Balancing Equations

Redox reactions involve the simultaneous transfer of electrons between chemical species — one species is oxidised (loses electrons) while another is reduced (gains electrons). Because mass and charge must both be conserved, writing a balanced redox equation requires more care than balancing a simple precipitation or neutralisation reaction. Two systematic methods have been developed to handle this: the ion-electron method (also called the half-reaction method) and the oxidation number method. Both produce the same final balanced equation but approach the problem from different angles. The ion-electron method splits the overall reaction into two half-reactions — one for oxidation and one for reduction — balances each separately, then combines them so that the electrons cancel. This approach is especially powerful because it keeps track of the ionic species actually present in solution and makes it easy to add water molecules or protons (in acid) or hydroxide ions (in base) to balance oxygen and hydrogen atoms. The oxidation number method, by contrast, works with the full molecular equation and uses changes in oxidation number to determine how many electrons are transferred, adjusting coefficients accordingly. Mastering both methods is essential for A-level and IB chemistry because redox equations appear throughout electrochemistry, organic oxidation, industrial processes (such as the extraction of metals), and analytical techniques like titrations with potassium manganate(VII). The rules are universal: atoms must balance, charge must balance, and the electrons gained must equal the electrons lost.

Key takeaways
  • Identify oxidation and reduction
  • Write the two unbalanced half-reactions
  • Balance atoms other than O and H
  • Balance oxygen by adding H₂O, then hydrogen by adding H⁺ (acid medium)
  • Balance charge by adding electrons
  • Equalise electrons and add half-reactions
  • Verify
  • Oxidation Number Method (overview)

Identify oxidation and reduction

Determine which species is oxidised and which is reduced by assigning oxidation numbers to all atoms. The species whose oxidation number increases is oxidised; the one that decreases is reduced. For example, in the reaction between manganate(VII) ions and iron(II) ions in acid solution: Mn goes from +7 to +2 (reduction), Fe goes from +2 to +3 (oxidation).

Write the two unbalanced half-reactions

Oxidation half-reaction: Fe²⁺ → Fe³⁺. Reduction half-reaction: MnO₄⁻ → Mn²⁺. These show the core chemical change without yet worrying about oxygen, hydrogen, or charge.

Balance atoms other than O and H

In most inorganic redox problems the non-oxygen, non-hydrogen atoms are already balanced at this stage. If not, insert stoichiometric coefficients now. Here both half-reactions already have one metal atom each.

Balance oxygen by adding H₂O, then hydrogen by adding H⁺ (acid medium)

Reduction half-reaction has 4 oxygen atoms on the left from MnO₄⁻, so add 4 H₂O to the right: MnO₄⁻ → Mn²⁺ + 4H₂O. Now 8 hydrogen atoms are on the right, so add 8H⁺ to the left: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O. In alkaline medium, instead add OH⁻ to balance H and H₂O to the other side.

Balance charge by adding electrons

Count the total charge on each side of each half-reaction and add electrons (e⁻) to the more positive side. Reduction: left side charge = (−1 + 8) = +7; right side = +2. Add 5e⁻ to the left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Oxidation: left = +2; right = +3. Add 1e⁻ to the right: Fe²⁺ → Fe³⁺ + e⁻.

Equalise electrons and add half-reactions

Multiply the oxidation half-reaction by 5 so both half-reactions involve 5 electrons: 5Fe²⁺ → 5Fe³⁺ + 5e⁻. Add both half-reactions and cancel the 5e⁻ from each side: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. This is the balanced ionic equation.

Verify

Check: Mn: 1=1 ✓, Fe: 5=5 ✓, O: 4=4 ✓, H: 8=8 ✓. Charge: left = −1+8+10 = +17; right = +2+0+15 = +17 ✓. The equation is balanced for both mass and charge.

Oxidation Number Method (overview)

Write the full molecular equation. Assign oxidation numbers to every atom. Identify atoms that change oxidation number. Calculate the increase (oxidised atom) and the decrease (reduced atom) per formula unit. Find the lowest common multiple of these changes and use it to assign coefficients so that total electrons lost = total electrons gained. Finally balance H and O using H₂O and H⁺ (or OH⁻) by inspection. For the same reaction: Mn changes by 5 (gain), Fe changes by 1 (loss). LCM = 5, so coefficient of Fe²⁺ is 5, coefficient of MnO₄⁻ is 1. This gives the same result.

The Key Equation

\ceMnO4−+8H++5Fe2+−>Mn2++4H2O+5Fe3+\ce{MnO4^{-} + 8H^{+} + 5Fe^{2+} -> Mn^{2+} + 4H2O + 5Fe^{3+}}
  • \ce{MnO4^{-}} — Manganate(VII) / Permanganate ion (mol): The oxidising agent; Mn is reduced from oxidation state +7 to +2, gaining 5 electrons.
  • \ce{H^{+}} — Hydrogen ion (proton) (mol): Provided by the acidic medium (usually dilute H₂SO₄); needed to balance oxygen atoms that leave as water.
  • \ce{Fe^{2+}} — Iron(II) ion (mol): The reducing agent; Fe is oxidised from +2 to +3, losing 1 electron each. Five Fe²⁺ ions supply the 5 electrons needed by one MnO₄⁻.
  • \ce{Mn^{2+}} — Manganese(II) ion (mol): Product of reduction; the pale pink colour of Mn²⁺ replaces the deep purple of MnO₄⁻, providing a visible end-point in titrations.
  • \ce{H2O} — Water (mol): Formed when the 4 oxygen atoms from MnO₄⁻ combine with the 8 H⁺ ions; balances both O and H atoms.

Worked Example

Problem. Balance the following redox reaction in alkaline (basic) medium using the ion-electron method: the dichromate ion (Cr₂O₇²⁻) oxidises iodide ions (I⁻) to iodine (I₂), and is itself reduced to Cr³⁺.

\ceCr2O72−+7H2O+6I−−>2Cr3++3I2+14OH−\ce{Cr2O7^{2-} + 7H2O + 6I^{-} -> 2Cr^{3+} + 3I2 + 14OH^{-}}

Answer. The balanced ionic equation in alkaline medium is: Cr₂O₇²⁻ + 7H₂O + 6I⁻ → 2Cr³⁺ + 3I₂ + 14OH⁻

Figure
Diagram
Ion-Electron Method: Electron Transfer in MnO₄⁻ / Fe²⁺ ReactionOXIDATION HALF-REACTIONFe²⁺ → Fe³⁺ + e⁻Oxidation state: +2 → +3Loses 1 electron×5 to equaliseREDUCTION HALF-REACTIONMnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂OOxidation state: +7 → +2Gains 5 electrons×15e⁻COMBINED (BALANCED) EQUATIONMnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺Acid MediumAdd H₂O for OAdd H⁺ for HBase MediumAdd H₂O for OAdd OH⁻ for HAlways CheckAtoms balanceCharge balance
Origins · 1 minWendell Mitchell Latimer and how this was discovered
WM
Wendell Mitchell Latimer
1893–1955 · American

Latimer systematised the use of electrode potentials and half-reactions in his landmark 1938 textbook 'The Oxidation States of the Elements and Their Potentials in Aqueous Solutions'. He introduced the Latimer diagram, which displays the standard reduction potentials between the successive oxidation states of an element, making it straightforward to determine which redox reactions are spontaneous and to write balanced half-reactions. His work provided the quantitative framework that underpins the half-reaction balancing method taught in schools worldwide. Latimer was also a pioneering low-temperature chemist and contributed to the Manhattan Project during World War II, studying the chemistry of uranium compounds.

  • Oxidation Numbers — Assigning correct oxidation numbers to every atom is the essential first step in both balancing methods; errors here propagate through the entire calculation.
  • Standard Electrode Potentials — Balanced half-reactions are the building blocks for calculating cell EMF using standard electrode (reduction) potentials via E°cell = E°cathode − E°anode.
  • Electrolytic Cells — The half-reactions at cathode and anode in electrolysis must be balanced using the same ion-electron method, and stoichiometry links charge passed (via Faraday's law) to moles of product.
  • Redox Titrations — Calculations in permanganate and dichromate titrations rely directly on the molar ratios in the balanced redox equation, so balancing must be correct to obtain accurate analytical results.
  • Acids, Bases, and pH — The medium (acid or base) determines whether H⁺ or OH⁻ is added to balance hydrogen in the half-reaction method, linking redox balancing directly to acid–base chemistry.