How to Balance a Permanganate Redox Reaction Using Half-Equations

A purple drop of potassium permanganate disappears when it enters a warm, acidified oxalate solution. The first few drops react slowly, but later drops seem to disappear almost immediately. Eventually, one drop leaves a faint pink colour that stays.

To use this experiment to calculate a concentration, you need to know exactly how many oxalate ions react with each permanganate ion. The ratio is not 1:1.

The key is to match the electrons transferred by the two supplied half-equations, then add the equations and cancel anything appearing on both sides.

For permanganate reduced to manganese(II) in acidic solution, the reduction half-equation consumes five electrons:

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

If the other half-equation produces two electrons, you need to multiply the permanganate equation by 2 and the oxidation equation by 5. Both processes then involve ten electrons, which cancel when you combine them.

Let’s work through why this method works, how the reaction medium changes the answer, and how to use the balanced equation in a titration calculation.

01What the experiment is measuring

One practical application of permanganate redox reactions is determining the concentration of a solution using a redox titration.

Consider a standardisation experiment. You have a potassium permanganate solution, but its exact concentration is unknown. You want to determine that concentration using a sodium oxalate solution of known concentration.

The apparatus is arranged as follows:

• A burette is clamped vertically above a conical flask. It contains the purple potassium permanganate solution.
• A pipette transfers a measured volume of sodium oxalate solution into the conical flask.
• Excess dilute sulfuric acid is added to the flask.
• The flask is kept warm, usually around 60-70 °C, while the permanganate is added gradually.
• A white tile beneath the flask makes the faint colour change easier to see.

What should happen as the permanganate enters the flask?

In acidic solution, permanganate ions are reduced to manganese(II) ions, which are nearly colourless at the concentrations used.

As long as oxalate ions remain, the purple permanganate is consumed. Once all the oxalate has reacted, the next small excess of permanganate produces a faint pink colour that persists.

This is a self-indicating titration. You do not need a separate indicator.

There is an interesting complication, though.

Why the solution is warmed

At room temperature, the reaction between permanganate and oxalate is relatively slow. This can make the endpoint difficult to judge.

As the reaction proceeds, manganese(II) ions are produced. These ions catalyse the reaction, increasing its rate. This is called autocatalysis: one of the products helps speed up the reaction that produces it.

The initial reaction can therefore be sluggish, while later additions react much faster.

Warming the solution makes the reaction practical to carry out. It changes the reaction rate, not the balanced equation or the mole ratio.

The experiment also needs sufficient acid. If the solution does not contain enough \(\mathrm{H}^{+}\), permanganate may be reduced to brown manganese dioxide rather than manganese(II).

That would mean a different half-equation and a different electron ratio.

Use dilute sulfuric acid for this titration. Hydrochloric acid is unsuitable because permanganate can oxidise chloride ions, interfering with the result.

02Step 1: Check the reaction medium

Before combining supplied half-equations, identify the conditions.

This is particularly important with permanganate because manganese can finish in different oxidation states.

The reduction product determines how many electrons each permanganate ion accepts.

In acidic solution, manganese changes from oxidation state \(+7\) to \(+2\).

It gains five electrons.

\[
+7 \longrightarrow +2
\]

The supplied reduction half-equation is:

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

In neutral or alkaline conditions where manganese dioxide forms, manganese changes from \(+7\) to \(+4\).

It gains only three electrons.

A suitable reduction half-equation is:

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

The important point is that you cannot automatically use the five-electron equation every time you see \(\mathrm{MnO_{4}}^{-}\).

Read the supplied equation and check the conditions. If the question specifies \(\ce{MnO2}\) as the product, you must use the three-electron reduction pathway.

03Step 2: Match the electrons

Once you have confirmed the correct reduction half-equation, look at the oxidation half-equation.

Remember:

• Reduction involves gaining electrons. Electrons appear on the left.
• Oxidation involves losing electrons. Electrons appear on the right.

The number of electrons gained must equal the number lost.

Think of it as finding a common denominator in fractions. You need to scale both equations until the electron quantities match.

For example, suppose your supplied half-equations contain:

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

and

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

The permanganate half-equation consumes five electrons.

The oxalate half-equation produces two electrons.

The lowest common multiple of 5 and 2 is 10.

Therefore:

Notice that you multiply the entire half-equation, not just the electrons.

For example, multiplying the oxalate half-equation by five gives:

\[
\ce{5C2O4^2- -> 10CO2 + 10e^-}
\]

All coefficients change together. The chemical formulas and their subscripts stay exactly the same.

04Step 3: Add the equations and cancel common species

Now place the two multiplied half-equations together.

Using the oxalate example:

Reduction:

\[
\ce{2MnO4^- + 16H+ + 10e^- -> 2Mn^2+ + 8H2O}
\]

Oxidation:

\[
\ce{5C2O4^2- -> 10CO2 + 10e^-}
\]

Add the reactants together and the products together:

\[
\begin{aligned}
&\ce{2MnO4^- + 16H+ + 10e^-}\\
&\ce{+ 5C2O4^2-}\\
&\ce{-> 2Mn^2+ + 8H2O}\\
&\ce{+ 10CO2 + 10e^-}
\end{aligned}
\]

The ten electrons appear on both sides, so they cancel.

The balanced overall ionic equation is:

\[
\ce{2MnO4^- + 16H+ + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2}
\]

This is not quite the end of the job. You still need to check that the equation conserves atoms and charge.

Check the atoms

Every element has the same number of atoms on both sides.

Check the charge

The total charge on the reactant side is:

\[
2(-1)+16(+1)+5(-2)=+4
\]

The total charge on the product side is:

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

Both sides have a net charge of \(+4\).

The equation is balanced.

Notice something important: a balanced ionic equation does not have to have zero charge on each side. It needs to have the same total charge on both sides.

07A quick diagnostic before you submit

Use this table to catch the mistakes that commonly appear when combining supplied half-equations.

If you need to construct the half-equations yourself before combining them, see Redox Half-Equations: Balancing HSC Chemistry Reactions.

The following questions test whether you can apply the method when the reducing agent, reaction medium, or available information changes.

All numerical values are illustrative.

08Questions and solutions

Question 1

In acidic solution, permanganate ions are reduced to manganese(II) ions, while iodide ions are oxidised to iodine.

The supplied half-equations are:

Reduction:

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

Oxidation:

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

Write the balanced overall ionic equation.

Identify the oxidising agent and the reducing agent.

Solution 1

The balanced overall ionic equation is:

\[
\ce{2MnO4^- + 16H+ + 10I^- -> 2Mn^2+ + 8H2O + 5I2}
\]

Permanganate is the oxidising agent, and iodide is the reducing agent.

The reduction half-equation consumes five electrons, while the oxidation half-equation produces two.

Step 1: Find the lowest common multiple.

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

Step 2: Multiply the reduction equation by 2.

\[
\ce{2MnO4^- + 16H+ + 10e^- -> 2Mn^2+ + 8H2O}
\]

Step 3: Multiply the oxidation equation by 5.

\[
\ce{10I^- -> 5I2 + 10e^-}
\]

Step 4: Add the equations and cancel ten electrons.

\[
\ce{2MnO4^- + 16H+ + 10I^- -> 2Mn^2+ + 8H2O + 5I2}
\]

Step 5: Check the charge.

Reactants:

\[
-2+16-10=+4
\]

Products:

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

The atoms and total charge are conserved.

Permanganate gains electrons, so it causes iodide to be oxidised. That makes permanganate the oxidising agent.

Iodide loses electrons, so it causes permanganate to be reduced. That makes iodide the reducing agent.

Question 2

A 10.00 mL sample of hydrogen peroxide is titrated with \(0.0200\ \mathrm{mol\,dm^{-3}}\) potassium permanganate in excess dilute sulfuric acid.

The average titre is 24.00 mL.

The supplied half-equations are:

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

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

Calculate the concentration of the hydrogen peroxide solution.

Assume the reaction proceeds to completion and all hydrogen peroxide is oxidised to oxygen.

Solution 2

The hydrogen peroxide concentration is:

\[
\boxed{0.120\ \mathrm{mol\,dm^{-3}}}
\]

The crucial step is recognising that the balanced equation gives a 2:5 ratio between permanganate and hydrogen peroxide.

Step 1: Balance the half-equations.

The lowest common multiple of five and two electrons is ten.

Multiply the permanganate reduction equation by 2:

\[
\ce{2MnO4^- + 16H+ + 10e^- -> 2Mn^2+ + 8H2O}
\]

Multiply the hydrogen peroxide oxidation equation by 5:

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

Step 2: Combine and simplify.

Cancel the electrons and ten hydrogen ions from each side.

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

The mole ratio is:

\[
\ce{MnO4^- : H2O2}=2:5
\]

Step 3: Calculate the amount of permanganate used.

Convert the titre into litres:

\[
24.00\ \mathrm{mL}=0.02400\ \mathrm{L}
\]

Using \(n=cV\):

\[
\begin{aligned}
n(\mathrm{MnO_{4}}^{-})
&=0.0200\times0.02400\\
&=4.80\times10^{-4}\ \mathrm{mol}
\end{aligned}
\]

Step 4: Calculate the amount of hydrogen peroxide.

From the 2:5 ratio:

\[
\begin{aligned}
n(\ce{H2O2})
&=\frac{5}{2}\times4.80\times10^{-4}\\
&=1.20\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

Step 5: Calculate the concentration.

The hydrogen peroxide sample has a volume of:

\[
10.00\ \mathrm{mL}=0.01000\ \mathrm{L}
\]

Therefore:

\[
\begin{aligned}
c&=\frac{n}{V}\\
&=\frac{1.20\times10^{-3}}{0.01000}\\
&=0.120\ \mathrm{mol\,dm^{-3}}
\end{aligned}
\]

The concentration is \(0.120\ \mathrm{mol\,dm^{-3}}\).

A common mistake is to reverse the 2:5 ratio and multiply by \(2/5\) when calculating hydrogen peroxide from permanganate. Always read the coefficients in the order of the substances you are calculating.

Question 3

In acidic solution, permanganate ions oxidise sulfite ions, \(\mathrm{SO_{3}}^{2-}\), to sulfate ions, \(\mathrm{SO_{4}}^{2-}\).

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

The supplied half-equations are:

Reduction:

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

Oxidation:

\[
\ce{SO3^2- + H2O -> SO4^2- + 2H+ + 2e^-}
\]

Construct the balanced overall ionic equation.

Explain which species is oxidised and which is reduced.

Solution 3

The balanced overall ionic equation is:

\[
\ce{2MnO4^- + 6H+ + 5SO3^2- -> 2Mn^2+ + 5SO4^2- + 3H2O}
\]

Sulfite is oxidised to sulfate, while permanganate is reduced to manganese(II).

The important complication is that both hydrogen ions and water appear on opposite sides after the equations are combined.

Step 1: Match the electrons.

The reduction half-equation consumes five electrons.

The oxidation half-equation produces two electrons.

The lowest common multiple is ten.

Step 2: Multiply the reduction equation by 2.

\[
\ce{2MnO4^- + 16H+ + 10e^- -> 2Mn^2+ + 8H2O}
\]

Step 3: Multiply the oxidation equation by 5.

\[
\ce{5SO3^2- + 5H2O -> 5SO4^2- + 10H+ + 10e^-}
\]

Step 4: Add the equations.

\[
\begin{aligned}
&\ce{2MnO4^- + 16H+ + 10e^-}\\
&\ce{+ 5SO3^2- + 5H2O}\\
&\ce{-> 2Mn^2+ + 8H2O}\\
&\ce{+ 5SO4^2- + 10H+ + 10e^-}
\end{aligned}
\]

Cancel ten electrons.

There are 16 hydrogen ions on the left and 10 on the right. Cancel ten, leaving six on the left.

There are eight water molecules on the right and five on the left. Cancel five, leaving three on the right.

This produces:

\[
\ce{2MnO4^- + 6H+ + 5SO3^2- -> 2Mn^2+ + 5SO4^2- + 3H2O}
\]

Step 5: Verify the charge.

Reactants:

\[
-2+6-10=-6
\]

Products:

\[
+4-10=-6
\]

The charges match.

There are also 2 Mn, 5 S, 23 O, and 6 H atoms on each side.

The oxidation state of sulfur increases from \(+4\) in sulfite to \(+6\) in sulfate. It loses two electrons per sulfur atom.

Manganese decreases from \(+7\) to \(+2\), gaining five electrons per manganese atom.

The tempting mistake is to leave all 16 hydrogen ions on the left and all 10 on the right. Those are not separate reactants and products in the final net ionic equation. They must be cancelled.

Question 4

A reaction takes place in alkaline solution.

Permanganate ions are reduced to solid manganese dioxide, while hydrogen peroxide is oxidised to oxygen gas.

The supplied half-equations are:

Reduction:

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

Oxidation:

\[
\ce{H2O2 + 2OH^- -> O2 + 2H2O + 2e^-}
\]

(a) Construct the balanced overall ionic equation.

(b) If \(0.400\ \mathrm{mmol}\) of permanganate reacts completely, calculate the amount of oxygen produced.

(c) Explain why the acidic permanganate reduction half-equation cannot be used for this reaction.

Solution 4

(a) The balanced overall equation is:

\[
\ce{2MnO4^- + 3H2O2 -> 2MnO2(s) + 3O2 + 2OH^- + 2H2O}
\]

(b) The amount of oxygen produced is:

\[
\boxed{0.600\ \mathrm{mmol}}
\]

(c) The acidic reduction equation describes manganese being reduced to \(+2\), whereas this reaction produces manganese dioxide, in which manganese has oxidation state \(+4\).

Step 1: Identify the electron numbers.

The permanganate reduction half-equation consumes three electrons.

The hydrogen peroxide oxidation half-equation produces two electrons.

The lowest common multiple is six.

Step 2: Multiply the reduction half-equation by 2.

\[
\ce{2MnO4^- + 4H2O + 6e^- -> 2MnO2 + 8OH^-}
\]

Step 3: Multiply the oxidation half-equation by 3.

\[
\ce{3H2O2 + 6OH^- -> 3O2 + 6H2O + 6e^-}
\]

Step 4: Add and cancel common species.

The six electrons cancel.

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

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

The final equation is:

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

Step 5: Calculate the amount of oxygen.

The balanced equation gives:

\[
\ce{MnO4^- : O2}=2:3
\]

Therefore:

\[
\begin{aligned}
n(\ce{O2})
&=\frac{3}{2}\times0.400\\
&=0.600\ \mathrm{mmol}
\end{aligned}
\]

Step 6: Explain the effect of the reaction medium.

In the acidic reaction, manganese changes from \(+7\) to \(+2\), gaining five electrons.

Here, manganese changes from \(+7\) to \(+4\), gaining only three electrons.

The two supplied reduction half-equations therefore describe different chemical pathways.

Using the acidic equation would give the wrong electron balance, the wrong overall equation, and the wrong mole ratio.

This is why identifying the reaction medium is not just a formality.

Question 5

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

Both substances react completely with permanganate:

• Iron(II) is oxidised to iron(III).
• Oxalate is oxidised to carbon dioxide.
• Permanganate is reduced to manganese(II).

A 20.00 mL aliquot of the mixture is titrated with \(0.0500\ \mathrm{mol\,dm^{-3}}\) potassium permanganate.

The titre is 12.00 mL.

An independent preparation record establishes that the aliquot initially contains a total of 1.80 mmol of \(\mathrm{Fe}^{2+}\) and \(\mathrm{C_{2}O_{4}}^{2-}\) ions combined.

A separate acid-balance measurement indicates that 4.80 mmol of \(\mathrm{H}^{+}\) was consumed.

Determine the amount, in mmol, of:

(a) \(\mathrm{Fe}^{2+}\) in the original aliquot.

(b) \(\mathrm{C_{2}O_{4}}^{2-}\) in the original aliquot.

(c) Explain whether the measured hydrogen ion consumption provides an independent equation that helps determine the mixture composition.

Assume that both oxidation reactions proceed to completion, sulfuric acid is present in excess, and no other reducing agents are present.

Solution 5

(a) The original aliquot contains:

\[
\boxed{0.60\ \mathrm{mmol\ Fe^{2+}}}
\]

(b) It contains:

\[
\boxed{1.20\ \mathrm{mmol\ C2O4^{2-}}}
\]

(c) The hydrogen ion measurement does not provide an independent equation. Both oxidation reactions consume eight moles of hydrogen ions per mole of permanganate, so the acid consumption depends only on the total amount of permanganate reacted.

This is the important conceptual distinction: a permanganate titration measures the total electron transfer, not simply the total number of reducing particles.

Step 1: Calculate the amount of permanganate used.

Convert the titre to litres:

\[
12.00\ \mathrm{mL}=0.01200\ \mathrm{L}
\]

\[
\begin{aligned}
n(\mathrm{MnO_{4}}^{-})
&=cV\\
&=0.0500\times0.01200\\
&=0.000600\ \mathrm{mol}
\end{aligned}
\]

Therefore:

\[
n(\mathrm{MnO_{4}}^{-})=0.600\ \mathrm{mmol}
\]

Step 2: Calculate the total number of electrons accepted.

Each permanganate ion accepts five electrons in acidic solution.

Therefore:

\[
n(e^-)=5\times0.600
\]

\[
n(e^-)=3.00\ \mathrm{mmol}
\]

The two reducing species must collectively supply 3.00 mmol of electrons.

Step 3: Work out how many electrons each reducing species supplies.

Iron(II) is oxidised according to:

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

Each mole of iron(II) releases one mole of electrons.

Oxalate is oxidised according to:

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

Each mole of oxalate releases two moles of electrons.

This is the detail that makes the problem different from an ordinary 1:5 permanganate titration.

Step 4: Set up two simultaneous equations.

Let:

• \(x\) = amount of \(\mathrm{Fe}^{2+}\), in mmol.
• \(y\) = amount of \(\mathrm{C_{2}O_{4}}^{2-}\), in mmol.

The independent preparation record gives the total number of reducing particles:

\[
x+y=1.80
\]

The electron balance gives:

\[
x+2y=3.00
\]

Notice that oxalate has a coefficient of 2 in the electron balance because each ion releases two electrons.

Step 5: Solve the simultaneous equations.

Subtract the first equation from the second:

\[
\begin{aligned}
(x+2y)-(x+y)&=3.00-1.80\\
y&=1.20
\end{aligned}
\]

Substitute into the first equation:

\[
\begin{aligned}
x+1.20&=1.80\\
x&=0.60
\end{aligned}
\]

Therefore, the original mixture contained:

\[
n(\mathrm{Fe}^{2+})=0.60\ \mathrm{mmol}
\]

\[
n(\mathrm{C_{2}O_{4}}^{2-})=1.20\ \mathrm{mmol}
\]

Step 6: Investigate the hydrogen ion measurement.

This is where the problem becomes more interesting.

For iron(II), the balanced overall equation is:

\[
\ce{MnO4^- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+}
\]

One mole of permanganate consumes eight moles of hydrogen ions.

For oxalate, the balanced equation is:

\[
\ce{2MnO4^- + 16H+ + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2}
\]

Here, two moles of permanganate consume 16 moles of hydrogen ions.

That is still eight moles of hydrogen ions per mole of permanganate.

Therefore, regardless of how much of the reducing capacity comes from iron(II) or oxalate:

\[
n(\mathrm{H}^{+})=8n(\mathrm{MnO_{4}}^{-})
\]

Using the measured amount of permanganate:

\[
\begin{aligned}
n(\mathrm{H}^{+})
&=8\times0.600\\
&=4.80\ \mathrm{mmol}
\end{aligned}
\]

This agrees exactly with the supplied acid-balance measurement.

However, it does not tell us anything new about the relative amounts of iron(II) and oxalate.

The acid measurement is consistent with the reaction, but it is mathematically redundant.

The tempting approach is to treat the electron balance and acid balance as two independent equations. They are not independent because both are controlled by the same amount of permanganate.

The actual second independent equation comes from the known total amount of reducing particles.

This is a useful distinction in more advanced redox problems: two measurements do not necessarily provide two independent pieces of information.

09Where to go next

Once you can combine supplied half-equations reliably, the next skill is using their coefficients to solve more involved stoichiometry problems.

The balanced equation gives you the mole ratios needed for titrations, limiting-reagent calculations, and unknown concentration problems.

For a broader comparison of the oxidation-number method and the half-equation method, see How to Balance Redox Equations Without Guessing.

The key habit to carry into your next question is simple: check the medium, match the electrons, cancel common species, and verify both atoms and charge before using the equation in a calculation.