Redox Chemistry and Groups 1, 2 and 7

Edexcel International A-Level Chemistry· Specification sections 8.1–8.28· 40 min read

1. Redox Chemistry Fundamentals★★☆☆☆⏱ 20 min

📘 Definition

Oxidation Number

A number assigned to an atom in a compound to represent the number of electrons it has lost or gained relative to its neutral state. Core rules: F=-1, O=-2 (except peroxides = -1, OF₂ = +2), H=+1 (except metal hydrides = -1), sum of oxidation numbers equals the charge on the species.

Example:

Oxidation number of S in H₂SO₄ = +6

Redox reactions involve a transfer of electrons: the reducing agent donates electrons (is oxidised, oxidation number increases), while the oxidising agent accepts electrons (is reduced, oxidation number decreases).

📘 Definition

Disproportionation

A redox reaction where the same element is both oxidised and reduced in the same reaction.

Example:

$C l_{2} + H_{2} O \to H C l + H C l O$ : Cl changes from 0 to -1 (reduction) and +1 (oxidation)

📐 Worked Example

Calculate the oxidation number of chlorine in $C l O_{3}^{-}$ , and write the half-equation for its reduction to $C l^{-}$ in acidic conditions.

1
Let $x$ = oxidation number of Cl. Sum of oxidation numbers = charge on ion: $x + 3 (- 2) = - 1$ so $x = + 5$ .
2
Balance Cl atoms: $C l O_{3}^{-} \to C l^{-}$
3
Balance O atoms by adding $H_{2} O$ : $C l O_{3}^{-} \to C l^{-} + 3 H_{2} O$
4
Balance H atoms by adding $H^{+}$ : $C l O_{3}^{-} + 6 H^{+} \to C l^{-} + 3 H_{2} O$
5
Balance charge by adding electrons: +5 charge on left, -1 on right, so add 6 $e^{-}$ to left: $C l O_{3}^{-} + 6 H^{+} + 6 e^{-} \to C l^{-} + 3 H_{2} O$

Exam tip:

Always show full working for oxidation number calculations, as method marks are awarded even if your final answer is incorrect.

2. Group 1 and 2 Chemistry★★★☆☆⏱ 25 min

Reactivity increases down Groups 1 and 2 because first ionisation energy decreases: larger atomic radius and more electron shielding reduce the attraction between the nucleus and outer electrons, making electrons easier to lose during reactions.

📘 Definition

Thermal Stability

The ability of a compound to resist decomposition when heated. For Group 1/2 carbonates and nitrates, stability increases down the group: larger cations have lower polarising power, so they cause less distortion of the anion electron cloud, making decomposition require more energy.

📐 Worked Example

Write balanced equations for (a) the reaction of strontium (Group 2) with water, and (b) the thermal decomposition of strontium nitrate.

1
Group 2 metal + water → metal hydroxide + hydrogen: $S r + 2 H_{2} O \to S r (O H)_{2} + H_{2}$
2
Group 2 nitrate decomposition produces metal oxide, nitrogen dioxide and oxygen: $2 S r (N O_{3})_{2} \to 2 S r O + 4 N O_{2} + O_{2}$

Key solubility trends: Group 2 hydroxides increase in solubility down the group (Mg(OH)₂ is almost insoluble, Ba(OH)₂ is highly soluble). Group 2 sulfates decrease in solubility down the group (MgSO₄ is soluble, BaSO₄ is almost insoluble).

Exam tip:

Solubility trends are often tested with sulfate ion test questions: the ionic equation for a positive sulfate test is $B a^{2 +} (a q) + S O_{4}^{2 -} (a q) \to B a S O_{4} (s)$ (white precipitate).

3. Group 7 (Halogen) Chemistry★★★☆☆⏱ 25 min

Key trends down Group 7: melting/boiling point increases (stronger London dispersion forces as molecule size increases), electronegativity decreases (larger atomic radius reduces attraction for bonding electrons), and reactivity decreases (harder to gain an outer electron).

📘 Definition

Halogen Displacement Reaction

A redox reaction where a more reactive halogen (higher up Group 7) displaces a less reactive halide ion from its compound, producing a characteristic colour change in aqueous or non-polar solvent.

Halogen	Aqueous solution colour	Hexane (non-polar) layer colour
$C l_{2}$	Pale green	Pale green
$B r_{2}$	Orange/yellow	Orange/red
$I_{2}$	Brown	Purple/violet

📐 Worked Example

Write the full ionic equation for the reaction between chlorine water and potassium iodide solution, and state the colour change in the hexane layer after shaking.

1
Cl₂ is more reactive than I₂, so it oxidises I⁻ to I₂ and is reduced to Cl⁻. Half equations: $C l_{2} + 2 e^{-} \to 2 C l^{-}$ and $2 I^{-} \to I_{2} + 2 e^{-}$
2
Combine half equations to cancel electrons: $C l_{2} + 2 I^{-} \to 2 C l^{-} + I_{2}$
3
Colour change in hexane: from colourless to purple/violet

Common disproportionation reactions of chlorine: (1) Cold dilute NaOH: $C l_{2} + 2 N a O H \to N a C l + N a C l O + H_{2} O$ (bleach production), (2) Water: $C l_{2} + H_{2} O ⇌ H C l + H C l O$ (water treatment, HClO kills bacteria), (3) Hot concentrated NaOH: $3 C l_{2} + 6 N a O H \to 5 N a C l + N a C l O_{3} + 3 H_{2} O$ .

Exam tip:

Always state the oxidation number changes for the element undergoing disproportionation to gain full marks in exam questions.

4. Volumetric Analysis and Ion Tests★★★★☆⏱ 30 min

📘 Definition

Standard Solution

A solution of accurately known concentration, prepared by dissolving a measured mass of primary standard solid in distilled water and making up to a fixed volume in a volumetric flask.

Core Practical 4 uses a standard solution of a solid acid (e.g. ethanedioic acid) to find the concentration of a NaOH solution via titration. Indicators: phenolphthalein (pink in alkali, colourless in acid) or methyl orange (yellow in alkali, red in acid).

📐 Worked Example

25.0 cm³ of NaOH solution is titrated against 0.100 mol dm⁻³ ethanedioic acid ( $H_{2} C_{2} O_{4}$ , diprotic). The mean titre is 22.4 cm³. Calculate the concentration of NaOH.

1
Balanced equation: $H_{2} C_{2} O_{4} + 2 N a O H \to N a_{2} C_{2} O_{4} + 2 H_{2} O$
2
Moles of ethanedioic acid = $0.100 \times 0.0224 = 0.00224$ mol
3
Mole ratio NaOH: $H_{2} C_{2} O_{4}$ = 2:1, so moles NaOH = $0.00224 \times 2 = 0.00448$ mol
4
Concentration of NaOH = $0.00448/0.025 = 0.179$ mol dm⁻³ (3 s.f.)

Carbonate test: Add dilute acid, bubble gas through limewater (turns cloudy). Ionic equation: $C O_{3}^{2 -} + 2 H^{+} \to C O_{2} + H_{2} O$
Ammonium test: Add NaOH, warm, gas turns damp red litmus blue. Ionic equation: $N H_{4}^{+} + O H^{-} \to N H_{3} + H_{2} O$

Exam tip:

Give titration calculation answers to the same number of significant figures as the least precise value given in the question, usually 3 s.f.

5. Common Pitfalls

Wrong move:

Writing oxidation numbers without the sign first, e.g. writing 2 instead of +2 for Mg²⁺

Why:

Oxidation numbers represent electron gain/loss, so the sign is essential to distinguish oxidation from reduction

Correct move:

Always write the sign before the magnitude for oxidation numbers, e.g. +2, -1

Wrong move:

Forgetting that Group 2 nitrate decomposition produces O₂ as well as NO₂ and metal oxide

Why:

Unbalanced equations lose marks, and O₂ is a required product of full Group 2 nitrate decomposition

Correct move:

Use the general equation: $2 M (N O_{3})_{2} \to 2 M O + 4 N O_{2} + O_{2}$ for Group 2 nitrates

Wrong move:

Using HCl instead of HNO₃ before adding AgNO₃ for halide tests

Why:

HCl contains Cl⁻ ions that produce a false positive white AgCl precipitate

Correct move:

Always use dilute nitric acid to acidify halide test samples to remove impurities without introducing Cl⁻

Wrong move:

Stating Group 2 sulfate solubility increases down the group

Why:

This is a common confusion with the Group 2 hydroxide solubility trend, which does increase down the group

Correct move:

Recall: Group 2 hydroxides = more soluble down, sulfates = less soluble down (BaSO₄ is highly insoluble)

Wrong move:

Rounding titration calculation results to 2 significant figures when given data is to 3 s.f.

Why:

Mark schemes explicitly penalise incorrect significant figures for quantitative calculations

Correct move:

Give answers to the same number of significant figures as the least precise value provided in the question

6. Quick Reference Cheatsheet

Concept	Key Exam Fact
Oxidation number rules	F=-1, O=-2 (peroxides=-1, OF₂=+2), H=+1 (metal hydrides=-1)
Disproportionation	Same element is both oxidised and reduced in one reaction
Group 2 solubility	Hydroxides ↑ down group, sulfates ↓ down group
Halide test results	AgCl=white (soluble dilute NH₃), AgBr=cream (conc NH₃), AgI=yellow (insoluble)
Titration indicators	Phenolphthalein: pink→colourless, Methyl orange: yellow→red for acid→base titrations
Flame colours	Li=red, Na=yellow, K=lilac, Ca=brick red, Ba=green

7. Frequently Asked

What is the difference between oxidation numbers and Roman numerals in compound names?

Oxidation numbers are written with a sign first (e.g., +2) to show electron loss or gain. Roman numerals (e.g., (II)) in names state the oxidation state of a metal without a sign, used for elements with variable oxidation states.

Why do Group 2 hydroxides get more soluble down the group?

Larger Group 2 cations have lower lattice enthalpy (weaker ionic bonds) as ionic radius increases, making it easier for water molecules to break the lattice and dissolve the hydroxide.

How do I distinguish between Cl⁻, Br⁻ and I⁻ ions?

Add acidified silver nitrate: AgCl (white, soluble in dilute NH₃), AgBr (cream, soluble only in concentrated NH₃), AgI (yellow, insoluble in both dilute and concentrated NH₃).

Going deeper

syllabusEdexcel IAL Chemistry SpecificationOfficial Unit 2 specification reference
core_practicalCore Practical 3 & 4 GuideStep-by-step for titration and group test procedures

What's Next

You have now mastered all core content for Redox Chemistry and Groups 1, 2, and 7 for Edexcel IAL Chemistry Unit 2. This topic forms the foundation for more advanced redox content in Unit 4, including electrode potentials and electrochemical cells, as well as periodicity extensions in higher units. To consolidate your learning, practise past paper questions focused on group trend explanations, titration calculations, and ion identification tests, making sure you follow the mark scheme wording for command terms like 'explain' and 'describe'. Pay special attention to core practical write-ups for titrations and flame tests, as these make up ~15% of Unit 2 exam marks.