Edexcel International A Level Chemistry: Redox chemistry (YCH11)

This guide teaches Topic 8A: Redox chemistry from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers learning outcomes 8.1 to 8.10. This is International AS content, examined in the IAS unit called Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. The rest of Topic 8 (Groups 1, 2 and 7) uses these ideas constantly, so learn this part first.

Course links: the course hub for Edexcel International A Level Chemistry and the printable Edexcel Chemistry checklist. After this guide, condense it with the redox revision notes and test yourself with the redox practice questions. Ionic equations were first met in formulae, equations and amount of substance. A free 10-minute diagnostic will point you to the outcomes you are least sure of.

Topic 8A at a glance

Outcome	What you must be able to do
8.1	Say what an oxidation number is and apply the rules for assigning one
8.2	Calculate oxidation numbers in compounds and ions, including peroxides and metal hydrides
8.3	Show an oxidation number in a name with a Roman numeral
8.4	Write a formula from a name that contains oxidation numbers
8.5	Describe oxidation and reduction as electron transfer and as oxidation number change, for s-block and p-block reactions
8.6	Know that oxidising agents gain electrons and reducing agents lose electrons
8.7	Recognise disproportionation: one element in a single species is oxidised and reduced at the same time
8.8	Use oxidation numbers to classify reactions as redox, and as disproportionation
8.9	Explain why metals generally form positive ions and non-metals negative ions
8.10	Write ionic half-equations and combine them into full ionic equations

8.1 and 8.2 Oxidation numbers

An oxidation number is the charge an atom would carry if every bond in the species were fully ionic, with the bonding electrons given to the more electronegative atom. It is a bookkeeping number. It lets you track electrons even in covalent molecules where no ions exist.

The rules, in priority order

1. An uncombined element has oxidation number 0 (Na, Cl₂, S₈, P₄).
2. The oxidation numbers in a neutral compound add up to 0.
3. The oxidation numbers in an ion add up to the charge on the ion. A simple ion such as Mg²⁺ is +2.
4. Fluorine is always −1 in compounds.
5. Group 1 metals are +1 and Group 2 metals are +2 in compounds.
6. Hydrogen is +1, except in metal hydrides (such as CsH or NaBH₄), where it is −1.
7. Oxygen takes −2 in most compounds. Peroxides (containing an O–O unit, as in CaO₂ or H₂O₂) give it −1, and it is positive when bonded to fluorine.
8. Chlorine is usually −1, but takes positive values when combined with oxygen or fluorine.

If two rules give different answers, follow the earlier rule. In a metal hydride the Group 1 or 2 rule (rule 5) beats the hydrogen rule, which is why hydrogen ends up negative.

Convention: oxidation numbers put the sign before the digit (+3, −2), while ionic charges put the digit first (3+, 2−).

Worked example 1: unknown oxidation numbers

Work out the oxidation number of each element named below.

(a) S in H₂S₂O₇ (disulfuric acid).

• H: 2 × (+1) = +2. O: 7 × (−2) = −14.
• Let S = x. Then 2 + 2x − 14 = 0, so 2x = +12.
• S = +6.

(b) B in NaBH₄ (sodium tetrahydridoborate).

• Na is +1. This is a metal hydride, so H is −1: 4 × (−1) = −4.
• 1 + x − 4 = 0, so B = +3.

(c) Quick ones. N in Mg₃N₂: 3(+2) + 2x = 0, so N = −3. O in CaO₂: +2 + 2x = 0, so O = −1 (a peroxide). H in CsH: −1. I in KIO₄: +1 + x − 8 = 0, so I = +7.

Check (b) against rule priority: if you had used H = +1, boron would come out as −5, a value that makes no chemical sense for boron. A strange answer is a signal that you have missed a peroxide or hydride.

8.3 and 8.4 Roman numerals in names

When an element can have more than one oxidation number, the name shows it as a Roman numeral in brackets, straight after the element it belongs to, with no space.

• For an oxyanion, the numeral belongs to the central atom: in NaClO₃, Cl is +5, so the name is sodium chlorate(V).
• In KNO₂, N is +3: potassium nitrate(III).
• In Na₂SO₃, S is +4: sodium sulfate(IV).
• In MnO₂, Mn is +4: manganese(IV) oxide.

Worked example 2: formulae from names

Work out the charge on each part from the oxidation numbers, then balance the charges.

Name	Reasoning	Formula
potassium bromate(V)	Br +5 with three O (−6) gives BrO₃⁻; K⁺ balances it	KBrO₃
copper(I) oxide	Cu⁺ and O²⁻; two Cu⁺ per O²⁻	Cu₂O
phosphorus(V) chloride	P +5, each Cl −1	PCl₅
barium chlorate(VII)	Cl +7 with four O (−8) gives ClO₄⁻; Ba²⁺ needs two	Ba(ClO₄)₂

The common oxyanion names are worth learning: sulfate(VI) SO₄²⁻, sulfate(IV) SO₃²⁻, nitrate(V) NO₃⁻, nitrate(III) NO₂⁻, chlorate(I) ClO⁻, chlorate(V) ClO₃⁻, chlorate(VII) ClO₄⁻.

8.5 and 8.6 Oxidation and reduction

There are two linked definitions.

	Electron view	Oxidation number view
Oxidation	electrons are lost	oxidation number goes up
Reduction	electrons are gained	oxidation number goes down

Oxidation and reduction always happen together. The electrons lost by one species are gained by another.

• An oxidising agent gains electrons. It is itself reduced.
• A reducing agent loses electrons. It is itself oxidised.

Worked example 3: an s-block metal with a p-block element

Magnesium burns in nitrogen: 3Mg + N₂ → Mg₃N₂.

1. Mg: 0 → +2. Each Mg atom loses 2 electrons. Magnesium is oxidised and is the reducing agent.
2. N: 0 → −3. Each N atom gains 3 electrons. Nitrogen is reduced and is the oxidising agent.
3. Electron check: 3 Mg lose 3 × 2 = 6 electrons; 2 N gain 2 × 3 = 6 electrons. The totals match, as they must.

The same reasoning works for 2Al + 3Cl₂ → 2AlCl₃ (Al 0 → +3, Cl 0 → −1) and for 2Na + 2H₂O → 2NaOH + H₂, where sodium goes 0 → +1 and hydrogen goes +1 → 0. In the water reaction, only the hydrogen atoms change; oxygen stays at −2 throughout.

8.9 Metals and non-metals

Metals have few electrons in their outer shell. In general they lose electrons to form positive ions, so their oxidation number increases (Na → Na⁺, 0 → +1). Non-metals have nearly full outer shells. In general they gain electrons to form negative ions, so their oxidation number decreases (Cl₂ → 2Cl⁻, 0 → −1).

This is why a reacting metal is usually the reducing agent and a reacting non-metal usually the oxidising agent. The words “in general” matter: non-metals can also show positive oxidation numbers, for example Cl in ClO₃⁻ or N in NO₂, when bonded to more electronegative atoms.

8.7 and 8.8 Disproportionation and classifying reactions

A disproportionation reaction is one in which the same element in a single species is both oxidised and reduced.

Worked example 4: spotting disproportionation

(a) Copper(I) oxide with dilute sulfuric acid: Cu₂O + 2H⁺ → Cu + Cu²⁺ + H₂O.

• Cu in Cu₂O is +1. In Cu it is 0 (reduced). In Cu²⁺ it is +2 (oxidised).
• One element, copper, in one species, Cu₂O, goes both up and down. This is disproportionation.
• H stays +1 and O stays −2.

(b) Nitrogen dioxide with water: 2NO₂ + H₂O → HNO₃ + HNO₂.

• N in NO₂ is +4. In HNO₃ it is +5 (oxidised). In HNO₂ it is +3 (reduced).
• This is disproportionation, of nitrogen.

Using oxidation numbers to classify

Assign oxidation numbers to every element on both sides.

• No changes at all: not redox. Examples are precipitation, neutralisation and many thermal decompositions, such as BaCO₃ → BaO + CO₂ (Ba +2, C +4, O −2 throughout).
• Changes in two different elements, or in the same element in two different species: redox.
• One element in one species goes both up and down: redox, and specifically disproportionation.

Watch the reverse case. If an element in two different species ends up at one in-between value, that is redox, but it is not disproportionation.

8.10 Ionic half-equations

Each half-equation follows a single species and includes the electrons it gives or takes. For example:

• Mg → Mg²⁺ + 2e⁻ (oxidation)
• Cl₂ + 2e⁻ → 2Cl⁻ (reduction)

Balancing a half-equation in acid

1. Get the right number of atoms of the redox element on each side.
2. Make up any oxygen shortfall with water molecules.
3. Make up any hydrogen shortfall with hydrogen ions.
4. Finally add electrons to the more positive side until the charges agree. Their number should equal the total oxidation number change.

Worked example 5: thiosulfate removing excess chlorine

Thiosulfate ions, S₂O₃²⁻, react with excess chlorine in water to form sulfate ions and chloride ions. Build the full ionic equation.

Oxidation half-equation

1. Sulfur: S₂O₃²⁻ → 2SO₄²⁻
2. Oxygen: 3 on the left, 8 on the right. Add 5H₂O to the left: S₂O₃²⁻ + 5H₂O → 2SO₄²⁻
3. Hydrogen: 10 on the left. Add 10H⁺ to the right.
4. Charge: left −2; right 2(−2) + 10 = +6. Add 8e⁻ to the right.

S₂O₃²⁻ + 5H₂O → 2SO₄²⁻ + 10H⁺ + 8e⁻

Check: S goes +2 → +6, a rise of 4 for each of two atoms = 8 electrons.

Reduction half-equation: Cl₂ + 2e⁻ → 2Cl⁻

Combine. Multiply the chlorine half-equation by 4 so both involve 8 electrons, then add and cancel the electrons:

S₂O₃²⁻ + 4Cl₂ + 5H₂O → 2SO₄²⁻ + 8Cl⁻ + 10H⁺

Final check: charge on the left −2; on the right 2(−2) + 8(−1) + 10 = −2. Balanced.

Worked example 6: finding a product from electron ratios

A solution containing 0.00250 mol of chlorate(V) ions, ClO₃⁻, reacts in acid with exactly 0.0150 mol of iodide ions, which are oxidised to iodine. Deduce the oxidation number of chlorine in the product.

1. 2I⁻ → I₂ + 2e⁻, so each I⁻ gives one electron: 0.0150 mol of electrons.
2. Electrons gained per ClO₃⁻ = 0.0150 ÷ 0.00250 = 6.
3. Cl starts at +5, so it ends at +5 − 6 = −1. The product is Cl⁻.
4. Half-equation: ClO₃⁻ + 6H⁺ + 6e⁻ → Cl⁻ + 3H₂O.
5. Full equation (multiply the iodide half-equation by 3): ClO₃⁻ + 6I⁻ + 6H⁺ → Cl⁻ + 3I₂ + 3H₂O.

Common errors

• Using H = +1 in a metal hydride, or O = −2 in a peroxide.
• Writing a charge as an oxidation number (2+ instead of +2), or leaving off the + sign.
• Putting the Roman numeral on the metal when it belongs to the central atom of the anion (sodium chlorate(V), not sodium(V) chlorate).
• Calling a reaction disproportionation when the element that changes comes from two different species.
• Saying “the oxidising agent is oxidised”. It is reduced.
• Adding electrons to the wrong side: electrons appear on the right in an oxidation half-equation.
• Combining half-equations without first making the electron numbers equal, or leaving electrons in the final equation.
• Forgetting to cancel H⁺ or H₂O that appear on both sides.

Where next

Condense this guide with the redox revision notes, then work through the redox practice questions, which have full worked answers. Tick off outcomes 8.1 to 8.10 on the checklist. Then apply these ideas to the elements of Groups 1 and 2 and Group 7.

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.