Redox Reactions · Lesson 5 of 8

Balancing Redox Equations by Oxidation Numbers

“Balance electron-equivalent changes first, then conserve charge and atoms in acidic and basic solutions.”

Learning Objectives

• Find the atoms whose oxidation numbers change in a skeletal equation. • Count changes per atom and per formula unit correctly. • Choose coefficients that equalise total oxidation and reduction. • Complete ionic equations using the stated acidic or basic medium. • Check the conservation of every element and of total charge.

A balanced equation must account for more than atoms

A skeletal equation tells us which substances react and which products form, but it does not yet state their correct proportions. In redox reactions, those proportions must match the electrons lost and gained. The oxidation-number method finds that match without first writing separate half reactions. It then uses the medium to finish the atom and charge balance.

Begin with correct chemical formulas and known products. Balancing can change coefficients in front of formulas; it cannot change a sulphite ion into a sulphate ion merely to fix an oxygen count. Nor does balancing predict which product a reaction should form under an unspecified set of conditions. Product identification comes before coefficient calculation.

Definition

Skeletal ionic equation

An equation containing the correct reactant and product ions or molecules before their coefficients have all been balanced. Ions that do not change can be omitted as spectators.

There are three linked checks: each element must be conserved, total charge must be conserved, and the total increase in oxidation number must equal the total decrease. The last condition expresses electron balance. If two atoms of the same element occur in an ion, include both in the electron-equivalent count.

Electron-equivalent balanceLaTeX

total increase in oxidation number = total decrease in oxidation number

Multiply the change per atom by the number of changing atoms and by the coefficient of the species.

Build the coefficients before balancing the medium

The method is easiest to use in a fixed order. Assign oxidation numbers and identify the changed atoms. Count each total increase and decrease, choose multipliers to make them equal, and balance the changing elements. For the ionic reactions below, then balance charge using hydrogen ions in acid or hydroxide ions in base; add water to complete hydrogen and oxygen. Finish by checking all atoms and charges.

Common mistake

Check how many atoms change within one formula unit. Chromium falls by 3 per atom when dichromate becomes chromium(III), but one dichromate ion contains two chromium atoms. Its total decrease is 6, not 3.

An acidic reaction: dichromate and sulphite

Potassium dichromate and sodium sulphite provide reacting ions in water. Potassium and sodium remain spectator ions. In acidic solution, dichromate is reduced to chromium(III), and sulphite is oxidised to sulphate. These specified products let us write and balance the net ionic equation.

Example — Dichromate oxidises sulphite

Problem
Balance $C r_{2} O_{7}^{2 -} + S O_{3}^{2 -} \to C r^{3 +} + S O_{4}^{2 -}$ in acid.

1.Assign states: chromium is +6 in dichromate and +3 in $C r^{3 +}$ ; sulphur is +4 in sulphite and +6 in sulphate.
2.Each chromium decreases by 3, so the two chromium atoms in one dichromate give a total decrease of 6. Each sulphur increases by 2. Three sulphite ions therefore supply a total increase of 6.
3.Balance these elements: $C r_{2} O_{7}^{2 -} + 3S O_{3}^{2 -} \to 2C r^{3 +} + 3S O_{4}^{2 -}$ .
4.Left charge is $- 2 + 3 (- 2) = - 8$ ; right charge is $2 (+ 3) + 3 (- 2) = 0$ . Add $8 H^{+}$ to the left, making its charge zero.
5.Eight hydrogen ions require four water molecules on the right. The result is $C r_{2} O_{7}^{2 -} + 3S O_{3}^{2 -} + 8 H^{+} \to 2C r^{3 +} + 3S O_{4}^{2 -} + 4 H_{2} O$ .
6.Check: Cr 2, S 3, O 16, and H 8 on each side; both total charges are zero. Dichromate is the oxidant and sulphite the reductant.

Notice why water is added only after the redox ratio is established. The three-to-one sulphite:dichromate ratio follows from oxidation-state changes. Water and hydrogen ions complete the environmental balance; they do not replace that electron accounting. This separates the essential redox proportion from the details of acidic solution.

A basic reaction: permanganate and bromide

A different medium can produce a different reduction product. In the next specified reaction, permanganate becomes manganese dioxide rather than manganese(II), and bromide becomes bromate. The manganese change is therefore +7 to +4, not +7 to +2. Always read the stated products before counting the electron equivalents.

Example — Permanganate produces bromate

Problem
Balance $Mn O_{4}^{-} + B r^{-} \to Mn O_{2} + Br O_{3}^{-}$ in basic medium.

1.Manganese decreases from +7 to +4, a decrease of 3 per Mn atom. Bromine increases from -1 to +5, an increase of 6 per Br atom.
2.Two permanganate ions match one bromide ion: $2Mn O_{4}^{-} + B r^{-} \to 2Mn O_{2} + Br O_{3}^{-}$ . Both total changes are six.
3.The left charge is -3 and the right charge -1. Since the medium is basic, add $2 O H^{-}$ to the right to make both sides -3.
4.The right side now has two H atoms. Add one $H_{2} O$ to the left: $2Mn O_{4}^{-} + B r^{-} + H_{2} O \to 2Mn O_{2} + Br O_{3}^{-} + 2O H^{-}$ .
5.Check oxygen: left 8 + 1 = 9, right 4 + 3 + 2 = 9. Hydrogen is 2 on each side, and Mn, Br, and total charge also match.

A basic equation can contain hydroxide as a product even though the reaction occurs in a basic solution. “Basic medium” tells us which balancing species are appropriate; it does not require all hydroxide to appear on the reactant side. What matters is the balanced net change.

When one reactant supplies both redox partners

The same electron-equivalent reasoning works in disproportionation. We follow the atoms ending in the lower state and the atoms ending in the higher state. Their numbers must make the total increase and decrease equal. This is a useful way to derive product ratios rather than memorise them.

Example — Deriving the phosphorus product ratio

Problem
Balance the alkaline disproportionation $P_{4} \to P H_{3} + H_{2} P O_{2}^{-}$ .

1.Phosphorus begins at 0. It becomes -3 in phosphine and +1 in hypophosphite. One phosphorus atom reduced to -3 requires three other phosphorus atoms each oxidised to +1.
2.The four phosphorus atoms therefore give $P_{4} \to P H_{3} + 3 H_{2} P O_{2}^{-}$ . The product ratio follows from electron balance.
3.The right charge is -3. Add $3 O H^{-}$ to the left, giving $P_{4} + 3O H^{-} \to P H_{3} + 3 H_{2} P O_{2}^{-}$ .
4.The right side has nine hydrogen atoms and six oxygen atoms; the left has three of each. Three water molecules on the left supply the missing six H and three O.
5.Thus $P_{4} + 3O H^{-} + 3 H_{2} O \to P H_{3} + 3 H_{2} P O_{2}^{-}$ . P is 4, H is 9, O is 6, and charge is -3 on each side.

Apply the method to a useful treatment process

After chlorination, excess dissolved chlorine can be removed by a reducing agent such as sulphur dioxide. The reaction links a practical water-treatment context to familiar oxidation-number changes. We can represent the net change using sulphate and chloride in water.

Example — Removing excess chlorine

Problem
Balance $C l_{2} + S O_{2} + H_{2} O \to C l^{-} + S O_{4}^{2 -} + H^{+}$ .

1.Each chlorine changes from 0 to -1, so one chlorine molecule accepts two electron equivalents. Sulphur changes from +4 to +6 and supplies two electron equivalents. Their ratio is therefore one to one.
2.Balance the chlorine atoms: $C l_{2} + S O_{2} \to 2C l^{-} + S O_{4}^{2 -}$ . The right charge is -4, so add $4 H^{+}$ to the right.
3.Add $2 H_{2} O$ to the left to supply four H atoms and complete oxygen: $C l_{2} + S O_{2} + 2 H_{2} O \to 2C l^{-} + S O_{4}^{2 -} + 4 H^{+}$ .
4.The total charge is zero on both sides. This equation describes the net redox change; control of treatment quantities is a separate practical requirement.

Common mistake

Do not stop when the oxidation-number changes match. An equation can have the correct redox ratio and still have unequal charge or unequal hydrogen and oxygen counts. Check every element and the signed charge sum before accepting it.

Quick check

What is the total chromium oxidation-number decrease for one dichromate ion becoming two chromium(III) ions?

Quick check

Why are three sulphite ions needed per dichromate ion in the acidic example?

Quick check

For $Mn O_{4}^{-} \to Mn O_{2}$ , how many electron equivalents are gained per Mn?

Quick check

Which change is never an acceptable way to balance an equation?

Quick check

In phosphorus disproportionation to states -3 and +1, what is the ratio of atoms reduced to atoms oxidised?

Practice Problems

Balance $Mn O_{4}^{-} + F e^{2 +} \to M n^{2 +} + F e^{3 +}$ in acid using oxidation numbers.
Balance $C r_{2} O_{7}^{2 -} + S O_{2} \to C r^{3 +} + S O_{4}^{2 -}$ in acid and identify both agents.
Rederive the basic permanganate–bromide equation, explaining why its manganese coefficient is 2.
Balance $C l_{2} + O H^{-} \to C l^{-} + Cl O^{-} + H_{2} O$ and explain its one-to-one chlorine-atom product ratio.
A student writes $C r_{2} O_{7}^{2 -} + 3S O_{3}^{2 -} \to 2C r^{3 +} + 3S O_{4}^{2 -}$ and stops. Identify every remaining imbalance.
Explain why changing a product from manganese dioxide to manganese(II) requires a new electron-equivalent calculation.

Key Takeaways

Start with correct reactant and product formulas. Count oxidation-number changes per atom and then per formula unit. Choose coefficients so the total increase equals the total decrease. Use the stated medium to balance charge, hydrogen, and oxygen. Disproportionation product ratios follow from the same electron-equivalent balance. Accept the final equation only after checking every element and the total charge.

Previous · Lesson 4

Recognising Redox Reaction Patterns

Next · Lesson 6

Balancing Redox Equations by Half Reactions

A balanced equation must account for more than atoms

Build the coefficients before balancing the medium

An acidic reaction: dichromate and sulphite

A basic reaction: permanganate and bromide

When one reactant supplies both redox partners

Apply the method to a useful treatment process

Quiz

Practice Problems

Key Takeaways