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S6 Chemistry: Redox Titration Calculations (Paper 2: intermediate)
S6 Chemistry: Redox Titration Calculations (Paper 2: intermediate)
Ten intermediate items on the oxidation number and electrons of dichromate, the dichromate to iron(II) ratio, moles of electrons converted to charge, concentration from a dichromate titre, iodometric analysis of chlorine water, percentage of copper in brass, standardising permanganate with ethanedioic acid, percentage of hydrogen peroxide, a back titration on pyrolusite and the misuse of starch.
Direct answers Answers in order: +6 and three electrons, 1 to 6, 5.00 x 10 to the minus 3 mol and 482.5 C, 0.0720 M, 0.0360 M and 2.56 g per dm3, 15.2 percent, 0.0250 M, 3.40 percent, 34.8 percent, and no starch is needed.
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)
- FoundationWhat is the oxidation number of chromium in dichromate, and how many electrons does each chromium gain on reduction to chromium(III)?
- FoundationIn the reaction of acidified dichromate with iron(II), what is the mole ratio of dichromate to iron(II)?
- FoundationWhen 5.00 x 10 to the minus 3 mol of iron(II) is oxidised to iron(III), how many moles of electrons are transferred and what charge is that, taking one mole of electrons as 96500 C?
- IntermediateTitrating 25.0 cm3 of an iron(II) solution needs 20.0 cm3 of 0.0150 M potassium dichromate. Find the iron(II) concentration.
- IntermediateExcess potassium iodide is added to 25.0 cm3 of chlorine water and the iodine released needs 18.0 cm3 of 0.100 M thiosulphate. Find the concentration of chlorine water and the mass of chlorine per dm3, taking chlorine as 71.0.
- IntermediateA 1.000 g brass sample is dissolved and titrated iodometrically, needing 24.0 cm3 of 0.100 M thiosulphate. Find the percentage of copper, taking Cu as 63.5.
- ChallengeStandardising permanganate against ethanedioic acid, 25.0 cm3 of 0.0500 M acid needs 20.0 cm3 of permanganate, with a ratio of 2 permanganate to 5 acid. Find the concentration of the permanganate.
- Challenge10.0 cm3 of hydrogen peroxide solution is diluted to 250 cm3 and a 25.0 cm3 portion needs 20.0 cm3 of 0.0200 M permanganate, with five peroxide to two permanganate. Find the percentage w/v of the original solution.
- Challenge1.00 g of pyrolusite is treated with 50.0 cm3 of 0.200 M iron(II) solution and the excess iron(II) needs 20.0 cm3 of 0.0200 M permanganate for back titration, with manganese dioxide to iron(II) being 1 to 2. Find the percentage of manganese dioxide, taking its molar mass as 86.9.
- Error redoA student adds starch indicator before a permanganate titration. Is that correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | +6, and each chromium gains three electrons | Two chromium atoms share +12 and the seven oxygen atoms contribute minus 14, giving minus 2 for the ion, so each chromium is +6 and 6 minus 3 is 3. |
| 2 | 1 to 6 | The balanced equation is dichromate plus fourteen hydrogen ions plus six iron(II) giving two chromium(III) plus six iron(III) plus seven water, so one dichromate oxidises six iron(II). |
| 3 | 5.00 x 10 to the minus 3 mol of electrons, which is 482.5 C | Each iron(II) ion releases one electron, so the moles of electrons are the same, and multiplying by 96500 gives 482.5 C. |
| 4 | 0.0720 M | The dichromate is 3.00 x 10 to the minus 4 mol, six times that is 1.80 x 10 to the minus 3 mol of iron(II), and dividing by 0.0250 dm3 gives 0.0720 M. |
| 5 | 0.0360 M, which is 2.56 g per dm3 | The 1.80 x 10 to the minus 3 mol of thiosulphate corresponds to 9.00 x 10 to the minus 4 mol of iodine, and chlorine is 1 to 1 with iodine; divide by 0.0250 dm3 and multiply by 71.0. |
| 6 | 15.2 percent | The 2.40 x 10 to the minus 3 mol of thiosulphate equals the moles of copper, giving 0.1524 g, and dividing by 1.000 g gives 15.24 percent which rounds to 15.2 percent. |
| 7 | 0.0250 M | The acid is 1.25 x 10 to the minus 3 mol, so the permanganate is two fifths of that which is 5.00 x 10 to the minus 4 mol, and dividing by 0.0200 dm3 gives 0.0250 M. |
| 8 | 3.40 percent w/v | The 4.00 x 10 to the minus 4 mol of permanganate corresponds to 1.00 x 10 to the minus 3 mol of peroxide in the portion, so the whole 250 cm3 holds 1.00 x 10 to the minus 2 mol which is 0.340 g, giving 34.0 g per dm3 and therefore 3.40 percent. |
| 9 | 34.8 percent | The iron(II) added is 1.00 x 10 to the minus 2 mol and the back titration uses 2.00 x 10 to the minus 3 mol, so 8.00 x 10 to the minus 3 mol reacted and the manganese dioxide is 4.00 x 10 to the minus 3 mol which weighs 0.3476 g. |
| 10 | No, a permanganate titration needs no starch because the purple of the permanganate ion is the indicator itself | Starch is used only in iodine titrations, while the end point of a permanganate titration is the first permanent faint purple colour. |
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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