Ch 3 · Some reactions of the halide ions · AS & A Level Chemistry 9701
Topic 3 of 4 in Group 17 — AS & A Level Chemistry 9701 — 48 lessons.
Introduction
If the halogens are oxidising agents that grow weaker down the group, the halide ions they form are reducing agents that grow stronger down the group. This single reversal explains two of the most heavily examined practicals in this subtopic: testing for halide ions with silver nitrate and ammonia, and observing what concentrated sulfuric acid does to each solid halide.
Key ideas — Halide ions as reducing agents
A halide ion reduces another species by losing an electron and being oxidised back to the halogen: 2X^- arrow X_2 + 2e^-. Reducing power increases down the group from Cl^- to I^-: down the group, the halide ion is larger, so its outer electrons are further from the nucleus and more shielded, and are therefore held less strongly and given up more readily. F^- and Cl^- are weak reducing agents; Br^- is a moderate reducing agent; I^- is a strong reducing agent, readily oxidised by species that cannot oxidise Cl^- or Br^-.
🧑🏫 Teacher's explanation
Think of a halide ion as a person carrying a heavy backpack (the extra electron). A 'reducing agent' is someone willing to throw that backpack away to help someone else. The bigger the person (the larger the ion), the more loosely they hold onto that backpack because the nucleus's grip is spread over a much larger area.
📖 Definitions
🔍 Worked example — Easy
Predict the relative order of reducing strength for F^-, Cl^-, Br^-, and I^-.
🔍 Worked example — Standard
Write the balanced equation for the reaction between I^- and Cl_2 in aqueous solution.
🔍 Worked example — Exam-style
A sample of NaBr(aq) is reacted with concentrated H_2SO_4. Identify the gas evolved and the colour of the observed fumes.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Silver nitrate test for halide ions
Aqueous silver nitrate reacts with each aqueous halide to give a precipitate of silver halide, distinguishable by colour and by its solubility in ammonia:
🧑🏫 Teacher's explanation
Think of the silver halide precipitates as a crowd of people holding hands very tightly. The silver ions (Ag^+) and halide ions (X^-) form a crystal lattice. When you add silver nitrate, you are essentially forcing these ions to stick together into a solid 'clump' (a precipitate). The color changes because as you go down Group 17, the ions get larger and the electron cloud becomes more 'squishy' (polarizable), which changes how they interact with light.
📖 Definitions
🔍 Worked example — Easy
A student adds AgNO_3(aq) to a solution containing Cl^-(aq). State the color of the precipitate formed.
🔍 Worked example — Standard
Write the balanced equation for the reaction between silver chloride and aqueous ammonia to form a soluble complex.
🔍 Worked example — Exam-style
A solution contains Br^-(aq). It forms a cream precipitate with AgNO_3(aq). This precipitate dissolves in dilute NH_3(aq). How would the precipitate behave in concentrated NH_3(aq)?
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
The student then adds dilute aqueous ammonia to the precipitate formed from X. The precipitate dissolves. The student then adds concentrated aqueous ammonia to the precipitate formed from Y. The precipitate dissolves. The student then adds concentrated aqueous ammonia to the precipitate formed from Z. The precipitate does not dissolve.
Key ideas — Reactions with concentrated sulfuric acid: NaCl and NaBr
Solid sodium halides react with concentrated sulfuric acid. With NaCl, only an acid–base (non-redox) reaction occurs, since Cl^- is too weak a reducing agent to be oxidised by sulfuric acid: NaCl(s) + H_2SO_4(l) arrow NaHSO_4(s) + HCl(g), giving steamy white fumes of HCl. With NaBr, the initial acid–base step still occurs, NaBr(s) + H_2SO_4(l) arrow NaHSO_4(s) + HBr(g), but Br^- is a strong enough reducing agent to be partly oxidised by the sulfuric acid, so some HBr is further oxidised to orange bromine vapour while H_2SO_4 is reduced to SO_2: 2HBr(g) + H_2SO_4(l) arrow Br_2(g) + SO_2(g) + 2H_2O(l).
🧑🏫 Teacher's explanation
Think of concentrated sulfuric acid as a 'chemical bully'. It is a very strong oxidizing agent, meaning it wants to steal electrons from other substances. When it meets a halide ion, a competition begins: can the halide ion resist being oxidized, or will it give up its electrons to the acid?
📖 Definitions
🔍 Worked example — Easy
Predict the products and state of the gas evolved when solid NaCl reacts with concentrated H_2SO_4.
🔍 Worked example — Standard
Write the balanced equation for the oxidation of HBr by concentrated H_2SO_4.
🔍 Worked example — Exam-style
A student reacts solid NaBr with concentrated H_2SO_4. Describe the observations made.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Write a balanced chemical equation for the reaction between NaBr and H_2SO_4 that shows the formation of Br_2 and SO_2. [2] (b) State the color of the gas evolved during this reaction. [1] [3 marks]
Key ideas — Reactions with concentrated sulfuric acid: NaI
I^- is the strongest reducing agent among these halide ions, so with NaI the redox reaction dominates and proceeds further, giving a vivid sequence of observations: steamy fumes of HI initially, then a colour change to brown as I_2 vapour forms, then a yellow solid deposit of solid iodine, alongside black solid and the choking, rotten-egg smell of hydrogen sulfide gas as reduction proceeds all the way to sulfur and even H_2S:
🧑🏫 Teacher's explanation
Think of concentrated sulfuric acid as a 'chemical bully' that wants to be reduced. It is a strong oxidizing agent, meaning it wants to grab electrons from other substances. However, the halide ions are also fighting for control. A 'reducing agent' is like a person giving away money; the more 'generous' (stronger) the agent, the more it can force the bully to change its state completely.
📖 Definitions
🔍 Worked example — Easy
Predict the products and state of the sulfur species when NaI reacts with concentrated H_2SO_4 if the reduction is only partial (to SO_2). Write the balanced equation for the reduction of HI to I_2 and SO_2.
🔍 Worked example — Standard
Calculate the number of moles of I_2 produced when 0.10 moles of HI react completely with concentrated H_2SO_4 to produce only I_2, SO_2, and H_2O.
🔍 Worked example — Exam-style
A student observes a reaction between NaI and conc. H_2SO_4. They see purple/brown vapors and a black solid. They conclude the sulfur has been reduced to H_2S. Which equation represents the reaction where sulfur is reduced to the -2 oxidation state?
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) State the observation regarding the color of the gas evolved if the reaction proceeds to produce sulfur dioxide. [1] (b) Write a balanced chemical equation for the reaction between HI and concentrated H_2SO_4 that results in the formation of solid sulfur and water. [2] [3 marks]
Worked example
Question. Solid sodium iodide is added to concentrated sulfuric acid.
🔍 Worked example — Easy
Predict the relative order of reactivity of Cl^-, Br^-, and I^- as reducing agents.
🔍 Worked example — Standard
Write the balanced equation for the reaction between NaBr and concentrated H_2SO_4 that produces Br_2 and SO_2.
🔍 Worked example — Exam-style
Solid KI reacts with concentrated H_2SO_4. Write the balanced equation for the formation of I_2 and H_2S, and state the change in oxidation state of sulfur.
⚠️ Common errors
Common mistakes
Practice
Answers