The shape of the lesson
By the end of the lesson, learners can
- Testing for halides: Run the silver nitrate test on each halide in the simulator, then add dilute and concentrated ammonia to the precipitates.
- With concentrated sulfuric acid: Each solid halide first forms the hydrogen halide.
- Reducing power: I− > Br− > Cl−.
- Conc. H2SO4: Cl− HCl only; Br− to SO2; I− to H2S.
- AgNO3: white, cream, yellow.
- Ammonia: dilute dissolves AgCl; conc. dissolves AgBr; AgI never.
One picture every idea returns to
Halide ions act as reducing agents, losing electrons to form halogens. Reducing power increases down the group (I− > Br− > Cl−) because the outer electrons of a larger ion are further from the nucleus and more easily lost. With concentrated sulfuric acid, chloride gives only HCl; bromide also reduces the acid to SO2 (S: +6 → +4); iodide reduces it as far as S and H2S (S: +6 → −2). To test for halides, acidify with dilute nitric acid and add AgNO3(aq): AgCl is white, AgBr cream, AgI yellow, distinguished further by their solubility in ammonia.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Iodide gives electrons away.. Halide ions are reducing agents, and the bigger they are the more readily they hand over electrons. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Testing for halides | Run the silver nitrate test on each halide in the simulator, then add dilute and concentrated ammonia to the precipitates. Project tab 5 of the lab (Group 17 lab, halide tests) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: With concentrated sulfuric acid | Each solid halide first forms the hydrogen halide. | Mix-Pair-Share |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in With concentrated sulfuric acid, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Chloride ions are too weak a reducing agent to reduce concentrated sulfuric acid, so no chlorine forms; the only product is HCl. | The further the sulfur is reduced, the stronger the reducing agent: iodide takes sulfur all the way to −2 in H 2 S. In the silver nitrate test, always acidify with nitric acid, never hydrochloric. |
Reactions of the halide ions: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Describe the test for halide ions in solution, including the results for chloride, bromide and iodide.
Write equations for the reactions of concentrated sulfuric acid with (a) sodium chloride and (b) hydrogen bromide.
Explain why concentrated sulfuric acid is reduced by bromide and iodide ions but not by chloride ions.
Write an equation for the reduction of concentrated sulfuric acid to hydrogen sulfide by hydrogen iodide, and give the change in oxidation number of sulfur.
Excess AgNO3(aq) is added to a solution containing 0.585 g of NaCl (Mr 58.5). Calculate the mass of AgCl (Mr 143.4) formed.
Explain why dilute nitric acid is added before silver nitrate in the halide test.
Answer key and marking notes
- Add dilute nitric acid, then aqueous silver nitrate.
- Chloride: white precipitate; bromide: cream; iodide: yellow.
- Confirm with ammonia: AgCl dissolves in dilute NH3, AgBr only in concentrated NH3, AgI in neither.
- (a) NaCl + H2SO4 → NaHSO4 + HCl.
- (b) 2HBr + H2SO4 → Br2 + SO2 + 2H2O.
- Reducing power increases down the group as the ions get larger.
- The outer electrons of Br− and I− are further from the nucleus and more shielded, so they are lost more easily.
- Cl− holds its electrons too strongly to reduce the sulfuric acid.
- 8HI + H2SO4 → 4I2 + H2S + 4H2O.
- Sulfur changes from +6 to −2.
- n(NaCl) = 0.585 ÷ 58.5 = 0.0100 mol = n(AgCl).
- Mass = 0.0100 × 143.4 = 1.43 g.
- It reacts with and removes ions such as carbonate that would also form precipitates with Ag+.
- Nitric acid is used because it adds no halide ions.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which list puts the halide ions in order of DECREASING strength as reducing agents?
- AI− > Br− > Cl−
- BCl− > Br− > I−
- CBr− > I− > Cl−
- Dthey are all equally strong, because each loses one electron
Concentrated sulfuric acid is added to solid sodium chloride. What is observed and what is formed?
- Asteamy white fumes of hydrogen chloride: NaCl + H2SO4 → NaHSO4 + HCl
- Bgreen chlorine gas and sulfur dioxide: 2HCl + H2SO4 → Cl2 + SO2 + 2H2O
- Cpurple fumes and a smell of rotten eggs
- Dorange-brown fumes of chlorine
The diagram shows how far each halide ion reduces the sulfur in concentrated sulfuric acid. Which reduction product, formed only with iodide, has a smell of rotten eggs, and what is the oxidation number of sulfur in it?
- AH2S, −2
- BSO2, +4
- CS, 0
- DH2SO4, +6
Before testing a solution for halide ions with aqueous silver nitrate, dilute nitric acid is added. Why is nitric acid used?
- Ato remove ions such as carbonate that would also give a precipitate with Ag+, without adding any halide or sulfate ions
- Bto supply chloride ions so that the precipitate forms faster
- Cto dissolve the silver halide precipitate so its colour can be seen
- Dany dilute acid would do; hydrochloric acid works equally well
A solution of a sodium halide is acidified and aqueous silver nitrate is added, giving a cream precipitate. The precipitate does not dissolve in dilute ammonia but dissolves in concentrated ammonia. Using the diagram, identify the halide and the precipitation equation.
- Abromide; Ag+(aq) + Br−(aq) → AgBr(s)
- Biodide; Ag+(aq) + I−(aq) → AgI(s)
- Cchloride; Ag+(aq) + Cl−(aq) → AgCl(s)
- Dbromide; Na+(aq) + NO3−(aq) → NaNO3(s)
Solid sodium bromide reacts with concentrated sulfuric acid. HBr forms first, then some of it reacts further. Which equation shows the second step, and how does the oxidation number of sulfur change?
- A2HBr + H2SO4 → Br2 + SO2 + 2H2O; sulfur is reduced from +6 to +4
- B2HBr + H2SO4 → Br2 + SO2 + 2H2O; sulfur is oxidised from +4 to +6
- C8HBr + H2SO4 → 4Br2 + H2S + 4H2O; sulfur is reduced from +6 to −2
- DNaBr + H2SO4 → NaHSO4 + HBr; sulfur is reduced from +6 to +4
A solution contains both chloride and iodide ions. Excess acidified silver nitrate is added, and then excess concentrated ammonia solution. What is seen at the end?
- Aa yellow precipitate remains, because AgI is insoluble in concentrated ammonia while AgCl dissolves
- Ba clear colourless solution, because both silver halides dissolve in concentrated ammonia
- Ca white precipitate remains, because AgCl is insoluble in ammonia
- Da mixture of white and yellow precipitates remains, because neither dissolves in ammonia
Excess acidified silver nitrate is added to 25.0 cm3 of aqueous sodium chloride. The dried precipitate of AgCl has a mass of 0.287 g. What was the concentration of the sodium chloride solution? (Mr AgCl = 143.4)
- A0.0800 mol dm−3
- B0.160 mol dm−3
- C0.00200 mol dm−3
- D8.00 × 10−5 mol dm−3
Quiz key
A halide ion acts as a reducing agent by losing an electron (2X− → X2 + 2e−). I− is the largest ion, so its outer electrons are furthest from the nucleus and most shielded, and it loses them most easily: reducing power increases down the group, I− > Br− > Cl−. The reverse order makes I− the weakest reducing agent, the classic error (it mirrors the oxidising power of the halogens, which does decrease down the group). Bromide lies between chloride and iodide, so it cannot be the strongest. Each loses one electron, but with different ease, so the strengths differ.
Concentrated H2SO4 acts as an acid and displaces the volatile hydrogen halide: NaCl + H2SO4 → NaHSO4 + HCl. HCl forms steamy fumes in moist air. Cl− (and HCl) is too weak a reducing agent to reduce the sulfuric acid, so no Cl2 or SO2 forms; claiming it does is the classic error. Purple fumes and rotten eggs (I2 and H2S) come from sodium iodide. Orange-brown fumes are bromine, from sodium bromide; chlorine is pale green, and is not formed here anyway.
Iodide is the strongest reducing agent, so it reduces sulfur all the way from +6 in H2SO4 to −2 in H2S, which smells of rotten eggs (8HI + H2SO4 → 4I2 + H2S + 4H2O). SO2 (+4) is formed by both bromide and iodide and has a choking smell, not a rotten-egg smell. Sulfur (0) is a yellow solid, also formed by iodide, but it is odourless. H2SO4 (+6) is the starting oxidising agent, not a reduction product.
Carbonate (and hydroxide) ions would also precipitate with Ag+ and confuse the result, so they are removed by acid first: CO32− + 2H+ → CO2 + H2O. Nitric acid is used because silver nitrate is soluble, so the nitrate ion adds nothing to the test. Nitric acid contains no chloride ions. It does not dissolve silver halides; the precipitate must stay to be seen. Hydrochloric acid would add Cl− ions, giving a white AgCl precipitate whatever the sample contained (and sulfuric acid could precipitate silver sulfate).
A cream precipitate that dissolves only in concentrated ammonia is silver bromide: Ag+(aq) + Br−(aq) → AgBr(s). Silver iodide is yellow and insoluble even in concentrated ammonia. Silver chloride is white and dissolves in dilute ammonia. Sodium nitrate is very soluble and stays in solution as spectator ions; the precipitate is the silver halide.
Bromide is strong enough to reduce sulfuric acid, but only as far as SO2: 2HBr + H2SO4 → Br2 + SO2 + 2H2O. Sulfur goes from +6 to +4 (reduced) while bromine goes from −1 to 0 (oxidised), giving orange-brown Br2 fumes. Sulfur starts at +6 in H2SO4, so it is reduced, not oxidised. Reduction all the way to H2S needs the stronger reducing agent iodide. The acid-base step forming HBr involves no change in oxidation number, so it is not the redox step.
Both AgCl (white) and AgI (yellow) precipitate. Concentrated ammonia dissolves AgCl by forming the complex ion [Ag(NH3)2]+, but AgI is so insoluble that even concentrated ammonia cannot dissolve it, so a yellow precipitate remains. Saying AgI dissolves in concentrated ammonia is the classic error; only AgCl and AgBr do. AgCl dissolves in ammonia, even dilute, so a white precipitate does not remain. Since AgCl does dissolve, the mixture of both colours does not survive.
n(AgCl) = 0.287 / 143.4 = 0.00200 mol. The ratio NaCl : AgCl is 1 : 1, so n(NaCl) = 0.00200 mol in 25.0 cm3 = 0.0250 dm3, and the concentration = 0.00200 / 0.0250 = 0.0800 mol dm−3. 0.160 mol dm−3 doubles the moles, as if two chloride ions formed each AgCl. 0.00200 mol dm−3 is the number of moles, not divided by the volume. 8.00 × 10−5 mol dm−3 divides the moles by 25.0, the volume in cm3 instead of dm3.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.