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Secondary 3 Chemistry Redox Electrochemistry Quiz
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Questions
Secondary 3 Chemistry Quiz - Redox Electrochemistry
Name: _________________________________ Class: _________________
Date: _________________ Score: _______ / 40
Duration: 50 minutes Total Marks: 40 marks
Instructions:
- Answer all questions.
- Write your answers in the spaces provided.
- For calculation questions, show all working clearly.
- Use the periodic table where necessary.
- Calculators are allowed.
Section A: Multiple Choice and Short Response (Questions 1–5)
[10 marks]
1. Which of the following is the correct oxidation state of sulfur in sulfuric acid, H₂SO₄?
| A | +4 |
| B | +6 |
| C | +2 |
| D | +8 |
[1 mark]
Answer: _________________
2. In the reaction between zinc metal and copper(II) sulfate solution, zinc displaces copper. Which statement correctly describes the change in oxidation state of zinc?
| A | Zinc is reduced from 0 to +2 |
| B | Zinc is oxidised from 0 to +2 |
| C | Zinc is reduced from +2 to 0 |
| D | Zinc does not change oxidation state |
[1 mark]
Answer: _________________
3. Define oxidation in terms of electron transfer. [1 mark]
4. (a) State the oxidation state of oxygen in hydrogen peroxide, H₂O₂. [1 mark]
(b) Explain why hydrogen peroxide can act as both an oxidising agent and a reducing agent. [2 marks]
5. The following half-equation represents the reduction of chlorine gas:
Use this to write the overall ionic equation for the reaction between chlorine gas and potassium bromide solution, given that bromide ions are oxidised to bromine. [2 marks]
[2 marks]
Section B: Structured Questions (Questions 6–15)
[20 marks]
6. Consider the redox reaction between magnesium and hydrochloric acid:
(a) Identify which element is oxidised and explain your answer in terms of electron transfer. [2 marks]
(b) Write the half-equation for the reduction process occurring in this reaction. [1 mark]
7. A student sets up the following electrochemical cell:
<image_placeholder> id: Q7-fig1 type: diagram linked_question: Q7 description: A simple galvanic cell with zinc electrode in zinc sulfate solution and copper electrode in copper sulfate solution, connected by a voltmeter and salt bridge. labels: Zinc electrode (Zn), Copper electrode (Cu), ZnSO₄(aq), CuSO₄(aq), Voltmeter (V), Salt bridge (labelled with KNO₃), wires connecting electrodes to voltmeter. values: No numerical values required. must_show: Direction of electron flow indicated with arrow, labelling of all components, salt bridge positioned to complete circuit, two separate half-cells. </image_placeholder>
(a) State the function of the salt bridge in this electrochemical cell. [1 mark]
(b) On the diagram, indicate the direction of electron flow in the external circuit. [1 mark]
(c) Write the half-equation occurring at the copper electrode. [1 mark]
8. The standard electrode potential of the Fe³⁺/Fe²⁺ half-cell is +0.77 V. When connected to a standard hydrogen electrode, the voltmeter reads +0.77 V.
(a) Explain what is meant by a "standard hydrogen electrode." [2 marks]
(b) Predict whether Fe³⁺ ions will oxidise iodide ions (I⁻) to iodine (I₂), given that E° for I₂/I⁻ = +0.54 V. Explain your answer. [2 marks]
9. Manganese(IV) oxide, MnO₂, reacts with concentrated hydrochloric acid to produce chlorine gas, manganese(II) chloride and water:
(a) Calculate the change in oxidation state of manganese in this reaction. [1 mark]
(b) Using oxidation states, explain why this is a redox reaction. [2 marks]
10. Electrolysis of molten lead(II) bromide is carried out using inert electrodes.
(a) State why graphite electrodes are described as "inert" in this context. [1 mark]
(b) Write the half-equations for the reactions occurring at: (i) the cathode (negative electrode) [1 mark]
(ii) the anode (positive electrode) [1 mark]
11. A metal rod is connected to the negative terminal of a power supply and immersed in copper(II) sulfate solution. Copper metal deposits on the rod.
(a) Explain whether this process is electrolysis or electroplating. [1 mark]
(b) If the metal rod is made of iron, explain why the copper coating may be poorly adherent. [2 marks]
12. The following table shows the results of experiments where metals are added to solutions containing ions of other metals:
| Experiment | Metal added | Solution | Observations |
|---|---|---|---|
| 1 | Magnesium | Copper(II) sulfate | Grey solid forms, blue colour fades |
| 2 | Copper | Zinc nitrate | No reaction |
| 3 | Zinc | Iron(II) sulfate | Grey solid forms, pale green fades |
(a) Use the results to arrange Mg, Zn, Cu and Fe in order of reactivity, most reactive first. [1 mark]
(b) Explain why no reaction occurs in Experiment 2. [1 mark]
13. In the electrolysis of dilute sulfuric acid using inert electrodes:
(a) State the ratio of volumes of hydrogen to oxygen produced at the electrodes. [1 mark]
(b) Explain how this ratio arises from the relevant half-equations. [2 marks]
14. Chromium plating is used to protect steel bicycle handlebars from corrosion.
(a) State whether the bicycle handlebar (steel) is made the anode or cathode during electroplating. [1 mark]
(b) Suggest why a low current is used for a longer time rather than a high current for a shorter time, to achieve a smooth chromium coating. [2 marks]
15. Hydroxylamine (NH₂OH) can reduce iron(III) ions to iron(II) ions. In this reaction, hydroxylamine is oxidised to nitrous oxide (N₂O).
(a) Determine the oxidation state of nitrogen in: (i) NH₂OH [1 mark]
(ii) N₂O [1 mark]
(b) By considering the change in oxidation state of nitrogen, explain why hydroxylamine acts as a reducing agent in this reaction. [2 marks]
Section C: Extended Response (Questions 16–20)
[10 marks]
16. The rusting of iron involves the oxidation of iron in the presence of oxygen and water. The overall process can be represented by:
This is an example of electrochemical corrosion. Explain the redox processes occurring at different regions on the iron surface, and explain how sacrificial protection using zinc blocks prevents rusting of steel structures such as bridges. [3 marks]
17. A student sets up an electrolytic cell to electroplate a copper spoon with silver.
<image_placeholder> id: Q17-fig1 type: experimental_setup linked_question: Q17 description: Electroplating setup with silver anode, copper spoon cathode, silver nitrate electrolyte, connected to DC power supply with ammeter. labels: Silver anode, Copper spoon cathode, Silver nitrate solution (AgNO₃(aq)), DC power supply, Ammeter, Rheostat (variable resistor). values: Current = 0.50 A, time = 20 minutes, mass of silver deposited = 0.67 g. must_show: Clear labelling of anode and cathode, direction of conventional current flow, complete circuit, electrolyte in beaker. </image_placeholder>
(a) Explain why the silver anode needs to be replaced periodically during this process. [1 mark]
(b) Given that the mass of silver deposited is 0.67 g, calculate the amount of silver deposited in moles. [Aᵣ: Ag = 108] [1 mark]
18. The following reaction occurs when sulfur dioxide gas is passed through acidified potassium dichromate(VI) solution:
(a) Determine the oxidation state of sulfur in SO₂ and in SO₄²⁻. [1 mark]
(b) Identify the oxidising agent in this reaction and explain your choice with reference to oxidation states. [2 marks]
19. The reaction between acidified potassium manganate(VII) and ethanedioic acid (oxalic acid, H₂C₂O₄) is used to study reaction rates. The relevant half-equations are:
(a) Construct the overall balanced ionic equation for this reaction. [2 marks]
(b) What colour change would be observed as this reaction proceeds? [1 mark]
20. Ethanol can be oxidised to ethanoic acid using acidified potassium dichromate(VI). In industry, this oxidation is often carried out using oxygen in the presence of certain bacteria (vinegar production).
Compare and contrast these two methods of oxidation (chemical vs biochemical) in terms of:
- the oxidising agent used
- reaction conditions required
- atom economy and environmental impact [3 marks]
END OF QUIZ
Answers
Secondary 3 Chemistry Quiz - Redox Electrochemistry: Answer Key
Total Marks: 40 marks
Section A: Multiple Choice and Short Response
Question 1 [1 mark]
Answer: B (+6)
Explanation: To find the oxidation state of sulfur in H₂SO₄:
- Hydrogen has oxidation state +1 (rule: +1 in compounds, except with metals)
- Oxygen has oxidation state −2 (rule: −2 in compounds, except peroxides)
- The compound is neutral, so sum of oxidation states = 0
Let oxidation state of S = x:
Marking note: Deduct mark if working not shown or arithmetic error made.
Question 2 [1 mark]
Answer: B (Zinc is oxidised from 0 to +2)
Explanation: In the reaction: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)
- Zinc starts as elemental metal: oxidation state = 0
- In zinc sulfate, Zn²⁺ ion has oxidation state = +2
- Increase in oxidation state = oxidation (loss of electrons)
- Zinc loses 2 electrons: Zn → Zn²⁺ + 2e⁻
This is a displacement reaction where more reactive zinc displaces less reactive copper. The oxidation state increases because electrons are lost to copper ions.
Common mistake: Students confuse oxidation and reduction. Use OILRIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).
Question 3 [1 mark]
Answer: Oxidation is the loss of electrons.
Explanation: Oxidation and reduction can be defined in three ways:
- Electron transfer: Oxidation is loss of electrons (OIL)
- Oxygen gain: Oxidation is gain of oxygen
- Hydrogen loss: Oxidation is loss of hydrogen
For ionic/electrochemical contexts, the electron transfer definition is most precise and universally applicable. When a species loses electrons, it becomes more positively charged (or less negatively charged), which we describe as an increase in oxidation state.
Question 4 [3 marks]
(a) [1 mark] Answer: −1
Explanation: In H₂O₂, using normal rules would give oxygen as −2, but this is a peroxide (contains O−O single bond). For peroxides, the special rule applies: oxygen has oxidation state −1.
Calculation: 2(+1) + 2(x) = 0, so x = −1.
(b) [2 marks] Answer: In H₂O₂, oxygen has an intermediate oxidation state of −1, between the values in O₂ (0) and typical oxides (−2). Therefore oxygen in H₂O₂ can be:
- Reduced to −2 (acting as oxidising agent, gaining electrons)
- Oxidised to 0 (acting as reducing agent, losing electrons)
Explanation for oxidation half-equation: H₂O₂ → O₂ + 2H⁺ + 2e⁻ (oxygen oxidised from −1 to 0)
Explanation for reduction half-equation: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O (oxygen reduced from −1 to −2)
Marking breakdown:
- [1 mark] for explaining that −1 is an intermediate oxidation state
- [1 mark] for identifying both directions of change with correct species
Question 5 [2 marks]
Answer:
Working:
- Given reduction: Cl₂ + 2e⁻ → 2Cl⁻ (chlorine gains electrons, is reduced)
- Oxidation of bromide: 2Br⁻ → Br₂ + 2e⁻ (bromide loses electrons, is oxidised)
- Electrons cancel: add the half-equations directly
Explanation: This is a displacement reaction where more reactive chlorine (stronger oxidising agent) displaces less reactive bromine from its compound. The E° value for Cl₂/Cl⁻ (+1.36 V) is more positive than Br₂/Br⁻ (+1.07 V), so chlorine has greater tendency to be reduced.
Marking breakdown:
- [1 mark] for correct formula of products and reactants
- [1 mark] for correctly balanced equation (species and charge)
Section B: Structured Questions
Question 6 [3 marks]
(a) [2 marks] Answer: Magnesium is oxidised. It loses two electrons to form Mg²⁺ ions: Mg → Mg²⁺ + 2e⁻. The oxidation state increases from 0 to +2.
Explanation:
- In elemental Mg, oxidation state = 0
- In MgCl₂, Mg has oxidation state = +2
- Increase in oxidation state indicates oxidation
- Magnesium is more reactive than hydrogen, so it displaces H⁺ from the acid
Marking breakdown:
- [1 mark] for identifying Mg as oxidised
- [1 mark] for correct explanation with electron loss/oxidation state change
(b) [1 mark] Answer: 2H⁺ + 2e⁻ → H₂
Explanation: Hydrogen ions from the acid gain electrons (reduction). This is the reduction half-equation. Two protons combine with two electrons to form one molecule of hydrogen gas.
Question 7 [3 marks]
(a) [1 mark] Answer: The salt bridge completes the circuit by allowing ions to flow between the two half-cells, maintaining electrical neutrality.
Explanation: Without the salt bridge, positive charge would build up in the anode compartment (as Zn²⁺ enters solution) and negative charge would build up in the cathode compartment (as Cu²⁺ is removed). The salt bridge allows NO₃⁻ ions to flow toward the anode and K⁺ ions to flow toward the cathode, balancing charge without mixing solutions.
(b) [1 mark] Answer: Electrons flow from zinc electrode (negative) to copper electrode (positive), or left to right as drawn.
Explanation: Zinc is more reactive than copper, so zinc metal is oxidised (loses electrons). Electrons travel through the external wire from the zinc half-cell to the copper half-cell, where Cu²⁺ ions are reduced to Cu metal.
(c) [1 mark] Answer: Cu²⁺ + 2e⁻ → Cu
Explanation: At the copper electrode (cathode), copper ions gain electrons and are reduced to copper metal. This is where reduction occurs because copper has a more positive reduction potential than zinc.
Question 8 [4 marks]
(a) [2 marks] Answer: A standard hydrogen electrode (SHE) consists of hydrogen gas at 1 atm pressure bubbled over a platinum electrode coated with platinum black, immersed in a solution containing H⁺ ions at 1 mol/dm³ concentration, at 298 K (25°C). By convention, its electrode potential is defined as 0.00 V.
Marking breakdown:
- [1 mark] for correct physical description (H₂ gas, Pt electrode, H⁺ at 1 mol/dm³)
- [1 mark] for standard conditions (298 K, 1 atm, 1 mol/dm³) and E° = 0.00 V
(b) [2 marks] Answer: Yes, Fe³⁺ will oxidise I⁻ to I₂. E°cell = E°(cathode) − E°(anode) = +0.77 − (+0.54) = +0.23 V. Since E°cell is positive, the reaction is feasible.
Explanation:
- Fe³⁺/Fe²⁺ has more positive E° (+0.77 V) than I₂/I⁻ (+0.54 V)
- Therefore Fe³⁺ is the stronger oxidising agent (more tendency to be reduced)
- Fe³⁺ will accept electrons from I⁻, oxidising I⁻ to I₂ while being reduced to Fe²⁺
Marking breakdown:
- [1 mark] for correct prediction with reasoning about E° values
- [1 mark] for calculating or stating that E°cell is positive
Question 9 [3 marks]
(a) [1 mark] Answer: From +4 to +2, a decrease of 2.
Working:
- In MnO₂: x + 2(−2) = 0, so x = +4
- In MnCl₂: x + 2(−1) = 0, so x = +2
- Change: +4 → +2
(b) [2 marks] Answer:
- Manganese is reduced: oxidation state decreases from +4 to +2, gaining electrons
- Chlorine is oxidised: in HCl, Cl has oxidation state −1; in Cl₂, Cl has oxidation state 0. Some chloride ions lose electrons to form chlorine gas
Explanation: Not all HCl is oxidised—only the chloride ions that form Cl₂. The chloride in MnCl₂ remains at −1. This is a disproportionation-related observation where HCl acts as both acid source and reducing agent.
Marking breakdown:
- [1 mark] for identifying manganese reduction with oxidation state evidence
- [1 mark] for identifying chlorine oxidation with correct oxidation states
Question 10 [3 marks]
(a) [1 mark] Answer: Graphite electrodes are inert because they do not react with the molten electrolyte or products; they merely provide a surface for electron transfer without being consumed or altered.
(b) (i) [1 mark] Answer: Pb²⁺ + 2e⁻ → Pb
Explanation: Lead ions are attracted to the negative cathode, where they gain electrons and are reduced to molten lead metal.
(ii) [1 mark] Answer: 2Br⁻ → Br₂ + 2e⁻
Explanation: Bromide ions are attracted to the positive anode, where they lose electrons and are oxidised to bromine vapour.
Question 11 [3 marks]
(a) [1 mark] Answer: This is electroplating, because a thin layer of one metal (copper) is deposited onto another metal object (the rod) using electricity.
(b) [2 marks] Answer: Iron is more reactive than copper, so if the iron surface is not perfectly clean and oxide-free, or if the electrolyte composition is not carefully controlled, the copper deposit may be powdery, non-uniform, or poorly adherent. Iron may also form a thin oxide layer that prevents direct metal-to-metal bonding.
Marking breakdown:
- [1 mark] for noting iron's greater reactivity or oxide layer issues
- [1 mark] for explaining consequence (poor adhesion, powdery deposit)
Question 12 [2 marks]
(a) [1 mark] Answer: Mg > Zn > Fe > Cu (most reactive to least reactive)
Explanation from results:
- Mg displaces Cu (Mg most reactive so far)
- Cu cannot displace Zn (Zn more reactive than Cu)
- Zn displaces Fe (Zn more reactive than Fe, but we need to place Fe relative to Mg)
Order: Mg > Zn > Fe > Cu
(b) [1 mark] Answer: Copper is less reactive than zinc, so copper cannot displace zinc from zinc nitrate; zinc ions have greater tendency to remain as ions than copper ions have to displace them.
Question 13 [3 marks]
(a) [1 mark] Answer: 2:1 (hydrogen:oxygen)
(b) [2 marks] Answer: At cathode: 4H⁺ + 4e⁻ → 2H₂ (two molecules of H₂ from 4 electrons) At anode: 4OH⁻ → O₂ + 2H₂O + 4e⁻ (one molecule of O₂ from 4 electrons)
Since the same quantity of electricity (4 moles of electrons) produces 2 volumes of H₂ and 1 volume of O₂, the ratio is 2:1.
Alternative using water as source: Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻
To match electrons: multiply cathode equation by 2: 4H₂O + 4e⁻ → 2H₂ + 4OH⁻
Now 4 electrons produce 2H₂ and 1O₂, giving 2:1 ratio.
Marking breakdown:
- [1 mark] for correct relevant half-equations
- [1 mark] for electron matching leading to 2:1 ratio
Question 14 [3 marks]
(a) [1 mark] Answer: The bicycle handlebar (steel) is made the cathode (negative electrode).
(b) [2 marks] Answer: A low current for longer time allows controlled, gradual deposition of chromium ions, permitting them to arrange regularly in the metallic lattice and settle smoothly on the surface. A high current causes rapid deposition, producing a rough, brittle, nodular coating with internal stress and poor adhesion.
Marking breakdown:
- [1 mark] for explaining that low current allows ordered/gradual deposition
- [1 mark] for explaining that high current causes rapid, uneven, rough deposition
Question 15 [4 marks]
(a) (i) [1 mark] Answer: −1
Working: Let oxidation state of N = x. In NH₂OH: x + 2(+1) + (−2) + (+1) = 0, so x + 2 − 2 + 1 = 0, thus x = −1.
(Alternatively: x + 2 + (−2) + 1 = 0 for H₂N-O-H structure)
(ii) [1 mark] Answer: +1
Working: In N₂O: 2x + (−2) = 0, so 2x = +2, x = +1.
(b) [2 marks] Answer: Nitrogen's oxidation state increases from −1 in NH₂OH to +1 in N₂O (increase of 2 per N atom, or 4 overall for N₂O formation). An increase in oxidation state indicates loss of electrons, which is oxidation. The species that is oxidised causes reduction in another species, so hydroxylamine acts as a reducing agent.
Marking breakdown:
- [1 mark] for correct oxidation state change (−1 to +1)
- [1 mark] for linking oxidation to reducing agent function
Section C: Extended Response
Question 16 [3 marks]
Answer structure and marking guidance:
Electrochemical corrosion process [2 marks]:
- Anodic region (iron surface impurities/stressed areas): Fe → Fe²⁺ + 2e⁻ (oxidation, iron loses electrons) [1 mark]
- Cathodic region (oxygen-rich areas): O₂ + 2H₂O + 4e⁻ → 4OH⁻ (reduction) [1 mark]
- Fe²⁺ and OH⁻ combine, further oxidise to Fe(OH)₃, which dehydrates to rust (Fe₂O₃·nH₂O)
Sacrificial protection with zinc [1 mark]:
- Zinc is more reactive than iron (higher in reactivity series), so zinc is oxidised preferentially: Zn → Zn²⁺ + 2e⁻
- Zinc sacrifices itself, protecting iron from oxidation; electrons flow from zinc to iron, keeping iron as the cathode in local cells
- Zinc blocks must be replaced periodically as they corrode away
Common mistakes to flag:
- Students describe zinc as "coating" rather than understanding electrochemical protection
- Confusing anode and cathode in rusting mechanism
- Not mentioning that zinc is oxidised preferentially due to higher reactivity
Question 17 [2 marks]
(a) [1 mark] Answer: The silver anode dissolves as Ag → Ag⁺ + e⁻, releasing silver ions into the solution to replace those deposited on the cathode, maintaining constant Ag⁺ concentration in the electrolyte.
(b) [1 mark] Answer:
Or approximately 6.204 × 10⁻³ mol
Calculation:
Question 18 [3 marks]
(a) [1 mark] Answer: In SO₂: +4; In SO₄²⁻: +6
Working for SO₂: x + 2(−2) = 0, x = +4 Working for SO₄²⁻: x + 4(−2) = −2, x − 8 = −2, x = +6
(b) [2 marks] Answer: The oxidising agent is Cr₂O₇²⁻ (potassium dichromate(VI) / dichromate ion).
Explanation: Chromium's oxidation state decreases from +6 in Cr₂O₇²⁻ to +3 in Cr³⁺. A decrease in oxidation state indicates gain of electrons (reduction). The species that is reduced is the oxidising agent because it causes oxidation in another species (SO₂ to SO₄²⁻, where sulfur increases from +4 to +6).
Marking breakdown:
- [1 mark] for correct identification with oxidation state evidence for Cr
- [1 mark] for explaining the oxidising agent is the species being reduced, linking to oxidation of sulfur
Question 19 [3 marks]
(a) [2 marks] Answer:
Working:
- MnO₄⁻ equation needs 5e⁻, H₂C₂O₄ equation produces 2e⁻
- LCM of 5 and 2 = 10 electrons
- Multiply MnO₄⁻ equation by 2 (10 electrons, 2 MnO₄⁻, 16 H⁺, 2 Mn²⁺, 8 H₂O)
- Multiply H₂C₂O₄ equation by 5 (10 electrons, 5 H₂C₂O₄, 10 CO₂, 10 H⁺)
- Add and simplify: 16H⁺ − 10H⁺ (from products side when reversing) = 6H⁺ remaining, but check carefully:
Actually: 2 × [MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O] gives 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O
5 × [H₂C₂O₄ → 2CO₂ + 2H⁺ + 2e⁻] gives 5H₂C₂O₄ → 10CO₂ + 10H⁺ + 10e⁻
Adding: 2MnO₄⁻ + 16H⁺ + 5H₂C₂O₄ → 2Mn²⁺ + 8H₂O + 10CO₂ + 10H⁺
Simplify: 2MnO₄⁻ + 6H⁺ + 5H₂C₂O₄ → 2Mn²⁺ + 8H₂O + 10CO₂
Check atoms and charges balance on both sides.
Marking breakdown:
- [1 mark] for correct multiplication to balance electrons
- [1 mark] for correctly simplified final equation with all species balanced
(b) [1 mark] Answer: Purple/pink to colourless (very pale pink) / or decolourised
Explanation: MnO₄⁻ ions are intensely purple. As they are reduced to nearly colourless Mn²⁺ ions (very pale pink, appears colourless in dilute solution), the purple colour disappears. This colour change makes the reaction self-indicating.
Question 20 [3 marks]
Answer structure with marking guidance:
| Aspect | Chemical oxidation (acidified K₂Cr₂O₇) | Biochemical oxidation (bacteria + O₂) |
|---|---|---|
| Oxidising agent | Dichromate(VI) ions (Cr₂O₇²⁻), inorganic compound | Oxygen (O₂) from air, catalysed by enzymes in Acetobacter bacteria |
| Reaction conditions | Strongly acidic (H₂SO₄), elevated temperature, controlled concentration | Room temperature, neutral/slightly acidic pH, requires bacterial culture, longer time |
| Atom economy / Environmental impact | Produces Cr³⁺ waste; chromium compounds are toxic and require disposal; lower atom economy due to by-products | High atom economy; O₂ as oxidant becomes H₂O; biodegradable, environmentally friendly; renewable process |
Marking breakdown [3 marks]:
- [1 mark] for correct identification and comparison of oxidising agents
- [1 mark] for clear contrast of reaction conditions (temperature, pH, time, catalyst/enzyme)
- [1 mark] for environmental comparison: chemical method produces toxic waste vs biochemical method being green/sustainable with oxygen/water as by-products
Teaching note: The biochemical method is the basis of traditional vinegar production. Acetobacter bacteria oxidise ethanol to ethanoic acid aerobically. Industrially, this is done in aerated tanks (acetifiers) at ~30°C, with the bacteria living as a biofilm on wood shavings or synthetic packing material.
END OF ANSWER KEY