How to Distinguish Oxidation from Adding Oxygen

A clean zinc strip placed in copper(II) sulfate solution develops a reddish-brown coating. The blue solution gradually fades, and the zinc starts dissolving.

Here’s the interesting part: no oxygen gas is added. Yet zinc is being oxidised.

This exposes a common mistake in Chemistry. Oxidation does not simply mean adding oxygen. In a general chemical reaction, oxidation means losing electrons, which corresponds to an increase in oxidation number. Adding oxygen is one way oxidation can occur, but it is not the definition.

To distinguish the two reliably, track the oxidation numbers of the elements before and after the reaction, then check that the electrons lost by one species are gained by another. You must also conserve every atom and the total electrical charge.

01Why adding oxygen is not a complete definition

The oxygen-based description of oxidation makes sense when you first look at reactions such as magnesium burning in oxygen.

\[ \ce{2Mg(s) + O2(g) -> 2MgO(s)} \]

Magnesium combines with oxygen to form magnesium oxide. Oxygen has been added to magnesium, so the reaction fits the traditional description of oxidation.

But something else is happening at the atomic level.

• Magnesium starts with an oxidation number of 0 and finishes at +2.
• Oxygen starts at 0 and finishes at -2.

Magnesium loses electrons, while oxygen gains them.

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

\[ \ce{O2 + 4e^- -> 2O^2-} \]

Magnesium is oxidised because it loses electrons. Oxygen is reduced because it gains electrons.

This is where the oxygen-based definition becomes incomplete.

Consider the zinc and copper(II) sulfate reaction from the opening example:

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

There is no oxygen in the net ionic equation.

Nevertheless, zinc loses electrons and becomes positively charged zinc ions. Copper ions gain those electrons and become copper metal.

Zinc has been oxidised without gaining oxygen.

The more general definition therefore focuses on electron transfer rather than oxygen transfer.

A useful memory aid is OIL RIG:

• Oxidation Is Loss (of electrons).
• Reduction Is Gain (of electrons).

02Following the experiment: from oxygen to electron transfer

The difference becomes clearer when you follow how the reaction can be investigated.

First experiment: zinc in copper(II) sulfate

Imagine a clean zinc strip suspended in a beaker containing blue aqueous copper(II) sulfate.

For a controlled demonstration, the solution can be deoxygenated so that oxygen gas is not involved in the reaction.

Before starting, make a prediction.

If oxidation requires the addition of oxygen, what should happen?

The oxygen-based model predicts that zinc cannot be oxidised because no oxygen is being supplied.

The actual observations challenge that prediction:

• A reddish-brown copper coating forms on the zinc.
• The zinc strip gradually loses material.
• The blue colour of the solution becomes less intense as copper(II) ions are consumed.

The chemical equation explains the observations:

\[ \ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)} \]

Sulfate ions do not participate in the actual electron-transfer reaction. Removing these spectator ions gives the net ionic equation:

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

Zinc atoms enter the solution as zinc ions. Copper ions leave the solution and form solid copper.

The electron-transfer explanation is:

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

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

Zinc loses two electrons. Copper(II) ions gain those same two electrons.

The total electron loss and gain are equal.

This is direct evidence that oxidation can occur without oxygen being added to the substance.

What the first experiment cannot show

The beaker experiment establishes that a redox reaction has occurred. However, it does not provide a separate pathway for observing electron flow.

Zinc and copper ions interact at the same metal-solution interface. You can infer electron transfer from the chemical changes, but you cannot use this arrangement to trace electrons through an external circuit.

A galvanic cell provides a better experimental design.

Revised experiment: separate the two half-reactions

Instead of placing zinc directly into copper sulfate, use two separate beakers.

• The first contains zinc sulfate solution, with a zinc electrode immersed in it.
• The second contains copper(II) sulfate solution, with a copper electrode immersed in it.
• A conducting wire connects the two electrodes.
• A salt bridge connects the solutions and allows ions to move between the half-cells.

An ammeter can be included in the external circuit to measure the current.

The two reactions can now occur separately.

At the zinc electrode:

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

At the copper electrode:

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

Electrons travel through the external wire from zinc to copper.

The salt bridge serves a different purpose. It allows ions to move so that electrical charge does not build up excessively in either solution.

The experiment makes the distinction between the two processes clear:

• Oxidation occurs at the zinc electrode, where electrons are released.
• Reduction occurs at the copper electrode, where electrons are accepted.
• The external wire carries electrons.
• The salt bridge carries ions, not electrons.

The revised design supports the electron-transfer model and explains how oxidation and reduction can occur in separate locations.

A small calculation: checking the electron transfer

Suppose a zinc electrode loses 0.0327 g of zinc.

The molar mass of zinc is 65.38 g mol\(^{-1}\).

The amount of zinc consumed is:

\[ \begin{aligned} n(\ce{Zn}) &= \frac{m}{M}\\ &= \frac{0.0327}{65.38}\\ &= 5.00\times10^{-4}\text{ mol} \end{aligned} \]

Each zinc atom releases two electrons.

Therefore:

\[ \begin{aligned} n(e^-) &= 2n(\ce{Zn})\\ &= 2(5.00\times10^{-4})\\ &= 1.00\times10^{-3}\text{ mol} \end{aligned} \]

Copper(II) ions also require two electrons each.

Therefore, \(1.00\times10^{-3}\) mol of copper can be deposited.

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

\[ m(\ce{Cu})=nM \]

\[ =1.00\times10^{-3}\times63.55 \]

\[ =0.0636\text{ g} \]

This is an illustrative calculation, not a reported experimental measurement. It assumes the reaction is complete and that no competing reactions occur.

The calculation demonstrates an important constraint: the electrons released by zinc must be accounted for by the reduction occurring at the other electrode.

03The reliable method: identify oxidation in any reaction

When an exam question gives you an unfamiliar chemical equation, don’t start by counting oxygen atoms.

Use the following method instead.

Step 1: Identify the reaction conditions

Check whether the reaction occurs in acidic, basic, neutral, aqueous, molten or gaseous conditions.

The medium can determine which products form and which species are available for balancing the equation.

For example:

• Acidic conditions allow \(\mathrm{H}^{+}\) and \(\ce{H2O}\) to participate in half-equations.
• Basic conditions allow \(\mathrm{OH}^{-}\) and \(\ce{H2O}\) to participate.
• Neutral aqueous conditions do not automatically justify adding \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\).

Never assume the reaction occurs in acidic solution simply because you are familiar with an acidic redox equation.

Step 2: Check that atoms and charge are conserved

Before identifying oxidation, make sure the equation is balanced.

Every element must have the same number of atoms on both sides.

For an ionic equation, the total electrical charge must also be equal on both sides.

For example:

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

Atom check:

• One zinc atom on each side.
• One copper atom on each side.

Charge check:

• Left side: \(+2\).
• Right side: \(+2\).

Both atoms and charge are conserved.

If an equation is unbalanced, don’t use its coefficients to calculate electron transfer.

Step 3: Assign oxidation numbers

Find the oxidation number of each element before and after the reaction.

The most useful rules are:

The sum of oxidation numbers must equal the overall charge of the substance.

For a neutral compound, the sum is zero.

For a polyatomic ion, the sum equals the ion’s charge.

If you need more practice with assigning oxidation numbers, see Oxidation numbers in redox reactions: HSC Chemistry Guide.

Step 4: Compare oxidation numbers

This is the decisive step.

Track the same element from reactants to products.

• If its oxidation number increases, it has been oxidised.
• If its oxidation number decreases, it has been reduced.
• If its oxidation number stays the same, that element has not undergone oxidation or reduction.

For example:

\[ \ce{Fe^2+ -> Fe^3+} \]

The oxidation number increases from +2 to +3.

Iron has lost one electron and has been oxidised.

By contrast:

\[ \ce{Fe^3+ -> Fe^2+} \]

The oxidation number decreases from +3 to +2.

Iron has gained one electron and has been reduced.

Step 5: Check electron balance

If one species loses two electrons, another species must gain two electrons, or an equivalent total amount.

Write the two half-equations to make the transfer clear.

For example:

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

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

The electrons cancel when the half-equations are added.

This provides a final check that the proposed redox interpretation is consistent.

Remember that oxidation numbers in covalent molecules are formal bookkeeping values. They do not always represent whole electrons physically moving between individual atoms. The oxidation-number method still provides a consistent way to identify oxidation and reduction.

04When adding oxygen gives the wrong prediction

There is another misconception worth addressing.

A student might see a reaction in which a molecule gains an oxygen atom and immediately conclude that its central element has been oxidised.

That conclusion is not always correct.

Consider the formation of carbonic acid:

\[ \ce{CO2(aq) + H2O(l) <=> H2CO3(aq)} \]

The product contains three oxygen atoms, compared with two in carbon dioxide.

A simple oxygen-counting approach might suggest that carbon has been oxidised.

But calculate the oxidation number of carbon.

In carbon dioxide:

\[ x+2(-2)=0 \]

\[ x=+4 \]

In carbonic acid:

\[ x+2(+1)+3(-2)=0 \]

\[ x=+4 \]

Carbon remains at +4.

Oxygen remains at -2, and hydrogen remains at +1.

No oxidation numbers change, so this is not a redox reaction.

The reaction is a hydration equilibrium. Water contributes an oxygen atom to the molecular formula, but carbon does not lose electrons.

This is why you must distinguish between two different ideas:

1. A substance gaining oxygen atoms.
2. An element increasing its oxidation number through electron loss.

They can occur together, but they are not interchangeable definitions.

Three reactions compared

Notice that oxygen addition is neither necessary for oxidation nor sufficient evidence of oxidation.

For more examples of reactions involving oxygen, see Metal Reactivity with Oxygen: HSC Chemistry Guide.

05Questions and solutions

The following questions progress from identifying straightforward electron transfer to analysing reactions where oxygen transfer, reaction conditions and electrical charge can lead to misleading conclusions.

Question 1

Magnesium reacts with dilute hydrochloric acid at room temperature.

The net ionic equation is:

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

Identify the species that is oxidised and the species that is reduced. Write the relevant half-equations and explain why oxidation occurs even though oxygen is not added.

Solution 1

Magnesium is oxidised, and hydrogen ions are reduced. Magnesium loses two electrons, which are gained by two hydrogen ions.

Start by comparing oxidation numbers.

Magnesium:

\[ 0\rightarrow +2 \]

Hydrogen:

\[ +1\rightarrow 0 \]

Magnesium’s oxidation number increases, so it undergoes oxidation.

Hydrogen’s oxidation number decreases, so it undergoes reduction.

The oxidation half-equation is:

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

The reduction half-equation is:

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

Now check the conservation requirements.

There is one magnesium atom and two hydrogen atoms on each side.

The total charge on the left is:

\[ 2(+1)=+2 \]

The total charge on the right is:

\[ +2 \]

Both atoms and charge are conserved.

Two electrons are lost by magnesium and gained by hydrogen ions.

This reaction is oxidation even though no oxygen atoms are added. The correct identification comes from electron transfer, not oxygen counting.

Question 2

Carbon dioxide dissolves in water and forms carbonic acid:

\[ \ce{CO2(aq) + H2O(l) <=> H2CO3(aq)} \]

A student claims that carbon must be oxidised because the product contains more oxygen atoms than carbon dioxide.

Evaluate this claim.

Determine the oxidation numbers of carbon, oxygen and hydrogen on both sides, and explain whether the reaction is a redox reaction.

Solution 2

The student is incorrect. This reaction is not a redox reaction because none of the elements changes oxidation number.

Start with carbon dioxide.

Oxygen has an oxidation number of -2, and the molecule is neutral.

Therefore:

\[ x+2(-2)=0 \]

\[ x=+4 \]

Carbon has an oxidation number of +4.

Now consider carbonic acid:

\[ \ce{H2CO3} \]

Hydrogen has an oxidation number of +1, and oxygen has an oxidation number of -2.

Let carbon’s oxidation number be \(x\):

\[ x+2(+1)+3(-2)=0 \]

\[ x+2-6=0 \]

\[ x=+4 \]

Carbon is still +4.

The oxidation numbers of all three elements are unchanged:

The reaction also conserves atoms.

• Carbon: 1 on each side.
• Hydrogen: 2 on each side.
• Oxygen: 3 on each side.

Both sides are electrically neutral.

The extra oxygen atom in the product comes from water. It does not cause carbon to lose electrons.

The important distinction is that a change in molecular composition does not automatically indicate a change in oxidation number.

Question 3

Aluminium reacts with excess aqueous copper(II) ions at room temperature:

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

A sample contains 0.0120 mol of aluminium, which reacts completely.

Calculate:

(a) The amount, in moles, of electrons transferred.

(b) The amount, in moles, of copper deposited.

(c) The mass of copper deposited.

Use a molar mass of 63.55 g mol\(^{-1}\) for copper.

Explain why the reaction is oxidation even though no oxygen is added.

Solution 3

The reaction transfers 0.0360 mol of electrons and deposits 0.0180 mol, or 1.14 g, of copper. Aluminium is oxidised because it loses electrons.

First, identify the oxidation-state changes.

Aluminium:

\[ 0\rightarrow +3 \]

Each aluminium atom loses three electrons.

Copper:

\[ +2\rightarrow 0 \]

Each copper(II) ion gains two electrons.

(a) Calculate the amount of electrons transferred.

The oxidation half-equation is:

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

Therefore:

\[ n(e^-)=3n(\ce{Al}) \]

Substitute the given amount:

\[ \begin{aligned} n(e^-)&=3(0.0120)\\ &=0.0360\text{ mol} \end{aligned} \]

The total amount of electrons transferred is 0.0360 mol.

(b) Calculate the amount of copper deposited.

The reduction half-equation is:

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

Two electrons are required to produce one copper atom.

Therefore:

\[ n(\ce{Cu})=\frac{n(e^-)}{2} \]

\[ =\frac{0.0360}{2} \]

\[ =0.0180\text{ mol} \]

(c) Calculate the mass of copper deposited.

Use:

\[ m=nM \]

Where:

• \(m\) is mass in grams.
• \(n\) is amount in moles.
• \(M\) is molar mass in g mol\(^{-1}\).

Substitute:

\[ \begin{aligned} m(\ce{Cu})&=0.0180\times63.55\\ &=1.1439\text{ g}\\ &=1.14\text{ g} \end{aligned} \]

The final mass is 1.14 g of copper.

The electron balance is also consistent with the overall equation.

Two aluminium atoms lose six electrons in total, while three copper(II) ions gain six electrons.

No oxygen is required. Aluminium undergoes oxidation through electron loss, while copper ions undergo reduction through electron gain.

Question 4

In neutral-to-alkaline aqueous solution, permanganate ions react with sulfite ions to form manganese dioxide and sulfate ions.

The balanced equation is:

\[ \ce{2MnO4^- + 3SO3^2- + H2O -> 2MnO2(s) + 3SO4^2- + 2OH^-} \]

(a) Identify the element that is oxidised and the element that is reduced.

(b) Calculate the total number of electrons lost and gained per reaction as written.

(c) A student argues that sulfur must be reduced because it gains oxygen when sulfite becomes sulfate. Explain why this conclusion is incorrect.

(d) Verify that the equation conserves atoms and charge. Explain why the reaction conditions matter.

Solution 4

Sulfur is oxidised, and manganese is reduced. Six electrons are transferred in each direction per reaction as written. Sulfur gains oxygen atoms, but its oxidation number increases, meaning it loses electrons.

(a) Identify oxidation and reduction.

Start with sulfite, \(\mathrm{SO_{3}}^{2-}\).

Let the oxidation number of sulfur be \(x\):

\[ x+3(-2)=-2 \]

\[ x=+4 \]

In sulfate, \(\mathrm{SO_{4}}^{2-}\):

\[ x+4(-2)=-2 \]

\[ x=+6 \]

Sulfur changes from +4 to +6.

Its oxidation number increases, so sulfur is oxidised.

Now examine manganese.

In permanganate, \(\mathrm{MnO_{4}}^{-}\):

\[ x+4(-2)=-1 \]

\[ x=+7 \]

In manganese dioxide, \(\ce{MnO2}\):

\[ x+2(-2)=0 \]

\[ x=+4 \]

Manganese changes from +7 to +4.

Its oxidation number decreases, so manganese is reduced.

(b) Calculate the electron transfer.

Each sulfur atom changes from +4 to +6.

It loses two electrons.

There are three sulfur atoms:

\[ 3\times2=6e^- \]

Each manganese atom changes from +7 to +4.

It gains three electrons.

There are two manganese atoms:

\[ 2\times3=6e^- \]

Therefore:

\[ \text{Electrons lost}=\text{Electrons gained}=6 \]

The electron transfer is balanced.

(c) Explain the oxygen-addition misconception.

The student has confused gaining oxygen atoms with gaining electrons.

Sulfite becomes sulfate:

\[ \ce{SO3^2- -> SO4^2-} \]

Sulfur gains an oxygen atom, but its oxidation number increases from +4 to +6.

An increase in oxidation number indicates electron loss.

Therefore, sulfur is oxidised, not reduced.

This is a particularly useful example because oxygen addition and oxidation occur together, but electron transfer is what establishes the correct classification.

(d) Check atom and charge conservation.

Count the atoms on each side.

Now check total charge.

Reactants:

\[ 2(-1)+3(-2)=-8 \]

Products:

\[ 3(-2)+2(-1)=-8 \]

The total charge is conserved.

The reaction conditions also matter.

In neutral-to-alkaline solution, manganese is reduced to manganese dioxide, and hydroxide ions appear in the balanced equation.

Under different conditions, permanganate can produce different reduction products. You cannot automatically use the same equation for an acidic reaction.

The stated medium determines which species and products are appropriate.

Question 5

A student investigates a galvanic cell containing two separate half-cells.

• A zinc electrode is immersed in aqueous zinc nitrate.
• A silver electrode is immersed in aqueous silver nitrate.
• The electrodes are connected by an external wire containing an ammeter.
• The solutions are connected by a potassium nitrate salt bridge.

The solutions are deoxygenated, and the experiment is conducted at room temperature.

Assume there are no competing reactions. All zinc lost from the electrode enters solution as \(\mathrm{Zn}^{2+}\), and all silver deposited comes from \(\mathrm{Ag}^{+}\) ions.

After the reaction, the following illustrative measurements are obtained:

Use the following molar masses:

• Zinc: 65.38 g mol\(^{-1}\).
• Silver: 107.87 g mol\(^{-1}\).

Answer the following:

(a) Write the oxidation and reduction half-equations and the overall net ionic equation. Identify the anode and cathode.

(b) Calculate the amount of electrons released by zinc.

(c) Use the silver electrode mass change to independently calculate the amount of electrons gained by silver ions. Determine whether the two measurements are consistent with the proposed electron-transfer reaction.

(d) A student argues that zinc is oxidised because it combines with oxygen from the nitrate ions. Another argues that silver is oxidised because the silver electrode gains mass.

Evaluate both explanations.

(e) Predict what happens if the salt bridge is removed while the external wire remains connected. Explain the charge changes in both half-cells and why the reaction cannot continue at its original rate.

Solution 5

Zinc is oxidised at the anode, and silver ions are reduced at the cathode. The two electrode mass changes independently support a transfer of approximately 0.00200 mol of electrons. Neither reaction requires oxygen transfer.

(a) Write the half-equations and overall reaction.

Zinc changes from oxidation number 0 to +2.

It loses two electrons:

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

This is oxidation, so the zinc electrode is the anode.

Silver changes from oxidation number +1 to 0.

It gains one electron per silver ion:

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

Multiply the reduction half-equation by two:

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

This is reduction, so the silver electrode is the cathode.

Add the half-equations and cancel the electrons:

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

The equation conserves atoms and charge.

There is one zinc atom and two silver atoms on each side.

The total charge is +2 on both sides.

(b) Calculate the electrons released by zinc.

The zinc electrode loses 0.0654 g.

Use:

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

Substitute:

\[ \begin{aligned} n(\ce{Zn})&=\frac{0.0654}{65.38}\\ &=1.00\times10^{-3}\text{ mol} \end{aligned} \]

Each zinc atom releases two electrons.

Therefore:

\[ \begin{aligned} n(e^-)&=2n(\ce{Zn})\\ &=2(1.00\times10^{-3})\\ &=2.00\times10^{-3}\text{ mol} \end{aligned} \]

The zinc electrode releases approximately 0.00200 mol of electrons.

(c) Calculate the electrons gained by silver ions.

The silver electrode gains 0.2157 g.

Calculate the amount of silver deposited:

\[ \begin{aligned} n(\ce{Ag})&=\frac{0.2157}{107.87}\\ &=2.00\times10^{-3}\text{ mol} \end{aligned} \]

Each silver ion gains one electron:

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

Therefore:

\[ n(e^-)=2.00\times10^{-3}\text{ mol} \]

Compare the two results:

The values agree within the precision of the supplied data.

This independently supports the proposed reaction stoichiometry.

The mass changes also make physical sense:

• Zinc atoms leave the solid electrode and enter the solution as ions, decreasing its mass.
• Silver ions gain electrons and form solid silver on the electrode, increasing its mass.

(d) Evaluate the two explanations.

Both explanations are incorrect.

The zinc electrode is not oxidised because it combines with oxygen from nitrate.

Nitrate ions are spectator ions in this reaction. They remain as \(\mathrm{NO_{3}}^{-}\), and their oxidation numbers do not change.

The net ionic equation contains no nitrate ions because they cancel when the complete ionic equation is simplified.

Zinc is oxidised because it loses electrons:

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

The second explanation confuses electrode mass gain with oxidation.

The silver electrode gains mass because silver ions are reduced to solid silver:

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

The silver atoms being deposited have gained electrons. They have been reduced, not oxidised.

An electrode’s mass change does not, by itself, tell you whether oxidation or reduction is occurring. You must identify the chemical species and track its electron transfer.

(e) Predict what happens when the salt bridge is removed.

The current rapidly decreases because charge begins accumulating in the two half-cells.

At the zinc electrode, oxidation continues to produce zinc ions:

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

The zinc solution becomes increasingly positively charged.

At the silver electrode, silver ions are consumed:

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

This leaves excess negative charge in the silver solution because nitrate ions remain.

Normally, the salt bridge prevents this charge separation from becoming excessive.

• Nitrate ions migrate towards the zinc half-cell to balance the increasing positive charge.
• Potassium ions migrate towards the silver half-cell to balance the excess negative charge.

When the salt bridge is removed, this ion migration is interrupted.

Charge separation creates an electrical potential that opposes further electron transfer through the external wire. A brief transient current may occur, but the current rapidly falls, and the reaction cannot continue at its original rate.

This final observation reinforces the distinction between electron transfer and ion movement. Electrons travel through the external circuit, while ions move through the salt bridge to maintain electrical neutrality.

06The key distinction to remember

When an exam question asks whether a substance has been oxidised, don’t rely on whether oxygen appears in the equation or whether the product contains more oxygen atoms.

Use the oxidation-number changes.

The most reliable approach is to conserve atoms and charge, identify the oxidation-state changes, and verify that the total electrons lost equal the total electrons gained.

Once you can do this confidently, the next step is balancing redox half-equations in the correct reaction medium. That is how you turn an identification of oxidation and reduction into a complete, balanced chemical equation.