Some reactions of the halide ions

1. Overview

Halide ions, such as chloride (Cl⁻), bromide (Br⁻) and iodide (I⁻), are the negative ions that the halogens form in salts like sodium chloride. This section compares how readily these ions give away electrons, which makes them reducing agents of different strengths, and uses that trend to explain why solid halides give different products with concentrated sulfuric acid. It also covers the laboratory test for a halide ion in solution: aqueous silver ions, followed by aqueous ammonia.

Key Definitions

• Halide ion: a negative ion with a charge of 1−, formed when a halogen atom gains one electron, for example chloride Cl⁻, bromide Br⁻ and iodide I⁻.
• Reducing agent: a substance that gives electrons to another substance and so reduces it; the reducing agent is itself oxidised, so the oxidation number of one of its elements increases.
• Oxidising agent: a substance that takes electrons from another substance and so oxidises it; the oxidising agent is itself reduced, so the oxidation number of one of its elements decreases.
• Redox reaction: a reaction in which oxidation and reduction happen together: one oxidation number increases (electrons are lost) while another decreases (electrons are gained); in a disproportionation reaction both changes happen to the same element.
• Brønsted-Lowry acid: a proton (H⁺ ion) donor; concentrated sulfuric acid acts as one when it gives a proton to a halide ion to form a hydrogen halide.
• Acid-base reaction: a reaction in which a proton is transferred from an acid to a base, with no change in the oxidation number of any element.
• Precipitate: an insoluble solid that forms in a solution, often when two solutions are mixed, for example silver chloride formed from aqueous silver ions and aqueous chloride ions.
• Steamy fumes: the misty white cloud seen when a hydrogen halide gas, such as HCl, meets moist air; the gas dissolves in the water vapour in the air, forming tiny droplets of acid.

Content

Halide ions can only be oxidised

A halogen atom gains one electron to form a halide ion with a charge of 1−. Keep the names apart: chlorine is the element, Cl₂, while chloride is the ion, Cl⁻. Topics 11.1 and 11.2 cover the elements; this page is about the ions.

In a halide ion the halogen has an oxidation number of −1, the lowest a halogen can have, so the ion cannot be reduced. It can only be oxidised, giving electrons away:

2X⁻ → X₂ + 2e⁻ (X = Cl, Br or I)

So whenever a halide ion is changed in a redox reaction, it is oxidised: it acts as the reducing agent, never as the oxidising agent.

Halide ions as reducing agents

The strength of the halide ions as reducing agents increases down the group:

F⁻ < Cl⁻ < Br⁻ < I⁻ (weakest to strongest reducing agent)

Why. Down the group the ions get larger, with more shells of electrons. The outer electrons of I⁻ are further from the nucleus, and more shielded by inner electrons, than those of Cl⁻. The extra nuclear charge is outweighed by the extra distance and shielding, so the outer electrons of I⁻ are held less strongly, and I⁻ gives an electron to an oxidising agent more readily.

This mirrors the halogens as oxidising agents (topic 11.2): chlorine gains electrons most readily of Cl₂, Br₂ and I₂, so chloride is the ion that gives them back least readily.

Evidence 1: halogen displacement. Among chlorine, bromine and iodine, a halide ion reduces any halogen above it in the group. For example, bromide ions reduce chlorine:

Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq)

Iodide ions reduce both chlorine and bromine. The reverse reactions do not happen: chloride ions cannot reduce bromine or iodine, and bromide ions cannot reduce iodine.

Evidence 2: concentrated sulfuric acid. How far each ion can reduce the sulfur in sulfuric acid follows the same order, as the next subsection shows.

The hydrogen halides follow the same order: HI is the strongest reducing agent of HCl, HBr and HI. Their H-X bonds also get weaker down the group (H-Cl 431, H-Br 366, H-I 299 kJ mol⁻¹), which makes HI the easiest of the three to break apart and oxidise.

Reactions with concentrated sulfuric acid

Concentrated sulfuric acid is about 98% H₂SO₄. It can behave as an acid (a proton donor) and as an oxidising agent. It is added to the solid halide at room temperature, in a fume cupboard, because several products are toxic.

Step 1, every halide: acid-base. The acid gives a proton to the halide ion, forming a hydrogen halide gas:

NaCl(s) + H₂SO₄(l) → NaHSO₄(s) + HCl(g)

Sodium bromide and sodium iodide react in the same way, giving HBr and HI. No oxidation number changes: sulfur is +6 in both H₂SO₄ and HSO₄⁻, and the halogen stays at −1. The hydrogen halide makes steamy fumes in moist air.

You may also see 2NaCl(s) + H₂SO₄(l) → Na₂SO₄(s) + 2HCl(g). It is balanced too, but forming sodium sulfate (Na₂SO₄, not NaSO₄) needs strong heating, so NaHSO₄ is the product at room temperature.

Chloride stops at step 1. Cl⁻ and HCl are too weak as reducing agents to reduce sulfuric acid, so you see only steamy fumes of HCl.

Bromide: step 1, then redox. HBr is a strong enough reducing agent to reduce some of the sulfuric acid to sulfur dioxide:

2HBr(g) + H₂SO₄(l) → Br₂(g) + SO₂(g) + 2H₂O(l)

Bromine is oxidised from −1 to 0, and sulfur is reduced from +6 to +4. In this step HBr is the reducing agent and sulfuric acid is the oxidising agent. The same step can be written with ions, the H⁺ ions coming from the acid: 2Br⁻ + 2H⁺ + H₂SO₄ → Br₂ + SO₂ + 2H₂O. This form shows the bromide ion as the reducing agent.

You see steamy fumes (some HBr escapes) and orange-brown fumes of bromine. Sulfur dioxide also forms: a colourless gas with a choking smell.

Iodide: step 1, then more redox. HI, the strongest reducing agent of the three, reduces sulfuric acid not only to sulfur dioxide but further, to sulfur and to hydrogen sulfide:

• To SO₂ (S from +6 to +4): 2HI(g) + H₂SO₄(l) → I₂(s) + SO₂(g) + 2H₂O(l)
• To S (S from +6 to 0): 6HI(g) + H₂SO₄(l) → 3I₂(s) + S(s) + 4H₂O(l)
• To H₂S (S from +6 to −2): 8HI(g) + H₂SO₄(l) → 4I₂(s) + H₂S(g) + 4H₂O(l)

In each, iodine is oxidised from −1 to 0. You see only a few steamy fumes (most of the HI reacts), a black solid that gives a purple vapour as the mixture warms (iodine), a yellow solid (sulfur) and a smell of bad eggs (hydrogen sulfide).

Halide	What you see	Sulfur reduced to
Cl⁻	steamy fumes	not reduced (stays +6)
Br⁻	steamy fumes; orange-brown fumes; choking gas	SO₂ (+4)
I⁻	black solid; purple vapour; yellow solid; bad-egg smell	SO₂ (+4), S (0), H₂S (−2)

Why the products differ. The stronger the reducing agent, the further it can lower the oxidation number of sulfur. Each halide ion gives one electron, so lowering it further takes more halide ions per sulfur atom. Chloride ions reduce no sulfur. Bromide ions take sulfur from +6 to +4 (2 Br⁻ per S atom). Iodide ions take it as far as −2 (8 I⁻ per S atom).

Overall equations from the salt. Combine the two steps so that the hydrogen halide cancels. Two NaBr make the two HBr that the redox step uses up, giving:

2NaBr(s) + 3H₂SO₄(l) → 2NaHSO₄(s) + Br₂(g) + SO₂(g) + 2H₂O(l)

This has a practical consequence. Concentrated sulfuric acid can make hydrogen chloride from sodium chloride, but not pure hydrogen bromide or hydrogen iodide, because they are oxidised as they form. A non-oxidising acid, concentrated phosphoric acid (H₃PO₄), is used instead; you will meet it again in topic 15.1, making halogenoalkanes from alcohols.

Worked example 1: balancing the equation that forms sulfur

Question: When concentrated sulfuric acid is added to solid potassium iodide, one product is a yellow solid, sulfur. Use changes in oxidation number to balance the equation for hydrogen iodide reducing sulfuric acid to sulfur. Then write the overall equation starting from potassium iodide.

HI + H₂SO₄ → I₂ + S + H₂O (not balanced)

Step 1: find the changes. Sulfur goes from +6 in H₂SO₄ to 0 in S, so each sulfur atom gains 6 electrons. Iodine goes from −1 in HI to 0 in I₂, so each iodine atom loses 1 electron.

Step 2: balance the electrons. Electrons lost must equal electrons gained, so 6 HI react with each H₂SO₄, and the 6 iodine atoms form 3 I₂:

6HI + H₂SO₄ → 3I₂ + S + H₂O

Step 3: balance oxygen, then check hydrogen. H₂SO₄ has 4 oxygen atoms, so 4 H₂O form. Hydrogen: 6 + 2 = 8 on the left and 4 × 2 = 8 on the right.

6HI(g) + H₂SO₄(l) → 3I₂(s) + S(s) + 4H₂O(l)

Step 4: start from the salt. Each HI comes from the acid-base step, KI(s) + H₂SO₄(l) → KHSO₄(s) + HI(g). Six HI need 6 KI and 6 H₂SO₄. Add this to the equation from Step 3 and cancel 6HI:

6KI(s) + 7H₂SO₄(l) → 6KHSO₄(s) + 3I₂(s) + S(s) + 4H₂O(l)

Check: K 6 and 6; I 6 and 6; S 7 and 6 + 1; O 28 and 24 + 4; H 14 and 6 + 8.

Answer: 6HI(g) + H₂SO₄(l) → 3I₂(s) + S(s) + 4H₂O(l); overall, 6KI(s) + 7H₂SO₄(l) → 6KHSO₄(s) + 3I₂(s) + S(s) + 4H₂O(l).

Reactions with aqueous silver ions, then aqueous ammonia

This is the test for a halide ion in solution.

1. Dissolve the sample in distilled water.
2. Add dilute nitric acid. It removes ions such as carbonate, which would also give a precipitate with silver ions. Do not use hydrochloric acid: it adds chloride ions of its own.
3. Add aqueous silver nitrate and note the colour of any precipitate.
4. Add aqueous ammonia in excess (some of it is used up neutralising the nitric acid) and note whether the precipitate dissolves.

Silver chloride, bromide and iodide are insoluble in water, so the ions combine as a solid:

Ag⁺(aq) + X⁻(aq) → AgX(s)

The other ions are spectator ions. As a full equation, for example: AgNO₃(aq) + NaBr(aq) → AgBr(s) + NaNO₃(aq).

Halide ion	Add Ag⁺(aq)	Then add NH₃(aq)
Cl⁻	white precipitate, AgCl	soluble: dissolves to give a colourless solution
Br⁻	cream (off-white) precipitate, AgBr	partially soluble: only some dissolves
I⁻	pale yellow precipitate, AgI	insoluble: the precipitate does not dissolve

Fluoride ions give no precipitate, because silver fluoride is soluble in water. The three colours are hard to tell apart, especially cream and pale yellow, so the ammonia result is what confirms the halide. In an answer, use the words in the table: soluble, partially soluble, insoluble.

Going further: concentrated ammonia. The table is for ordinary bench aqueous ammonia, which is dilute. If a question uses concentrated ammonia instead, AgCl and AgBr both dissolve, but AgI still does not.

Why ammonia dissolves some silver halides

Each of AgCl, AgBr and AgI dissolves only very slightly in water, setting up an equilibrium whose position lies far to the left:

AgX(s) ⇌ Ag⁺(aq) + X⁻(aq)

The solubility in water falls from AgCl to AgBr to AgI.

Ammonia molecules bond to silver ions, forming a soluble complex ion (you do not need its formula or how it forms). This removes free Ag⁺(aq) from the solution. By Le Chatelier's principle, the position of equilibrium shifts to the right to replace those ions, so solid dissolves.

• AgCl is the most soluble of the three. Aqueous ammonia lowers the concentration of Ag⁺(aq) enough for all of it to dissolve.
• AgBr is less soluble. Aqueous ammonia dissolves only part of it, so it is partially soluble. (Concentrated ammonia ties up more of the silver ions and dissolves it.)
• AgI is so insoluble that ammonia cannot lower the concentration of Ag⁺(aq) enough. Only a tiny amount dissolves, even in concentrated ammonia, so the precipitate stays.

Worked example 2: identifying a halide from the mass of precipitate

Question: A white solid is sodium chloride, sodium bromide or sodium iodide. A 0.720 g sample is dissolved in water and acidified with dilute nitric acid, then excess aqueous silver nitrate is added. The precipitate is filtered, washed, dried and weighed: 1.31 g. Identify the solid by calculation, and describe what you would see when aqueous ammonia is added to the precipitate. Use $A_{\text{r}}$: Na 23.0, Ag 107.9, Cl 35.5, Br 79.9, I 126.9.

Step 1: link the amounts. From Ag⁺(aq) + X⁻(aq) → AgX(s), one mole of NaX gives one mole of AgX.

Step 2: predict the mass of precipitate for each salt. For sodium bromide ($M$ = 102.9 g mol⁻¹; AgBr, $M$ = 187.8 g mol⁻¹): $$n(\text{NaBr}) = \frac{0.720}{102.9} = 6.997 \times 10^{-3}\ \text{mol}$$ $$m(\text{AgBr}) = 6.997 \times 10^{-3} \times 187.8 = 1.31\ \text{g}$$

The same working for the other two salts gives:

Salt	$M$ / g mol⁻¹	Predicted precipitate / g
NaCl	58.5	1.76 (AgCl)
NaBr	102.9	1.31 (AgBr)
NaI	149.9	1.13 (AgI)

Step 3: compare. Only sodium bromide predicts the measured 1.31 g.

Step 4: the ammonia test. The precipitate is cream AgBr, which is partially soluble in aqueous ammonia: some of the precipitate dissolves, but not all of it. (In concentrated ammonia it would dissolve.)

Answer: the solid is sodium bromide, NaBr; its cream precipitate partially dissolves in aqueous ammonia. The method relies on washing and drying the precipitate fully: water or dissolved salts left on it would make the mass too large.

Key Equations

Halide ion acting as a reducing agent (X = Cl, Br or I; X is oxidised from −1 to 0):

2X⁻ → X₂ + 2e⁻

Precipitation of a silver halide (X = Cl, Br or I):

Ag⁺(aq) + X⁻(aq) → AgX(s)

Concentrated sulfuric acid, step 1: acid-base (every halide; no oxidation number changes):

NaX(s) + H₂SO₄(l) → NaHSO₄(s) + HX(g)

Step 2: redox (bromide: S from +6 to +4; iodide also gives S and H₂S):

2HBr(g) + H₂SO₄(l) → Br₂(g) + SO₂(g) + 2H₂O(l)

8HI(g) + H₂SO₄(l) → 4I₂(s) + H₂S(g) + 4H₂O(l)

Electron balance (each halide ion loses one electron; use it to balance any redox step):

number of X⁻ oxidised = number of S atoms reduced × fall in the oxidation number of S

For example, H₂S (+6 to −2) needs 8 I⁻ for each S atom.

Amount from mass ($n$ = amount in mol; $m$ = mass in g; $M$ = molar mass in g mol⁻¹; one mole of NaX gives one mole of AgX): $$n = \frac{m}{M}$$

Common Mistakes to Avoid

• Mistake: writing a redox equation for sodium chloride with concentrated sulfuric acid, with chlorine as a product. Fix: chloride ions are too weak a reducing agent to reduce sulfuric acid, so only the acid-base step happens: NaCl(s) + H₂SO₄(l) → NaHSO₄(s) + HCl(g).
• Mistake: giving sulfur in HSO₄⁻ an oxidation number of +7 or +8. Fix: H is +1, each O is −2 and the ion's charge is −1, so S + 1 − 8 = −1 and S = +6, the same as in H₂SO₄. That is why step 1 is not a redox reaction.
• Mistake: calling halide ions or hydrogen halides oxidising agents, or writing "the oxidising power of the halides increases down the group". Fix: halide ions give away electrons, so they are reducing agents, and their reducing power increases from Cl⁻ to I⁻. In 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O, HBr is the reducing agent and H₂SO₄ the oxidising agent.
• Mistake: keeping HBr as a product in the overall equation for making bromine from sodium bromide. Fix: the HBr from step 1 is used up in step 2, so it cancels: 2NaBr(s) + 3H₂SO₄(l) → 2NaHSO₄(s) + Br₂(g) + SO₂(g) + 2H₂O(l). Check: 3 S, 12 O and 6 H on each side.
• Mistake: not recognising iodine from its appearance. Fix: in the iodide reaction, the black (dark grey) solid and the purple vapour are both iodine; the yellow solid is sulfur. Orange-brown fumes mean bromine, from a bromide.
• Mistake: writing "white solution" or "turns cream" for the silver nitrate test. Fix: name the solid: "white precipitate" or "cream precipitate". With aqueous ammonia, write "the precipitate dissolves to give a colourless solution" for AgCl and "the precipitate partially dissolves" for AgBr.
• Mistake: deciding the halide from the precipitate colour alone, then writing the ammonia result to match. Fix: let the ammonia test decide. With aqueous ammonia, a precipitate that dissolves is AgCl, one that partially dissolves is AgBr, and one that does not dissolve is AgI.

Exam Tips

• Describe observations with colour and state together: "steamy fumes", "orange-brown fumes", "cream precipitate", "black solid". When asked to identify, name the substance as well, for example "purple vapour: iodine, I₂".
• Write equations for the sulfuric acid reactions step by step, with state symbols. Balance each redox step with oxidation numbers: the electrons lost by the halide ions equal those gained by sulfur (2 per S for SO₂, 6 for S, 8 for H₂S).
• Explain why the products differ as a chain: the ions get larger from Cl⁻ to I⁻, so the outer electrons are held less strongly, so the ion is a stronger reducing agent, so it reduces sulfur further (to +4 with Br⁻, as far as −2 with I⁻). State that Cl⁻ is too weak to reduce sulfuric acid at all.
• State the role of sulfuric acid in each step: an acid (proton donor) in step 1; an oxidising agent in step 2, quoting the change in the oxidation number of sulfur.
• Identify or distinguish halides by giving each reagent in full and the result for each ion: "add dilute nitric acid, then aqueous silver nitrate, then aqueous ammonia". Give a whole name or formula, such as silver nitrate or AgNO₃(aq), never just "Ag⁺" or "silver". An ionic equation for the test needs state symbols: Ag⁺(aq) + Br⁻(aq) → AgBr(s).
• Explain the ammonia results with the equilibrium AgX(s) ⇌ Ag⁺(aq) + X⁻(aq): ammonia removes Ag⁺ ions, so the equilibrium shifts to the right, and how much dissolves depends on how soluble the silver halide is.