Redox I
Notes for every section, eight new exam-style questions with full worked solutions, and online recall cards for your review days. Nothing here is copied from the book.
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Read one section at a time, then answer its check question before you open the answer.
3.1 Oxidation numbers
An oxidation number is the charge an atom would have if all its bonds were ionic. It lets you track electrons in any reaction.
| Rule | Example |
|---|---|
| Uncombined element = 0 | O2, Fe, S8 |
| Simple ion = its charge | Na+ +1, Cl− −1 |
| Oxidation numbers in a compound add up to 0 | H2O: 2(+1) + (−2) = 0 |
| In an ion they add up to the charge | SO42−: S + 4(−2) = −2, so S = +6 |
| F is always −1; O is −2 | except O in peroxides (−1) and in OF2 (+2) |
| H is +1 | except in metal hydrides such as NaH (−1) |
| Group 1 is +1, Group 2 is +2 | in compounds |
Work out the unknown last: write what you know, then balance to the overall charge.
Check yourself: What is the oxidation number of Cr in Cr2O72−?
2Cr + 7(−2) = −2, so Cr = +6
3.2 Oxidation and reduction
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3.3 Roman numerals in names and formulae
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3.4 Ionic half-equations
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3.5 Disproportionation and classifying reactions
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Extra worked questions
Cover the solution, answer on paper, then mark yourself against each step. Each step shows the mark it earns.
All 8 worked solutions are free for members. The first one is open to everyone.
Working out oxidation numbers
Give the oxidation number of (a) sulfur in H2SO4, (b) nitrogen in NH4+ and (c) chlorine in ClO3−.
Try it first, then show the worked solution
| (a) | 2(+1) + S + 4(−2) = 0, so S = +6 | 1 |
|---|---|---|
| (b) | N + 4(+1) = +1, so N = −3 | 1 |
| (c) | Cl + 3(−2) = −1, so Cl = +5 | 1 |
Answer: (a) +6 (b) −3 (c) +5
Examiner tip: Always write the sign. "6" is not the same as "+6" to an examiner.
Names with Roman numerals
(a) Name the compound Fe2O3. (b) Write the formula of potassium manganate(VII).
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Chlorine and bromide ions
Chlorine reacts with bromide ions: Cl2 + 2Br− → 2Cl− + Br2. Identify the oxidising agent, and explain your answer using oxidation numbers.
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Balancing a manganate(VII) half-equation
Write the half-equation for MnO4− being reduced to Mn2+ in acid solution.
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Disproportionation of chlorine
Chlorine reacts with cold, dilute sodium hydroxide: Cl2 + 2NaOH → NaCl + NaClO + H2O. Explain why this is a disproportionation reaction.
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Hydrogen peroxide and iron(II)
In acid solution, hydrogen peroxide oxidises Fe2+ to Fe3+ and is itself reduced to water. Write the two half-equations and the overall ionic equation.
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Dichromate and iodide
Acidified dichromate(VI) ions oxidise iodide ions to iodine and are reduced to Cr3+. Write the two half-equations and the overall ionic equation.
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Chlorine in hot alkali
Chlorine reacts with hot, concentrated sodium hydroxide to form sodium chloride and sodium chlorate(V). (a) Give the oxidation numbers of chlorine in each species. (b) Use them to balance the equation. (c) Name this type of reaction.
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Video walkthroughs
Narrated walkthroughs of exam-style questions from this topic of the book, showing where each mark comes from. Try the question first, then watch. Captions are available in the player.
Oxidation numbers in a hydride and a peroxide
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Copper and nitric acid half-equations
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Manganate(VII) and ethanedioate
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Iodate and iodide: not disproportionation
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Recall cards
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