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S6 Chemistry: Redox Titration Calculations (Paper 1: basics)
S6 Chemistry: Redox Titration Calculations (Paper 1: basics)
Ten basics items on moles of permanganate and iron(II) oxidised, the oxidation number change of manganese, the meaning of molarity, concentration from a titre, the iodine to thiosulphate ratio, mass of iron, percentage of iron in a sample, vitamin C concentration, mean titre and the permanganate to iron(II) ratio.
Direct answers Answers in order: 5.00 x 10 to the minus 4 mol and 2.50 x 10 to the minus 3 mol, from +7 to +2, 0.100 mol per dm3 and 2.50 x 10 to the minus 3 mol, 0.0960 M, 1.25 x 10 to the minus 3 mol, 0.1395 g, 31.2 percent, 4.40 g per dm3, 23.90 cm3 and 0.0956 M, and 1 to 5.
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 moles of permanganate ion are in 25.0 cm3 of 0.0200 M potassium permanganate, and how many moles of iron(II) ion can it oxidise?
- FoundationWhen permanganate is reduced to manganese(II) in acid solution, how does the oxidation number of manganese change?
- FoundationWhat does 0.100 M mean, and how many moles of solute are in 25.0 cm3 of such a solution?
- IntermediateTitrating 25.0 cm3 of an iron(II) solution needs 24.0 cm3 of 0.0200 M permanganate to reach the end point. Find the concentration of iron(II).
- IntermediateIn an iodine titration, 25.0 cm3 of 0.100 M sodium thiosulphate is used. How many moles of iodine does it reduce?
- IntermediateIf 2.50 x 10 to the minus 3 mol of iron(II) came from an iron sample, what mass of iron does the sample contain? Take Fe as 55.8.
- ChallengeA 0.500 g iron sample needs 28.0 cm3 of 0.0200 M permanganate. Find the percentage of iron in the sample.
- Challenge25.0 cm3 of a vitamin C sample needs 12.5 cm3 of 0.0500 M iodine solution, with vitamin C reacting with iodine in a 1 to 1 ratio, M equals 176.1. Find the concentration in g per dm3.
- ChallengeThree titres read 23.80, 23.90 and 24.00 cm3. What is the mean volume, and if that volume of 0.0200 M permanganate titrates 25.0 cm3 of an iron(II) sample, what is the iron(II) concentration?
- Error redoA student writes that one mole of permanganate reacts exactly with one mole of iron(II). What is the correct ratio?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 5.00 x 10 to the minus 4 mol of permanganate, which can oxidise 2.50 x 10 to the minus 3 mol of iron(II) | Moles equal concentration times volume in dm3, so 0.0200 times 0.0250 gives 5.00 x 10 to the minus 4, and the ratio of permanganate to iron(II) is 1 to 5. |
| 2 | It falls from +7 to +2, so each permanganate ion gains five electrons | A fall in oxidation number means reduction, and 7 minus 2 is 5, so each permanganate ion accepts five electrons. |
| 3 | It means 0.100 mol of solute per dm3, so 25.0 cm3 contains 2.50 x 10 to the minus 3 mol | The volume must first be changed from cm3 to dm3, and 0.0250 dm3 times 0.100 gives 2.50 x 10 to the minus 3 mol. |
| 4 | 0.0960 M | The permanganate is 4.80 x 10 to the minus 4 mol, five times that gives 2.40 x 10 to the minus 3 mol of iron(II), and dividing by 0.0250 dm3 gives 0.0960 M. |
| 5 | 1.25 x 10 to the minus 3 mol of iodine | The thiosulphate is 2.50 x 10 to the minus 3 mol, and the ratio of iodine to thiosulphate is 1 to 2. |
| 6 | 0.1395 g, about 0.140 g | Mass equals moles times molar mass, so 2.50 x 10 to the minus 3 times 55.8 gives 0.1395 g. |
| 7 | 31.2 percent | The permanganate is 5.60 x 10 to the minus 4 mol, so the iron(II) is 2.80 x 10 to the minus 3 mol which is 0.15624 g, and dividing by 0.500 g gives 31.2 percent. |
| 8 | 4.40 g per dm3 | The iodine is 6.25 x 10 to the minus 4 mol and so is the vitamin C, which weighs 0.110 g, and dividing by 0.0250 dm3 gives 4.40 g per dm3. |
| 9 | Mean 23.90 cm3, giving 0.0956 M iron(II) | The mean of 23.90 cm3 is 0.02390 dm3, so 4.78 x 10 to the minus 4 mol of permanganate times five divided by 0.0250 gives 0.0956 M. |
| 10 | One mole of permanganate reacts with five moles of iron(II) | The balanced ionic equation is permanganate plus eight hydrogen ions plus five iron(II) ions giving manganese(II) plus five iron(III) plus four water, so the ratio is 1 to 5. |
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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