What are the general reactions of halide ions with silver nitrate in aqueous solution?

Halide ions react with silver nitrate to form precipitates of silver halides. Chloride ions form a white precipitate of silver chloride, bromide ions form a cream precipitate of silver bromide, and iodide ions form a yellow precipitate of silver iodide. These reactions are commonly used in qualitative analysis to identify halide ions.

How do halide ions react with concentrated sulfuric acid?

When halide ions react with concentrated sulfuric acid, the reactions vary depending on the halide. Chloride ions produce hydrogen chloride gas, bromide ions produce hydrogen bromide and bromine, and iodide ions produce hydrogen iodide, iodine, and sulfur dioxide. These reactions demonstrate the increasing reducing power of halide ions down the group.

Why do halide ions show different reactivity with oxidizing agents?

Halide ions show different reactivity with oxidizing agents due to their varying reducing abilities. As you move down Group 17, the reducing power of halide ions increases. This means iodide ions are more easily oxidized compared to bromide and chloride ions, making them more reactive with oxidizing agents like sulfuric acid.

What is the role of halide ions in displacement reactions?

In displacement reactions, a more reactive halogen can displace a less reactive halide ion from its compound. For example, chlorine can displace bromide and iodide ions from their salts, while bromine can displace iodide ions. This reactivity trend is due to the decreasing reactivity of halogens down the group.

How do halide ions participate in redox reactions?

Halide ions participate in redox reactions by acting as reducing agents. They donate electrons to oxidizing agents, resulting in the formation of halogen molecules. For instance, iodide ions can be oxidized to iodine by agents like chlorine or potassium permanganate, highlighting their role in redox chemistry.

Why is "Acidifying with HCl before the silver nitrate test." a common mistake in Some reactions of the halide ions?

Why it happens: Habit of using any acid. How to avoid it: Use dilute HNO₃ — HCl would add chloride and give a false positive. Source: 9701 Examiner Reports 2022-2024

Why is "Thinking NaCl + conc H₂SO₄ is a redox reaction." a common mistake in Some reactions of the halide ions?

Why it happens: Assuming all halides reduce sulfuric acid. How to avoid it: Cl⁻ is too weak a reducing agent — it only gives HCl (no change in sulfur).

Why is "Incomplete products in the conc. H₂SO₄ reactions." a common mistake in Some reactions of the halide ions?

Why it happens: Forgetting the reduced sulfur species. How to avoid it: Br⁻ → SO₂; I⁻ → H₂S (plus the halogen X₂ in each).

Why is "Saying Cl⁻ is the strongest reducing agent." a common mistake in Some reactions of the halide ions?

Why it happens: Confusing with chlorine's oxidising power. How to avoid it: Halide reducing power increases down the group (I⁻ strongest).

Group 17(AS-Level Inorganic Chemistry)Cambridge International A Level Chemistry 9701

Some reactions of the halide ions

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Some Reactions of the Halide Ions — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

Halide ions as reducing agents, the silver nitrate / ammonia test for distinguishing chloride, bromide and iodide, and their reactions with concentrated sulfuric acid that reveal the reducing-power trend.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

11.3.1 — Describe the relative reactivity of halide ions as reducing agents.
11.3.2 — Describe and explain the reactions of halide ions with (a) aqueous silver ions followed by aqueous ammonia, and (b) concentrated sulfuric acid (with balanced equations).

Halide ions as reducing agents

Reducing power increases down the group — the bigger the ion, the more easily it loses an electron.

A halide ion acts as a reducing agent by donating (losing) an electron ( $\text{X}^- \rightarrow \tfrac{1}{2}\text{X}_2 + e^-$ ).

Reducing power increases down the group: $\text{I}^- > \text{Br}^- > \text{Cl}^-$ Going down, the ion is larger, so the outer electron is further from the nucleus and more shielded — it is lost more easily, making the ion a better reducing agent.

(Note this is the opposite trend to the oxidising power of the elements, which decreases down the group.)

Halide ion reduces by losing an electron.
Reducing power: I⁻ > Br⁻ > Cl⁻.
Larger ion loses electron more easily.

The silver nitrate / ammonia test

Add acidified AgNO₃ to see the precipitate colour, then ammonia to confirm which halide.

To identify a halide, add acidified silver nitrate solution (AgNO₃ with dilute nitric acid — the acid removes carbonate/hydroxide that would also give precipitates):

Halide	Precipitate	Colour	Solubility in NH₃
Cl⁻	AgCl	white	dissolves in dilute NH₃
Br⁻	AgBr	cream	dissolves in concentrated NH₃
I⁻	AgI	yellow	insoluble even in concentrated NH₃

Ionic equation: $\text{Ag}^+(aq) + \text{X}^-(aq) \rightarrow \text{AgX}(s)$ .

Because the colours can be hard to tell apart, ammonia is used to confirm: the more soluble the silver halide, the more reactive (smaller) the halide.

Acidified AgNO₃: AgCl white, AgBr cream, AgI yellow.
NH₃ confirms: AgCl dilute, AgBr conc, AgI insoluble.
Acidify first to remove interfering ions.

Reactions with concentrated sulfuric acid

How far the halide reduces H₂SO₄ shows its reducing power: Cl⁻ none, Br⁻ to SO₂, I⁻ to H₂S.

Solid halides react with concentrated sulfuric acid. How far the halide reduces the sulfur in H₂SO₄ reveals its reducing power.

Chloride (weakest reducer) — no redox; just an acid–base reaction giving HCl: $\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}$ (steamy white fumes of HCl).

Bromide — first gives HBr, then HBr reduces H₂SO₄ (S: +6 → +4) to SO₂, and Br⁻ is oxidised to Br₂ (red-brown fumes): $2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}$

Iodide (strongest reducer) — reduces H₂SO₄ all the way to H₂S (S: +6 → −2), and I⁻ is oxidised to I₂ (purple vapour / black solid; also some S and SO₂): $8\text{HI} + \text{H}_2\text{SO}_4 \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{H}_2\text{O}$

The further the reduction (SO₂ → S → H₂S), the stronger the reducing agent — confirming $\text{I}^- > \text{Br}^- > \text{Cl}^-$ .

Cl⁻: HCl only (no redox).
Br⁻: Br₂ + SO₂ (reduces S to +4).
I⁻: I₂ + H₂S (reduces S to −2) — strongest reducer.

How it’s examined

The silver-nitrate/ammonia test is a guaranteed practical and Paper 2 question; the concentrated-sulfuric-acid reactions test reducing power with full equations. Examiner reports flag forgetting to acidify with HNO₃, mixing up the precipitate colours, and incomplete H₂SO₄ products (omitting SO₂/H₂S).

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 11.3; 9701 Examiner Reports 2022-2024; 9701/22 & 9701/33 May/Jun 2024 question papers and mark schemes. Last reviewed 2026-05-20.

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Step-by-step worked examples — Some reactions of the halide ions

Step-by-step solutions to past-paper-style questions on some reactions of the halide ions, written exactly the way a tutor would explain them at the board.

1Identifying halides (3 marks)
Getting started• halide test
Question
Three solutions contain Cl⁻, Br⁻ and I⁻. Describe how acidified silver nitrate and ammonia distinguish them.
Step-by-step solution
1. Step 1
  Add acidified AgNO₃: Cl⁻ → white AgCl; Br⁻ → cream AgBr; I⁻ → yellow AgI.
2. Step 2
  Add dilute NH₃: only AgCl dissolves.
3. Step 3
  Add concentrated NH₃: AgBr dissolves; AgI stays insoluble.
Answer
Colours (white/cream/yellow) + NH₃ solubility (dilute/conc/insoluble) distinguish Cl⁻, Br⁻, I⁻.
Examiner tip
Acidify with dilute HNO₃ first — not HCl.
2Explaining the reducing-power trend (2 marks)
Getting started• reducing power
Question
Explain why I⁻ is a better reducing agent than Cl⁻.
Step-by-step solution
1. Step 1
  I⁻ is larger; its outer electron is further from the nucleus and more shielded.
2. Step 2
  So I⁻ loses an electron more easily → better reducing agent.
Answer
Larger I⁻ loses its outer electron more easily than Cl⁻.
3Reducing power with conc. H₂SO₄ (4 marks)
Building confidence• reducing power
Question
Explain, with equations, how the reactions of solid NaCl and NaI with concentrated sulfuric acid show that I⁻ is a stronger reducing agent than Cl⁻.
Step-by-step solution
1. Step 1
  NaCl: no redox, only HCl produced.
  $\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}$
2. Step 2
  NaI: I⁻ reduces H₂SO₄ all the way to H₂S (S +6 → −2), itself oxidised to I₂.
  $8\text{HI} + \text{H}_2\text{SO}_4 \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{H}_2\text{O}$
3. Step 3
  I⁻ reduces sulfur far more than Cl⁻ (which can't) → I⁻ is the stronger reducing agent.
Answer
Cl⁻ gives only HCl; I⁻ reduces H₂SO₄ to H₂S (forming I₂) → I⁻ is the stronger reducing agent.
4The intermediate case: NaBr (3 marks)
Building confidence• reducing power
Question
Write the equations for solid NaBr with concentrated sulfuric acid and name the reduced sulfur product.
Step-by-step solution
1. Step 1
  First HBr is displaced.
  $\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}$
2. Step 2
  Then HBr reduces H₂SO₄ to SO₂ (S +6 → +4), forming brown Br₂.
  $2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}$
Answer
NaBr + H₂SO₄ → NaHSO₄ + HBr, then 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O (reduced product SO₂).
Examiner tip
Br⁻ reduces sulfur only to SO₂ — between Cl⁻ (no reduction) and I⁻ (down to H₂S).
5Ranking reducing power from the products (4 marks)
Stretch• reducing power, oxidation number
Question
Using the sulfur products with concentrated H₂SO₄ (Cl⁻ → none; Br⁻ → SO₂; I⁻ → H₂S), rank the reducing power of the halides and justify with oxidation numbers.
Step-by-step solution
1. Step 1
  Cl⁻: does not reduce S (stays +6) → weakest reducing agent.
2. Step 2
  Br⁻: reduces S from +6 → +4 (SO₂) → intermediate.
3. Step 3
  I⁻: reduces S all the way from +6 → −2 (H₂S) → strongest.
4. Step 4
  The bigger the oxidation-number drop it can cause, the stronger the reducing agent → I⁻ > Br⁻ > Cl⁻.
Answer
I⁻ > Br⁻ > Cl⁻: the further it reduces sulfur (+6 → −2 vs +6 → +4 vs none), the stronger the reducing agent.
Examiner tip
Quantify with the sulfur oxidation-number change — that is the evidence for the ranking.
6Deducing an unknown halide (4 marks)
Stretch• deduction
Question
A solid sodium halide gives a cream precipitate with acidified AgNO₃ that dissolves only in concentrated NH₃, and with concentrated H₂SO₄ produces a brown vapour and a choking gas. Identify the halide and justify.
Step-by-step solution
1. Step 1
  Cream AgX, soluble only in concentrated NH₃ → AgBr → the halide is bromide.
2. Step 2
  Brown vapour = Br₂; choking gas = SO₂ (the H₂SO₄ reduction product for Br⁻).
3. Step 3
  Both tests agree → the halide is bromide, Br⁻.
Answer
Bromide (Br⁻): cream AgBr (dissolves in conc NH₃) and Br₂ + SO₂ with conc H₂SO₄.
Examiner tip
Cross-check: the AgX colour/NH₃ solubility and the H₂SO₄ products must point to the same halide.

Key Formulae — Some reactions of the halide ions

The formulae you need to memorise for some reactions of the halide ions on the Cambridge International A Level 9701 paper, with every variable defined in plain English and a note on when to use it.

Reducing power of halide ions
$\text{reducing power} \propto \text{ionic size (ease of losing an electron)}$
$ionic size$
larger ions (I⁻) lose the outer electron more easily
When to use
Explaining why reducing power increases down the group: Cl⁻ < Br⁻ < I⁻. (Sulfur products: Cl⁻ none, Br⁻ → SO₂, I⁻ → H₂S.)

Key Definitions and Keywords — Some reactions of the halide ions

Definitions to memorise and the exact keywords mark schemes credit for some reactions of the halide ions answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

Reducing power (halide ions)
Examiner keyword
The ability of a halide ion to donate an electron; increases down the group (I⁻ > Br⁻ > Cl⁻).
Silver nitrate halide test
Examiner keyword
Acidified AgNO₃ gives AgCl (white), AgBr (cream) or AgI (yellow); ammonia solubility (dilute/conc/insoluble) confirms which halide.
Conc. H₂SO₄ reduction products
Examiner keyword
Cl⁻ → no reduction (HCl only); Br⁻ → SO₂ (S +6 → +4); I⁻ → H₂S (S +6 → −2). The more it reduces, the stronger the reducing agent.

Common Mistakes and Misconceptions — Some reactions of the halide ions

The traps other students keep falling into on some reactions of the halide ions questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Acidifying with HCl before the silver nitrate test.
9701 Examiner Reports 2022-2024
Why it happens
Habit of using any acid.
How to avoid it
Use dilute HNO₃ — HCl would add chloride and give a false positive.
✕Thinking NaCl + conc H₂SO₄ is a redox reaction.
Why it happens
Assuming all halides reduce sulfuric acid.
How to avoid it
Cl⁻ is too weak a reducing agent — it only gives HCl (no change in sulfur).
✕Incomplete products in the conc. H₂SO₄ reactions.
Why it happens
Forgetting the reduced sulfur species.
How to avoid it
Br⁻ → SO₂; I⁻ → H₂S (plus the halogen X₂ in each).
✕Saying Cl⁻ is the strongest reducing agent.
Why it happens
Confusing with chlorine's oxidising power.
How to avoid it
Halide reducing power increases down the group (I⁻ strongest).

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Some reactions of the halide ions — frequently asked questions

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Some reactions of the halide ions

Short Study Notes in the form of Slides

Short Study Notes — Some reactions of the halide ions

Detailed Notes

What you’ll learn

How it’s examined

1Identifying halides (3 marks)

2Explaining the reducing-power trend (2 marks)

3Reducing power with conc. H₂SO₄ (4 marks)

4The intermediate case: NaBr (3 marks)

5Ranking reducing power from the products (4 marks)

6Deducing an unknown halide (4 marks)

Reducing power of halide ions

Reducing power (halide ions)

Silver nitrate halide test

Conc. H₂SO₄ reduction products

✕Acidifying with HCl before the silver nitrate test.

✕Thinking NaCl + conc H₂SO₄ is a redox reaction.

✕Incomplete products in the conc. H₂SO₄ reactions.

✕Saying Cl⁻ is the strongest reducing agent.

Practice questions

Past paper style quiz

Assess your understanding

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Some reactions of the halide ions — frequently asked questions