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8873 / 2027
The Mole Concept and Stoichiometry

Lesson 2 of 3

Chemical accounting

Balance atoms, charge and electron transfer.

A-Level 8873, revised syllabus (2026-2027)

  • I can construct balanced equations.Syllabus

    Syllabus 8873, 5(f). Construct balanced equations.

  • I can explain redox through electrons and oxidation numbers.Syllabus

    Syllabus 8873, 5(g). Explain redox through electrons and oxidation numbers.

  • I can combine relevant redox half-equations.Syllabus

    Syllabus 8873, 5(h). Combine relevant redox half-equations.

Balance atoms and charge separately

Oxidation numbers identify electron transfer; half-equations make the accounting visible.

Balance an equation by changing coefficients, never subscripts inside a correct formula. A coefficient multiplies every atom in the formula. In an ionic equation, conserve total signed charge as well as every element. Include state symbols when they matter, especially for precipitation, gases and aqueous ions.

Oxidation is electron loss and an increase in oxidation number; reduction is electron gain and a decrease. A reducing agent loses electrons and is oxidised; an oxidising agent gains electrons and is reduced. An element in its elemental form has oxidation number zero; a monatomic ion has oxidation number equal to its charge. Oxygen is usually -2 and hydrogen usually +1 in the compounds used here; the sum equals the species' charge.

Worked exampleLocate the changing atom

What happens to iron and manganese when Fe2+ reacts with acidified MnO4-?

Try it first. Then show the worked stepsHide the worked steps
  1. Fe2+ becomes Fe3+: oxidation number rises from +2 to +3, so each Fe loses one electron.
  2. In MnO4-, four oxygens total -8 and the ion totals -1, so Mn is +7.
  3. Mn becomes Mn2+: +7 to +2 means gain of five electrons.
Answer

Fe2+ is the reducing agent and is oxidised. MnO4- is the oxidising agent and is reduced.

Construct a half-equation in acidic solution
  1. Balance the changing element

    Start MnO4- → Mn2+.

  2. Balance oxygen with water

    Add 4H2O on the right.

  3. Balance hydrogen with H+

    Add 8H+ on the left.

  4. Balance charge with electrons

    Left is +7 before electrons, right +2: add five electrons on the left.

Reduction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O.
Oxidation: Fe2+ → Fe3+ + e-.
Multiply the iron half-equation by five, add, then cancel electrons.

Overall: MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq). The charge totals +17 on each side. The 1:5 mole ratio follows from the electrons transferred, not from comparing ion charges.

Check your understanding

In 2Fe3+ + 2I- -> 2Fe2+ + I2, which species is the oxidising agent?

  1. Fe3+, because it gains electrons and is reduced to Fe2+
  2. I-, because it loses electrons and is oxidised to I2
  3. I2, because it is the element formed in the reaction
Show the answer

Fe3+, because it gains electrons and is reduced to Fe2+

Iron goes from +3 to +2, gaining an electron. An oxidising agent accepts electrons and is itself reduced. Iodide is oxidised from -1 to 0.