When inspection gets stuck
Lesson 1.6.3 balanced equations by inspection: start with the most complicated substance and adjust one coefficient at a time. That works well for simple reactions, but try copper with dilute nitric acid:
N appears in HNO₃, Cu(NO₃)₂, and NO, and O appears in four substances, so trial and error goes round in circles. Redox reactions offer an extra clue: the electrons.
The principle: electrons lost = electrons gained
Lesson 2.13.2 showed that in every redox reaction the electrons lost by the reducing agent equal the electrons gained by the oxidizing agent. Electrons can’t appear or vanish. So you first make electrons lost equal electrons gained, which fixes the coefficients of the species that change oxidation number, and then balance everything else by inspection. This is the oxidation-number method化合价升降法.
The oxidation-number method, step by step
Using Cu + dilute HNO₃:
- Mark the changes. Cu: 0 → +2, up 2. N: +5 → +2 (for the N that becomes NO), down 3.
- Count electrons per formula. Each Cu loses 2 electrons; the N in each NO gains 3. If a formula contains several atoms that change, multiply by the number of atoms.
- Make lost = gained. The least common multiple of 2 and 3 is 6: put 3 in front of Cu (6 lost) and 2 in front of NO (6 gained).
- Balance the rest by inspection. 3 Cu end up in 3 Cu(NO₃)₂, which contain 6 N that don’t change; add the 2 N that become NO, and 8 HNO₃ are needed. The 8 HNO₃ bring 8 H, so H₂O gets 4.
- Check. Count every atom (O: 24 on the left, 18 + 2 + 4 = 24 on the right).
Notice that step 3 puts the coefficient on NO, not on HNO₃: only part of the HNO₃ is reduced; the rest just supplies NO₃⁻.
Try a few reactions below. The first two steps are done for you. In step 3, enter two coefficients that make electrons lost equal electrons gained, then see how the rest is balanced:
Half-reactions: writing the electrons in
You can also split a redox reaction into two half-reactions半反应: one for the oxidation (electrons lost) and one for the reduction (electrons gained), with the electrons written into the equations. For copper displacing silver:
To make electrons lost equal electrons gained, multiply the reduction by 2 and add the two equations; the 2e⁻ cancel:
A balanced ionic equation must conserve charge as well as atoms: 2 Ag⁺ on the left total +2, and Cu²⁺ on the right is +2. Lesson 2.13.4 uses half-reactions to describe what happens at each electrode of a battery.
How Chinese textbooks differ
- Chinese textbooks lean on the oxidation-number method (化合价升降法); US textbooks use it too but more often use the half-reaction method.
- In AP and US textbooks, half-reactions are often balanced in acidic solution: balance the atoms that change, add H₂O to balance O and H⁺ to balance H, then add e⁻ to balance charge. That step isn’t covered here.
- Chinese textbooks often write an electrode half-reaction as Cu − 2e⁻ = Cu²⁺, subtracting electrons on the left; US textbooks write Cu → Cu²⁺ + 2e⁻.
Example 1: oxidizing ammonia
Balance NH₃ + O₂ → NO + H₂O (catalyst, heat).
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Mark the changes. N: −3 → +2, up 5. O: 0 → −2, down 2. H stays +1.
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Electrons per molecule. Each NH₃ has one N and loses 5 electrons; each O₂ has 2 O atoms and gains 2 × 2 = 4.
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Make lost = gained. The LCM of 5 and 4 is 20:
so NH₃ gets 4 and O₂ gets 5.
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Inspection for the rest. 4 N become 4 NO; the left has 4 × 3 = 12 H, so H₂O gets 6.
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Check O. 5 × 2 = 10 on the left, 4 + 6 = 10 on the right.
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A common mistake. Counting only one O and saying O₂ gains 2 electrons. Count per molecule: O₂ has 2 O atoms, so it gains 4.
Example 2: balancing an ionic equation with half-reactions
A strip of aluminum in copper(II) sulfate solution becomes coated with copper. Write the half-reactions and the balanced net ionic equation.
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Oxidation. Al loses 3 electrons:
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Reduction. Cu²⁺ gains 2 electrons:
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Make lost = gained. The LCM of 3 and 2 is 6: multiply the oxidation by 2 and the reduction by 3:
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Add. The 6e⁻ cancel:
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Check. Atoms: 2 Al and 3 Cu on each side. Charge: left 3(+2) = +6; right 2(+3) = +6. Charge is conserved.
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A common mistake. Writing Al + Cu²⁺ → Al³⁺ + Cu. The atoms balance, but the charges don’t (+2 vs. +3). Always check both.
- Al in Al + Fe₂O₃ → Al₂O₃ + Fe (high temperature)
Al loses 3 electrons and Fe₂O₃ gains 2 × 3 = 6: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe.
- Cl₂ in Fe + Cl₂ → FeCl₃
Fe loses 3 and Cl₂ gains 2; the LCM is 6: 2Fe + 3Cl₂ → 2FeCl₃.
- O₂ in NH₃ + O₂ → NO + H₂O
See Example 1: 4NH₃ + 5O₂ → 4NO + 6H₂O.
- H₂O in the same reaction
4 NH₃ hold 12 H, so H₂O gets 6.
- HNO₃ in Cu + HNO₃ (dilute) → Cu(NO₃)₂ + NO + H₂O
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O.
- Cu in the same reaction
Cu loses 2 and NO gains 3; the LCM is 6, so Cu gets 3.
- H₂O in the same reaction
8 HNO₃ hold 8 H, so H₂O gets 4.
- Cl₂ in MnO₂ + HCl → MnCl₂ + Cl₂ + H₂O
Mn gains 2 electrons and forming one Cl₂ loses 2: MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O.
- Ag⁺ in Zn + Ag⁺ → Zn²⁺ + Ag
Zn loses 2 electrons and each Ag⁺ gains 1: Zn + 2Ag⁺ → Zn²⁺ + 2Ag, with +2 charge on each side.
Common mistakes
Mistake 1: “If the atoms balance, the ionic equation is balanced.” Cu + Ag⁺ → Cu²⁺ + Ag has the same atoms on both sides, but the charge is +1 on the left and +2 on the right. An ionic equation must also conserve charge: Cu + 2Ag⁺ → Cu²⁺ + 2Ag.
Mistake 2: “Electrons per formula = the change in oxidation number.” Multiply by the number of atoms that change. Each Fe₂O₃ has 2 Fe going from +3 to 0, so it gains 2 × 3 = 6 electrons, not 3.
Mistake 3: “The coefficient of HNO₃ is set by the part that is reduced.” In copper with dilute nitric acid, only 2 of the 8 HNO₃ are reduced to NO; the other 6 supply the NO₃⁻ in Cu(NO₃)₂. So the electron count goes on NO, and HNO₃ comes last, from counting N atoms.
Quick check
After balancing □Al + □O₂ → □Al₂O₃, what is the coefficient of O₂?
Al loses 3 electrons and O₂ gains 4; the LCM is 12, so Al gets 4 and O₂ gets 3: 4Al + 3O₂ → 2Al₂O₃.
How many electrons are transferred in Cu + 2Ag⁺ → Cu²⁺ + 2Ag?
One Cu loses 2 electrons, and the two Ag⁺ gain 1 each: 2 electrons in total.
Summary
- Redox equations are balanced on the principle that electrons lost equal electrons gained, along with conservation of atoms (and of charge, for ionic equations).
- Oxidation-number method: mark the changes → count electrons per formula (times the number of atoms) → use the LCM to make lost = gained → balance the rest by inspection → check.
- Half-reaction method: write the oxidation and reduction half-reactions, multiply so the electrons match, and add.
On to AP (for later)
AP topic 4.9 asks you to write and balance half-reactions and the overall redox equation, often in acidic solution, where H₂O and H⁺ balance O and H. In Unit 9, both cell potentials and electrolysis calculations start from the number of electrons transferred in the half-reactions.