How to Balance a Redox Half-Equation in Acidic Solution

Add hydrogen peroxide to acidified potassium iodide, then add a few drops of starch. The initially colourless mixture turns blue-black. You can use the reaction to identify iodine, but how do you write the half-equation that explains where the electrons went?

The difficulty is that balancing atoms is only half the job. A redox half-equation must conserve both the number of atoms and the total electric charge, including the electrons transferred.

For an acidic aqueous solution, there is a reliable method: balance the other atoms first, use water to balance oxygen, use hydrogen ions to balance hydrogen, and finally use electrons to balance charge.

01Start with the experimental evidence

Imagine a test tube containing aqueous potassium iodide, \(\ce{KI}\), and dilute sulfuric acid. Hydrogen peroxide, \(\ce{H2O2}\), is added, followed by a few drops of starch solution.

The solution turns blue-black because iodine has formed. In the presence of excess iodide ions, some of this iodine exists as triiodide ions, which also produce the characteristic starch colour.

Before writing an equation, predict which substance loses electrons.

Iodide ions are converted into iodine molecules:

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

Each iodide ion starts with a charge of \(-1\). In an iodine molecule, each iodine atom has an oxidation number of \(0\).

The iodide ions therefore lose electrons. This is oxidation.

But those electrons must be accepted by another substance. Hydrogen peroxide is reduced to water.

The corresponding reduction half-equation is:

\[ \ce{H2O2 -> H2O} \]

There is a problem, though. This equation is not balanced. It contains different numbers of hydrogen and oxygen atoms on each side, and it does not show the electron transfer.

We need to balance it properly.

What the experiment tells us

The two half-equations describe separate parts of the same reaction:

Oxidation of iodide:

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

Reduction of hydrogen peroxide:

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

The electrons released by iodide are accepted by hydrogen peroxide.

Adding the half-equations gives the overall ionic equation:

\[ \ce{H2O2 + 2H+ + 2I^- -> I2 + 2H2O} \]

This explains the experimental observation: iodine forms, producing the blue-black starch colour.

Notice that the acid is important. The hydrogen ions provide the hydrogen needed to form water.

The colour change is evidence that iodine has formed. It does not directly show electrons moving between individual particles. The half-equations provide the mathematical model that accounts for the electron transfer.

02How to balance a redox half-equation in acidic solution

For an aqueous reaction in acidic conditions, use the following sequence.

The order matters. Water introduces both hydrogen and oxygen, so balance oxygen before hydrogen. Electrons come last because adding hydrogen ions changes the total charge.

Let’s examine why each step works.

Step 1: Balance all elements except hydrogen and oxygen

Start with the chemical species given in the question.

Balance elements other than hydrogen and oxygen using coefficients. A coefficient changes the number of particles, not the identity of the substance.

For example, if you need two chlorine atoms, write \(\ce{2Cl}\), not a different chemical formula.

If the element is already balanced, leave it alone.

Step 2: Balance oxygen using water

In acidic aqueous solutions, water molecules can be added to either side of the half-equation.

Each water molecule contains one oxygen atom and two hydrogen atoms.

Add enough \(\ce{H2O}\) to make the oxygen atom counts equal.

Do not use hydrogen ions to balance oxygen. Hydrogen ions contain no oxygen.

Step 3: Balance hydrogen using hydrogen ions

Once oxygen is balanced, count the hydrogen atoms on each side.

Add \(\mathrm{H}^{+}\) to whichever side needs more hydrogen atoms.

Remember that each hydrogen ion contributes:

• One hydrogen atom.
• A charge of \(+1\).

This second point will matter in the next step.

Step 4: Balance charge using electrons

Now calculate the total charge on each side.

An electron has a charge of \(-1\). Add electrons to the side with the greater positive charge until both sides have equal net charge.

For example, suppose the left side has a total charge of \(+5\), while the right side has a total charge of \(+2\).

The left side is three units more positive, so add three electrons to the left:

\[ +5-3=+2 \]

Both sides now have a total charge of \(+2\).

The position of the electrons also identifies the type of reaction:

• Electrons on the left: reduction, because electrons are gained.
• Electrons on the right: oxidation, because electrons are lost.

Step 5: Check atoms and charge independently

A balanced half-equation must satisfy two separate conditions.

1. Every element must have the same number of atoms on both sides.
2. The total electric charge must be identical on both sides.

Do not assume that balancing the atoms automatically balances the charge.

04An important contrast: Hydrogen peroxide can also be oxidised

Here is a misconception worth correcting before it becomes a habit.

You might assume that hydrogen peroxide always gains electrons because it is commonly used as an oxidising agent.

That is not correct. Whether a substance is oxidised or reduced depends on the reaction it undergoes.

Consider a different half-equation in which hydrogen peroxide forms oxygen gas:

\[ \ce{H2O2 -> O2} \]

Balance it in acidic solution.

The oxygen atoms are already balanced: there are two on each side.

There are two hydrogen atoms on the left and none on the right, so add two hydrogen ions to the right:

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

The right side now has a charge of \(+2\), while the left side is neutral.

Add two electrons to the right:

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

This time, hydrogen peroxide releases two electrons. It is oxidised.

Compare the two possibilities:

The key point is that the starting substance alone does not determine the half-equation. You must know what product it forms in the reaction being considered.

This is particularly important in redox questions where hydrogen peroxide reacts with different oxidising or reducing agents.

06What a quantitative experiment adds

The starch test tells us that iodine has formed, but it does not tell us how much hydrogen peroxide reacted.

To measure the amount, the iodine can be collected in a known reaction mixture and titrated using a standard sodium thiosulfate solution.

The apparatus is arranged with a burette containing sodium thiosulfate above a conical flask containing the iodine solution. The titrant is added gradually while the flask is swirled. Starch is used near the endpoint, and the blue-black colour disappears when the iodine has been consumed.

Representing iodine as \(\ce{I2}\), the titration reaction is:

\[ \ce{I2 + 2S2O3^2- -> 2I^- + S4O6^2-} \]

This establishes a mole ratio of:

\[ 2\text{ mol }\mathrm{S_{2}O_{3}}^{2-}: 1\text{ mol }\ce{I2} \]

From the original redox reaction, one mole of hydrogen peroxide produces one mole of iodine.

Therefore, the amount of iodine measured by titration can be used to calculate the original amount of hydrogen peroxide.

Worked measurement: Calculate the hydrogen peroxide concentration

Illustrative data:

A \(25.00\text{ mL}\) aliquot of hydrogen peroxide solution is reacted with excess acidified potassium iodide. The iodine produced requires \(24.60\text{ mL}\) of \(0.1000\text{ mol L}^{-1}\) sodium thiosulfate to reach the endpoint.

Calculate the hydrogen peroxide concentration.

Step 1: Calculate the moles of thiosulfate used.

Use:

\[ n=cV \]

where \(n\) is the amount in moles, \(c\) is concentration in \(\text{mol L}^{-1}\), and \(V\) is volume in litres.

Convert the titre:

\[ 24.60\text{ mL}=0.02460\text{ L} \]

Then:

\[ \begin{aligned} n(\mathrm{S_{2}O_{3}}^{2-}) &=0.1000\times0.02460\\ &=0.002460\text{ mol} \end{aligned} \]

Step 2: Calculate the moles of iodine.

The thiosulfate-to-iodine ratio is \(2:1\).

\[ n(\ce{I2})=\frac{0.002460}{2} \]

\[ n(\ce{I2})=0.001230\text{ mol} \]

Step 3: Calculate the moles of hydrogen peroxide.

The original redox reaction has a \(1:1\) mole ratio between hydrogen peroxide and iodine.

Therefore:

\[ n(\ce{H2O2})=0.001230\text{ mol} \]

Step 4: Calculate the concentration of hydrogen peroxide.

The original aliquot has a volume of \(0.02500\text{ L}\).

\[ \begin{aligned} c&=\frac{n}{V}\\ &=\frac{0.001230}{0.02500}\\ &=0.04920\text{ mol L}^{-1} \end{aligned} \]

The hydrogen peroxide concentration is:

\[ \boxed{0.04920\text{ mol L}^{-1}} \]

This calculation assumes that iodide and acid are in excess, all hydrogen peroxide reacts, and no iodine is lost before titration. The numerical data are illustrative, not measurements from an actual experiment.

The important connection is that a correctly balanced half-equation establishes the electron and mole relationships needed to interpret experimental measurements.

07Questions and solutions

Question 1

Balance the following reduction half-equation in acidic aqueous solution:

\[ \ce{NO2^- -> NO} \]

Solution 1

The balanced half-equation is:

\[ \boxed{\ce{NO2^- + 2H+ + e^- -> NO + H2O}} \]

Start by balancing oxygen. There are two oxygen atoms on the left and one on the right, so add one water molecule to the right:

\[ \ce{NO2^- -> NO + H2O} \]

The right side now contains two hydrogen atoms. Add two hydrogen ions to the left:

\[ \ce{NO2^- + 2H+ -> NO + H2O} \]

The left-hand charge is:

\[ -1+2=+1 \]

The right-hand charge is zero.

Add one electron to the left to reduce its total charge to zero.

Both sides now contain one nitrogen atom, two oxygen atoms, and two hydrogen atoms. Their total charges are also equal.

Nitrogen changes from \(+3\) in nitrite to \(+2\) in nitric oxide, gaining one electron. This confirms the result.

Question 2

A student is asked to balance the following oxidation half-equation in acidic solution:

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

They add water and hydrogen ions because the reaction occurs in acid.

Write the correctly balanced half-equation and explain whether water or hydrogen ions are needed.

Solution 2

The correctly balanced half-equation is:

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

First, balance carbon. There are two carbon atoms in the oxalate ion, so add a coefficient of two to carbon dioxide:

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

There are now four oxygen atoms on each side. Hydrogen does not appear in either species, so there is no need to introduce water or hydrogen ions.

Next, balance charge.

The left side has a charge of \(-2\), while the right side has a charge of zero.

Add two electrons to the right:

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

The total charge is now \(-2\) on both sides.

The student has made a common mistake: assuming that acidic conditions require hydrogen ions to appear in every half-equation.

They do not. Acidic conditions allow hydrogen ions and water to be used when needed to balance atoms. They should never be added without a reason.

The electrons appear on the right, so oxalate is oxidised.

Question 3

Balance the following reduction half-equation in acidic aqueous solution:

\[ \ce{BrO3^- -> Br^-} \]

Use oxidation numbers to independently verify the number of electrons required.

Solution 3

The balanced half-equation is:

\[ \boxed{\ce{BrO3^- + 6H+ + 6e^- -> Br^- + 3H2O}} \]

Bromine is already balanced, with one atom on each side.

There are three oxygen atoms on the left and none on the right. Add three water molecules to the right:

\[ \ce{BrO3^- -> Br^- + 3H2O} \]

These water molecules contain six hydrogen atoms, so add six hydrogen ions to the left:

\[ \ce{BrO3^- + 6H+ -> Br^- + 3H2O} \]

Now calculate the charge.

Left:

\[ -1+6=+5 \]

Right:

\[ -1 \]

The left side is six units more positive, so add six electrons to the left.

The final charge on each side is \(-1\).

To verify the electron coefficient, calculate bromine’s oxidation number in bromate.

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

\[ \begin{aligned} x+3(-2)&=-1\\ x-6&=-1\\ x&=+5 \end{aligned} \]

Bromine changes from \(+5\) to \(-1\), a decrease of six.

It therefore gains six electrons, confirming the half-equation.

Question 4

Hydrogen sulfide, \(\ce{H2S}\), is oxidised to sulfate ions, \(\mathrm{SO_{4}}^{2-}\), in acidic aqueous solution.

Write the balanced oxidation half-equation.

Explain why hydrogen ions appear on the product side, even though the reaction occurs in acidic solution.

Solution 4

The balanced oxidation half-equation is:

\[ \boxed{\ce{H2S + 4H2O -> SO4^2- + 10H+ + 8e^-}} \]

Start by balancing sulfur. There is one sulfur atom on each side.

Next, balance oxygen. Sulfate contains four oxygen atoms, while hydrogen sulfide contains none.

Add four water molecules to the left:

\[ \ce{H2S + 4H2O -> SO4^2-} \]

The left side now contains ten hydrogen atoms: two from hydrogen sulfide and eight from the four water molecules.

Add ten hydrogen ions to the right:

\[ \ce{H2S + 4H2O -> SO4^2- + 10H+} \]

Now balance charge.

Left-hand charge:

\[ 0 \]

Right-hand charge:

\[ -2+10=+8 \]

The right side is eight units more positive, so add eight electrons to the right:

\[ \ce{H2S + 4H2O -> SO4^2- + 10H+ + 8e^-} \]

The final equation conserves all atoms and charge.

Hydrogen ions appear on the product side because the hydrogen atoms must balance there. The acidic medium does not dictate that hydrogen ions must always be reactants.

There is also an independent electron check. Sulfur has an oxidation number of \(-2\) in hydrogen sulfide and \(+6\) in sulfate.

Its oxidation number increases by eight, meaning it loses eight electrons.

The eight electrons on the product side are therefore correct.

Question 5

An acidic reaction mixture contains:

• \(0.00400\text{ mol}\) of \(\mathrm{MnO_{4}}^{-}\)
• \(0.0120\text{ mol}\) of \(\ce{H2O2}\)

The permanganate ions are reduced to \(\mathrm{Mn}^{2+}\), and all the permanganate reacts.

A gas syringe collects \(0.248\text{ L}\) of dry oxygen gas at \(100\text{ kPa}\) and \(298\text{ K}\).

Assume:

• Acid is present in excess.
• Hydrogen peroxide is the only source of the oxygen gas.
• The gas behaves ideally.
• The reaction proceeds according to the specified products.

Determine:

(a) Whether hydrogen peroxide is oxidised or reduced.

(b) The balanced overall ionic equation.

(c) The amount of hydrogen peroxide remaining after the reaction.

(d) Whether the measured oxygen volume is consistent with the balanced equation.

Solution 5

Hydrogen peroxide is oxidised to oxygen gas, the balanced overall ionic equation is:

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

The reaction consumes \(0.0100\text{ mol}\) of hydrogen peroxide, leaving \(0.0020\text{ mol}\). The measured oxygen volume is consistent with the equation.

The important part is recognising which half-equation hydrogen peroxide must follow.

(a) Identify the electron transfer

Oxygen gas is an observed product. Therefore, hydrogen peroxide must be oxidised to oxygen.

Its oxidation half-equation is:

\[ \ce{H2O2 -> O2} \]

Balance oxygen. There are already two oxygen atoms on each side.

Balance hydrogen by adding two hydrogen ions to the right:

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

The right-hand charge is now \(+2\), so add two electrons to the right:

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

Hydrogen peroxide releases two electrons.

(b) Construct the balanced overall ionic equation

First, balance the permanganate reduction half-equation.

Start with:

\[ \ce{MnO4^- -> Mn^2+} \]

Balance oxygen by adding four water molecules to the right:

\[ \ce{MnO4^- -> Mn^2+ + 4H2O} \]

Balance hydrogen by adding eight hydrogen ions to the left:

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

The left-hand charge is \(+7\), while the right-hand charge is \(+2\).

Add five electrons to the left:

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

Now we have two balanced half-equations:

Reduction:

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

Oxidation:

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

The first half-equation consumes five electrons, while the second produces two.

The lowest common multiple of five and two is ten.

Multiply the permanganate half-equation by two:

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

Multiply the hydrogen peroxide half-equation by five:

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

Add the equations and cancel the ten electrons.

There are sixteen hydrogen ions on the left and ten on the right, leaving six on the left.

The resulting equation is:

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

This equation conserves all atoms and charge.

The left-hand charge is:

\[ 2(-1)+6(+1)=+4 \]

The right-hand charge is:

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

Both sides have the same total charge.

(c) Calculate the hydrogen peroxide remaining

From the balanced equation:

\[ 2\text{ mol }\mathrm{MnO_{4}}^{-} : 5\text{ mol }\ce{H2O2} \]

Therefore, \(0.00400\text{ mol}\) of permanganate requires:

\[ \begin{aligned} n(\ce{H2O2}) &=0.00400\times\frac{5}{2}\\ &=0.0100\text{ mol} \end{aligned} \]

The initial amount of hydrogen peroxide was \(0.0120\text{ mol}\).

Subtract the amount consumed:

\[ \begin{aligned} n_{\text{remaining}} &=0.0120-0.0100\\ &=0.0020\text{ mol} \end{aligned} \]

Therefore, \(0.0020\text{ mol}\) of hydrogen peroxide remains.

Permanganate is the limiting reactant because the available hydrogen peroxide exceeds the amount required by the balanced equation.

(d) Check the oxygen gas measurement

The balanced equation predicts that two moles of permanganate produce five moles of oxygen.

Therefore:

\[ \begin{aligned} n(\ce{O2}) &=0.00400\times\frac{5}{2}\\ &=0.0100\text{ mol} \end{aligned} \]

Now calculate the amount of oxygen using the ideal gas equation:

\[ PV=nRT \]

Rearrange:

\[ n=\frac{PV}{RT} \]

Use the supplied measurements:

• \(P=100\,000\text{ Pa}\)
• \(V=0.248\text{ L}=2.48\times10^{-4}\text{ m}^3\)
• \(R=8.314\text{ J mol}^{-1}\text{ K}^{-1}\)
• \(T=298\text{ K}\)

Substitute:

\[ \begin{aligned} n&=\frac{PV}{RT}\\ &=\frac{100000\times2.48\times10^{-4}} {8.314\times298}\\ &=0.0100\text{ mol} \end{aligned} \]

The measured amount of oxygen is \(0.0100\text{ mol}\), matching the amount predicted by the balanced equation.

The experimental data are therefore consistent with the proposed reaction.

There is a tempting alternative here. A student might remember the earlier reduction of hydrogen peroxide to water and use:

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

But permanganate is also accepting electrons. Both half-equations would place electrons on the reactant side, meaning there would be no electrons available to cancel.

More importantly, that alternative predicts water rather than oxygen as the hydrogen peroxide product.

The observed oxygen gas resolves the ambiguity. Hydrogen peroxide must be the electron donor in this reaction, not the electron acceptor.

This is why identifying the actual products is just as important as knowing the balancing sequence.

Once you can balance each half-equation independently, the next challenge is combining oxidation and reduction so that the electrons cancel completely.

For the broader method of identifying oxidation and reduction before balancing, read How to Balance Redox Equations Without Guessing.

When the two half-equations contain different numbers of electrons, continue with How to Combine Redox Half-Equations with Different Electron Numbers.