The shape of the lesson
By the end of the lesson, learners can
- Oxidation states and half-equations: Work out the oxidation number of the metal in each species in the simulator, then build the half-equations by balancing O with H 2 O, H with H + and charge with electrons.
- Titration calculations: Practise combining half-equations into full ionic equations in the simulator, then use the mole ratios in the worked titrations below.
- MnO4− : Fe2+ 1 : 5; MnO4− : C2O42− 2 : 5.
- Cr2O72− : Fe2+ 1 : 6.
- Cu2+ : S2O32− 1 : 1 (via I2, starch indicator).
- Acidify with H2SO4; E⊖cell positive means feasible.
One picture every idea returns to
Transition elements have variable oxidation states because the 3d and 4s electrons are close in energy. Acidified manganate(VII), MnO4−, is reduced to Mn2+ (purple to colourless) and is its own indicator: the end point is the first permanent pale pink. It reacts with Fe2+ in a 1 : 5 ratio and with ethanedioate in a 2 : 5 ratio. Acidified dichromate(VI), Cr2O72−, is reduced to Cr3+ (orange to green) and reacts with Fe2+ in a 1 : 6 ratio. E⊖ values predict feasibility: a reaction is feasible if E⊖cell is positive. Cu2+ oxidises I− to I2, which is titrated with thiosulfate: overall n(Cu2+) = n(S2O32−).
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, Titrate with a colour change.. Because transition metals have several stable oxidation states, their compounds make excellent oxidising and reducing agents. 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: Oxidation states and half-equations | Work out the oxidation number of the metal in each species in the simulator, then build the half-equations by balancing O with H 2 O, H with H + and charge with electrons. Project tab 6 of the lab (Transition elements lab, redox titrations) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Titration calculations | Practise combining half-equations into full ionic equations in the simulator, then use the mole ratios in the worked titrations below. Project tab 6 of the lab (Transition elements lab, redox titrations) and let a student drive it while the class predicts. | Round Table |
| 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, 16 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 Titration calculations, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| Get the ratios right: MnO 4 −: Fe 2+ = 1: 5, MnO 4 −: C 2 O 4 2− = 2: 5, Cr 2 O 7 2−: Fe 2+ = 1: 6, Cu 2+: S 2 O 3 2− = 1: 1. | Acidify manganate(VII) with sulfuric acid, never hydrochloric. The manganate(VII) end point is the first permanent pale pink, not colourless. |
Variable oxidation states and redox: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Write the ionic equation for the reaction of acidified manganate(VII) ions with iron(II) ions, and give the colour change at the end point.
6.00 g of impure iron(II) sulfate crystals, FeSO4.7H2O (Mr 278.0), is dissolved in dilute H2SO4 and made up to 250 cm3. 25.0 cm3 portions need 20.00 cm3 of 0.0200 mol dm−3 KMnO4. Calculate the percentage purity.
25.0 cm3 of 0.0500 mol dm−3 sodium ethanedioate is acidified and warmed. Calculate the volume of 0.0200 mol dm−3 KMnO4 needed.
Excess KI is added to 25.0 cm3 of copper(II) sulfate solution. The iodine released needs 21.20 cm3 of 0.100 mol dm−3 sodium thiosulfate. Calculate the concentration of Cu2+.
Use E⊖ values to explain why potassium manganate(VII) titrations are acidified with sulfuric acid rather than hydrochloric acid.
Write the half-equation for the reduction of dichromate(VI) ions in acid, and use it to deduce the ratio Cr2O72− : Fe2+.
Answer key and marking notes
- MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O.
- Colourless (very pale yellow) to the first permanent pale pink.
- n(MnO4−) = 0.0200 × 0.02000 = 4.00 × 10−4 mol.
- n(Fe2+) in 25.0 cm3 = 5 × 4.00 × 10−4 = 2.00 × 10−3 mol; in 250 cm3 = 0.0200 mol.
- Mass = 0.0200 × 278.0 = 5.56 g.
- Purity = 5.56 ÷ 6.00 × 100 = 92.7%.
- n(C2O42−) = 0.0500 × 0.0250 = 1.25 × 10−3 mol.
- 2MnO4− : 5C2O42−, so n(MnO4−) = 1.25 × 10−3 × 2/5 = 5.00 × 10−4 mol.
- V = 5.00 × 10−4 ÷ 0.0200 = 0.0250 dm3 = 25.0 cm3.
- 2Cu2+ + 4I− → 2CuI + I2; I2 + 2S2O32− → 2I− + S4O62−, so Cu2+ : S2O32− = 1 : 1.
- n(Cu2+) = 0.100 × 0.02120 = 2.12 × 10−3 mol.
- [Cu2+] = 2.12 × 10−3 ÷ 0.0250 = 0.0848 mol dm−3.
- E⊖(MnO4−/Mn2+) = +1.52 V is more positive than E⊖(Cl2/Cl−) = +1.36 V, so MnO4− oxidises Cl− to Cl2.
- Extra manganate(VII) would be used, giving too large a titre; sulfate ions are not oxidised.
- Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O (each Cr goes +6 to +3).
- Six electrons are gained and each Fe2+ gives one, so the ratio is 1 : 6.
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.
25.0 cm3 of acidified iron(II) sulfate solution reacts exactly with 22.40 cm3 of 0.0200 mol dm−3 KMnO4. MnO4− + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O. What is the concentration of Fe2+?
- A0.0896 mol dm−3
- B0.0179 mol dm−3
- C0.00358 mol dm−3
- D8.96 × 10−5 mol dm−3
Manganate(VII) titrations are acidified with dilute sulfuric acid, never hydrochloric acid. Why? (E⊖: MnO4−/Mn2+ +1.52 V; Cl2/Cl− +1.36 V)
- AMnO4− would oxidise Cl− to chlorine, so extra manganate(VII) would be used and the titre would be too high
- Bhydrochloric acid is a weak acid, so it cannot supply enough H+
- Cchloride ions form a complex with Mn2+ that hides the end-point colour
- Dsulfuric acid is needed to oxidise Fe2+ before the titration starts
25.0 cm3 of acidified Fe2+ solution reacts exactly with 20.0 cm3 of 0.0200 mol dm−3 Cr2O72−. Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O. What is the concentration of Fe2+?
- A0.0960 mol dm−3
- B0.0160 mol dm−3
- C0.00267 mol dm−3
- D0.0480 mol dm−3
Why is no separate indicator needed when acidified KMnO4 is titrated into iron(II) solution?
- AMnO4− is deep purple and Mn2+ is almost colourless, so the first permanent pale pink colour shows excess MnO4−
- Bthe Fe3+ formed turns the solution purple at the end-point
- Cmanganate(VII) is colourless until all the Fe2+ has reacted
- Dthe solution turns green at the end-point as Cr3+ forms
Can aqueous Fe3+ ions oxidise iodide ions to iodine under standard conditions? (E⊖: Fe3+/Fe2+ +0.77 V; I2/I− +0.54 V)
- Ayes; Ecell = +0.23 V, which is positive
- Bno; Ecell = −0.23 V, which is negative
- Cyes; Ecell = +1.31 V, which is positive
- Dno; Ecell = −0.31 V, because the iodine half-equation is doubled
25.0 cm3 of 0.0600 mol dm−3 sodium ethanedioate is warmed with dilute H2SO4 and titrated with 0.0200 mol dm−3 KMnO4. 2MnO4− + 16H+ + 5C2O42− → 2Mn2+ + 10CO2 + 8H2O. What volume of KMnO4 is needed?
- A30.0 cm3
- B75.0 cm3
- C188 cm3
- D15.0 cm3
Excess KI is added to 25.0 cm3 of a copper(II) solution: 2Cu2+ + 4I− → 2CuI + I2. The iodine needs 21.0 cm3 of 0.100 mol dm−3 Na2S2O3: I2 + 2S2O32− → 2I− + S4O62−. What is the concentration of Cu2+?
- A0.0840 mol dm−3
- B0.0420 mol dm−3
- C0.168 mol dm−3
- D0.00210 mol dm−3
1.00 g of brass (copper and zinc) is dissolved in nitric acid, neutralised and made up to 250 cm3. A 25.0 cm3 portion is treated with excess KI, and the iodine released needs 12.6 cm3 of 0.100 mol dm−3 thiosulfate. Zinc ions do not react. What is the percentage by mass of copper in the brass? (Ar: Cu 63.5, Zn 65.4)
- A80.0%
- B8.00%
- C40.0%
- D82.4%
Quiz key
n(MnO4−) = 0.0200 × 0.02240 = 4.48 × 10−4 mol. n(Fe2+) = 5 × 4.48 × 10−4 = 2.24 × 10−3 mol. Concentration = 2.24 × 10−3 / 0.0250 = 0.0896 mol dm−3. 0.0179 uses a 1:1 ratio. 0.00358 divides by 5 instead of multiplying, putting the ratio the wrong way round. 8.96 × 10−5 divides by 25.0 instead of 0.0250, forgetting to convert cm3 to dm3.
MnO4− (+1.52 V) is more positive than Cl2/Cl− (+1.36 V), so Ecell = 1.52 − 1.36 = +0.16 V and MnO4− can oxidise Cl− to Cl2. Some manganate(VII) would be used up on the acid rather than on the analyte, giving a titre that is too large. Hydrochloric acid is a strong acid, so supplying H+ is not the problem. Chloride does not mask the end-point colour; the issue is a side reaction. Sulfuric acid is not an oxidising agent here: it only supplies the H+ needed in the half-equation MnO4− + 8H+ + 5e− → Mn2+ + 4H2O.
n(Cr2O72−) = 0.0200 × 0.0200 = 4.00 × 10−4 mol. n(Fe2+) = 6 × 4.00 × 10−4 = 2.40 × 10−3 mol. Concentration = 2.40 × 10−3 / 0.0250 = 0.0960 mol dm−3. Each Cr goes from +6 to +3 (3 electrons) and there are two Cr per ion, hence 1:6. 0.0160 uses a 1:1 ratio. 0.00267 divides by 6 instead of multiplying. 0.0480 uses 1:3, counting the electrons for only one chromium atom.
While Fe2+ remains, each drop of purple MnO4− is reduced to very pale Mn2+ and decolourised. One drop past the end-point stays unreacted, giving a permanent pale pink colour, so KMnO4 is self-indicating. Fe3+(aq) is pale yellow, not purple. Manganate(VII) is purple from the start; it is its reduction product that is nearly colourless. The orange to green change belongs to dichromate(VI) titrations, which need an added redox indicator.
Fe3+ is reduced and I− is oxidised, so Ecell = 0.77 − 0.54 = +0.23 V. A positive value means the reaction 2Fe3+ + 2I− → 2Fe2+ + I2 is feasible. −0.23 V subtracts the wrong way round. +1.31 V adds the two values. Doubling a half-equation never changes its E⊖ value, so −0.31 V is wrong. (Feasible does not guarantee a fast rate.)
n(C2O42−) = 0.0600 × 0.0250 = 1.50 × 10−3 mol. n(MnO4−) = 1.50 × 10−3 × 2/5 = 6.00 × 10−4 mol. Volume = 6.00 × 10−4 / 0.0200 = 0.0300 dm3 = 30.0 cm3. 75.0 cm3 uses a 1:1 ratio. 188 cm3 (187.5) inverts the ratio, multiplying by 5/2. 15.0 cm3 uses the 1:5 ratio from the iron(II) titration, which does not apply: each C2O42− releases 2 electrons, not 1.
n(S2O32−) = 0.100 × 0.0210 = 2.10 × 10−3 mol. n(I2) = 1.05 × 10−3 mol, and n(Cu2+) = 2 × 1.05 × 10−3 = 2.10 × 10−3 mol, so Cu2+:S2O32− is 1:1. Concentration = 2.10 × 10−3 / 0.0250 = 0.0840 mol dm−3. 0.0420 stops at the moles of iodine (it uses Cu2+:S2O32− as 1:2). 0.168 doubles instead of halving at one step. 0.00210 is the number of moles, not a concentration.
n(S2O32−) = 0.100 × 0.0126 = 1.26 × 10−3 mol = n(Cu2+) in 25.0 cm3. In 250 cm3: 1.26 × 10−2 mol. Mass of Cu = 0.0126 × 63.5 = 0.800 g, so 0.800 / 1.00 × 100 = 80.0%. 8.00% forgets to scale up from the 25.0 cm3 portion to the whole 250 cm3. 40.0% halves the moles, using a 2:1 ratio of thiosulfate to copper. 82.4% uses the Ar of zinc instead of copper.
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.