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S6 Chemistry: Redox Titration Calculations (Paper 4: misconception drill)
S6 Chemistry: Redox Titration Calculations (Paper 4: misconception drill)
Ten misconception drill items on the oxidation number of manganese, a mole calculation, the dichromate ratio, concentration from a titre, the mass of permanganate to weigh out, moles of iodine, percentage of iron, iron content of the hydrated salt, the mean titre and the rough titration that must be excluded.
Direct answers Answers in order: reduction from +7 to +2, 1.00 x 10 to the minus 3 mol, 1 to 6, 0.0780 M, 0.790 g, 5.00 x 10 to the minus 4 mol, 20.9 percent, 20.1 percent, 23.60 cm3 and 0.0944 M, and the rough titration is excluded.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S6 Chemistry: Redox Titration Calculations |
| Level | S6 |
02 Questions (10 items)
- FoundationWhen manganese in permanganate is reduced to manganese(II) in acid, is that oxidation or reduction, and how does the oxidation number change?
- FoundationHow many moles of solute are in 20.0 cm3 of a 0.0500 M solution?
- FoundationIn the titration of iron(II) with acidified dichromate, what is the mole ratio of dichromate to iron(II) and why?
- IntermediateTitrating 25.0 cm3 of an iron(II) solution needs 26.0 cm3 of 0.0150 M permanganate. Find the iron(II) concentration.
- IntermediateWhat mass of potassium permanganate is needed to make 250.0 cm3 of 0.0200 M solution, taking its molar mass as 158.0?
- IntermediateHow many moles of iodine can 20.0 cm3 of 0.0500 M thiosulphate reduce?
- ChallengeA 0.800 g iron sample needs 30.0 cm3 of 0.0200 M permanganate. Find the percentage of iron, taking Fe as 55.8.
- ChallengeWhat is the mass percentage of iron in pure hydrated iron(II) sulphate, taking Fe as 55.8 and the molar mass as 278.0?
- ChallengeThree titres read 23.60, 23.65 and 23.55 cm3 and are used for 25.0 cm3 of an iron(II) sample. Find the mean volume and the iron(II) concentration, with 0.0200 M permanganate as titrant.
- Error redoA student includes the rough titration volume in the mean. What is wrong with that?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Reduction, with the oxidation number falling from +7 to +2, so each permanganate ion gains five electrons | A fall in oxidation number is reduction, and 7 minus 2 is 5, so each permanganate ion gains five electrons. |
| 2 | 1.00 x 10 to the minus 3 mol | First change 20.0 cm3 to 0.0200 dm3, then multiply by 0.0500 to get 1.00 x 10 to the minus 3 mol. |
| 3 | 1 to 6, because each dichromate ion accepts six electrons while each iron(II) gives up only one | Each of the two chromium atoms falls from +6 to +3, giving six electrons in total, which oxidises exactly six iron(II) ions. |
| 4 | 0.0780 M | The permanganate is 3.90 x 10 to the minus 4 mol, five times that is 1.95 x 10 to the minus 3 mol of iron(II), and dividing by 0.0250 dm3 gives 0.0780 M. |
| 5 | 0.790 g | The moles needed are 0.0200 times 0.2500 which is 5.00 x 10 to the minus 3 mol, and multiplying by 158.0 gives 0.790 g. |
| 6 | 5.00 x 10 to the minus 4 mol | The thiosulphate is 1.00 x 10 to the minus 3 mol and the ratio of iodine to thiosulphate is 1 to 2, so the iodine is 5.00 x 10 to the minus 4 mol. |
| 7 | 20.9 percent | The permanganate is 6.00 x 10 to the minus 4 mol, the iron(II) is 3.00 x 10 to the minus 3 mol which weighs 0.1674 g, and dividing by 0.800 g gives 20.9 percent. |
| 8 | 20.1 percent | Dividing 55.8 by 278.0 and multiplying by 100 gives 20.07 percent, which is 20.1 percent. |
| 9 | Mean 23.60 cm3, giving 0.0944 M iron(II) | The mean is 23.60 cm3, and 4.72 x 10 to the minus 4 mol of permanganate times five divided by 0.0250 dm3 gives 0.0944 M. |
| 10 | The rough titration only estimates the end point and is usually too large, so it must be left out and only agreeing titres within 0.10 cm3 should be averaged | The rough titration is run quickly, so its volume is usually too large and imprecise, and including it makes the calculated concentration too high. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Write the half equations and the balanced equation first, then convert with the mole ratio to concentration, mass or percentage.Tutor Registration →
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