Follow the reversal from halogen oxidising power to halide reducing power, distinguish acid–base and redox stages, and identify halides with controlled precipitation tests.
Larger halide ions lose electrons more readily
A halide ion is a potential reducing agent because it can lose an electron. Down the group, its outer electrons are farther from the nucleus and more shielded, so electron loss becomes easier. Reducing ability increases from chloride to bromide to iodide; this is the opposite direction from halogen oxidising power.
With a solid Group 1 halide and concentrated sulfuric acid, first consider proton transfer to form a hydrogen halide. Bromide and iodide can then reduce sulfuric acid, so additional products appear. The acid’s role changes from proton donor in the initial stage to oxidising agent in the redox stage.
Chloride gives an acid–base reaction; bromide also gives redox
For sodium chloride, NaCl(s) + H₂SO₄(l) → NaHSO₄(s) + HCl(g). Steamy acidic fumes appear as HCl encounters moisture. Chloride is not a strong enough reducing agent to reduce sulfuric acid under these conditions, so this is an acid–base reaction without oxidation-number changes. Fluoride gives the analogous initial acid–base pattern, but HF requires specialist hazard controls and is not a routine student preparation.
Sodium bromide initially gives HBr. Some HBr/bromide then reduces sulfur in H₂SO₄ from +6 to +4 in SO₂ while bromide goes from −1 to 0 in Br₂. Orange-brown bromine vapour appears as well as colourless sulfur dioxide; steamy HBr fumes may also be present. Bromide does not normally reduce the sulfur as far as elemental sulfur in this standard comparison.
Use the two stages to organise an answer. The molecular redox equation below uses HBr, while a valid acidic ionic version can use sulfate, H⁺ and Br⁻. Do not mix half of each representation without checking atoms and charge.
Iodide can reduce sulfur to three different products
Iodide is the stronger reducing agent. After HI forms in the initial acid–base step, sulfuric acid may be reduced to SO₂, sulfur or H₂S, where sulfur has oxidation numbers +4, 0 and −2 respectively. Iodide is oxidised to iodine, seen as a dark solid and purple vapour. Sulfur can appear as a yellow solid; SO₂ and H₂S are colourless gases.
From sulfur’s initial +6 state, the electron gains are two to form SO₂, six to form S and eight to form H₂S. Each iodide loses one electron, so the corresponding equations consume 2, 6 or 8 HI. The equations describe possible product channels, not a claim that every experiment gives fixed equal amounts of all products.
Hydrogen sulfide is highly toxic and sulfur dioxide is harmful. Although characteristic odours are sometimes described in data, never use deliberate smelling as the identification method. These concentrated-acid reactions require a fume cupboard and a prescribed microscale demonstration with suitable protection.
Hydrogen halides form acids in water and ammonium salts with ammonia
Hydrogen chloride, bromide and iodide are molecular substances as dry gases. They react with water to form aqueous hydrogen/hydroxonium ions and halide ions, giving acidic conducting solutions. For HCl, HCl(g) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq). Dry HCl is not already a bottle of aqueous H⁺ ions.
HF is highly soluble but only partially ionises in dilute water, so do not infer that high solubility means complete ionisation or that every hydrogen halide is equally strong as an aqueous acid. Aqueous acid strength is distinct from the reducing power of the halide ion discussed above.
A hydrogen halide reacts with ammonia to give fine white solid particles of the ammonium halide. HCl + NH₃ → NH₄Cl; analogous equations give NH₄Br and NH₄I. Describe a white smoke/fume of solid salt, rather than a white gas. These are acid–base reactions, not changes in halogen oxidation number.
If a preparation of HBr or HI is required, concentrated sulfuric acid is unsuitable when its oxidising action introduces halogen and sulfur products. A non-oxidising acid such as phosphoric acid can supply H⁺ without that competing redox chemistry. Any practical method must account for the gases’ corrosiveness and very high solubility in water.
Acidify with nitric acid before adding silver nitrate
For an aqueous unknown, add dilute nitric acid, then aqueous silver nitrate. Acid removes carbonate or hydroxide interference that could otherwise produce silver precipitates. Nitric acid is used because nitrate does not introduce a competing insoluble silver salt. Hydrochloric acid would add Cl⁻ and create AgCl, confusing the test.
A precipitate’s colour suggests the halide, but similar colours are better distinguished by its solubility in aqueous ammonia. Test the precipitate with dilute ammonia first, then concentrated ammonia if needed using the approved procedure. Silver nitrate can damage/stain tissue and ammonia fumes are irritating; use small quantities and appropriate eye protection and ventilation.
Pearson 9CH0/01 June 2023 Q3(b)(ii–iii) required identifying the interference and why an unsuitable acid would form another precipitate. The report explains why ‘removes impurities’ or ‘prevents a false positive’ alone is too vague in that context. State the chemical species and unwanted observation.
| Halide | AgNO₃ observation after HNO₃ | With aqueous NH₃ |
|---|---|---|
| Cl⁻ | White AgCl precipitate | Dissolves in dilute ammonia |
| Br⁻ | Cream AgBr precipitate | Does not dissolve in dilute; dissolves in concentrated ammonia |
| I⁻ | Yellow AgI precipitate | Remains in dilute and concentrated ammonia |
| F⁻ | No silver-halide precipitate | No analogous precipitate test; AgF is soluble |
Ammonia changes the silver-ion equilibrium
Ammonia binds to Ag⁺ to form the soluble complex [Ag(NH₃)₂]⁺. This lowers the concentration of free Ag⁺ and can draw more silver halide into solution. AgCl dissolves readily; AgBr needs more concentrated ammonia; AgI is too sparingly soluble to dissolve appreciably under the test conditions.
At AS, learn the observed distinctions and correct ionic precipitation equations. The complex equation provides a useful link to later transition-metal/equilibrium chemistry: AgCl(s) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq) + Cl⁻(aq). Ammonia is not reducing Ag⁺ to silver metal in this test.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.
Q1. Why does NaCl with concentrated sulfuric acid produce HCl without the redox chemistry seen for NaBr?Show answer
Chloride accepts a proton in the acid–base step but is not a sufficiently strong reducing agent to reduce sulfuric acid under these conditions. Bromide can lose electrons to reduce sulfur from +6 to +4 while forming Br₂.
Q2. How many iodide ions lose electrons for each sulfur atom reduced from +6 to −2, and how many I₂ molecules form?Show answer
Sulfur gains eight electrons. Eight I⁻ each lose one, giving four I₂ molecules. This is consistent with 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O.
Q3. What is observed when HBr gas reacts with NH₃ gas, and what is the equation?Show answer
White smoke/fume of fine solid ammonium bromide: HBr(g) + NH₃(g) → NH₄Br(s). The visible material is a solid salt aerosol, not a white gas.
Q4. An acidified unknown gives a cream precipitate with AgNO₃, insoluble in dilute but soluble in concentrated ammonia. Identify the anion.Show answer
Br⁻. The precipitate is AgBr: Ag⁺(aq) + Br⁻(aq) → AgBr(s). The two-stage ammonia result supports the cream-colour identification.
Q5. Explain chemically why HCl is unsuitable for acidifying a halide test.Show answer
It supplies Cl⁻, which reacts with Ag⁺ to produce a white AgCl precipitate even if the original sample contained no chloride. Nitric acid avoids introducing this precipitating anion.
Sources
Sources and examiner guidance (reviewed 9 October 2026)
- Pearson Edexcel 9CH0 specification, Issue 3 — Topic 4, printed pp. 13–14, checked against 8CH0 Topic 4, printed pp. 11–12. Reviewed 9 October 2026.
- Chemrevise — Edexcel Inorganic Chemistry and the Periodic Table — All nine pages reviewed as a secondary coverage check; explanations and questions here are original.
- Pearson 8CH0/01 June 2023 mark scheme — Q2(b–c), Q3(b)(ii), Q6 and Q9(b); PDF pp. 8–9, 13, 20–21, 28–30. Guidance is specific to these questions.
- Pearson 8CH0/01 June 2023 examiner report — Q2–3, Q6 and Q9; PDF pp. 4–8. Reviewed with question context on 9 October 2026.
- Pearson 8CH0/01 June 2023 question paper — Context for ion tests, displacement, redox and Group 2 reasoning. Original exercises below do not copy these questions.
- Pearson 9CH0/01 June 2023 mark scheme — Q3(a–b), PDF pp. 8–9: bromide/sulfuric acid and nitric-acid/silver-nitrate reasoning.
- Pearson 9CH0/01 June 2023 examiner report — Q3(b)(ii–iii), PDF p. 8: identify interfering ions and the unwanted precipitate, not just a vague 'false positive'.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.
