Chemistry HL

1. 1.

   Define oxidation and reduction in terms of electron transfer.

   [2 marks] · no calculator

   Worked explanation

   1. An electron lost appears on the product side: Zn -> Zn2+ + 2e- is oxidation.
   2. An electron gained appears on the reactant side: Cu2+ + 2e- -> Cu is reduction. Equal electron transfer links the two in an overall redox reaction.

   Common mistake: Defining reduction only as oxygen loss excludes redox reactions with no oxygen.

   Marking points

   • States that oxidation is the loss of electrons by a species.
   • States that reduction is the gain of electrons by a species.

   Examiner tip: The mnemonic OIL RIG (Oxidation Is Loss, Reduction Is Gain) applies to electrons, not oxygen — do not confuse this with the older, narrower definition based on oxygen transfer.

2. 2.

   Marking analysis: A learner attempts the following task: “Define oxidation and reduction in terms of electron transfer.” Their response addresses only this point: “States that oxidation is the loss of electrons by a species.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [2 marks] · no calculator

   Marking points

   • Recognises credit for the stated point: States that oxidation is the loss of electrons by a species.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that reduction is the gain of electrons by a species.

   Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

3. 3.

   Using standard electrode potentials E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, calculate E°cell for the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), stating which half-cell is the cathode.

   [4 marks]

   Worked explanation

   1. The more positive reduction potential is Cu2+/Cu, so copper reduction occurs at the cathode; zinc oxidation occurs at the anode.
   2. Ecell standard = Ecathode standard - Eanode standard = +0.34-(-0.76) = +1.10 V. Potentials are not multiplied by electron-balancing coefficients.

   Common mistake: Reversing the zinc sign and then subtracting reverses it twice.

   Marking points

   • Identifies the Cu²⁺/Cu half-cell as the cathode, since it has the more positive (less negative) standard electrode potential.
   • Identifies the Zn²⁺/Zn half-cell as the anode.
   • Uses E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76).
   • Obtains E°cell = +1.10 V.

   Examiner tip: Always use cathode minus anode with the tabulated reduction potentials exactly as given — do not flip the sign of the anode's value before subtracting.

4. 4.

   Marking analysis: A learner attempts the following task: “Using standard electrode potentials E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, calculate E°cell for the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), stating which half-cell is the cathode.” Their response addresses only this point: “Identifies the Cu²⁺/Cu half-cell as the cathode, since it has the more positive (less negative) standard electrode potential.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [4 marks]

   Marking points

   • Recognises credit for the stated point: Identifies the Cu²⁺/Cu half-cell as the cathode, since it has the more positive (less negative) standard electrode potential.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Identifies the Zn²⁺/Zn half-cell as the anode.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Uses E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76).
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Obtains E°cell = +1.10 V.

   Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

5. 5.

   State how the sign of E°cell indicates whether a redox reaction is spontaneous under standard conditions, and hence state whether the reaction between Zn(s) and Cu²⁺(aq) is spontaneous, given E°cell = +1.10 V.

   [2 marks] · no calculator

   Worked explanation

   1. Delta G standard = -nF Ecell standard, with positive n and F, makes positive cell potential correspond to negative standard reaction Gibbs energy.
   2. The stated +1.10 V therefore favours Zn + Cu2+ -> Zn2+ + Cu under standard conditions. It predicts thermodynamic direction, not reaction speed or behaviour at every concentration.

   Common mistake: Interpreting positive voltage as positive Gibbs energy misses the minus sign.

   Marking points

   • States that a positive E°cell indicates a spontaneous (feasible) reaction under standard conditions, while a negative E°cell indicates a non-spontaneous one.
   • States that since E°cell = +1.10 V is positive, the reaction between Zn(s) and Cu²⁺(aq) is spontaneous.

   Examiner tip: This links to ΔG° = −nFE°cell: a positive E°cell gives a negative ΔG°, which is the general thermodynamic condition for spontaneity.

6. 6.

   Marking analysis: A learner attempts the following task: “State how the sign of E°cell indicates whether a redox reaction is spontaneous under standard conditions, and hence state whether the reaction between Zn(s) and Cu²⁺(aq) is spontaneous, given E°cell = +1.10 V.” Their response addresses only this point: “States that a positive E°cell indicates a spontaneous (feasible) reaction under standard conditions, while a negative E°cell indicates a non-spontaneous one.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [2 marks] · no calculator

   Marking points

   • Recognises credit for the stated point: States that a positive E°cell indicates a spontaneous (feasible) reaction under standard conditions, while a negative E°cell indicates a non-spontaneous one.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that since E°cell = +1.10 V is positive, the reaction between Zn(s) and Cu²⁺(aq) is spontaneous.

   Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

7. 7.

   A current of 2.00 A is passed through molten lead(II) bromide, PbBr₂(l), for 1930 seconds using inert electrodes. Calculate the number of moles of electrons transferred. (F = 96500 C mol⁻¹)

   [4 marks]

   Course scope: Optional enrichment

   Syllabus placement evidence: First assessment 2025, updated March 2024; Reactivity 3.2.1-3.2.16, pp.59-61, especially 3.2.8 and 3.2.14-3.2.16. Requires Q=It and n(e)=Q/F quantitative electrolysis, beyond specified conduction/products and standard Gibbs-potential use of F.

   Worked explanation

   1. Current is charge per second: Q = It = 2.00 C s^-1 x 1930 s = 3860 C.
   2. n(e-) = Q/F = 3860 C/(96500 C mol^-1) = 0.0400 mol electrons. This is electron amount, not yet lead amount.

   Common mistake: Dividing by two here answers moles of Pb rather than the requested electrons.

   Marking points

   • States the formula for charge: Q = It.
   • Calculates Q = 2.00 × 1930 = 3860 C.
   • Uses n(e⁻) = Q / F.
   • Obtains n(e⁻) = 3860 / 96500 = 0.0400 mol.

   Examiner tip: Faraday's constant, F, is the charge on one mole of electrons — always convert charge to moles of electrons before relating it to moles of product.

8. 8.

   Marking analysis: A learner attempts the following task: “A current of 2.00 A is passed through molten lead(II) bromide, PbBr₂(l), for 1930 seconds using inert electrodes. Calculate the number of moles of electrons transferred. (F = 96500 C mol⁻¹)” Their response addresses only this point: “States the formula for charge: Q = It.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [4 marks]

   Marking points

   • Recognises credit for the stated point: States the formula for charge: Q = It.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates Q = 2.00 × 1930 = 3860 C.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Uses n(e⁻) = Q / F.
   • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Obtains n(e⁻) = 3860 / 96500 = 0.0400 mol.

   Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

9. 9.

   Using n(e⁻) = 0.0400 mol from the electrolysis of molten PbBr₂, calculate the mass of lead deposited at the cathode. (Pb²⁺ + 2e⁻ → Pb; Ar(Pb) = 207)

   [3 marks]

   Syllabus placement evidence: First assessment 2025, updated March 2024; Structure 1.4.1,1.4.3, p.36; Reactivity 2.1.1-2.1.2, p.52; 3.2.2, p.59. Electron amount and half-equation are supplied; only prescribed mole ratio and molar-mass conversion are required.

   Worked explanation

   1. Pb2+ + 2e- -> Pb requires two moles electrons per mole lead, so n(Pb) = 0.0400/2 = 0.0200 mol.
   2. m(Pb) = 0.0200 mol x 207 g mol^-1 = 4.14 g, assuming all the specified electrons deposit lead and the metal is recovered.

   Common mistake: Multiplying electron moles by 207 directly doubles the deposited mass.

   Marking points

   • States that 2 moles of electrons are required per mole of Pb deposited, from Pb²⁺ + 2e⁻ → Pb.
   • Uses n(Pb) = n(e⁻) / 2 = 0.0400 / 2 = 0.0200 mol.
   • Calculates mass = n × Ar = 0.0200 × 207 = 4.14 g.

   Examiner tip: Always check the number of electrons in the half-equation for the specific ion being discharged — it is not always 1, and using the wrong ratio is the most common error in these calculations.

10. 10.

    Marking analysis: A learner attempts the following task: “Using n(e⁻) = 0.0400 mol from the electrolysis of molten PbBr₂, calculate the mass of lead deposited at the cathode. (Pb²⁺ + 2e⁻ → Pb; Ar(Pb) = 207)” Their response addresses only this point: “States that 2 moles of electrons are required per mole of Pb deposited, from Pb²⁺ + 2e⁻ → Pb.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks]

    Marking points

    • Recognises credit for the stated point: States that 2 moles of electrons are required per mole of Pb deposited, from Pb²⁺ + 2e⁻ → Pb.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Uses n(Pb) = n(e⁻) / 2 = 0.0400 / 2 = 0.0200 mol.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates mass = n × Ar = 0.0200 × 207 = 4.14 g.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

11. 11.

    Explain why the standard hydrogen electrode (SHE) is assigned a standard electrode potential of exactly 0.00 V, and outline why this matters for tabulating other electrode potentials.

    [3 marks] · no calculator

    Worked explanation

    1. Only differences between electrode potentials are experimentally accessible in a complete cell, so a shared reference is necessary.
    2. Assigning SHE exactly 0.00 V fixes that reference, allowing other standard reduction potentials to be compared on one scale. The zero is conventional, not a measured claim that hydrogen has no redox tendency.

    Common mistake: Treating the reference zero as a chemical inertness measurement misconstrues the scale.

    Marking points

    • States that the SHE is an arbitrary but universally agreed reference electrode.
    • States that its standard electrode potential is defined as exactly 0.00 V by convention.
    • States that this common reference allows every other standard electrode potential to be measured and tabulated on the same scale, so they can be meaningfully compared with one another.

    Examiner tip: Only the relative values of electrode potentials matter for predicting reactions — the SHE's 0.00 V is a reference point, not a claim that hydrogen has no tendency to be reduced.

12. 12.

    Marking analysis: A learner attempts the following task: “Explain why the standard hydrogen electrode (SHE) is assigned a standard electrode potential of exactly 0.00 V, and outline why this matters for tabulating other electrode potentials.” Their response addresses only this point: “States that the SHE is an arbitrary but universally agreed reference electrode.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that the SHE is an arbitrary but universally agreed reference electrode.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that its standard electrode potential is defined as exactly 0.00 V by convention.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this common reference allows every other standard electrode potential to be measured and tabulated on the same scale, so they can be meaningfully compared with one another.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

13. 13.

    Aqueous copper(II) sulfate is electrolysed using inert platinum electrodes under the usual laboratory conditions. Identify the cathode and anode products, explain why water rather than sulfate is oxidised at the anode, and write a balanced anode half-equation.

    [4 marks] · no calculator

    Worked explanation

    1. At the cathode, Cu2+ + 2e- -> Cu deposits copper under usual aqueous CuSO4/Pt conditions rather than evolving hydrogen.
    2. At the anode, 2H2O -> O2 + 4H+ + 4e- evolves oxygen; sulfate stays mainly in solution. Water oxidation is favoured over sulfate oxidation under these conditions, with electrode kinetics and overpotential also relevant.

    Common mistake: Discharging sulfate simply because it is the named anion ignores competing water chemistry.

    Marking points

    • States that copper metal, Cu(s), is deposited at the cathode, since Cu²⁺ is preferentially reduced over H⁺ (Cu²⁺/Cu has the more positive standard electrode potential).
    • States that oxygen gas, O₂(g), is produced at the anode.
    • Explains that oxygen evolution from water is favoured over sulfate oxidation under these conditions; sulfate generally remains in solution rather than being discharged.
    • Writes 2H2O(l) -> O2(g) + 4H+(aq) + 4e-, or the equivalent alkaline form 4OH-(aq) -> O2(g) + 2H2O(l) + 4e-.

    Examiner tip: Compare competing aqueous electrode processes under the stated conditions. Concentrations, electrode material and overpotential can matter; standard potentials alone do not universally determine the observed electrolysis products.

14. 14.

    Marking analysis: A learner attempts the following task: “Aqueous copper(II) sulfate is electrolysed using inert platinum electrodes under the usual laboratory conditions. Identify the cathode and anode products, explain why water rather than sulfate is oxidised at the anode, and write a balanced anode half-equation.” Their response addresses only this point: “States that copper metal, Cu(s), is deposited at the cathode, since Cu²⁺ is preferentially reduced over H⁺ (Cu²⁺/Cu has the more positive standard electrode potential).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that copper metal, Cu(s), is deposited at the cathode, since Cu²⁺ is preferentially reduced over H⁺ (Cu²⁺/Cu has the more positive standard electrode potential).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that oxygen gas, O₂(g), is produced at the anode.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Explains that oxygen evolution from water is favoured over sulfate oxidation under these conditions; sulfate generally remains in solution rather than being discharged.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Writes 2H2O(l) -> O2(g) + 4H+(aq) + 4e-, or the equivalent alkaline form 4OH-(aq) -> O2(g) + 2H2O(l) + 4e-.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

15. 15.

    Two electrolytic cells are connected in series: one contains molten PbBr2 and the other aluminium oxide dissolved in molten cryolite. Assume 100% current efficiency for metal deposition in each. Deduce which cell deposits more moles of metal for the same charge and explain why. Use Pb2+ + 2e- -> Pb and Al3+ + 3e- -> Al.

    [3 marks] · no calculator

    Syllabus placement evidence: First assessment 2025, updated March 2024; Structure 1.4.1, p.36; Reactivity 2.1.1-2.1.2, p.52; 3.2.2,3.2.8, pp.59-60; Tool 3, p.29. At the stated same charge, compares supplied half-equation ratios without quantitative current-time or Faraday calculation.

    Worked explanation

    1. Series cells receive the same Q and hence the same electron amount ne = Q/F.
    2. At 100% current efficiency, n(Pb)=ne/2 and n(Al)=ne/3, so lead amount is 3/2 of aluminium amount. The comparison is moles, not masses or metal ion charges alone.

    Common mistake: Giving equal metal amounts confuses shared electrons with different deposition stoichiometries.

    Marking points

    • States that cells connected in series carry the same current, so the same quantity of charge (and hence the same number of moles of electrons) passes through both.
    • States that depositing 1 mole of Pb requires 2 moles of electrons, while depositing 1 mole of Al requires 3 moles of electrons.
    • Deduces that for the same number of moles of electrons, more moles of Pb are deposited than Al, since fewer electrons are needed per mole of Pb.

    Examiner tip: In cells connected in series, moles of electrons are the shared quantity across cells — always convert to moles of electrons first, then apply each cell's own half-equation stoichiometry separately.

16. 16.

    Marking analysis: A learner attempts the following task: “Two electrolytic cells are connected in series: one contains molten PbBr2 and the other aluminium oxide dissolved in molten cryolite. Assume 100% current efficiency for metal deposition in each. Deduce which cell deposits more moles of metal for the same charge and explain why. Use Pb2+ + 2e- -> Pb and Al3+ + 3e- -> Al.” Their response addresses only this point: “States that cells connected in series carry the same current, so the same quantity of charge (and hence the same number of moles of electrons) passes through both.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that cells connected in series carry the same current, so the same quantity of charge (and hence the same number of moles of electrons) passes through both.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that depositing 1 mole of Pb requires 2 moles of electrons, while depositing 1 mole of Al requires 3 moles of electrons.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Deduces that for the same number of moles of electrons, more moles of Pb are deposited than Al, since fewer electrons are needed per mole of Pb.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

17. 17.

    State the Nernst equation. For Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s) at fixed temperature, explain how increasing only [Cu2+] above 1 mol dm^-3 affects cell potential when [Zn2+] remains at 1 mol dm^-3. Treat ion activities as proportional to concentration.

    [3 marks] · no calculator

    Course scope: Optional enrichment

    Syllabus placement evidence: First assessment 2025, updated March 2024; Reactivity 3.2.12-3.2.16, p.61; 1.4.4, p.51; 2.3.5-2.3.7, pp.55-56. Requires recalling Nernst and nonstandard potential correction beyond the specified standard-potential and Gibbs/quotient equations.

    Worked explanation

    1. E = E standard - (RT/nF) ln Q; for Zn + Cu2+ -> Zn2+ + Cu, Q = a(Zn2+)/a(Cu2+) and n=2, with pure solid activities unity.
    2. Increasing only copper-ion concentration lowers Q below 1 in the stated activity approximation. ln Q becomes negative, so E exceeds E standard. At high concentration the proportional-activity approximation becomes less reliable.

    Common mistake: Putting solid zinc concentration into Q contradicts its unit activity.

    Marking points

    • States the Nernst equation, E = E° − (RT/nF) ln Q, relating cell potential under non-standard conditions to the standard cell potential and the reaction quotient Q.
    • States that increasing [Cu²⁺] above standard (1 mol/dm³) decreases the reaction quotient Q for the reaction as written, since Cu²⁺ is a reactant.
    • States that this makes the cell potential E slightly greater (more positive) than the standard value E°, since the reaction is driven further from equilibrium in the forward direction.

    Examiner tip: For a given reaction at fixed temperature, E equals E standard whenever Q = 1. Standard-state activities ensure Q = 1, but are not necessary: for this cell equal nonstandard Zn2+ and Cu2+ activities also give Q = 1.

18. 18.

    Marking analysis: A learner attempts the following task: “State the Nernst equation. For Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s) at fixed temperature, explain how increasing only [Cu2+] above 1 mol dm^-3 affects cell potential when [Zn2+] remains at 1 mol dm^-3. Treat ion activities as proportional to concentration.” Their response addresses only this point: “States the Nernst equation, E = E° − (RT/nF) ln Q, relating cell potential under non-standard conditions to the standard cell potential and the reaction quotient Q.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States the Nernst equation, E = E° − (RT/nF) ln Q, relating cell potential under non-standard conditions to the standard cell potential and the reaction quotient Q.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that increasing [Cu²⁺] above standard (1 mol/dm³) decreases the reaction quotient Q for the reaction as written, since Cu²⁺ is a reactant.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that this makes the cell potential E slightly greater (more positive) than the standard value E°, since the reaction is driven further from equilibrium in the forward direction.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

19. 19.

    A current of 1.5 A is passed through molten aluminium oxide to deposit aluminium metal at the cathode: Al³⁺ + 3e⁻ → Al. Calculate the time required to deposit 2.70 g of aluminium. (Ar(Al) = 27, F = 96500 C mol⁻¹)

    [5 marks]

    Course scope: Optional enrichment

    Syllabus placement evidence: First assessment 2025, updated March 2024; Reactivity 3.2.8,3.2.12-3.2.16, pp.60-61. Requires quantitative current-charge-duration electrolysis beyond the complete specified electron-transfer requirements.

    Worked explanation

    1. 2.70 g/(27 g mol^-1)=0.100 mol Al. Al3+ requires 3e-, so electrons needed = 0.300 mol and Q = 0.300(96500)=28950 C.
    2. t=Q/I=28950 C/(1.5 C s^-1)=19300 s, about 5.36 h. This assumes steady current and 100% current efficiency for aluminium deposition.

    Common mistake: Using one electron per Al underestimates time by a factor of three.

    Marking points

    • Calculates moles of Al = 2.70 / 27 = 0.100 mol.
    • Uses the half-equation ratio to find moles of electrons: 0.100 × 3 = 0.300 mol.
    • Calculates charge Q = n(e⁻) × F = 0.300 × 96500 = 28 950 C.
    • Uses t = Q / I.
    • Obtains t = 28 950 / 1.5 = 19 300 s.

    Examiner tip: This is the same Faraday's law relationship as before, simply rearranged to solve for time instead of mass — always identify which single quantity is unknown before rearranging.

20. 20.

    Marking analysis: A learner attempts the following task: “A current of 1.5 A is passed through molten aluminium oxide to deposit aluminium metal at the cathode: Al³⁺ + 3e⁻ → Al. Calculate the time required to deposit 2.70 g of aluminium. (Ar(Al) = 27, F = 96500 C mol⁻¹)” Their response addresses only this point: “Calculates moles of Al = 2.70 / 27 = 0.100 mol.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Marking points

    • Recognises credit for the stated point: Calculates moles of Al = 2.70 / 27 = 0.100 mol.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Uses the half-equation ratio to find moles of electrons: 0.100 × 3 = 0.300 mol.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Calculates charge Q = n(e⁻) × F = 0.300 × 96500 = 28 950 C.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Uses t = Q / I.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Obtains t = 28 950 / 1.5 = 19 300 s.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

21. 21.

    State one practical reason for electroplating an object with a metal such as chromium or silver, and outline the role of the object being plated in the electrolytic cell.

    [3 marks] · no calculator

    Worked explanation

    1. Silver plating can improve appearance; a suitable chromium coating can provide a protective surface. These are practical purposes, not changes to the bulk object's composition.
    2. Make the object the cathode in an electrolyte supplying the plating-metal ions. Reduction, for example Ag+ + e- -> Ag, deposits metal onto its surface.

    Common mistake: Making the object the anode promotes oxidation rather than the desired deposition.

    Marking points

    • States a valid reason, e.g. to protect the underlying metal from corrosion, or to improve its appearance.
    • States that the object to be plated is made the cathode of the electrolytic cell.
    • States that the electrolyte contains ions of the plating metal, which are reduced and deposited onto the cathode (object).

    Examiner tip: The object being plated must always be the cathode, since reduction (electron gain, depositing neutral metal atoms) occurs at the cathode — this is the same principle as any electrolytic metal deposition.

22. 22.

    Marking analysis: A learner attempts the following task: “State one practical reason for electroplating an object with a metal such as chromium or silver, and outline the role of the object being plated in the electrolytic cell.” Their response addresses only this point: “States a valid reason, e.g. to protect the underlying metal from corrosion, or to improve its appearance.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States a valid reason, e.g. to protect the underlying metal from corrosion, or to improve its appearance.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the object to be plated is made the cathode of the electrolytic cell.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the electrolyte contains ions of the plating metal, which are reduced and deposited onto the cathode (object).

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

23. 23.

    Outline how a hydrogen-oxygen fuel cell generates electrical energy, including the half-equations at each electrode in alkaline conditions, and state one advantage of a fuel cell compared with burning hydrogen directly as a fuel.

    [4 marks] · no calculator

    Worked explanation

    1. Alkaline anode: 2H2+4OH- -> 4H2O+4e-. Cathode: O2+2H2O+4e- -> 4OH-. Adding and cancelling gives 2H2+O2 -> 2H2O.
    2. Electrons travel through the external load from anode to cathode, directly generating electricity. This can be more efficient than combustion followed by a heat engine; water-only product is not a distinguishing advantage over burning pure hydrogen.

    Common mistake: Using no CO2 as the sole comparison ignores that hydrogen combustion also has no direct CO2 product.

    Marking points

    • States the anode (negative electrode) half-equation: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻.
    • States the cathode (positive electrode) half-equation: O₂ + 2H₂O + 4e⁻ → 4OH⁻.
    • States that electrons flow through an external circuit from the anode to the cathode, generating an electric current, while the only product formed overall is water.
    • States a valid comparative advantage, e.g. direct conversion to electricity can be more efficient than a combustion-based heat engine, or avoids combustion-generated NOx when hydrogen is burned in air. Do not credit 'no direct CO2' or 'water is produced' alone as advantages over burning pure hydrogen: both routes share those features.

    Examiner tip: A fuel cell is essentially a voltaic cell that is continuously supplied with fresh reactants (hydrogen and oxygen) rather than relying on a fixed internal store of chemicals that becomes depleted.

24. 24.

    Marking analysis: A learner attempts the following task: “Outline how a hydrogen-oxygen fuel cell generates electrical energy, including the half-equations at each electrode in alkaline conditions, and state one advantage of a fuel cell compared with burning hydrogen directly as a fuel.” Their response addresses only this point: “States the anode (negative electrode) half-equation: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States the anode (negative electrode) half-equation: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States the cathode (positive electrode) half-equation: O₂ + 2H₂O + 4e⁻ → 4OH⁻.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that electrons flow through an external circuit from the anode to the cathode, generating an electric current, while the only product formed overall is water.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States a valid comparative advantage, e.g. direct conversion to electricity can be more efficient than a combustion-based heat engine, or avoids combustion-generated NOx when hydrogen is burned in air. Do not credit 'no direct CO2' or 'water is produced' alone as advantages over burning pure hydrogen: both routes share those features.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

25. 25.

    Distinguish between a primary cell and a secondary cell, giving one example of each.

    [2 marks] · no calculator

    Worked explanation

    1. A standard alkaline primary cell is designed for one discharge; its chemistry and construction do not support safe practical reversal for charging.
    2. A lead-acid or lithium-ion secondary cell uses reactions and electrode structures that can be restored by externally driven reverse processes. Rechargeability is a design-and-chemistry property, not a feature of every voltaic cell.

    Common mistake: Assuming connecting any used primary cell to a charger makes it secondary is chemically and practically unsound.

    Marking points

    • States that a primary cell cannot be recharged, since its electrode reactions are not easily reversible, e.g. a standard alkaline battery.
    • States that a secondary cell can be recharged by passing a current through it in the reverse direction, reversing the chemical reaction, e.g. a lithium-ion or lead-acid battery.

    Examiner tip: Secondary cells use electrode reactions and electrode structures that permit practical repeated restoration by external charging. Rechargeability depends on chemistry and cell design, not merely whether any reverse reaction can be forced. Do not attempt to recharge a primary cell.

26. 26.

    Marking analysis: A learner attempts the following task: “Distinguish between a primary cell and a secondary cell, giving one example of each.” Their response addresses only this point: “States that a primary cell cannot be recharged, since its electrode reactions are not easily reversible, e.g. a standard alkaline battery.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [2 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that a primary cell cannot be recharged, since its electrode reactions are not easily reversible, e.g. a standard alkaline battery.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a secondary cell can be recharged by passing a current through it in the reverse direction, reversing the chemical reaction, e.g. a lithium-ion or lead-acid battery.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

27. 27.

    Write the standard cell notation (cell diagram) for a voltaic cell made from a Zn(s)/Zn²⁺(aq) half-cell and a Cu(s)/Cu²⁺(aq) half-cell, given that zinc is the more easily oxidised metal.

    [3 marks] · no calculator

    Course scope: Optional enrichment

    Syllabus placement evidence: First assessment 2025, updated March 2024; Reactivity 3.2.5-3.2.6, p.60; 3.2.12-3.2.16, p.61. Requires unsupplied vertical-bar line-notation conventions beyond prescribed cell construction, electron flow and potentials.

    Worked explanation

    1. Put zinc oxidation on the left: Zn(s) | Zn2+(aq). Put copper reduction on the right: Cu2+(aq) | Cu(s). Single bars mark phase boundaries.
    2. Join them with the salt-bridge symbol: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s). For a standard cell, specify unit ion activities at the stated temperature.

    Common mistake: Placing both metals next to the double bar obscures the actual phase sequence.

    Marking points

    • States that the anode (oxidation, the more easily oxidised metal) is written on the left: Zn(s) | Zn²⁺(aq).
    • States that a double vertical line represents the salt bridge, separating the two half-cells: ||.
    • Writes the complete notation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), with the cathode (reduction) on the right.

    Examiner tip: Cell notation always follows the convention: anode (oxidation) on the left, salt bridge (||) in the middle, cathode (reduction) on the right — the same order as reading 'left to right, oxidation to reduction'.

28. 28.

    Marking analysis: A learner attempts the following task: “Write the standard cell notation (cell diagram) for a voltaic cell made from a Zn(s)/Zn²⁺(aq) half-cell and a Cu(s)/Cu²⁺(aq) half-cell, given that zinc is the more easily oxidised metal.” Their response addresses only this point: “States that the anode (oxidation, the more easily oxidised metal) is written on the left: Zn(s) | Zn²⁺(aq).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that the anode (oxidation, the more easily oxidised metal) is written on the left: Zn(s) | Zn²⁺(aq).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that a double vertical line represents the salt bridge, separating the two half-cells: ||.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: Writes the complete notation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), with the cathode (reduction) on the right.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

29. 29.

    Explain, in terms of electron transfer, why iron rusts when exposed to both oxygen and water, and state one method of preventing rusting based on this mechanism.

    [4 marks] · no calculator

    Worked explanation

    1. Anodic regions oxidise Fe -> Fe2+ + 2e-; cathodic regions reduce dissolved O2 + 2H2O + 4e- -> 4OH-. Further reactions yield hydrated iron(III) oxide rust.
    2. Paint blocks oxygen/water contact. Alternatively attached zinc oxidises preferentially, supplying electrons and protecting iron as the cathodic material even at small coating scratches while zinc remains available.

    Common mistake: Calling rust simply Fe(OH)2 stops before further oxidation to iron(III) corrosion products.

    Marking points

    • States that iron atoms are oxidised (lose electrons) to form Fe²⁺ ions: Fe → Fe²⁺ + 2e⁻, acting as an anode region.
    • States that oxygen dissolved in water is reduced at a different point on the iron surface, acting as a cathode region: O₂ + 2H₂O + 4e⁻ → 4OH⁻.
    • States that the Fe²⁺ and OH⁻ ions produced react further and are oxidised by more oxygen to eventually form hydrated iron(III) oxide, Fe₂O₃·xH₂O (rust).
    • States a valid prevention method, e.g. coating the iron with a barrier such as paint or grease to exclude oxygen and water, or attaching a more reactive metal (e.g. zinc, in galvanisation) which is oxidised preferentially instead of the iron.

    Examiner tip: Rusting is a genuine electrochemical process with separate anode and cathode regions on the same piece of metal — sacrificial protection works by making a more reactive metal act as the anode instead, so the iron remains as the (protected) cathode.

30. 30.

    Marking analysis: A learner attempts the following task: “Explain, in terms of electron transfer, why iron rusts when exposed to both oxygen and water, and state one method of preventing rusting based on this mechanism.” Their response addresses only this point: “States that iron atoms are oxidised (lose electrons) to form Fe²⁺ ions: Fe → Fe²⁺ + 2e⁻, acting as an anode region.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Marking points

    • Recognises credit for the stated point: States that iron atoms are oxidised (lose electrons) to form Fe²⁺ ions: Fe → Fe²⁺ + 2e⁻, acting as an anode region.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that oxygen dissolved in water is reduced at a different point on the iron surface, acting as a cathode region: O₂ + 2H₂O + 4e⁻ → 4OH⁻.
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States that the Fe²⁺ and OH⁻ ions produced react further and are oxidised by more oxygen to eventually form hydrated iron(III) oxide, Fe₂O₃·xH₂O (rust).
    • Checks this criterion, crediting equivalent work already shown or identifying what is missing: States a valid prevention method, e.g. coating the iron with a barrier such as paint or grease to exclude oxygen and water, or attaching a more reactive metal (e.g. zinc, in galvanisation) which is oxidised preferentially instead of the iron.

    Examiner tip: Check each criterion, including equivalent work already shown. Do not demand an extra step merely to rewrite an equivalent answer. Do not award the same idea twice or infer work the learner did not show.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.