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S5 Chemistry Paper: Redox, Electrochemical Cells and Electrolysis (100 marks)
S5 Chemistry – redox and electrolysis
S5 Chemistry paper: six questions on oxidation numbers, half equations, electrochemical cells, electrolysis and Faraday calculations, with answers and marking notes.
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Paper details S5 Chemistry practice paper with six questions on oxidation numbers, half equations and electron transfer, electrochemical cells and standard EMF, electrolysis and Faraday calculations, brine electrolysis and a manganate(VII) titration. Complete answers with marking notes.
Section A: Oxidation Numbers and Identifying Redox Changes (16 marks)
An oxidation number is the charge an atom carries after the bonding electrons are assigned to the more electronegative atom. Elements in their standard state have zero, oxygen is usually minus two and hydrogen usually plus one, and the oxidation numbers in a compound must sum to zero. A rise in oxidation number means oxidation, and a fall means reduction.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Find the oxidation number of manganese in potassium manganate(VII), KMnO4. | Answer: potassium is plus one and oxygen is minus two, so plus one plus x plus four times minus two equals zero, giving x equals plus seven. (Marking: 1 mark for the oxidation numbers of potassium and oxygen, 1 mark for the equation, 1 mark for the algebra, 1 mark for the value with sign.) |
| 2 | Find the oxidation number of chromium in K2Cr2O7. | Answer: plus two plus two x plus seven times minus two equals zero, so two x equals plus twelve and x equals plus six. (Marking: 2 marks for the equation, 1 mark for the working, 1 mark for the value.) |
| 3 | The oxygen in hydrogen peroxide, H2O2, has an oxidation number of minus one. Explain how this differs from ordinary oxides and find the oxidation number of sulphur in Na2S2O3. | Answer: in H2O2 the two oxygen atoms are bonded to each other, so the electrons are not fully transferred and each oxygen is minus one instead of minus two. For Na2S2O3, two times plus one plus two x plus three times minus two equals zero, so two x equals plus four and sulphur is plus two. (Marking: 2 marks for the oxygen to oxygen bond, 1 mark for the equation, 1 mark for the value.) |
| 4 | In the reaction Zn plus Cu2+ giving Zn2+ plus Cu, state what is oxidised, what is reduced and which species is the oxidising agent. | Answer: zinc rises from zero to plus two, so it is oxidised and acts as the reducing agent, while the copper ion falls from plus two to zero, so it is reduced and is the oxidising agent. (Marking: 1 mark each for oxidation and reduction, 1 mark for the oxidising agent, 1 mark for the reason.) |
Section B: Half Equations and Electron Transfer (16 marks)
A half equation must balance both atoms and charge. In acidic solution, hydrogen ions and water molecules are used to supply the missing atoms. Each electron carries 1.6 times 10^-19 C, so one mole of electrons carries the Faraday constant, 96485 C.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Write the half equation for the reduction of MnO4- to Mn2+ in acidic solution, and state the change in oxidation number of manganese and the number of electrons involved. | Answer: MnO4- plus eight H+ plus five e- gives Mn2+ plus four H2O, with manganese falling from plus seven to plus two and five electrons taken up. (Marking: 1 mark for species, 2 marks for balancing, 1 mark for the electron count.) |
| 6 | Write the half equation for the oxidation of Fe2+ to Fe3+ and combine it with the manganese half equation into an overall ionic equation. | Answer: Fe2+ gives Fe3+ plus one e-, and the overall equation is MnO4- plus five Fe2+ plus eight H+ giving Mn2+ plus five Fe3+ plus four H2O. (Marking: 1 mark for the half equation, 2 marks for matching the electrons, 1 mark for the balanced overall equation.) |
| 7 | If 0.020 mol of MnO4- reacts completely, find the amount of Fe2+ oxidised and the amount of electrons transferred. | Answer: the ratio is one to five, so Fe2+ equals five times 0.020, that is 0.100 mol, and one mole of Fe2+ releases one mole of electrons, so 0.100 mol of electrons is transferred. (Marking: 2 marks for the ratio, 1 mark for Fe2+, 1 mark for the electrons.) |
| 8 | Calculate the quantity of charge carried by that amount of electrons, in C. | Answer: Q equals n times F, that is 0.100 times 96485, giving 9648.5 C, about 9.65 times 10^3 C. (Marking: 1 mark for the formula, 1 mark for substitution, 1 mark for the value, 1 mark for significant figures.) |
Section C: Electrochemical Cells and Standard EMF (16 marks)
A chemical cell converts chemical energy into electrical energy. Electrons leave the electrode that loses them more readily and travel through the external circuit to the other electrode, so the first is the anode and the negative terminal, while a salt bridge keeps both half cells electrically neutral. The standard EMF equals the cathode potential minus the anode potential.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A zinc strip in zinc sulphate solution and a copper strip in copper sulphate solution are joined by a salt bridge. Identify the anode and the cathode, and state which electrode is the negative terminal. | Answer: zinc is the anode and copper is the cathode, and because zinc loses electrons more readily it is the negative terminal. (Marking: 1 mark for the anode, 1 mark for the cathode, 1 mark for the negative terminal with reason.) |
| 10 | Given E of Zn2+ over Zn is minus 0.76 V and E of Cu2+ over Cu is plus 0.34 V, find the standard EMF of the cell. | Answer: the standard EMF equals 0.34 minus minus 0.76, which is 1.10 V. (Marking: 1 mark for the formula, 1 mark for correct signs, 1 mark for the value with unit.) |
| 11 | Describe the direction of electron flow in the external circuit and explain the function of the salt bridge. | Answer: electrons flow from the zinc electrode through the external circuit to the copper electrode, and ions in the salt bridge migrate into the two half cells to cancel the charge that builds up, keeping both sides neutral so the current keeps flowing. (Marking: 2 marks for the electron direction, 2 marks for ion migration, 1 mark for neutrality.) |
| 12 | If the zinc ion concentration is 1.0 mol dm^-3 and the copper ion concentration is 0.010 mol dm^-3, use the Nernst equation to estimate the cell EMF at 298 K, taking the constant as 0.0592 V. | Answer: the concentration ratio is 100 and its logarithm is two, so the EMF is 1.10 minus 0.0592, giving 1.04 V, which is smaller than the standard value. (Marking: 1 mark for the ratio, 1 mark for the logarithm, 2 marks for substitution, 1 mark for the answer.) |
Section D: Electrolysis and Faraday Calculations (20 marks)
Electrolysis uses an external supply to drive a non spontaneous redox reaction. Charge equals current times time, and the amount of electrons equals charge divided by the Faraday constant. Reduction occurs at the cathode and oxidation at the anode. With graphite electrodes in copper sulphate solution, copper is deposited at the cathode and oxygen is released at the anode.
| # | Question | Answer and marking notes |
|---|---|---|
| 13 | A current of 2.00 A passes for 30.0 minutes. Find the charge that flows. | Answer: the time is 1800 s, so the charge is 2.00 times 1800, that is 3600 C. (Marking: 1 mark for the time, 1 mark for the formula, 2 marks for the value with unit.) |
| 14 | Find the amount of electrons transferred in that process. | Answer: the amount equals 3600 divided by 96485, which is 0.0373 mol, about 3.73 times 10^-2 mol. (Marking: 1 mark for the formula, 1 mark for substitution, 1 mark for the value, 1 mark for significant figures.) |
| 15 | Find the mass of copper deposited at the cathode, taking the molar mass of copper as 63.5 g mol^-1. | Answer: two moles of electrons give one mole of copper, so the amount of copper is 0.0187 mol and the mass is 0.0187 times 63.5, that is 1.18 g. (Marking: 2 marks for the two to one ratio, 1 mark for the amount, 1 mark for the mass with unit.) |
| 16 | Find the volume of oxygen released at the anode at room temperature and pressure, in cm3, taking the molar volume as 24.0 dm3 mol^-1. | Answer: four electrons release one oxygen molecule, so the amount is 9.33 times 10^-3 mol and the volume is 0.224 dm3, that is 224 cm3. (Marking: 2 marks for the four electron ratio, 1 mark for the amount, 1 mark for the unit conversion.) |
| 17 | To deposit 2.00 g of copper, find the time needed, in minutes. | Answer: the amount of copper is 0.0315 mol, needing 0.0630 mol of electrons and 6079 C, so the time is 3039 s, about 50.7 minutes. (Marking: 1 mark for the amount, 1 mark for the electrons, 1 mark for the charge, 1 mark for the time with unit.) |
Section E: Electrolysis of Brine and Industrial Use (16 marks)
In the electrolysis of concentrated brine, chlorine is released at the anode while hydrogen and sodium hydroxide are produced at the cathode, giving the overall change 2NaCl plus 2H2O to 2NaOH plus H2 plus Cl2. Industry uses an ion exchange membrane to keep the products apart.
| # | Question | Answer and marking notes |
|---|---|---|
| 18 | Write the electrode reactions at the anode and at the cathode, and state the gas produced at each. | Answer: at the anode, two Cl- give Cl2 plus two e-, releasing chlorine, and at the cathode, two H2O plus two e- give H2 plus two OH-, releasing hydrogen. (Marking: 2 marks for the anode equation, 1 mark for the cathode, 1 mark for matching the gases.) |
| 19 | If 1.00 mol of chlorine is formed, find the mass of sodium hydroxide produced at the same time. | Answer: the equation gives two moles of sodium hydroxide for each mole of chlorine, so the mass is two times 40.0, that is 80.0 g. (Marking: 2 marks for the ratio, 1 mark for the molar mass, 1 mark for the value with unit.) |
| 20 | Find the volume of the hydrogen formed at room temperature and pressure, taking the molar volume as 24.0 dm3 mol^-1. | Answer: one mole of chlorine goes with one mole of hydrogen, so the volume is 24.0 dm3. (Marking: 2 marks for the ratio, 1 mark for the molar volume, 1 mark for the value with unit.) |
| 21 | State the change in oxidation number of chlorine and explain why an ion exchange membrane is used in industry. | Answer: chlorine rises from minus one to zero, so it is oxidised. The membrane lets sodium ions through but blocks hydroxide and chloride ions, which stops chlorine from reacting with the alkali and keeps hydrogen apart from chlorine. (Marking: 2 marks for the oxidation number change, 2 marks for the role of the membrane.) |
Section F: Potassium Manganate(VII) Titration (16 marks)
Acidified potassium manganate(VII) is a strong oxidising agent that turns iron(II) into iron(III) while being reduced to manganese(II). At the end point the purple colour persists, so the colour change acts as its own indicator. The calculation starts from concentration and volume, then uses the equation ratio.
| # | Question | Answer and marking notes |
|---|---|---|
| 22 | A 25.0 cm3 sample of iron(II) solution is titrated with 0.0200 mol dm^-3 potassium manganate(VII), and the mean titre is 24.0 cm3. Find the amount of manganate(VII) used. | Answer: the volume is 0.0240 dm3, so the amount is 0.0200 times 0.0240, that is 4.80 times 10^-4 mol. (Marking: 2 marks for the volume conversion, 1 mark for the formula, 1 mark for substitution, 1 mark for the value.) |
| 23 | Find the amount of iron(II) oxidised in the 25.0 cm3 sample. | Answer: the equation shows one manganate ion for five iron(II) ions, so the amount is five times 4.80 times 10^-4, that is 2.40 times 10^-3 mol. (Marking: 3 marks for the ratio, 1 mark for substitution, 1 mark for the value.) |
| 24 | Find the concentration of the original iron(II) solution, and predict the titre if 0.0100 mol dm^-3 potassium manganate(VII) were used instead. | Answer: the concentration is 2.40 times 10^-3 divided by 0.0250, that is 0.0960 mol dm^-3, and since halving the concentration doubles the volume needed, the expected titre is 48.0 cm3. (Marking: 2 marks for the concentration formula, 2 marks for its value, 2 marks for the doubled volume.) |
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