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F2.3 · Balance redox equations using oxidation numbers and half-reactions

Learn to balance redox equations using oxidation numbers and half-reactions through clear examples and targeted practice.

Ontario Grade 12 Chemistry

Electrochemistry

Oxidation numbers and half-reactions | Study topic F2.3

A metal surface can change colour or develop a new coating when it reacts with another substance. Such an observation can result from particles transferring electrons. A redox reaction is a reaction in which oxidation and reduction occur together: one reacting species loses electrons, and another gains them. The equation must represent this transfer while conserving atoms and total charge. This lesson reviews oxidation numbers, then uses two Grade 12 methods to balance redox equations.

What you will learn

  • Use oxidation numbers to identify oxidation and reduction.
  • Balance electron changes when balancing a redox equation.
  • Balance a redox equation in acidic solution using half-reactions.
  • Check that a balanced ionic equation conserves atoms and net charge.

1. Prerequisite bridge: atoms, charge, and oxidation numbers

A chemical equation shows reactants changing into products. A balanced equation has the same number of each kind of atom on both sides. An ionic equation must also have the same total charge on both sides. A coefficient is a number placed before a formula. It changes how many units of that substance are present, but it does not change the substance’s formula or the charge of one ion.
An oxidation number is a bookkeeping value used to track changes linked to electron transfer. It is not always the actual charge on an atom. An element by itself has oxidation number zero. For a one-atom ion, its oxidation number is the ion’s charge. In a neutral compound, oxidation numbers add to zero; in a polyatomic ion, they add to the ion’s overall charge.
In the reactions in this lesson, oxygen usually has oxidation number −2-2 and hydrogen usually has +1+1. Use the sum rule to find an unknown value. In extMnO4− ext{MnO}_4^-, four oxygen atoms contribute a total of −8-8. The manganese oxidation number must be +7+7, so that the oxidation numbers add to the ion’s charge of −1-1. A coefficient before the ion would not change these oxidation numbers.
Oxidation is an increase in oxidation number and represents loss of electrons. Reduction is a decrease in oxidation number and represents gain of electrons. These always occur together in a redox reaction. The total electrons lost must equal the total electrons gained.
∑oxidation numbers=overall charge\sum \text{oxidation numbers} = \text{overall charge}
  • Use the overall charge to find an unknown oxidation number.
  • An increase in oxidation number indicates oxidation; a decrease indicates reduction.
  • Change coefficients when balancing, not subscripts.

2. Balance by comparing oxidation-number changes

Start with the correct reactant and product formulas. Find oxidation numbers for the elements that change. Compare each element’s value before and after the reaction. The size of the increase tells how many electrons each atom loses; the size of the decrease tells how many electrons each atom gains.
Choose coefficients so that the total increase matches the total decrease. This matches the electrons lost with the electrons gained. Then balance the remaining atoms. In an acidic solution, water can be used to balance oxygen atoms, and hydrogen ions can be used to balance hydrogen atoms. These are balancing species appropriate to the stated conditions.
Do not change a formula’s subscripts to force a match. A subscript is part of a substance’s identity. After completing the atom balance, count the total charge on both sides. If the charges differ, revisit the balancing rather than accepting an equation that only conserves atoms.
total increase=total decrease\text{total increase} = \text{total decrease}
  • Match the total oxidation-number increase and decrease.
  • Use water and hydrogen ions when balancing a reaction in acidic solution.
  • Finish by checking atoms and net charge.

3. Balance by half-reactions

A half-reaction shows only the oxidation part or only the reduction part of a redox reaction. Separating the two parts makes the electron transfer easier to track. Write one half-reaction for the species oxidized and one for the species reduced.
For each half-reaction, balance atoms other than hydrogen and oxygen first. Balance oxygen by adding water. In acidic solution, balance hydrogen by adding hydrogen ions. Then balance charge by adding electrons to the more positive side. Electrons carry a negative charge, so adding them changes the total charge on that side.
Multiply the half-reactions as needed so that the number of electrons lost equals the number gained. Add the equations, then cancel electrons and any identical species on both sides. The result is the overall ionic equation. Check atoms and total charge in that final equation.
The oxidation-number method and the half-reaction method describe the same electron transfer. Oxidation numbers help find the electron ratio. Half-reactions show how to balance each side’s atoms and charge before combining the changes.
electrons lost=electrons gained\text{electrons lost} = \text{electrons gained}
  • Balance oxygen with water and hydrogen with hydrogen ions in acidic solution.
  • Use electrons to balance charge in each half-reaction.
  • Match electron counts before adding the half-reactions.

4. Check the finished equation

Apply every coefficient, then count each element on the left and right. Count atoms in water and hydrogen ions as well as in the other formulas. Next, add the charges of all ions on each side, taking coefficients into account. The equation is balanced only when every atom count and the total charge match.
Keep the conditions in the question in view. Hydrogen ions are used here for reactions in acidic solution. Do not add a species just because it makes one part of the equation easier; it must fit the stated reaction and conditions. A final check also confirms that the formulas have not been altered.
  • Check formulas, atom counts, and net charge.
  • Use coefficients to balance equations and keep subscripts fixed.
  • The electron ratio from both methods should agree.

Worked example

Permanganate oxidizes iron(II) in acidic solution

Balance the net ionic equation for permanganate ions reacting with iron(II) ions in acidic solution to form manganese(II) ions and iron(III) ions. Use oxidation numbers, then verify using half-reactions.
  1. Write the unbalanced equation
    Use the stated formulas for the reactants and products. Leave coefficients at one while identifying which elements change oxidation number.
    MnO4−+Fe2+→Mn2++Fe3+\mathrm{MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}}
  2. Compare oxidation numbers
    Oxygen contributes 4(−2)=−84(-2)=-8 in permanganate. Since the ion’s overall charge is −1-1, manganese is +7+7. It becomes +2+2, so each manganese gains five electrons. Iron changes from +2+2 to +3+3, so each iron ion loses one electron. Five iron ions are therefore needed for each permanganate ion.
    Mn:+7→+2Fe:+2→+3\mathrm{Mn: +7 \rightarrow +2 \qquad Fe: +2 \rightarrow +3}
  3. Balance atoms and charge
    The electron match gives one permanganate ion for five iron(II) ions. Four oxygen atoms on the left require four water molecules on the right. Those water molecules contain eight hydrogen atoms, so add eight hydrogen ions to the left. The resulting charges are equal.
    MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}
  4. Verify with half-reactions
    In the reduction half-reaction, balance oxygen with water, hydrogen with hydrogen ions, and charge with electrons. The oxidation half-reaction releases one electron per iron ion, so multiply it by five before adding. The electrons cancel, giving the same overall equation.
    MnO4−+8H++5e−→Mn2++4H2O5Fe2+→5Fe3++5e−\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}\qquad 5\mathrm{Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-}
Answer: The balanced net ionic equation is MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}.
Check: Both sides have one manganese atom, five iron atoms, four oxygen atoms, and eight hydrogen atoms. The left-side charge is −1+10+8=+17-1+10+8=+17. The right-side charge is +2+15=+17+2+15=+17. Manganese gains five electrons, matching the five electrons lost by the iron ions.

Common mistakes and how to avoid them

Changing a subscript to balance an equation.
Correction: Keep each formula fixed and change coefficients. Changing a subscript changes the substance.
Balancing atoms but ignoring charge in an ionic equation.
Correction: Count the total charge on each side after balancing atoms. Both atoms and net charge must be conserved.
Adding half-reactions before matching their electron counts.
Correction: Multiply the half-reactions until electrons lost and gained are equal. Then add and cancel electrons.
Calling a decrease in oxidation number oxidation.
Correction: A decrease is reduction and represents gaining electrons. An increase is oxidation and represents losing electrons.

Lesson summary

  • Oxidation numbers help track changes linked to electron transfer.
  • In the oxidation-number method, match the total increase and decrease before completing the atom balance.
  • In the half-reaction method, balance atoms, balance charge with electrons, then match and cancel electrons.
  • Check every atom and the total charge in the final equation.

Check your understanding

Question 1

An atom’s oxidation number changes from +3+3 to +1+1. What happens to it?
  1. It is oxidized and loses two electrons.
  2. It is reduced and gains two electrons.
  3. It is oxidized and gains two electrons.
  4. It is reduced and loses one electron.
Show answer and explanation
It is reduced and gains two electrons.
The oxidation number decreases by two, so the atom is reduced and gains two electrons.

Question 2

When balancing a half-reaction in acidic solution, what is used to balance oxygen atoms?
  1. Electrons
  2. Hydrogen ions
  3. Water
  4. A changed subscript
Show answer and explanation
Water
Add water to balance oxygen. Then balance hydrogen with hydrogen ions and charge with electrons.

Question 3

In the worked reaction, how many iron(II) ions are needed per permanganate ion to match electron transfer?
  1. One
  2. Two
  3. Five
  4. Eight
Show answer and explanation
Five
Manganese gains five electrons as its oxidation number decreases from +7+7 to +2+2. Each iron ion loses one electron, so five iron(II) ions are needed.

Key terms

Redox reaction
A reaction in which oxidation and reduction occur together through electron transfer.
Oxidation number
A bookkeeping value used to track changes linked to electron transfer.
Half-reaction
An equation showing only the oxidation part or only the reduction part of a redox reaction.
Net ionic equation
An equation showing the reacting ions and other species that change, without spectator ions.

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Chemistry (SCH4U), expectation F2.3. It is a study resource, not an official curriculum publication.

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