Edexcel International A Level Chemistry: Redox chemistry (YCH11) – Revision Notes

These notes condense Topic 8A: Redox chemistry of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, learning outcomes 8.1 to 8.10. The topic sits in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols, an IAS unit. For full explanations and longer worked examples, read the redox study guide first.

Then test yourself with the redox practice questions. Track outcomes on the printable checklist and find more on the course hub. A free diagnostic will show which topics need the most work.

8.1 Oxidation number: definition

The oxidation number of an atom is the charge it would have if all its bonds were ionic, with bonding electrons assigned to the more electronegative atom. Sign first: +2, −1.

8.1 and 8.2 Rules (higher rule wins a clash)

Rule	Value
Uncombined element (Fe, O₂, Br₂, P₄)	0
All oxidation numbers in an uncharged compound, added	0
All oxidation numbers in an ion, added	equals the ion’s charge
Fluorine in compounds	−1
Group 1 / Group 2 metals in compounds	+1 / +2
Hydrogen	+1; −1 in metal hydrides (RbH, CaH₂)
Oxygen	−2; −1 in peroxides (H₂O₂, Na₂O₂); positive with F
Chlorine	−1 in chlorides; positive with O or F

Method in steps: unknown oxidation number

1. Write down every known value, with peroxide and hydride exceptions checked first.
2. Multiply each by the number of atoms.
3. Make the total zero for a compound, or the ion’s charge for an ion.
4. Solve for the unknown and write it with its sign.

Reminder 1. Se in H₂SeO₄: 2(+1) + x + 4(−2) = 0, so x = +6. Reminder 2. N in N₂H₄: 2x + 4(+1) = 0, so x = −2. Reminder 3. O in H₂O₂: 2(+1) + 2x = 0, so x = −1 (peroxide).

8.3 and 8.4 Names and formulae

• The Roman numeral follows the element it describes: iron(III) chloride, nitrogen(IV) oxide.
• In an oxyanion it describes the central atom: chlorate(I) ClO⁻, chlorate(V) ClO₃⁻, chlorate(VII) ClO₄⁻, nitrate(III) NO₂⁻, nitrate(V) NO₃⁻, sulfate(IV) SO₃²⁻, sulfate(VI) SO₄²⁻.
• Name to formula: find each ion’s charge from its oxidation numbers, then balance the charges. Sodium chlorate(I) is NaClO; magnesium nitrate(V) is Mg(NO₃)₂.

8.5 and 8.6 Definitions

Term	Electrons	Oxidation number
Oxidation	lost	increases
Reduction	gained	decreases
Oxidising agent	gains electrons (is reduced)	its oxidation number falls
Reducing agent	loses electrons (is oxidised)	its oxidation number rises

s-block and p-block examples

• Ca + Cl₂ → CaCl₂: Ca 0 → +2 (oxidised, reducing agent); Cl 0 → −1 (reduced, oxidising agent).
• 2Li + 2H₂O → 2LiOH + H₂: Li 0 → +1; H +1 → 0; O unchanged at −2.
• N₂H₄ + O₂ → N₂ + 2H₂O: N −2 → 0 (oxidised); O 0 → −2 (reduced). No ions involved, but oxidation numbers still track the change.

8.9 Metals and non-metals

• Metals, in general, lose electrons → positive ions → oxidation number rises. Al → Al³⁺ + 3e⁻.
• Non-metals, in general, gain electrons → negative ions → oxidation number falls. O₂ + 4e⁻ → 2O²⁻.
• So in 4Al + 3O₂ → 2Al₂O₃, aluminium is the reducing agent and oxygen the oxidising agent.
• “In general” matters: a non-metal bonded to a more electronegative atom has a positive oxidation number (Cl in ClO⁻ is +1).

8.7 and 8.8 Classifying reactions

What the oxidation numbers do	Classification
Nothing changes	Not redox
One element rises and another falls	Redox
The same element in one species both rises and falls	Redox and disproportionation
The same element in two species meets at one value	Redox, not disproportionation

Reminder 4: disproportionation. 2H₂O₂ → 2H₂O + O₂. O in H₂O₂ is −1. It falls to −2 in H₂O and rises to 0 in O₂. One element, one species: disproportionation.

Reminder 5: not redox. Precipitation of a sulfate, or a neutralisation such as H⁺ + OH⁻ → H₂O: no element changes oxidation number.

8.10 Half-equations

Method in steps (acid conditions)

1. Match the redox element first.
2. Water fixes oxygen.
3. Hydrogen ions fix hydrogen.
4. Electrons fix charge, added on the more positive side.
5. Cross-check the electron count against the oxidation number change.

Method in steps: combining

1. Multiply one or both half-equations so the electron numbers are equal.
2. Add them.
3. Cancel the electrons, plus any H⁺ or H₂O that appears on both sides.
4. Confirm that every element and the overall charge balance.

Reminder 6. Chlorate(I) oxidising iodide in acid:

• ClO⁻ + 2H⁺ + 2e⁻ → Cl⁻ + H₂O
• 2I⁻ → I₂ + 2e⁻
• Overall: ClO⁻ + 2I⁻ + 2H⁺ → Cl⁻ + I₂ + H₂O (charge −1 on both sides).

Reminder 7. Peroxide as an oxidising agent: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O (O −1 → −2).

Electron bookkeeping

Electrons transferred = change in oxidation number × number of atoms that change. The total lost must equal the total gained.

Reminder 8. In 4Al + 3O₂ → 2Al₂O₃, four Al atoms each lose 3 electrons (12 in total). Six O atoms each gain 2 electrons (12 in total). The totals match, so the equation is consistent.

Method in steps: deducing a product from amounts

1. Find the moles of electrons lost by the reducing agent (moles × electrons per particle).
2. Divide by the moles of oxidising agent to get electrons gained per particle.
3. Subtract that number from the starting oxidation number of the element being reduced.
4. Name the product with that oxidation number, then write the half-equation and check it.

The same steps work the other way round, starting from the oxidising agent, to find what a reducing agent becomes.

Typical oxidation numbers in s-block and p-block chemistry

Element	Usual state in compounds	Usual role when the element reacts
Li, Na, K, Rb	+1	reducing agent (loses 1 electron)
Mg, Ca, Sr, Ba	+2	reducing agent (loses 2 electrons)
Al	+3	reducing agent (loses 3 electrons)
O	−2 in oxides; −1 in peroxides	oxidising agent
S	−2 in sulfides; +4 and +6 with oxygen	oxidising agent with metals
N	−3 in nitrides; +1 to +5 with oxygen	oxidising agent with reactive metals
Cl, Br, I	−1 in halides; positive with oxygen	oxidising agent
H	+1 with non-metals; −1 in metal hydrides	either, depending on partner

Method in steps: classifying a reaction

1. Write the oxidation number above every element on both sides.
2. If none change, it is not redox.
3. If they change, find which element goes up (oxidised) and which goes down (reduced).
4. If both changes belong to one element that started in one species, it is disproportionation.

Must-know distinctions

• Charge versus oxidation number: Fe³⁺ has charge 3+; Fe in it has oxidation number +3.
• Oxidising agent versus oxidised: the oxidising agent is the species that gets reduced.
• Peroxide versus oxide: Na₂O has O at −2; Na₂O₂ has O at −1.
• Metal hydride versus non-metal hydride: H is −1 in NaH but +1 in HCl and NH₃.
• Disproportionation versus the reverse: one species splitting two ways, not two species meeting.
• Ionic half-equation versus full ionic equation: electrons appear only in the half-equation.

Quick self-test

1. State the oxidation number of Xe in XeO₃.
2. State the oxidation number of P in PH₄⁺.
3. State the oxidation number of Br in BrF₃.
4. State the oxidation number of H in RbH, and explain why.
5. State the oxidation number of Sn in SnO₃²⁻.
6. Name KClO₄ using a Roman numeral.
7. Write the formula of calcium nitrate(III).
8. In 2Rb + Br₂ → 2RbBr, identify the oxidising agent.
9. Is Mg + 2H⁺ → Mg²⁺ + H₂ a disproportionation? Explain.
10. Write the half-equation for bromine forming bromide ions.
11. Balance in acid: H₂SO₃ → SO₄²⁻ (oxidation).
12. State what happens to the oxidation number of a reducing agent.

Answers

1. +6: x + 3(−2) = 0.
2. −3: x + 4(+1) = +1.
3. +3: F is −1, so x + 3(−1) = 0.
4. −1: rubidium is a Group 1 metal at +1, and that rule outranks the hydrogen rule, so this is a metal hydride.
5. +4: x + 3(−2) = −2.
6. Potassium chlorate(VII): Cl is +7.
7. Ca(NO₂)₂: nitrate(III) is NO₂⁻, and Ca²⁺ needs two.
8. Br₂: Br goes 0 → −1, gaining electrons.
9. No: Mg is oxidised (0 → +2) and H is reduced (+1 → 0). Two different elements change, so it is redox but not disproportionation.
10. Br₂ + 2e⁻ → 2Br⁻.
11. H₂SO₃ + H₂O → SO₄²⁻ + 4H⁺ + 2e⁻ (S +4 → +6; charge 0 on both sides).
12. It increases, because a reducing agent is oxidised.

Where marks are usually lost

• Giving H as +1 in a metal hydride such as CaH₂, or O as −2 in a peroxide.
• Writing “2+” when an oxidation number “+2” is asked for, or dropping the sign altogether.
• Placing the Roman numeral after the cation when it belongs to the anion’s central atom.
• Defining oxidation by electrons only when the question asks about oxidation number (or the reverse).
• Naming the oxidising agent as “chlorine is oxidised” instead of “chlorine is reduced”.
• Calling any reaction with two changes of one element disproportionation without checking it comes from a single species.
• Balancing atoms in a half-equation but not charge, or putting the electrons on the wrong side.
• Leaving electrons, or uncancelled H⁺ and H₂O, in a full ionic equation.
• Forgetting that oxygen in water often keeps its oxidation number of −2 while only hydrogen changes.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 2, Topic 8A: Redox chemistry, outcomes 8.1 to 8.10.