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S6 Chemistry: Redox Titration Calculations (Paper 3: challenge)
S6 Chemistry: Redox Titration Calculations (Paper 3: challenge)
Ten challenge items on electron transfer, iron in a whole flask, mass of iodine, percentage of copper in an alloy, purity of hydrated iron(II) sulphate, vitamin C in juice, available chlorine in bleach, iron(II) and iron(III) together, burette reading uncertainty and the timing of starch addition.
Direct answers Answers in order: five electrons and 0.100 mol, 1.26 g, 0.190 g, 14.0 percent, 89.0 percent, 35.2 mg, 44.4 g per dm3, iron(II) 0.0800 M and iron(III) 0.0600 M, about 0.41 percent, and starch only near the end point.
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)
- FoundationHow many electrons does each permanganate ion transfer when it becomes manganese(II) in acid, and how many moles of iron(II) can 0.0200 mol of permanganate oxidise?
- FoundationA 250.0 cm3 iron solution is sampled at 25.0 cm3 and needs 22.50 cm3 of 0.0200 M permanganate. Find the mass of iron in the whole flask, taking Fe as 55.8.
- FoundationWhat mass of iodine can 15.0 cm3 of 0.100 M thiosulphate reduce? Take iodine as 253.8.
- IntermediateA 1.20 g copper alloy needs 26.40 cm3 of 0.1000 M thiosulphate in an iodometric titration. Find the percentage of copper, taking Cu as 63.5.
- Intermediate2.50 g of impure hydrated iron(II) sulphate is made up to 250.0 cm3, and a 25.0 cm3 portion needs 8.00 cm3 of 0.0200 M permanganate. Find the purity of the sample, taking the hydrated salt as 278.0.
- Intermediate250 cm3 of fruit juice needs 25.0 cm3 of 0.0200 M iodine, with vitamin C reacting 1 to 1 with iodine and having a molar mass of 176.1. Find the milligrams of vitamin C per 100 cm3.
- Challenge5.00 cm3 of bleach is diluted to 250 cm3 and a 25.0 cm3 portion with potassium iodide needs 12.50 cm3 of 0.0500 M thiosulphate, with hypochlorite 1 to 1 with iodine. Find the available chlorine in g per dm3, counting it as chlorine of molar mass 71.0.
- Challenge25.0 cm3 of an iron solution needs 20.0 cm3 of 0.0200 M permanganate directly, while another 25.0 cm3 portion reduced with zinc needs 35.0 cm3 of the same permanganate. Find the concentrations of iron(II) and iron(III).
- ChallengeA titre is the difference of two burette readings, each with an uncertainty of plus or minus 0.05 cm3, and the measured volume is 24.30 cm3. What is the maximum percentage uncertainty in the volume?
- Error redoA student adds a lot of starch at the start of an iodine titration, so the solution turns deep blue and the end point is hard to see. What is the correct practice?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Five electrons, and it can oxidise 0.100 mol of iron(II) | The oxidation number falls from +7 to +2, which is five electrons, and since each iron(II) gives up only one electron, 0.0200 times 5 gives 0.100 mol. |
| 2 | 1.26 g | The portion holds 4.50 x 10 to the minus 4 mol of permanganate and 2.25 x 10 to the minus 3 mol of iron(II), so the whole flask holds 2.25 x 10 to the minus 2 mol which weighs 1.2555 g. |
| 3 | 0.190 g | The thiosulphate is 1.50 x 10 to the minus 3 mol, half of that is 7.50 x 10 to the minus 4 mol of iodine, and multiplying by 253.8 gives 0.19035 g. |
| 4 | 14.0 percent | The 2.640 x 10 to the minus 3 mol of thiosulphate equals the moles of copper, which weigh 0.16764 g, and dividing by 1.20 g gives 13.97 percent which rounds to 14.0 percent. |
| 5 | 89.0 percent | The portion contains 8.00 x 10 to the minus 4 mol of iron(II), so the whole flask holds 8.00 x 10 to the minus 3 mol which weighs 2.224 g, and dividing by 2.50 g gives 88.96 percent. |
| 6 | 35.2 mg | The 5.00 x 10 to the minus 4 mol of iodine means 5.00 x 10 to the minus 4 mol of vitamin C weighing 0.08805 g, which is 0.03522 g or 35.2 mg per 100 cm3. |
| 7 | 44.4 g per dm3 | The 6.25 x 10 to the minus 4 mol of thiosulphate means 3.125 x 10 to the minus 4 mol of iodine and the same of hypochlorite, so the flask holds 3.125 x 10 to the minus 3 mol, and dividing by 0.00500 dm3 gives 0.625 M which times 71.0 gives 44.375 g per dm3. |
| 8 | Iron(II) 0.0800 M and iron(III) 0.0600 M | The first titration measures only iron(II) and gives 2.00 x 10 to the minus 3 mol, which is 0.0800 M, while the second measures total iron and gives 3.50 x 10 to the minus 3 mol or 0.140 M, so the difference is 0.0600 M of iron(III). |
| 9 | About 0.41 percent | The two reading uncertainties add to plus or minus 0.10 cm3, and dividing by 24.30 cm3 and multiplying by 100 gives 0.412 percent, so about 0.41 percent. |
| 10 | Titrate until the solution is pale yellow first and add starch only near the end point, so that the loss of the blue colour marks the end point | The deep blue starch iodine complex holds on to iodine while much iodine is still present, which makes the end point sluggish, so starch should go in only near the end point. |
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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