How to Identify the Oxidant and Reductant in a Reaction

A clean strip of zinc is placed into blue copper(II) sulfate solution. A reddish-brown coating begins to form on the zinc, and the blue solution becomes paler.

Which reactant is the oxidant? Is it zinc, because the zinc strip appears to be reacting, or copper(II) ions, because they are being removed from solution?

The answer is copper(II) ions, \(\mathrm{Cu}^{2+}\). They accept electrons and are reduced. Zinc is the reductant because it donates electrons and is oxidised.

The reliable way to identify either substance is to compare oxidation states before and after the reaction. You don’t need to guess from the chemical formula, the colour change, or the charge of an ion.

01What the zinc and copper reaction actually tells us

Start with the apparatus. A beaker contains aqueous copper(II) sulfate, and a zinc strip is immersed in the solution.

Assume room-temperature conditions, with no added acid or alkali. The sulfate ions do not participate in the redox reaction.

Before looking at the chemical explanation, predict what should happen if zinc atoms donate electrons to copper(II) ions.

Two changes must occur:

• Zinc atoms lose electrons and enter the solution as \(\mathrm{Zn}^{2+}\) ions.
• Copper(II) ions gain electrons and become solid copper.

The balanced molecular equation is:

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

Removing the spectator sulfate ions gives the net ionic equation:

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

Now compare the oxidation states.

Zinc increases from 0 to +2. It loses two electrons.

Copper decreases from +2 to 0. It gains two electrons.

The corresponding half-equations are:

Oxidation:

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

Reduction:

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

The two electrons lost by zinc are exactly the two electrons gained by copper(II) ions.

This gives us the answer:

• Zinc is the reducing agent (reductant).
• Copper(II) ions are the oxidising agent (oxidant).

Does the experimental evidence support this?

Yes. Three observations support the proposed reaction.

1. A reddish-brown solid appears on the zinc. This is deposited copper metal.
2. The blue solution becomes paler as copper(II) ions are removed.
3. The zinc strip gradually dissolves as zinc atoms enter the solution.

The observations are consistent with the oxidation-state changes. However, the colour change alone does not directly demonstrate electrons moving between particles. That is an interpretation based on the reaction model.

We can also make a small quantitative prediction about the zinc strip.

Suppose \(0.0100\text{ mol}\) of zinc reacts completely, and all deposited copper remains attached to the strip.

Using:

• \(M(\ce{Zn})=65.38\text{ g mol}^{-1}\)
• \(M(\ce{Cu})=63.55\text{ g mol}^{-1}\)

The mass of zinc removed is:

\[ m(\ce{Zn})=0.0100\times65.38 =0.654\text{ g} \]

The mass of copper deposited is:

\[ m(\ce{Cu})=0.0100\times63.55 =0.636\text{ g} \]

Therefore:

\[ \Delta m=0.636-0.654 \]

\[ \boxed{\Delta m=-0.0183\text{ g}} \]

The strip becomes slightly lighter, despite gaining a copper coating.

This is a useful reminder that seeing a solid deposit does not automatically mean the object’s total mass increases.

02Separating the half-reactions: a better way to trace electron movement

The zinc and copper experiment has one limitation.

Both oxidation and reduction happen at the surface of the zinc strip. We can infer electron transfer from the chemical changes, but there is no separate external path through which to measure the electron flow.

A Daniell cell improves the design by separating the two half-reactions.

The apparatus contains:

• A zinc electrode immersed in aqueous zinc sulfate.
• A copper electrode immersed in aqueous copper(II) sulfate.
• An external wire connecting the two electrodes, with a voltmeter.
• A salt bridge containing an electrolyte such as potassium nitrate, connecting the two solutions.

The zinc electrode releases electrons:

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

These electrons travel through the external wire towards the copper electrode.

At the copper electrode, copper(II) ions accept the electrons:

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

The salt bridge has a different job. It allows ions to migrate between the half-cells, preventing charge from building up in either solution.

• Nitrate ions migrate towards the zinc half-cell, where positive zinc ions accumulate.
• Potassium ions migrate towards the copper half-cell, where copper(II) ions are being consumed.

Electrons travel through the external wire, not through the salt bridge.

The revised apparatus makes the electron pathway easier to investigate. The voltmeter measures the potential difference between the electrodes, while the electrode mass changes and solution composition provide further evidence of the reaction.

The chemical conclusion is the same: zinc supplies electrons, and copper(II) ions accept them.

03The reliable method: identify the oxidation-state changes

You can now turn the experiment into a method that works for unfamiliar HSC Chemistry reactions.

The key is to distinguish the substance that gives away electrons from the substance that accepts them.

Remember these two relationships:

Oxidant = electron acceptor = reduced.

Reductant = electron donor = oxidised.

The names can feel backwards at first. The oxidant causes another substance to be oxidised, but the oxidant itself is reduced.

Likewise, the reductant causes another substance to be reduced, but the reductant itself is oxidised.

Step 1: Identify the reaction conditions

Before assigning oxidation states, establish the reaction medium.

Is the reaction occurring in:

• Acidic solution?
• Alkaline solution?
• Neutral aqueous solution?

This matters because the medium can affect the products formed.

For example, permanganate ions can be reduced to \(\mathrm{Mn}^{2+}\) in acidic solution, whereas different manganese products can form under other conditions.

Do not assume the products or oxidation-state changes without considering the conditions supplied.

Step 2: Assign oxidation states

Work out the oxidation states of the relevant elements in the reactants and products.

You do not necessarily need to calculate every element. Focus on elements whose oxidation states might change.

Useful rules include:

• An element in its uncombined form has oxidation state 0.
• A monatomic ion has an oxidation state equal to its charge.
• Oxygen is usually −2.
• Hydrogen is usually +1.
• The sum of oxidation states equals the overall charge of the species.

For example, in \(\mathrm{SO_{4}}^{2-}\), let the sulfur oxidation state be \(x\).

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

\[ x=+6 \]

Sulfur has an oxidation state of +6.

Step 3: Compare the same element before and after

This is the most important step.

Do not compare the oxidation state of zinc with the oxidation state of copper. Follow each element independently.

For example:

\[ \mathrm{Fe}^{2+}\rightarrow\mathrm{Fe}^{3+} \]

The oxidation state increases from +2 to +3.

Iron loses one electron and is oxidised.

Now consider:

\[ \mathrm{Ag}^{+}\rightarrow\mathrm{Ag} \]

The oxidation state decreases from +1 to 0.

Silver gains one electron and is reduced.

Step 4: Name the oxidant and reductant

Once you know what is oxidised and reduced, the names follow.

• The reactant that is oxidised is the reductant.
• The reactant that is reduced is the oxidant.

Be precise about the chemical species.

For example, in the zinc and copper sulfate reaction, \(\mathrm{Cu}^{2+}\) is the oxidant. The sulfate ion is a spectator, not the oxidant.

Step 5: Check atoms, charge, and electrons

A correctly balanced redox equation must conserve atoms and total electrical charge.

Check:

1. Every element has the same number of atoms on both sides.
2. The total charge on the reactant side equals the total charge on the product side.
3. The total electrons lost equal the total electrons gained.

For ionic equations, calculate the charge using the coefficients as well as the individual ion charges.

For example, five \(\mathrm{Fe}^{2+}\) ions contribute a total charge of +10, not +2.

If the electron totals do not match, revisit your oxidation-state changes and coefficients.

04Worked examples

05A contrasting case: not every reaction has an oxidant and reductant

Consider the neutralisation reaction between hydrochloric acid and sodium hydroxide.

The molecular equation is:

\[ \ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)} \]

The net ionic equation is:

\[ \ce{H+(aq) + OH^-(aq) -> H2O(l)} \]

Has oxidation or reduction occurred?

No.

Check the oxidation states:

No element changes oxidation state.

There is no electron transfer, so there is no oxidant or reductant in this reaction.

This is a useful distinction because students sometimes assume that any reaction involving ions, acids, or changes in chemical formulae must be a redox reaction.

It doesn’t.

The test is always the oxidation-state change.

06Questions and solutions

Question 1

Iron metal is added to aqueous copper(II) sulfate at room temperature. No acid or alkali is added, and sulfate ions are spectators.

The net ionic equation is:

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

Identify:

(a) The oxidant.

(b) The reductant.

(c) The oxidation-state change of each element.

(d) The two half-equations.

Solution 1

Copper(II) ions are the oxidant, and iron metal is the reductant.

(a) Oxidant

Copper changes from +2 to 0:

\[ \mathrm{Cu}: \quad +2\rightarrow0 \]

It gains two electrons and is reduced.

Therefore, \(\mathrm{Cu}^{2+}\) is the oxidant.

(b) Reductant

Iron changes from 0 to +2:

\[ \mathrm{Fe}: \quad 0\rightarrow+2 \]

It loses two electrons and is oxidised.

Therefore, \(\ce{Fe}\) is the reductant.

(c) Oxidation-state changes

• Iron: 0 to +2, oxidation.
• Copper: +2 to 0, reduction.

(d) Half-equations

Oxidation:

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

Reduction:

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

The two electrons lost by iron are gained by copper(II) ions.

The total ionic charge is +2 on both sides, and both elements are conserved.

Question 2

A galvanic cell is constructed using two half-cells at \(25^\circ\text{C}\).

• A zinc strip is immersed in \(100.0\text{ mL}\) of \(0.50\text{ mol L}^{-1}\) aqueous zinc nitrate.
• A silver strip is immersed in \(100.0\text{ mL}\) of \(0.50\text{ mol L}^{-1}\) aqueous silver nitrate.
• A potassium nitrate salt bridge connects the solutions.
• An external wire and voltmeter connect the metal electrodes.

After the cell operates, the following mass changes are measured:

Assume the metal deposits remain attached to the electrodes, and no competing reactions occur.

Use \(M(\ce{Zn})=65.4\text{ g mol}^{-1}\) and \(M(\ce{Ag})=107.9\text{ g mol}^{-1}\).

Determine:

(a) The oxidant and reductant.

(b) Which electrode is the anode and which is the cathode.

(c) The direction of electron flow through the external wire.

(d) The direction of ion migration through the salt bridge.

(e) The amount of electrons transferred, calculated independently using both electrode mass changes. Explain whether the measurements are consistent.

Solution 2

Silver ions are the oxidant, and zinc is the reductant. The zinc electrode is the anode, and the silver electrode is the cathode.

(a) Identify the agents

Zinc changes from 0 to +2:

\[ \mathrm{Zn}: \quad 0\rightarrow+2 \]

It loses electrons and is oxidised.

Therefore, zinc is the reductant.

Silver changes from +1 to 0:

\[ \mathrm{Ag}: \quad +1\rightarrow0 \]

It gains electrons and is reduced.

Therefore, silver ions are the oxidant.

The net ionic equation is:

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

(b) Identify the electrodes

Oxidation occurs at the anode.

Therefore, the zinc electrode is the anode.

Reduction occurs at the cathode.

Therefore, the silver electrode is the cathode.

Remember:

• Anode = oxidation.
• Cathode = reduction.

(c) Determine electron flow

Electrons travel from the zinc electrode towards the silver electrode through the external wire.

This is because zinc releases electrons, while silver ions accept them.

(d) Determine salt bridge ion migration

Zinc oxidation produces \(\mathrm{Zn}^{2+}\), increasing the positive charge in the zinc half-cell.

Nitrate ions migrate towards this half-cell to maintain electrical neutrality.

At the silver electrode, \(\mathrm{Ag}^{+}\) ions are consumed, leaving a relative excess of negative charge in the solution.

Potassium ions migrate towards the silver half-cell to maintain electrical neutrality.

(e) Calculate the electron transfer independently

First, use the zinc mass change.

The amount of zinc consumed is:

\[ \begin{aligned} n(\ce{Zn})&=\frac{m}{M}\\ &=\frac{0.0654}{65.4}\\ &=0.00100\text{ mol} \end{aligned} \]

Each zinc atom releases two electrons.

Therefore:

\[ \begin{aligned} n(e^-)_{\text{Zn}}&=0.00100\times2\\ &=0.00200\text{ mol} \end{aligned} \]

Now use the silver mass change.

The amount of silver deposited is:

\[ \begin{aligned} n(\ce{Ag})&=\frac{0.2158}{107.9}\\ &=0.00200\text{ mol} \end{aligned} \]

Each silver ion accepts one electron.

Therefore:

\[ n(e^-)_{\text{Ag}}=0.00200\text{ mol} \]

Both calculations give:

\[ \boxed{n(e^-)=0.00200\text{ mol}} \]

The measurements are consistent with the proposed reaction.

This is an important experimental check: the electron loss calculated from the zinc electrode matches the electron gain calculated from the silver electrode.

Question 3

Dichromate ions react with iodide ions in acidic aqueous solution.

The acid is supplied by excess dilute sulfuric acid, and the reaction proceeds according to:

\[ \ce{Cr2O7^2- + 14H+ + 6I^- -> 2Cr^3+ + 3I2 + 7H2O} \]

A reaction mixture contains \(0.0150\text{ mol}\) of dichromate ions, which react completely.

Determine:

(a) The oxidation state of chromium before and after the reaction.

(b) The oxidation state of iodine before and after the reaction.

(c) The oxidant and reductant.

(d) The amount of electrons transferred in moles.

(e) Why hydrogen ions are not the oxidant or reductant.

Verify that the ionic equation conserves charge.

Solution 3

Dichromate ions are the oxidant, and iodide ions are the reductant. The reaction transfers \(0.0900\text{ mol}\) of electrons.

(a) Chromium oxidation states

In \(\mathrm{Cr_{2}O_{7}}^{2-}\), let the oxidation state of chromium be \(x\).

\[ 2x+7(-2)=-2 \]

\[ 2x=12 \]

\[ x=+6 \]

Chromium finishes as \(\mathrm{Cr}^{3+}\), so:

\[ \mathrm{Cr}: \quad +6\rightarrow+3 \]

Chromium gains three electrons per atom.

There are two chromium atoms in each dichromate ion.

Therefore, each dichromate ion accepts six electrons in total.

(b) Iodine oxidation states

Iodine begins in \(\mathrm{I}^{-}\), with oxidation state −1.

In elemental iodine, \(\ce{I2}\), the oxidation state is 0.

Therefore:

\[ \mathrm{I}: \quad -1\rightarrow0 \]

Iodine loses one electron per atom.

(c) Identify the agents

Chromium is reduced, so dichromate ions are the oxidant.

Iodine is oxidised, so iodide ions are the reductant.

(d) Calculate electron transfer

Each mole of dichromate ions accepts six moles of electrons.

\[ \begin{aligned} n(e^-)&=0.0150\times6\\ &=0.0900\text{ mol} \end{aligned} \]

Therefore:

\[ \boxed{n(e^-)=0.0900\text{ mol}} \]

The same amount of electrons must be released by the iodide ions.

There are six iodide ions per dichromate ion, so:

\[ n(\mathrm{I}^{-})=0.0150\times6 \]

\[ \boxed{n(\mathrm{I}^{-})=0.0900\text{ mol}} \]

(e) Why hydrogen ions are not the oxidant

Hydrogen begins at oxidation state +1 in \(\mathrm{H}^{+}\).

It remains at +1 in water.

Therefore, hydrogen is not oxidised or reduced.

The hydrogen ions help balance the hydrogen and oxygen atoms in the acidic reaction, but they do not act as the oxidant or reductant.

Finally, check charge conservation.

Reactant-side charge:

\[ -2+14(+1)+6(-1)=+6 \]

Product-side charge:

\[ 2(+3)=+6 \]

Both sides have a total charge of +6.

The atoms are also conserved.

Question 4

Manganate(VI) ions react in acidic aqueous solution according to:

\[ \ce{3MnO4^2- + 4H+ -> 2MnO4^- + MnO2(s) + 2H2O} \]

A student argues:

“Only one manganese-containing reactant is present, so there cannot be a separate oxidant and reductant. The manganate ions must all undergo the same process.”

Evaluate this claim.

Determine:

(a) The oxidation state of manganese in each manganese-containing species.

(b) Which manganese atoms are oxidised and which are reduced.

(c) Whether manganate(VI) acts as an oxidant, a reductant, or both.

(d) How many electrons are lost and gained per reaction as written.

(e) Whether the equation conserves atoms and charge.

Solution 4

The student is incorrect. Manganate(VI) ions act as both the oxidant and the reductant because different manganese atoms undergo oxidation and reduction.

(a) Calculate the oxidation states

In \(\mathrm{MnO_{4}}^{2-}\):

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

\[ x=+6 \]

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

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

\[ x=+7 \]

In \(\ce{MnO2}\):

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

\[ x=+4 \]

Therefore:

(b) Identify oxidation and reduction

Two manganese atoms change from +6 to +7.

\[ +6\rightarrow+7 \]

They each lose one electron.

This is oxidation.

One manganese atom changes from +6 to +4.

\[ +6\rightarrow+4 \]

It gains two electrons.

This is reduction.

(c) Identify the oxidant and reductant

The manganate ions that become \(\ce{MnO2}\) are reduced.

Therefore, the original manganate ions act as the oxidant.

The manganate ions that become permanganate are oxidised.

Therefore, the original manganate ions also act as the reductant.

The same reactant species performs both roles.

(d) Check electron transfer

Two manganese atoms are oxidised, each losing one electron:

\[ n(e^-)_{\text{lost}}=2\times1=2 \]

One manganese atom is reduced, gaining two electrons:

\[ n(e^-)_{\text{gained}}=1\times2=2 \]

Therefore:

\[ \boxed{2\text{ electrons lost}=2\text{ electrons gained}} \]

(e) Check atoms and charge

Count the manganese atoms:

• Reactants: 3.
• Products: 2 in permanganate and 1 in manganese dioxide, giving 3.

Count oxygen atoms:

• Reactants: \(3\times4=12\).
• Products: \(2\times4+2+2=12\).

Count hydrogen atoms:

• Reactants: 4.
• Products: 4 in two water molecules.

Now check charge.

Reactant side:

\[ 3(-2)+4(+1)=-2 \]

Product side:

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

Both sides have the same total charge.

The equation is balanced.

This reaction is an example of disproportionation: the same starting species undergoes both oxidation and reduction.

The important trap is to average the oxidation states of all the product manganese atoms. Their average is still +6, but that does not mean no redox has occurred. Different atoms have followed different pathways.

For a broader explanation of this reaction pattern, see Disproportionation: When One Substance Is Oxidised and Reduced.

Question 5

A chemist investigates a two-stage reaction sequence in aqueous solution at \(25^\circ\text{C}\).

The first reaction is:

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

The resulting iron(II) solution is transferred quantitatively into a second reaction vessel containing excess silver ions.

The second reaction is:

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

Assume both reactions proceed completely as written, no side reactions occur, and all iron ions produced in the first stage are transferred into the second stage.

A sample contains \(0.00400\text{ mol}\) of zinc, which reacts completely.

A student makes the following claim:

“Since \(\mathrm{Fe}^{3+}\) accepts electrons in the first reaction, it must be the overall oxidant. Since \(\mathrm{Fe}^{2+}\) donates electrons in the second reaction, it must be the overall reductant.”

Evaluate this claim.

Determine:

(a) The oxidant and reductant in the first reaction.

(b) The oxidant and reductant in the second reaction.

(c) The overall balanced net ionic equation after combining both stages.

(d) The oxidant and reductant in the overall reaction.

(e) The amount of silver deposited.

(f) The amount of electrons transferred in each stage, and whether the electron totals agree.

Solution 5

The student’s claim is incorrect for the overall reaction.

Zinc is the overall reductant, and silver ions are the overall oxidant. The iron(III)/iron(II) pair acts as an intermediate electron-transfer system and is regenerated by the end of the sequence.

(a) Identify the agents in Stage 1

The first reaction is:

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

Zinc changes from 0 to +2.

It loses two electrons and is oxidised.

Therefore, zinc is the reductant.

Iron changes from +3 to +2.

Each iron(III) ion gains one electron and is reduced.

Therefore, \(\mathrm{Fe}^{3+}\) is the oxidant.

Stage 1:

• Oxidant: \(\mathrm{Fe}^{3+}\).
• Reductant: \(\ce{Zn}\).

(b) Identify the agents in Stage 2

The second reaction is:

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

Iron changes from +2 to +3.

It loses one electron per iron ion and is oxidised.

Therefore, \(\mathrm{Fe}^{2+}\) is the reductant.

Silver changes from +1 to 0.

It gains one electron per silver ion and is reduced.

Therefore, \(\mathrm{Ag}^{+}\) is the oxidant.

Stage 2:

• Oxidant: \(\mathrm{Ag}^{+}\).
• Reductant: \(\mathrm{Fe}^{2+}\).

Notice that iron changes roles between the two reactions.

In Stage 1, iron(III) accepts electrons.

In Stage 2, iron(II) donates electrons.

(c) Combine the two reactions

Write the equations underneath each other:

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

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

Add the equations:

\[ \begin{aligned} \ce{Zn + 2Fe^3+ + 2Fe^2+ + 2Ag+}\\ \ce{-> Zn^2+ + 2Fe^2+ + 2Fe^3+ + 2Ag} \end{aligned} \]

The iron(III) and iron(II) ions appear on both sides.

Cancel them.

The overall net ionic equation is:

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

Check the charge:

Reactant side:

\[ 0+2(+1)=+2 \]

Product side:

\[ +2+0=+2 \]

The equation also conserves zinc and silver atoms.

(d) Identify the overall agents

Look at the overall equation rather than assigning labels from just one stage.

Zinc changes from 0 to +2.

It is oxidised, so it is the overall reductant.

Silver changes from +1 to 0.

It is reduced, so silver ions are the overall oxidant.

The iron ions have disappeared from the net equation because they are regenerated.

They are not consumed or produced overall.

This is the crucial distinction: an ion can act as an oxidant or reductant in an individual reaction without being an overall reactant or product in the complete reaction sequence.

(e) Calculate the amount of silver deposited

The amount of zinc reacting is:

\[ n(\ce{Zn})=0.00400\text{ mol} \]

From Stage 1:

\[ 1\text{ mol Zn}:2\text{ mol }\mathrm{Fe}^{2+} \]

Therefore:

\[ n(\mathrm{Fe}^{2+})=2\times0.00400 \]

\[ =0.00800\text{ mol} \]

From Stage 2, the mole ratio between iron(II) ions and silver ions is 1:1.

Therefore:

\[ n(\ce{Ag})=0.00800\text{ mol} \]

The amount of silver deposited is:

\[ \boxed{0.00800\text{ mol}} \]

(f) Check the electron transfer in both stages

In Stage 1, each zinc atom loses two electrons.

\[ \begin{aligned} n(e^-)_{\text{Stage 1}} &=0.00400\times2\\ &=0.00800\text{ mol} \end{aligned} \]

In Stage 2, each iron(II) ion loses one electron.

\[ \begin{aligned} n(e^-)_{\text{Stage 2}} &=0.00800\times1\\ &=0.00800\text{ mol} \end{aligned} \]

Therefore:

\[ \boxed{ n(e^-)_{\text{Stage 1}} = n(e^-)_{\text{Stage 2}} } \]

Both stages transfer \(0.00800\text{ mol}\) of electrons.

The electron balance is consistent.

The deeper lesson is that identifying an oxidant or reductant depends on the particular reaction you are analysing. If the question gives a sequence of reactions, you must distinguish the agents in each stage from the agents in the overall reaction.

07What to do next

Once you can identify the oxidant and reductant, the next step is to construct the oxidation and reduction half-equations and combine them so that electrons cancel.

For acidic reactions, see How to Balance a Redox Half-Equation in Acidic Solution.

The order matters: identify what is oxidised and reduced first, then use the half-equations to balance the atoms and charge.