Redox Reactions · Lesson 6 of 8

Balancing Redox Equations by Half Reactions

“Build balanced oxidation and reduction halves, then combine them so every transferred electron is accounted for.”

Learning Objectives

• Separate a skeletal redox equation into oxidation and reduction halves. • Balance atoms and charge using water, hydrogen ions, and electrons. • Equalise electron counts and cancel common species correctly. • Convert an acidic balance into the corresponding basic-medium form. • Explain why atom, charge, and electron checks are all necessary.

Separate the bookkeeping into two manageable parts

The half-reaction method makes electron transfer visible in the equation. We balance the oxidation and reduction separately, then multiply them so that their electrons cancel. The conservation rules are the same as in the oxidation-number method; the organisation of the work is different. This is why both methods should give the same final equation for the same specified reaction.

For an acidic half reaction, first balance atoms other than oxygen and hydrogen. Add water to balance oxygen, hydrogen ions to balance hydrogen, and electrons to balance charge. An oxidation half has electrons among its products; a reduction half has electrons among its reactants. If the electron appears on the wrong side, revisit the charge calculation.

Dichromate oxidises iron in acid

An iron(II) ion can lose one electron to become iron(III). Dichromate can accept electrons and become chromium(III) in acid. The difficulty is that these changes involve different numbers of electrons per formula unit. Balancing the halves separately makes the required ratio clear.

Example — From two halves to a complete equation

Problem
Balance $F e^{2 +} + C r_{2} O_{7}^{2 -} \to F e^{3 +} + C r^{3 +}$ in acid.

1.The oxidation half is $F e^{2 +} \to F e^{3 +} + e^{-}$ . Adding one electron to the right makes the charge +2 on both sides.
2.For reduction, balance chromium first: $C r_{2} O_{7}^{2 -} \to 2C r^{3 +}$ . Add seven waters to the right for oxygen, then fourteen hydrogen ions to the left for hydrogen.
3.This gives $C r_{2} O_{7}^{2 -} + 14 H^{+} \to 2C r^{3 +} + 7 H_{2} O$ . Charge is +12 on the left and +6 on the right, so add six electrons to the left.
4.The balanced reduction half is $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2C r^{3 +} + 7 H_{2} O$ .
5.Multiply the iron half by 6: $6F e^{2 +} \to 6F e^{3 +} + 6 e^{-}$ . Add the halves and cancel six electrons.
6.The final equation is $6F e^{2 +} + C r_{2} O_{7}^{2 -} + 14 H^{+} \to 6F e^{3 +} + 2C r^{3 +} + 7 H_{2} O$ .
7.Check the total charge: left +12 - 2 + 14 = +24; right +18 + 6 = +24. Iron, chromium, hydrogen, and oxygen are also balanced.

The six electrons do not survive in the overall equation because they are transferred internally between the reacting partners. Their cancellation is more than a notational convenience: it confirms that one half supplies exactly what the other consumes. Multiply every coefficient in a half reaction when equalising electrons, including water and hydrogen ions.

Matching the electron counts— One dichromate reduction consumes the electrons released by six iron(II) oxidations. The electrons cancel in the overall equation.

Convert an acidic balance into a basic one

A convenient way to balance in base is to first balance oxygen and hydrogen as though the medium were acidic. Then add the same number of hydroxide ions to both sides for every hydrogen ion present. Hydrogen and hydroxide on the same side form water. Finally cancel any water common to the two sides. This preserves the balance while removing the hydrogen ions that should not remain in the final basic equation.

Example — Permanganate oxidises iodide in base

Problem
Balance $Mn O_{4}^{-} + I^{-} \to Mn O_{2} + I_{2}$ in basic solution.

1.The oxidation half is $2 I^{-} \to I_{2} + 2 e^{-}$ . Both iodine atoms and charge are balanced.
2.Balance the reduction half in acid first: $Mn O_{4}^{-} + 4 H^{+} + 3 e^{-} \to Mn O_{2} + 2 H_{2} O$ . Manganese decreases from +7 to +4, so three electrons are accepted.
3.Add $4 O H^{-}$ to both sides. On the left, $4 H^{+} + 4O H^{-}$ becomes four waters. Cancel two waters common to both sides.
4.The basic reduction half is $Mn O_{4}^{-} + 2 H_{2} O + 3 e^{-} \to Mn O_{2} + 4O H^{-}$ . Its charge is -4 on each side.
5.The smallest common electron count is 6. Multiply the iodine half by 3 and the manganese half by 2.
6.Add and cancel electrons to obtain $2Mn O_{4}^{-} + 6 I^{-} + 4 H_{2} O \to 2Mn O_{2} + 3 I_{2} + 8O H^{-}$ .
7.Check: Mn 2, I 6, O 12, and H 8 on each side. Total charge is -8 on each side.

Compare this result with the earlier permanganate–bromide example. Manganese ends at +4 in both, but iodide ends as iodine while bromide in that example ended as bromate. The halogen electron changes are therefore different. A reagent name alone cannot determine the coefficients; the products and medium are part of the problem.

A peroxide can be the oxidant

Hydrogen peroxide contains oxygen at -1. It can accept electrons and become water, where oxygen is -2. This case is useful because the peroxide exception and the electron balance reinforce each other. Iron(II) provides the electrons.

Example — Hydrogen peroxide with iron(II)

Problem
Balance $H_{2} O_{2} + F e^{2 +} \to F e^{3 +} + H_{2} O$ in acid.

1.For reduction, balance oxygen with two waters: $H_{2} O_{2} \to 2 H_{2} O$ . Add two hydrogen ions to the left to balance hydrogen.
2.The left charge is +2 and the right charge zero. Add two electrons to the left: $H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O$ .
3.The oxidation half is $F e^{2 +} \to F e^{3 +} + e^{-}$ . Multiply it by 2.
4.Combine to obtain $H_{2} O_{2} + 2 H^{+} + 2F e^{2 +} \to 2F e^{3 +} + 2 H_{2} O$ .
5.The peroxide is the oxidant and iron(II) the reductant. Charge is +6 on each side, and every atom is conserved.

The product ion determines the hydrogen balance

Sometimes the specified sulphur product is hydrogen sulphate rather than sulphate. These are different chemical species, so the hydrogen-ion and water coefficients must reflect the actual product. We will balance the source’s acidic sulphur dioxide example with hydrogen sulphate explicitly present.

Example — Sulphur dioxide becomes hydrogen sulphate

Problem
Balance $Mn O_{4}^{-} + S O_{2} \to M n^{2 +} + HS O_{4}^{-}$ in acid.

1.The reduction half is $Mn O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ .
2.For sulphur oxidation, balance oxygen and hydrogen: $S O_{2} + 2 H_{2} O \to HS O_{4}^{-} + 3 H^{+}$ . The right charge is +2, so add two electrons to the right.
3.Thus $S O_{2} + 2 H_{2} O \to HS O_{4}^{-} + 3 H^{+} + 2 e^{-}$ . Sulphur increases from +4 to +6, matching two electrons released.
4.Use ten electrons: multiply the manganese half by 2 and the sulphur half by 5. Add them and cancel the electrons.
5.Cancel fifteen hydrogen ions from the sixteen on the left and eight waters from the ten on the left. The result is $2Mn O_{4}^{-} + 5S O_{2} + 2 H_{2} O + H^{+} \to 2M n^{2 +} + 5HS O_{4}^{-}$ .
6.Left and right charges are both -1. Mn is 2, S is 5, H is 5, and O is 20 on each side.

Common mistake

Never remove electrons by crossing out unequal numbers. First multiply complete half reactions to make the electron counts equal. Also, adding hydroxide only to the side containing hydrogen ions changes the charge balance; add it to both sides before combining hydrogen and hydroxide into water.

A final equation should use the smallest common whole-number coefficients. Cancelling common species is permitted because those species occur in equal quantities on both sides. It is not permitted to cancel atoms inside different compounds: an oxygen atom in permanganate is not a separate oxygen reactant that can simply be deleted.

Quick check

In an acidic half reaction, what is normally used to balance oxygen atoms?

Quick check

Why are electrons added to the left of a reduction half reaction?

Quick check

Two half reactions release 2 electrons and consume 3 electrons respectively. What is the smallest matched count?

Quick check

How is an acidic balance converted to basic conditions?

Quick check

How many iron(II) ions are oxidised per $H_{2} O_{2}$ reduced to water?

Practice Problems

Balance $Mn O_{4}^{-} + S O_{2} \to M n^{2 +} + S O_{4}^{2 -}$ in acid, then compare it with the hydrogen-sulphate version.
Balance $C r_{2} O_{7}^{2 -} + S O_{2} \to C r^{3 +} + S O_{4}^{2 -}$ in acid using half reactions.
Balance $M n^{3 +} \to M n^{2 +} + Mn O_{2}$ in water using half reactions and identify the source of the hydrogen ions.
Balance $N_{2} H_{4} + Cl O_{3}^{-} \to NO + C l^{-}$ in basic medium.
Balance $C l_{2} O_{7} + H_{2} O_{2} \to Cl O_{2}^{-} + O_{2}$ in basic medium. Determine whether peroxide is the oxidant or reductant.
Rework the phosphorus disproportionation equation by half reactions and compare its coefficients with the oxidation-number result.
For one equation above, show separate atom, charge, and electron checks rather than only writing a final answer.

Key Takeaways

Balance each half reaction independently before combining it with the other. In acid, use water for oxygen, hydrogen ions for hydrogen, and electrons for charge. Convert to base by adding hydroxide to both sides, forming water, and cancelling it where appropriate. Equalise complete half reactions so electrons cancel exactly. The final coefficients depend on the specified products and medium. Both balancing methods express conservation of atoms, charge, and transferred electrons.

Previous · Lesson 5

Balancing Redox Equations by Oxidation Numbers

Next · Lesson 7

Redox in Titrations and Electrochemical Cells

Separate the bookkeeping into two manageable parts

Dichromate oxidises iron in acid

Convert an acidic balance into a basic one

A peroxide can be the oxidant

The product ion determines the hydrogen balance

Quiz

Practice Problems

Key Takeaways