2018 AQA IGCSE Chemistry (9202) 模擬試題連答案詳解

01.1 The solution changes from colorless to orange/brown due to the displacement of bromine molecules.

01.2 Chlorine displaces bromine: Cl2(aq) + 2KBr(aq) -> 2KCl(aq) + Br2(aq).

01.3 Going down Group 7, atomic radius increases and shielding increases. The electrostatic attraction between the nucleus and the outer shell decreases, making it harder to gain an electron, so reactivity decreases.

01.4 Iodine is less reactive than bromine, so no displacement reaction occurs.

01.1 [2 marks]
- Colorless starting solution (1 mark)
- Turns orange/yellow/brown (1 mark)

01.2 [2 marks]
- Correct products and reactants: Cl2 + 2KBr -> 2KCl + Br2 (1 mark)
- Correct state symbols: (aq) for all (1 mark)

01.3 [4 marks]
- Reactivity decreases down the group (1 mark)
- Atomic size/shielding increases (1 mark)
- Weaker attraction for outer shell electrons / incoming electron (1 mark)
- Harder to gain/attract an electron (1 mark)

01.4 [2 marks]
- Iodine (1 mark)
- It is less reactive than bromine (1 mark)

02.1 Positive copper ions migrate to the negative cathode where they are reduced to copper metal. Negative chloride ions migrate to the positive anode where they are oxidized to chlorine gas.

02.2 The reduction half-equation at the cathode is Cu2+(aq) + 2e- -> Cu(s).

02.3 Damp litmus paper test: Damp blue litmus paper is placed in the gas, which turns red (as chlorine is acidic in water) and then bleaches white.

02.4 Copper ions (Cu2+) are positive, and the cathode is negative; opposite charges attract.

02.5 Copper ions give the solution its blue color. As electrolysis proceeds, Cu2+ ions are removed from the solution, reducing their concentration and causing the color to fade.

02.1 [2 marks]
- Cathode: Copper / Cu (1 mark)
- Anode: Chlorine / Cl2 (1 mark)

02.2 [2 marks]
- Reactants and products correct: Cu2+ + 2e- -> Cu (1 mark)
- Balanced equation (1 mark)

02.3 [2 marks]
- Damp blue litmus paper (1 mark)
- Turns red then bleaches white (1 mark)

02.4 [2 marks]
- Copper ions are positively charged (1 mark)
- Opposites attract / attracted to negative electrode (1 mark)

02.5 [2 marks]
- Blue color is due to Cu2+ ions (1 mark)
- Concentration of Cu2+ ions decreases as they turn into copper metal (1 mark)

03.1 Roasting ZnS: 2ZnS(s) + 3O2(g) -> 2ZnO(s) + 2SO2(g).

03.2 Reduction of ZnO: ZnO(s) + CO(g) -> Zn(g) + CO2(g). The reducing agent is carbon monoxide (CO) because it gains oxygen.

03.3 Carbon reduction only works for metals less reactive than carbon. Aluminium is highly reactive (more reactive than carbon), so it must be extracted using electrolysis.

03.4 Recycling zinc requires significantly less energy than extracting it from zinc blende, lowering production costs. Environmentally, it prevents the release of harmful mining byproducts and reduces landfill waste.

03.1 [2 marks]
- Correct formulas: ZnS, O2, ZnO, SO2 (1 mark)
- Correct balancing: 2ZnS + 3O2 -> 2ZnO + 2SO2 (1 mark)

03.2 [2 marks]
- Correct balanced equation: ZnO + CO -> Zn + CO2 (1 mark)
- Carbon monoxide identified as the reducing agent (1 mark)

03.3 [2 marks]
- Aluminium is more reactive than carbon (1 mark)
- Carbon cannot displace/remove oxygen from aluminium oxide (1 mark)

03.4 [4 marks]
- Environmental advantage: conserves ore reserves / reduces mining damage / less waste / lower CO2 emissions (any one, 2 marks)
- Economic advantage: uses less energy / cheaper than mining and processing raw ore (any one, 2 marks)

04.1 Mr of CaCO3 = 40 + 12 + (3 * 16) = 100. Mr of CaO = 40 + 16 = 56.

04.2 Moles of CaCO3 = mass / Mr = 15.0 g / 100 = 0.15 mol. From the equation, 1 mole of CaCO3 produces 1 mole of CaO. Therefore, moles of CaO = 0.15 mol. Mass of CaO = moles * Mr = 0.15 * 56 = 8.4 g.

04.3 From the equation, 1 mole of CaCO3 produces 1 mole of CO2. Moles of CO2 = 0.15 mol. Volume of CO2 = moles * 24.0 dm3 = 0.15 * 24.0 = 3.6 dm3.

04.4 In an open beaker, gaseous products are free to escape. The decrease in mass is due to the loss of carbon dioxide gas to the air. The total mass is conserved if the escaped gas is taken into account.

04.1 [2 marks]
- Mr of CaCO3 = 100 (1 mark)
- Mr of CaO = 56 (1 mark)

04.2 [3 marks]
- Moles of CaCO3 = 0.15 mol (1 mark)
- Moles of CaO = 0.15 mol (1 mark)
- Mass of CaO = 8.4 g (1 mark)

04.3 [3 marks]
- 1:1 molar ratio of CaCO3 to CO2 (1 mark)
- Moles of CO2 = 0.15 mol (1 mark)
- Volume of CO2 = 3.6 dm3 (1 mark)

04.4 [2 marks]
- Carbon dioxide is a gas and escapes into the atmosphere (1 mark)
- Total mass is conserved if the escaped gas is measured/included (1 mark)

05.1 The rate can be monitored by collecting gas in a gas syringe and recording the volume at regular time intervals, or by placing the flask on a digital balance and measuring the mass loss as CO2 gas escapes.

05.2 With higher concentration, there are more reacting particles packed into the same volume. Consequently, the frequency of successful collisions increases, increasing the reaction rate.

05.3 Breaking a solid into powder increases the total surface area. More reactant particles are exposed to collide with the acid, resulting in a higher rate of collisions per second.

05.4 The reaction stops when the limiting reactant (typically calcium carbonate or the acid) is entirely consumed, leaving no more particles to react.

05.1 [2 marks]
- Method 1: measure gas volume using a gas syringe over time (1 mark)
- Method 2: measure mass loss over time using a balance (1 mark)

05.2 [3 marks]
- More particles per unit volume / space (1 mark)
- Increased collision frequency / more collisions per second (1 mark)
- More successful collisions per unit time (1 mark)

05.3 [3 marks]
- Powder has a larger surface area (1 mark)
- More reactant particles are exposed to the acid (1 mark)
- Increases frequency of collisions / collisions per second (1 mark)

05.4 [2 marks]
- Reactants are used up / completely reacted (1 mark)
- Limiting reactant (1 mark)

06.1 Sodium is in Group 1 and has 1 outer electron, which it transfers to chlorine (Group 7, with 7 outer electrons). This forms Na+ and Cl- ions, which have stable noble gas configurations and attract each other strongly.

06.2 An oxygen atom shares 2 electrons (one with each of the two hydrogen atoms) to form two single covalent bonds. This leaves oxygen with 4 non-bonding (lone pair) electrons in its outer shell.

06.3 Ionic substances like NaCl require high temperatures to break the strong electrostatic attractions throughout the giant ionic lattice. Simple molecular substances like water only require enough thermal energy to overcome the weak intermolecular forces between the molecules; the strong covalent bonds within the water molecules are not broken during boiling.

06.1 [4 marks]
- Sodium loses one electron (1 mark)
- Chlorine gains one electron (1 mark)
- Stable/full outer shells formed (or Na+ and Cl- ions represented) (1 mark)
- Strong electrostatic attraction between oppositely charged ions (1 mark)

06.2 [2 marks]
- Two shared pairs of electrons (one per O-H bond) (1 mark)
- Four non-bonding outer electrons on oxygen (1 mark)

06.3 [4 marks]
- NaCl has a giant ionic structure / lattice (1 mark)
- Strong electrostatic forces / ionic bonds require large amounts of energy to overcome (1 mark)
- Water has simple molecules (1 mark)
- Weak intermolecular forces require very little energy to overcome (1 mark)
- Note: max 3 marks if student says covalent bonds are broken during boiling of water.

07.1 Energy to break reactant bonds: 1 * (N≡N) + 3 * (H-H) = 945 + 3 * 436 = 945 + 1308 = 2253 kJ.

07.2 Energy released when product bonds are formed: Ammonia has 3 N-H bonds. 2 moles of NH3 have 6 N-H bonds. 6 * 391 = 2346 kJ.

07.3 Overall energy change = Energy taken in - Energy given out = 2253 - 2346 = -93 kJ/mol. Since the energy change is negative, the reaction is exothermic.

07.4 Activation energy is the minimum quantity of energy that reacting species must possess to undergo a specified reaction.

07.1 [3 marks]
- 1 x 945 = 945 (1 mark)
- 3 x 436 = 1308 (1 mark)
- Sum = 2253 kJ (1 mark)

07.2 [3 marks]
- Identifies 6 N-H bonds in 2NH3 (1 mark)
- 6 x 391 (1 mark)
- Total = 2346 kJ (1 mark)

07.3 [3 marks]
- Energy change = Reactants - Products = 2253 - 2346 (1 mark)
- -93 kJ/mol (1 mark)
- Exothermic because more energy is released than absorbed / negative sign (1 mark)

07.4 [1 mark]
- Minimum energy required for a collision to result in a reaction (1 mark)

08.1 Ink is soluble in chromatography solvents, meaning it would separate into its component dyes and run up the paper, ruinous to the experiment. Graphite (pencil) is insoluble and remains fixed.

08.2 Keeping the spots above the solvent line ensures they migrate upwards through the capillary action of the paper rather than washing out into the bulk solvent container.

08.3 The Rf value is calculated by dividing the distance moved by the solute by the distance moved by the solvent front: Rf = 5.2 cm / 8.0 cm = 0.65.

08.4 A chemically pure substance contains only one substance. Since chromatography separates substances based on their solubility and retention, a pure substance produces one spot on the paper, whereas a mixture produces two or more distinct spots.

08.1 [2 marks]
- Pencil is insoluble / does not dissolve in the solvent (1 mark)
- Ink is soluble / would run or contaminate the chromatogram (1 mark)

08.2 [2 marks]
- Prevents the dye from washing off/dissolving directly into the solvent in the beaker (2 marks)

08.3 [2 marks]
- Working: 5.2 / 8.0 (1 mark)
- Rf = 0.65 (no units) (1 mark)

08.4 [4 marks]
- Pure substance definition: single substance / element or compound (not mixed with anything else) (1 mark)
- Pure substance on chromatogram: only one spot (1 mark)
- Mixture on chromatogram: more than one spot / multiple spots (1 mark)
- Separation based on different solubilities/affinities (1 mark)

**1.1** Group 7 elements all have 7 electrons in their outermost energy level. During chemical reactions, they all need to gain exactly one electron to achieve a stable full outer shell. This identical outer shell configuration gives them highly similar chemical properties.

**1.2** Moving down Group 7, the melting points increase, meaning the elements progress from gas to liquid to solid at room temperature. Since iodine is already a solid, astatine must also be a solid. The colours also become progressively darker (pale green \(\rightarrow\) red-brown \(\rightarrow\) dark grey). Therefore, astatine is predicted to be black or very dark grey.

**1.3** Chlorine is more reactive than bromine, so it displaces bromide ions from potassium bromide. The reaction produces bromine molecule (\(\text{Br}_2\)), which turns the colourless solution orange, yellow, or brown.
The balanced equation is:
\(\text{Cl}_2\text{ (aq)} + 2\text{KBr (aq)} \rightarrow \text{Br}_2\text{ (aq)} + 2\text{KCl (aq)}\)

**1.4** Chlorine is higher up in Group 7 than bromine, meaning a chlorine atom has fewer electron shells (a smaller atomic radius). Because the outer shell is closer to the nucleus, there is less shielding and a stronger electrostatic attraction between the positive nucleus and an incoming electron. Thus, chlorine gains an electron more easily than bromine, making it more reactive.

**1.1**
* all have 7 electrons in their outer shell / energy level (1)
* (so they all) need to gain one electron (to form a stable structure) (1)

**1.2**
* (state) solid (1)
* (colour) black / dark grey (1) *(allow very dark purple; do not accept purple or grey)*

**1.3**
* colourless to orange / yellow / brown (1) *(allow turns orange / yellow / brown)*
* \(\text{Br}_2 + 2\text{KCl}\) (2) *(1 mark for correct products, 1 mark for correct balancing)*

**1.4**
* chlorine atoms have fewer shells / energy levels (or chlorine is a smaller atom) (1)
* so there is a stronger attraction between the nucleus and the outer shell / incoming electron (or less shielding) (1)
* therefore chlorine gains an electron more easily (1)

01.1:
* State: Gas (1)
* Color: Pale yellow (1)

01.2:
* Boiling points increase down the group (1)
* because molecular size increases / there are more electrons (1)
* which leads to stronger intermolecular forces / attraction between molecules, requiring more thermal energy to break (1)

01.3:
* Correct formulas: \(\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2\) (1)
* Correct balancing (1)

01.4:
* Bromine atoms are larger / have more electron shells than chlorine atoms (1)
* The outer shell of bromine is further from the nucleus and experiences more shielding (1)
* Therefore, there is a weaker attraction between the nucleus and incoming electrons, meaning bromine is less reactive and cannot displace chlorine (1)

01.1: [2 marks] 1 mark for state (gas), 1 mark for color (pale yellow).
01.2: [3 marks] 1 mark for identifying the trend (increase), 1 mark for referencing larger molecules/more electrons, 1 mark for stating that stronger intermolecular forces require more energy to overcome.
01.3: [2 marks] 1 mark for correct reactants and products, 1 mark for correct balancing.
01.4: [3 marks] 1 mark for stating bromine has more shells/is larger, 1 mark for mentioning more shielding/greater distance from the nucleus, 1 mark for concluding there is a weaker attraction for incoming electrons making it less reactive.

02.1:
* Negative electrode (cathode): Zinc (1)
* Positive electrode (anode): Chlorine (1)

02.2:
* Correct species: \(\text{Zn}^{2+}(\text{l}) + 2\text{e}^- \rightarrow \text{Zn}(\text{s})\) (1)
* Correct state symbols (1)

02.3:
* Gas: Hydrogen (1)
* Reason: Aqueous solution contains hydrogen ions (from water) (1)
* Reason: Hydrogen is less reactive than zinc, so hydrogen ions are preferentially reduced / discharged at the cathode (1)

02.4:
* Test: Place damp blue litmus paper (1)
* into the gas (1)
* Observation: Litmus paper turns red then is bleached white (1)

02.1: [2 marks] 1 mark for zinc at the cathode, 1 mark for chlorine at the anode.
02.2: [2 marks] 1 mark for balanced half equation, 1 mark for correct state symbols (l and s).
02.3: [3 marks] 1 mark for identifying hydrogen gas, 1 mark for noting the presence of H+ ions, 1 mark for explaining that hydrogen is less reactive than zinc and is therefore discharged.
02.4: [3 marks] 1 mark for stating 'damp blue litmus paper', 1 mark for specifying putting it into the test tube/gas, 1 mark for the paper bleaching white (accept turns red then bleaches).

03.1:
* Correct reactants and products: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\) (1)
* Correct balancing (1)

03.2:
* Reduction is the loss of oxygen (by iron oxide) (1)

03.3:
* Sodium is more reactive than titanium (1)
* It is a displacement / redox reaction (1)

03.4:
* Correct species: \(\text{TiCl}_4 + 4\text{Na} \rightarrow \text{Ti} + 4\text{NaCl}\) (1)
* Correct balancing (1)

03.5:
* Any three from:
- Sodium is very expensive to obtain (requires electrolysis of molten salt) (1)
- High energy costs to maintain high temperatures (1)
- It is a batch process (rather than a continuous furnace process) which is slower and less efficient (1)
- Requires an expensive inert argon atmosphere to prevent oxidation of titanium/reaction of sodium with air (1)

03.1: [2 marks] 1 mark for correct formulas, 1 mark for correct balancing.
03.2: [1 mark] 1 mark for stating reduction is loss of oxygen.
03.3: [2 marks] 1 mark for stating sodium is more reactive, 1 mark for identifying it as displacement.
03.4: [2 marks] 1 mark for correct formulas, 1 mark for correct balancing.
03.5: [3 marks] 1 mark for each valid economic/procedural reason (max 3 marks).

04.1:
* \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\) (1)

04.2:
* \(M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100\) (1)
* \(M_r(\text{CaO}) = 40 + 16 = 56\) (1)

04.3:
* Moles of \(\text{CaCO}_3 = \frac{25.0}{100} = 0.25\text{ moles}\) (1)
* Moles of \(\text{CaO} = 0.25\text{ moles}\) (1)
* Theoretical mass of \(\text{CaO} = 0.25 \times 56 = 14.0\text{ g}\) (1)

04.4:
* Percentage yield formula: \(\frac{\text{Actual Mass}}{\text{Theoretical Mass}} \times 100 = \frac{11.2}{14.0} \times 100\) (1)
* Percentage yield = \(80.0\%\) (1)

04.5:
* Any two from:
- The reaction is reversible or may not have gone to completion (incomplete thermal decomposition) (1)
- Some product (calcium oxide) was left behind in the crucible / lost during transfer (1)
- Impurities in the starting material (calcium carbonate) (1)

04.1: [1 mark] 1 mark for correct balanced equation.
04.2: [2 marks] 1 mark for Mr of CaCO3 = 100, 1 mark for Mr of CaO = 56.
04.3: [3 marks] 1 mark for calculating moles of CaCO3 (0.25 mol), 1 mark for realizing molar ratio is 1:1, 1 mark for calculating correct mass of CaO (14.0 g).
04.4: [2 marks] 1 mark for correct calculation setup, 1 mark for correct percentage yield (80.0%). Allow error carried forward (ecf) from 04.3.
04.5: [2 marks] 1 mark for each distinct, valid practical reason (max 2 marks).

05.1:
* Increase the concentration of the acid (1)
* Increase the surface area / crush the calcium carbonate into a powder (1)

05.2:
* There are more reactant particles in the same volume / per unit volume (1)
* The particles are closer together, leading to more frequent collisions (1)
* This results in a higher frequency of successful collisions / more successful collisions per second (1)

05.3:
* The gradient of the curve decreases over time / becomes less steep and eventually horizontal (1)
* This is because the reactants are being used up, so their concentration decreases (1)
* Leading to less frequent collisions between reacting particles (1)

05.4:
* A catalyst increases the rate of reaction but is chemically unchanged / not used up at the end (1)
* It provides an alternative reaction pathway that has a lower activation energy (1)

05.1: [2 marks] 1 mark for each correct change listed (e.g., use powdered marble, use more concentrated acid).
05.2: [3 marks] 1 mark for more particles per unit volume, 1 mark for particles being closer together / more frequent collisions, 1 mark for more successful collisions per unit time / greater frequency of successful collisions.
05.3: [3 marks] 1 mark for describing gradient change (gets flatter/levels off), 1 mark for explaining concentration of reactants drops as they are used up, 1 mark for stating that collision frequency decreases.
05.4: [2 marks] 1 mark for defining catalyst (speeds up reaction without being consumed), 1 mark for explaining activation energy lowering.

06.1:
* Sodium loses its one outer shell electron (to form a positive sodium ion, \(\text{Na}^+\)) (1)
* Chlorine gains this electron into its outer shell (to form a negative chloride ion, \(\text{Cl}^-\)) (1)
* Strong electrostatic forces of attraction hold the oppositely charged ions together (ionic bonding) (1)

06.2:
* Sodium chloride has a giant ionic lattice structure with strong electrostatic forces of attraction between oppositely charged ions (1)
* Water consists of small molecules held together by weak intermolecular forces (1)
* Much more energy is needed to break the strong ionic bonds than to overcome the weak intermolecular forces in water (1)

06.3:
* Copper has a giant metallic structure made of regular layers of positive metal ions (cations) surrounded by a sea of delocalised electrons (1)
* Delocalised electrons are free to move throughout the structure and carry electrical charge (allowing conduction) (1)
* Layers of positive ions can slide past each other (1)
* without breaking the metallic bonds / sea of delocalised electrons holds the ions in place (allowing it to be shaped / malleable) (1)

06.1: [3 marks] 1 mark for sodium losing 1 electron, 1 mark for chlorine gaining 1 electron, 1 mark for electrostatic attraction between oppositely charged ions.
06.2: [3 marks] 1 mark for describing giant ionic structure of NaCl and its strong electrostatic forces, 1 mark for describing simple molecular structure of water with weak intermolecular forces, 1 mark for comparing the relative energy needed to break these bonds/forces.
06.3: [4 marks] 1 mark for describing metallic structure (positive ions in a sea of delocalised electrons), 1 mark for explaining conduction (free/delocalised electrons can move/carry charge), 1 mark for explaining shapeability (layers of ions can slide), 1 mark for mentioning that metallic bonding is preserved during sliding.

08.1:
* A pure substance melts at a single, sharp temperature (1)
* An impure mixture melts over a range of temperatures (and at a lower temperature than the pure substance) (1)

08.2:
* Ink would dissolve in the solvent and run/separate up the paper, interfering with the chromatogram (1)
* Pencil lead (graphite) is insoluble in the solvent and will not run (1)

08.3:
* If the solvent is above the start line, the spots of food coloring will dissolve directly into the solvent in the beaker (1)
* rather than traveling up the paper (1)

08.4:
* \(R_f = \frac{\text{Distance traveled by substance}}{\text{Distance traveled by solvent}} = \frac{4.5}{6.0}\) (1)
* \(R_f = 0.75\) (1)

08.5:
* The mobile phase (solvent) moves through the stationary phase (paper) (1)
* Different substances have different solubilities in the mobile phase and different strengths of attraction/adsorption to the stationary phase, causing them to travel at different rates (1)

08.1: [2 marks] 1 mark for pure substance sharp melting point, 1 mark for impure mixture melting over a range of temperatures.
08.2: [2 marks] 1 mark for identifying that ink dissolves/runs/contaminates the results, 1 mark for stating pencil graphite is insoluble/doesn't run.
08.3: [2 marks] 1 mark for noting the spots would wash off the paper into the beaker, 1 mark for stating they would not separate up the paper.
08.4: [2 marks] 1 mark for correct division setup (4.5 / 6.0), 1 mark for correct calculated value (0.75).
08.5: [2 marks] 1 mark for defining mobile phase moving through stationary phase, 1 mark for explaining separation due to different relative solubilities/attractions.

01.1 Two sodium atoms each lose one electron from their outer shell to form Na+ ions. One oxygen atom gains these two electrons into its outer shell to form an O2- ion. 01.2 Sodium oxide is a giant ionic lattice with strong electrostatic forces of attraction between oppositely charged ions, which require a large amount of energy to break. Carbon dioxide consists of simple molecules with weak intermolecular forces between them, which require very little energy to overcome. 01.3 Solid sodium oxide does not conduct electricity because the ions are fixed in a giant lattice and cannot move to carry charge.

01.1
- Sodium atoms lose one electron (1)
- Oxygen atoms gain two electrons (1)
- Two sodium atoms react with one oxygen atom (1)
- Forms Na+ and O2- ions (1)

01.2
- Sodium oxide has a giant ionic lattice / structure (1)
- Strong electrostatic forces of attraction between oppositely charged ions (1)
- Carbon dioxide consists of simple molecules / simple covalent structure with weak intermolecular forces (1)
- Giant structure requires much more energy to break bonds than simple molecular structure (1)

01.3
- Does not conduct (1)
- Ions are in fixed positions / cannot move in the solid state (1)

01.1 At the anode (positive electrode), bubbles of gas are observed because hydroxide ions from water are discharged to form oxygen gas: \( 4\text{OH}^{-}(\text{aq}) \rightarrow \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4e^{-} \).
01.2 At the cathode (negative electrode), copper ions gain electrons (reduction) to form copper metal: \( \text{Cu}^{2+}(\text{aq}) + 2e^{-} \rightarrow \text{Cu}(\text{s}) \).
01.3 The characteristic blue colour of copper(II) sulfate solution is caused by hydrated copper(II) ions, \( \text{Cu}^{2+} \). As the electrolysis proceeds, these ions are continually discharged and deposited as solid copper on the cathode, causing their concentration in the solution to decrease.
01.4 When active copper electrodes are used, copper from the anode dissolves into solution, so the mass of the anode decreases. Copper ions from the solution deposit onto the cathode, so the mass of the cathode increases.
01.5 At the anode, copper atoms in the electrode lose electrons (are oxidised) to form soluble copper(II) ions: \( \text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2e^{-} \), which dissolve into the electrolyte, reducing the anode's mass.

01.1
• Bubbles / effervescence / fizzing (1 mark)
• Oxygen (1 mark) (accept \( \text{O}_2 \))

01.2
• \( \text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu} \) (2 marks)
• Allow 1 mark for correct reactants and products with incorrect balancing.

01.3
• Blue colour is due to copper ions / \( \text{Cu}^{2+} \) (1 mark)
• Copper ions are removed/discharged from the solution (1 mark)

01.4
• (mass of) anode decreases (1 mark)
• (mass of) cathode increases (1 mark)

01.5
• copper atoms (at the anode) lose electrons / are oxidised (1 mark)
• to form copper ions / \( \text{Cu}^{2+} \) which go into solution / dissolve (1 mark)