How to Combine Redox Half-Equations with Different Electron Numbers

A burette delivers purple potassium permanganate solution into a flask containing colourless hydrogen peroxide. Each drop loses its colour as the flask is swirled. Eventually, a faint pink colour remains.

You know a redox reaction is taking place, but how many hydrogen peroxide molecules react with each permanganate ion? Is the ratio 1:1, or something different?

The answer comes from the electrons transferred. In acidic solution, each permanganate ion gains five electrons, while each hydrogen peroxide molecule loses two electrons when it forms oxygen.

The smallest number of electrons both reactions can exchange is 10. Therefore, you need two permanganate ions for every five hydrogen peroxide molecules.

The method is straightforward: find the lowest common multiple (LCM) of the electron numbers, multiply each entire half-equation by the appropriate factor, and then add the equations. Let’s work through why this works and how to use it in HSC Chemistry problems.

01Why the electrons must match

Start by picturing the titration apparatus.

A burette is clamped vertically above a conical flask. The burette contains potassium permanganate solution of known concentration. The flask contains a measured volume of hydrogen peroxide solution and excess dilute sulfuric acid.

As permanganate enters the flask, the purple \(\mathrm{MnO_{4}}^{-}\) ions react with hydrogen peroxide. In acidic conditions, they become \(\mathrm{Mn}^{2+}\) ions, which have a very pale colour.

What should happen as the titration proceeds?

If hydrogen peroxide is still present, newly added permanganate reacts and its purple colour disappears. Once the hydrogen peroxide has been consumed, the next slight excess of permanganate gives the solution a persistent pale pink colour.

This is the endpoint.

The colour change tells us when the reaction is complete. However, it doesn’t tell us the mole ratio directly. We need the balanced redox equation to convert the measured volume of permanganate into an amount of hydrogen peroxide.

The two half-equations

In acidic solution, permanganate is reduced to manganese(II):

\[ \ce{MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O} \]

Each permanganate ion gains five electrons.

Hydrogen peroxide is oxidised to oxygen:

\[ \ce{H2O2 -> O2 + 2H+ + 2e^-} \]

Each hydrogen peroxide molecule loses two electrons.

Notice the problem. One reaction consumes five electrons, while the other produces only two.

If you simply add these equations, three electrons would be left unmatched. That cannot represent a complete redox reaction.

The electrons released during oxidation must equal the electrons accepted during reduction.

Find the lowest common multiple

We need to make the electron numbers equal.

The lowest common multiple of 5 and 2 is 10.

Therefore:

• Multiply the entire permanganate half-equation by 2.
• Multiply the entire hydrogen peroxide half-equation by 5.

This produces:

\[ \begin{aligned} \ce{2MnO4^- + 16H+ + 10e^-} &\ce{-> 2Mn^2+ + 8H2O}\\ \ce{5H2O2} &\ce{-> 5O2 + 10H+ + 10e^-} \end{aligned} \]

Now the electrons cancel completely.

Add the two equations, then cancel the ten electrons and any identical species appearing on both sides.

There are 16 \(\mathrm{H}^{+}\) ions on the left and 10 on the right, leaving six on the left.

The final balanced equation is:

\[ \boxed{ \begin{gathered} \ce{2MnO4^- + 6H+ + 5H2O2}\\ \ce{-> 2Mn^2+ + 8H2O + 5O2} \end{gathered} } \]

The mole ratio is therefore:

\[ n(\mathrm{MnO_{4}}^{-}):n(\ce{H2O2})=2:5 \]

This is the key result. Two moles of permanganate react with five moles of hydrogen peroxide under these acidic conditions.

02A reliable method for combining half-equations

The permanganate example gives us a repeatable process for exam questions.

Step 1: Check that both half-equations are balanced.

Each half-equation must have equal numbers of every atom on both sides, and equal total charge.

Make sure both equations represent the same reaction conditions. An acidic half-equation cannot automatically be combined with one balanced for alkaline conditions.

Step 2: Identify the electron numbers.

Record how many electrons are gained and how many are lost.

Treat the electron numbers as positive magnitudes. The side on which the electrons appear tells you whether they are being gained or released.

Step 3: Find the lowest common multiple.

Find the smallest whole number that both electron counts divide into.

For example:

\[ \operatorname{LCM}(4,6)=12 \]

Step 4: Calculate the multipliers.

Divide the LCM by each original electron number.

For electron counts of 4 and 6:

\[ \frac{12}{4}=3 \]

\[ \frac{12}{6}=2 \]

Multiply the first entire half-equation by 3 and the second entire half-equation by 2.

Step 5: Add the half-equations.

Put reactants on the left and products on the right. Cancel the electrons, followed by any identical species appearing on both sides.

Step 6: Check the final equation.

Confirm that:

• Every element has the same number of atoms on both sides.
• Total charge is equal on both sides.
• No electrons remain in the overall equation.
• All coefficients are in the smallest whole-number ratio.
• The chemical species are appropriate for the stated conditions.

The most important detail is that you multiply the entire half-equation, not just the electron coefficient.

Why multiplying only the electrons is wrong

Consider the hydrogen peroxide half-equation:

\[ \ce{H2O2 -> O2 + 2H+ + 2e^-} \]

Suppose you want five times as many electrons.

An incorrect attempt would be:

\[ \ce{H2O2 -> O2 + 10H+ + 10e^-} \]

This is not balanced. The left side has a total charge of zero, while the right side has a charge of:

\[ +10-10=0 \]

The charge happens to match, but the hydrogen and oxygen atoms do not. There are two hydrogen atoms on the left and ten on the right, and two oxygen atoms on the left but two on the right only if the oxygen is counted without changing the products. The hydrogen balance has clearly failed.

The correct approach is to multiply every coefficient:

\[ \ce{5H2O2 -> 5O2 + 10H+ + 10e^-} \]

Now both atoms and charge remain balanced.

A coefficient changes the quantity of a substance. A subscript changes the identity of the substance. Never change a chemical formula just to force the electrons to match.

03Worked examples

What the experiment cannot tell you

A titration gives useful quantitative information, but it relies on some important assumptions.

The calculation above assumes:

• The solution is sufficiently acidic.
• Hydrogen peroxide is the only substance reacting with permanganate.
• The reaction proceeds completely.
• The endpoint is identified accurately.

If another reducing agent is present, it can also consume permanganate. The measured titre would then represent the combined electron transfer, not hydrogen peroxide alone.

A separate selective measurement or separation would be needed to determine the individual substances.

The reaction conditions also matter. In insufficiently acidic or alkaline conditions, permanganate can form manganese dioxide rather than manganese(II). That changes the reduction half-equation and therefore the electron number.

This is why you should always check the products and conditions before deciding on your electron multipliers.

04Questions and solutions

Question 1

Consider the following balanced half-equations:

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

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

Combine the half-equations to produce a balanced overall ionic equation.

State the mole ratio of tin(II) ions to silver ions.

Solution 1

The balanced overall equation is:

\[ \boxed{ \ce{Sn^2+ + 2Ag+ -> Sn^4+ + 2Ag} } \]

The mole ratio of tin(II) ions to silver ions is 1:2.

Step 1: Identify the electron numbers.

Tin(II) produces two electrons, while each silver ion consumes one electron.

Step 2: Find the LCM.

\[ \operatorname{LCM}(2,1)=2 \]

The tin half-equation remains unchanged, while the silver half-equation is multiplied by 2.

Step 3: Scale the entire silver half-equation.

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

Step 4: Add the equations.

\[ \ce{Sn^2+ + 2Ag+ + 2e^-} \]

\[ \ce{-> Sn^4+ + 2Ag + 2e^-} \]

Cancel the electrons.

Step 5: Check charge.

The left-hand charge is:

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

The right-hand charge is:

\[ +4 \]

The equation is balanced.

One mole of tin(II) ions supplies the two moles of electrons required to reduce two moles of silver ions.

Question 2

In acidic solution, iodide ions are oxidised to iodine, while nitrate ions are reduced to nitrogen monoxide, \(\ce{NO}\).

The balanced half-equations are:

\[ \ce{2I^- -> I2 + 2e^-} \]

\[ \ce{NO3^- + 4H+ + 3e^- -> NO + 2H2O} \]

(a) Determine the lowest common multiple of the electron numbers and state the multiplier for each half-equation.

(b) Write the overall balanced ionic equation.

(c) Explain why adding the original half-equations without scaling would produce an invalid overall equation.

Solution 2

The balanced overall equation is:

\[ \boxed{ \begin{gathered} \ce{6I^- + 2NO3^- + 8H+}\\ \ce{-> 3I2 + 2NO + 4H2O} \end{gathered} } \]

The multipliers are 3 for the iodide half-equation and 2 for the nitrate half-equation.

Step 1: Identify the electron numbers.

The iodide half-equation produces two electrons.

The nitrate half-equation consumes three electrons.

Step 2: Calculate the LCM.

\[ \operatorname{LCM}(2,3)=6 \]

Therefore:

\[ \text{Iodide multiplier}=\frac{6}{2}=3 \]

\[ \text{Nitrate multiplier}=\frac{6}{3}=2 \]

Step 3: Multiply both half-equations.

Oxidation:

\[ \ce{6I^- -> 3I2 + 6e^-} \]

Reduction:

\[ \ce{2NO3^- + 8H+ + 6e^-} \]

\[ \ce{-> 2NO + 4H2O} \]

Step 4: Add and cancel the electrons.

\[ \ce{6I^- + 2NO3^- + 8H+} \]

\[ \ce{-> 3I2 + 2NO + 4H2O} \]

Step 5: Check charge and atoms.

The total charge on the left is:

\[ -6-2+8=0 \]

The right side is neutral.

There are six iodine atoms, two nitrogen atoms, six oxygen atoms, and eight hydrogen atoms on each side.

The equation is balanced.

If you added the original equations without scaling, two electrons would be produced but only three consumed. The remaining electron would have no corresponding reaction to accept or supply it. This is why electron cancellation is essential, not optional.

Question 3

A student has a concentrated hydrogen peroxide solution that needs to be diluted before titration.

They pipette 10.00 mL of the original solution into a 100.0 mL volumetric flask and fill it to the calibration mark with distilled water.

A 20.00 mL aliquot of the diluted solution is then titrated with \(0.0200\ \mathrm{mol\,L^{-1}}\) acidified potassium permanganate.

The mean titre is 16.40 mL.

Assume the reaction produces \(\mathrm{Mn}^{2+}\) and \(\ce{O2}\).

Calculate:

(a) The concentration of hydrogen peroxide in the diluted solution.

(b) The concentration of hydrogen peroxide in the original solution.

(c) The mass concentration of hydrogen peroxide in the original solution, in \(\mathrm{g\,L^{-1}}\).

The molar mass of hydrogen peroxide is \(34.0\ \mathrm{g\,mol^{-1}}\).

Solution 3

The original hydrogen peroxide concentration is \(0.410\ \mathrm{mol\,L^{-1}}\), equivalent to \(13.9\ \mathrm{g\,L^{-1}}\).

Step 1: Identify the mole ratio.

The balanced acidic reaction gives:

\[ n(\mathrm{MnO_{4}}^{-}):n(\ce{H2O2})=2:5 \]

Therefore:

\[ n(\ce{H2O2})=\frac{5}{2}n(\mathrm{MnO_{4}}^{-}) \]

Step 2: Calculate the moles of permanganate used.

Convert the titre into litres:

\[ 16.40\ \mathrm{mL}=0.01640\ \mathrm{L} \]

Use \(n=cV\):

\[ \begin{aligned} n(\mathrm{MnO_{4}}^{-}) &=0.0200\times0.01640\\ &=3.28\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Step 3: Calculate the moles of hydrogen peroxide.

\[ \begin{aligned} n(\ce{H2O2}) &=\frac{5}{2}\times3.28\times10^{-4}\\ &=8.20\times10^{-4}\ \mathrm{mol} \end{aligned} \]

This is the amount in the 20.00 mL diluted aliquot.

Step 4: Calculate the diluted concentration.

The aliquot volume is:

\[ 20.00\ \mathrm{mL}=0.02000\ \mathrm{L} \]

Therefore:

\[ \begin{aligned} c_{\text{diluted}} &=\frac{8.20\times10^{-4}}{0.02000}\\ &=0.0410\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 5: Calculate the original concentration.

The dilution factor is:

\[ \frac{100.0}{10.00}=10.00 \]

The original solution is ten times as concentrated as the diluted solution.

\[ c_{\text{original}}=10.00\times0.0410 \]

\[ \boxed{c_{\text{original}}=0.410\ \mathrm{mol\,L^{-1}}} \]

Step 6: Convert to mass concentration.

Use:

\[ \text{Mass concentration}=c\times M \]

where \(M\) is the molar mass.

\[ \begin{aligned} &=0.410\times34.0\\ &=13.94\ \mathrm{g\,L^{-1}} \end{aligned} \]

To three significant figures:

\[ \boxed{13.9\ \mathrm{g\,L^{-1}}} \]

The important detail is that the mole ratio must be applied before calculating the concentration. Using a 1:1 ratio would produce an incorrect result, even if every volume conversion and concentration calculation were otherwise correct.

Question 4

An acidic solution contains both iron(II) ions, \(\mathrm{Fe}^{2+}\), and oxalate ions, \(\mathrm{C_{2}O_{4}}^{2-}\).

The solution reacts completely with acidified dichromate ions, \(\mathrm{Cr_{2}O_{7}}^{2-}\).

The products are:

• \(\mathrm{Fe}^{3+}\) from oxidation of iron(II),
• \(\ce{CO2}\) from oxidation of oxalate,
• \(\mathrm{Cr}^{3+}\) from reduction of dichromate.

A 25.00 mL aliquot consumes 0.900 mmol of dichromate ions.

An independent measurement establishes that the aliquot contains 0.600 mmol of oxalate ions.

Calculate the concentration of iron(II) ions in the original solution.

Assume both reducing agents react completely, and no other species transfer electrons.

Solution 4

The iron(II) concentration is:

\[ \boxed{0.168\ \mathrm{mol\,L^{-1}} } \]

The key is that dichromate accepts electrons from both reducing agents. You must account for both contributions before calculating the amount of iron(II).

Step 1: Write the relevant half-equations.

Dichromate reduction:

\[ \ce{Cr2O7^2- + 14H+ + 6e^-} \]

\[ \ce{-> 2Cr^3+ + 7H2O} \]

Iron(II) oxidation:

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

Oxalate oxidation:

\[ \ce{C2O4^2- -> 2CO2 + 2e^-} \]

Step 2: Calculate the total electrons accepted by dichromate.

Each dichromate ion accepts six electrons.

The amount of dichromate is:

\[ 0.900\ \mathrm{mmol} \]

Therefore:

\[ n(e^-)=6\times0.900 \]

\[ =5.400\ \mathrm{mmol} \]

The two reducing agents must supply a combined 5.400 mmol of electrons.

Step 3: Calculate the electrons supplied by oxalate.

Each oxalate ion releases two electrons.

\[ n(e^-)=2\times0.600 \]

\[ =1.200\ \mathrm{mmol} \]

Step 4: Calculate the electrons supplied by iron(II).

Subtract the contribution from oxalate:

\[ n(e^-)_{\mathrm{Fe}^{2+}} =5.400-1.200 \]

\[ =4.200\ \mathrm{mmol} \]

Each iron(II) ion releases one electron.

Therefore:

\[ n(\mathrm{Fe}^{2+})=4.200\ \mathrm{mmol} \]

Step 5: Calculate the iron(II) concentration.

The aliquot volume is 25.00 mL, or 0.02500 L.

\[ \begin{aligned} c(\mathrm{Fe}^{2+}) &=\frac{4.200\times10^{-3}}{0.02500}\\ &=0.168\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The final concentration is:

\[ \boxed{0.168\ \mathrm{mol\,L^{-1}} } \]

Why is the tempting alternative wrong?

You might initially assume that all 5.400 mmol of electrons came from iron(II).

That would give:

\[ \frac{5.400\times10^{-3}}{0.02500} =0.216\ \mathrm{mol\,L^{-1}} \]

But this ignores the electrons already supplied by oxalate.

The oxidising agent does not distinguish between the two sources of electrons. Its total electron requirement must be satisfied by the combined oxidation reactions.

This is an important extension of the usual two-half-equation method: when multiple oxidation processes occur, add their electron contributions together before matching them to the reduction process.

Question 5

A student investigates the reaction between potassium permanganate and hydrogen peroxide under two different conditions.

The starting reactants are the same in both experiments, and hydrogen peroxide is oxidised to oxygen in each case.

However, the manganese products differ.

Experiment A: Acidic conditions

Permanganate is reduced to \(\mathrm{Mn}^{2+}\).

Experiment B: Weakly alkaline conditions

Permanganate is reduced to solid \(\ce{MnO2}\).

Assume these are the only redox products in each experiment and that both reactions proceed completely.

(a) Balance the overall ionic equation for Experiment A.

(b) Balance the overall ionic equation for Experiment B.

(c) Determine the mole ratio of permanganate to hydrogen peroxide in each experiment.

(d) A student argues that both experiments must have the same mole ratio because the starting reactants are identical. Explain precisely why this reasoning is incorrect.

Solution 5

The mole ratios are different:

• Experiment A: \(2:5\).
• Experiment B: \(2:3\).

The difference occurs because manganese accepts a different number of electrons in each experiment. The reduction product, not just the identity of the starting reactant, determines the electron requirement.

Step 1: Balance the reduction half-equation for Experiment A.

In acidic solution:

\[ \ce{MnO4^- + 8H+ + 5e^-} \]

\[ \ce{-> Mn^2+ + 4H2O} \]

Manganese changes from oxidation state \(+7\) to \(+2\).

It therefore gains five electrons.

Step 2: Write the oxidation half-equation.

Hydrogen peroxide forms oxygen:

\[ \ce{H2O2 -> O2 + 2H+ + 2e^-} \]

Each hydrogen peroxide molecule loses two electrons.

Step 3: Find the LCM and combine the half-equations.

\[ \operatorname{LCM}(5,2)=10 \]

Multiply the reduction half-equation by 2:

\[ \ce{2MnO4^- + 16H+ + 10e^-} \]

\[ \ce{-> 2Mn^2+ + 8H2O} \]

Multiply the oxidation half-equation by 5:

\[ \ce{5H2O2 -> 5O2 + 10H+ + 10e^-} \]

Add and cancel the electrons and ten hydrogen ions.

The overall equation is:

\[ \boxed{ \begin{gathered} \ce{2MnO4^- + 6H+ + 5H2O2}\\ \ce{-> 2Mn^2+ + 8H2O + 5O2} \end{gathered} } \]

Therefore:

\[ \boxed{n(\mathrm{MnO_{4}}^{-}):n(\ce{H2O2})=2:5} \]

Step 4: Balance the reduction half-equation for Experiment B.

In weakly alkaline solution, manganese forms \(\ce{MnO2}\).

The balanced reduction half-equation is:

\[ \ce{MnO4^- + 2H2O + 3e^-} \]

\[ \ce{-> MnO2 + 4OH^-} \]

Manganese changes from \(+7\) to \(+4\).

It now gains only three electrons per permanganate ion.

Step 5: Balance the hydrogen peroxide oxidation half-equation for alkaline conditions.

The balanced equation is:

\[ \ce{H2O2 + 2OH^-} \]

\[ \ce{-> O2 + 2H2O + 2e^-} \]

Each hydrogen peroxide molecule still loses two electrons.

Step 6: Find the new LCM.

The electron numbers are now three and two.

\[ \operatorname{LCM}(3,2)=6 \]

Multiply the permanganate half-equation by 2:

\[ \ce{2MnO4^- + 4H2O + 6e^-} \]

\[ \ce{-> 2MnO2 + 8OH^-} \]

Multiply the hydrogen peroxide half-equation by 3:

\[ \ce{3H2O2 + 6OH^-} \]

\[ \ce{-> 3O2 + 6H2O + 6e^-} \]

Step 7: Add and cancel common species.

The six electrons cancel.

There are six hydroxide ions on the left and eight on the right, leaving two on the right.

There are four water molecules on the left and six on the right, leaving two on the right.

The overall equation becomes:

\[ \boxed{ \begin{gathered} \ce{2MnO4^- + 3H2O2}\\ \ce{-> 2MnO2 + 3O2 + 2OH^- + 2H2O} \end{gathered} } \]

The mole ratio is:

\[ \boxed{n(\mathrm{MnO_{4}}^{-}):n(\ce{H2O2})=2:3} \]

Step 8: Explain why the two experiments have different mole ratios.

The student’s mistake is assuming that the electron transfer depends only on the starting substances.

It actually depends on how far each substance changes during the reaction.

In acidic solution, manganese changes from \(+7\) to \(+2\), gaining five electrons per ion.

In weakly alkaline solution, manganese changes from \(+7\) to \(+4\), gaining only three electrons per ion.

Hydrogen peroxide loses two electrons in both experiments, but the number of electrons accepted by each permanganate ion changes.

Consequently, the LCM changes from 10 to 6, producing different mole ratios.

This is why you must identify the reaction products and conditions before combining half-equations. A familiar reactant does not guarantee a familiar electron count.

05Where to go next

Combining half-equations is the final stage of a process, not the first. You need to identify oxidation and reduction, balance each half-equation for the correct conditions, and only then match the electrons.

For more practice with the earlier stages, see these tutorgum resources:

• Redox Half-Equations: Balancing HSC Chemistry Reactions – constructing and balancing oxidation and reduction half-equations in acidic and alkaline solutions.
• Oxidation numbers in redox reactions: HSC Chemistry Guide – identifying which elements are oxidised and reduced by tracking oxidation-state changes.

When you next encounter two half-equations with different electron numbers, don’t try to make the equations fit by changing individual coefficients at random. Find the LCM, scale both complete equations, cancel common species, and check atoms and charge. That method works regardless of whether the electron numbers are 1 and 2, 2 and 3, or 5 and 6.