Balancing redox equations – the electron balance method
ChemistryReactions & RedoxAges 15–16
Use it with your class: students open it from a link or a QR code, with no account needed.
Next in Reactions & Redox
Step by step: assign oxidation numbers to each element, find which ones gain or lose electrons, write the oxidation and reduction half-reactions, then multiply them so the electrons lost equal the electrons gained (an animation shows electrons moving from the reducing agent to the oxidizing agent), and finally place the coefficients into the full equation. Students pick a textbook reaction and click Next step, or let it run automatically.
Lesson: Oxidation numbers, reducing and oxidizing agents, balancing redox equations
What it shows
In a redox reaction, electrons pass from a reducing agent, whose oxidation number increases, to an oxidizing agent, whose oxidation number decreases. The electron balance method builds on this: assign oxidation numbers, write the oxidation and reduction half-reactions, multiply them so that electrons lost equal electrons gained, then transfer the coefficients to the full equation and balance the remaining atoms. The seven reactions include cases where only part of a reactant is reduced, such as copper with dilute nitric acid, where most of the HNO₃ simply forms the nitrate salt.
How to use
Choose a Reaction and work through the five steps with Next step ▶, pausing at each one to ask students what comes next: which oxidation numbers change, what the two half-reactions are, and which multipliers make the electrons balance. For a whole-class demonstration, use Auto-run with a longer Time per step. Keep Label reducing / oxidizing agent on at first, then turn it off and let students identify the agents themselves.
Parameters you can change
- Reaction Fe + O₂ → Fe₂O₃, Mg + HCl → MgCl₂ + H₂, Cu + dilute HNO₃ → Cu(NO₃)₂ + NO + H₂O, Cu + concentrated H₂SO₄ → CuSO₄ + SO₂ + H₂O, KMnO₄ + HCl → KCl + MnCl₂ + Cl₂ + H₂O, NH₃ + O₂ → NO + H₂O, Fe₂O₃ + CO → Fe + CO₂
- Auto-advance steps
- Time per step 1–6 s
- Label reducing / oxidizing agent
Questions to explore
- In KMnO₄ + HCl, which element is oxidized, and why does only part of the HCl act as the reducing agent?
- Why is the coefficient of HNO₃ larger than the number of nitrogen atoms that are reduced?
- In Fe₂O₃ + CO, which substance is the reducing agent, and how many electrons does each molecule of it lose?
Common questions
How do I tell which substance is the oxidizing agent and which is the reducing agent?
Compare oxidation numbers on both sides of the equation: the substance containing the element whose number rises is the reducing agent, and the one containing the element whose number falls is the oxidizing agent. The reducing agent loses electrons and is oxidized while the oxidizing agent gains them and is reduced, so in Mg + HCl, Mg (0 to +2) is the reducing agent.
How do you assign oxidation numbers to the elements in a compound?
Start from fixed values, with an element in its free state at 0, H at +1 and O at −2 (except in metal hydrides, peroxides and OF₂), then use the rule that the numbers sum to 0 in a molecule and to the ion's charge in an ion. In HNO₃, for example, +1 + x + 3(−2) = 0 gives x = +5 for nitrogen.
Why must the electrons lost equal the electrons gained in a redox equation?
Electrons lost must equal electrons gained because every electron released by the reducing agent has to be taken up by the oxidizing agent, so the half-reactions are multiplied up to the lowest common multiple of their electrons. In Fe + O₂, each Fe loses 3 e⁻ and each O₂ gains 4 e⁻, so the multipliers 4 and 3 give 12 e⁻ on both sides.
