is there enough acid to support the oxygen-driven reaction?
(d) If the copper had dissolved through the proposed direct reaction with hydrogen ions, calculate the theoretical volume of hydrogen that would have formed. Explain why the absence of hydrogen is important.
(e) Calculate the standard cell potential for the oxygen-driven reaction and compare it with the standard cell potential for the proposed copper-hydrogen reaction.
Solution 5
The copper loss in Flask B is consistent with oxygen-driven corrosion, not the direct displacement of hydrogen from hydrochloric acid.
The key evidence is that copper dissolves in the oxygen-containing flask without detectable hydrogen production, while the deoxygenated flask shows negligible copper loss.
We can test whether the oxygen supply is sufficient to account for the observations.
(a) Write the net ionic equation.
Copper is oxidised:
\[ \ce{Cu(s) -> Cu^2+(aq) + 2e-} \]
Oxygen is reduced:
\[ \ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)} \]
Multiply the copper oxidation half-equation by two:
\[ \ce{2Cu(s) -> 2Cu^2+(aq) + 4e-} \]
Add the half-equations:
\[ \boxed{ \ce{2Cu(s) + O2(g) + 4H+(aq) -> 2Cu^2+(aq) + 2H2O(l)} } \]
Oxygen is the oxidising agent because it accepts electrons.
Hydrogen ions are consumed in the reaction, but they are not reduced to hydrogen gas.
(b) Calculate the oxygen available and required.
First, calculate the initial amount of air in the headspace.
\[ \begin{aligned} n(\text{air}) &=\frac{V}{V_m}\\ &=\frac{0.250}{24.8}\\ &=0.01008\ \mathrm{mol} \end{aligned} \]
Since air contains 21% oxygen:
\[ \begin{aligned} n(\ce{O2})_{\mathrm{available}} &=0.21\times0.01008\\ &=0.002117\ \mathrm{mol} \end{aligned} \]
Therefore, the initial oxygen supply is:
\[ \boxed{n(\ce{O2})=2.12\times10^{-3}\ \mathrm{mol}} \]
Now calculate the amount of copper lost.
\[ \begin{aligned} n(\ce{Cu}) &=\frac{m}{M}\\ &=\frac{0.127}{63.55}\\ &=0.001998\ \mathrm{mol} \end{aligned} \]
From the balanced ionic equation:
\[ \ce{2Cu:O2}=2:1 \]
Therefore:
\[ \begin{aligned} n(\ce{O2})_{\mathrm{required}} &=\frac{0.001998}{2}\\ &=0.000999\ \mathrm{mol} \end{aligned} \]
Compare the amounts:
\[ n(\ce{O2})_{\mathrm{available}} > n(\ce{O2})_{\mathrm{required}} \]
There is approximately \(0.002117\ \mathrm{mol}\) of oxygen available, while only \(0.000999\ \mathrm{mol}\) is required.
Therefore, the initial oxygen supply is sufficient to account for all the observed copper loss under the stated assumptions.
(c) Calculate the amount of hydrogen ions available.
Use:
\[ n=cV \]
Convert the acid volume into litres:
\[ 50.0\ \mathrm{mL}=0.0500\ \mathrm{L} \]
Therefore:
\[ \begin{aligned} n(\mathrm{H}^{+}) &=0.200\times0.0500\\ &=0.0100\ \mathrm{mol} \end{aligned} \]
The oxygen-driven reaction requires four moles of hydrogen ions for every two moles of copper.
Therefore:
\[ \begin{aligned} n(\mathrm{H}^{+})_{\mathrm{required}} &=2n(\ce{Cu})\\ &=2(0.001998)\\ &=0.003997\ \mathrm{mol} \end{aligned} \]
The solution initially contains \(0.0100\ \mathrm{mol}\) of hydrogen ions.
This exceeds the required amount, so hydrochloric acid is not the limiting reactant.
(d) Calculate the theoretical hydrogen volume if direct displacement had occurred.
The proposed direct reaction would be:
\[ \ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)} \]
The mole ratio between copper and hydrogen is 1:1.
Therefore:
\[ n(\ce{H2})=0.001998\ \mathrm{mol} \]
Use the given molar gas volume:
\[ \begin{aligned} V(\ce{H2}) &=nV_m\\ &=0.001998\times24.8\\ &=0.04956\ \mathrm{L} \end{aligned} \]
Convert to millilitres:
\[ \boxed{V(\ce{H2})=49.6\ \mathrm{mL}} \]
If the observed copper loss had occurred through direct hydrogen displacement, approximately \(49.6\ \mathrm{mL}\) of hydrogen should have been produced under the stated assumptions.
However, no hydrogen was detected.
Combined with the available oxygen and the balanced oxygen-reduction pathway, this supports oxygen-driven corrosion instead.
The tempting conclusion is that any loss of copper in hydrochloric acid proves copper reacts with hydrogen ions. The missing constraint is the identity of the electron acceptor. Copper can be oxidised by oxygen even when hydrogen ions cannot drive the proposed displacement reaction.
(e) Compare the standard cell potentials.
For oxygen-driven corrosion, oxygen is reduced at the cathode, while copper is oxidised at the anode.
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\ce{O2/H2O}} -E^\circ_{\ce{Cu^2+/Cu}}\\ &=+1.23-(+0.34)\\ &=\boxed{+0.89\ \mathrm{V}} \end{aligned} \]
The positive value indicates that the oxygen-driven reaction is thermodynamically favourable under standard conditions.
For direct copper oxidation by hydrogen ions:
\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\ce{H+/H2}} -E^\circ_{\ce{Cu^2+/Cu}}\\ &=0.00-(+0.34)\\ &=\boxed{-0.34\ \mathrm{V}} \end{aligned} \]
The negative value indicates that the direct hydrogen-producing reaction is thermodynamically unfavourable under standard conditions.
The two pathways therefore have different thermodynamic predictions:
| Proposed process | Standard cell potential | Prediction |
|---|---|---|
| Copper oxidised by hydrogen ions | -0.34 V | Unfavourable |
| Copper oxidised by oxygen in acidic solution | +0.89 V | Favourable |
The observations are consistent with the second pathway.
This is the crucial distinction: copper can undergo oxidation in an acidic environment without reacting directly with the acid’s hydrogen ions.
10Where to take this next
The copper-hydrochloric acid example connects directly to galvanic cells and electrochemical series.
In a galvanic cell, the same standard reduction potentials help predict which species will be oxidised, which will be reduced, and the direction electrons will flow through an external circuit.
The next useful step is learning how to combine two half-cells, calculate their cell potential, and use its sign to predict the spontaneous direction of a redox reaction.
NSW syllabus context: This explanation draws on the Chemistry Stage 6 Syllabus (2017), particularly Year 11 Module 3, which includes metal reactivity, redox reactions, standard reduction potentials, and predicting reaction spontaneity. NESA schedules implementation of the Chemistry 11-12 Syllabus (2025) from 2028, with its first HSC examination in 2029. This page does not claim that any particular detail is a specified requirement of the 2025 syllabus. See NESA’s current Chemistry syllabus information.