IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 4: Electrochemistry -- Challenge Paper

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0620

Instructions -- Challenge Level

Answer all questions in the spaces provided.
This paper is at challenge level -- expect multi-step reasoning and unfamiliar contexts.
Show all working for calculations.
Write balanced half-equations where required -- pay careful attention to electron balance.
Your answers will be automatically graded when you submit.

Question Navigation

Question 1 -- Writing and Balancing Half-Equations

Total: 12 marks

A research laboratory in Cambridge investigates the electrolysis of several different compounds. You are given the following electrolytes and must write balanced half-equations for each.

(a) [4]

Write the balanced half-equation for the reaction at:

(i) the cathode [2]

(ii) the anode [2]

Model Answer -- 1(a)

Cathode: Ca²⁺ + 2e⁻ → Ca [1 for species, 1 for balanced]

Ca²+ + 2e− → Ca

Anode: 2Cl⁻ → Cl₂ + 2e⁻ [1 for species, 1 for balanced]

2Cl− → Cl₂ + 2e−

⚠ If you missed marks here: The usual slip is putting electrons on the wrong side: at the cathode electrons are GAINED so they belong on the left (Ca²⁺ + 2e⁻ → Ca); at the anode they are LOST and appear on the right. Also check you wrote Cl₂ (a diatomic molecule), not Cl, and that 2Cl⁻ matches 2e⁻ so the charges balance.

1 mark -- cathode calcium ion Ca2+ gains electrons 1 mark -- cathode balanced 2 electrons correct 1 mark -- anode chloride ions lose electrons chlorine formed 1 mark -- anode balanced 2Cl- gives Cl2 plus 2e-

(b) [4]

Write the balanced half-equation for the reaction at:

(i) the cathode [2]

(ii) the anode [2]

Hint: consider the charge on the aluminium ion and the oxide ion carefully.

Model Answer -- 1(b)

Cathode: Al³⁺ + 3e⁻ → Al [1 for correct ion/product, 1 for 3 electrons]

Al³+ + 3e− → Al

Anode: 2O²⁻ → O₂ + 4e⁻ [1 for correct ion/product, 1 for 4 electrons and balanced charges]

2O²− → O₂ + 4e−

⚠ If you missed marks here: The classic error is writing Al²⁺ or only 2 electrons – aluminium is in Group III, so it is Al³⁺ + 3e⁻ → Al. At the anode most lost marks come from O²⁻ → O₂ + 2e⁻: you need TWO oxide ions, so 2O²⁻ releases 4e⁻ to balance both the atoms and the 4− of charge.

1 mark -- cathode Al3+ ion gains electrons forms aluminium 1 mark -- cathode 3 electrons correctly balanced 1 mark -- anode O2- oxide ions lose electrons oxygen formed 1 mark -- anode 4 electrons balanced 2O2- gives O2 plus 4e-

Dilute sodium sulfate solution is electrolysed using platinum electrodes.

(i) State the four ions present in the solution. [1]

(ii) Using your knowledge of selective discharge, state the product at the cathode and write the half-equation. [1]

(iii) State the product at the anode and write the half-equation. [2]

Model Answer -- 1(c)

Na⁺, SO₄²⁻ (from sodium sulfate) and H⁺, OH⁻ (from water) [1]

Cathode: hydrogen gas; 2H⁺ + 2e⁻ → H₂ (H⁺ discharged because Na is more reactive than H) [1]

2H+ + 2e− → H₂

Anode: oxygen gas; 4OH⁻ → 2H₂O + O₂ + 4e⁻ (OH⁻ discharged because SO₄²⁻ is too stable to be discharged) [1]

4OH− → 2H₂O + O₂ + 4e−

This is effectively the electrolysis of water [1]

⚠ If you missed marks here: If you only listed Na⁺ and SO₄²⁻, you forgot that WATER also supplies H⁺ and OH⁻ ions – and those are the ones discharged. Answers saying 'sodium at the cathode' miss that Na is more reactive than hydrogen, so H⁺ is discharged; and sulfate is never discharged at the anode – OH⁻ gives oxygen via 4OH⁻ → 2H₂O + O₂ + 4e⁻.

1 mark -- Na+ SO42- H+ OH- four ions identified 1 mark -- cathode hydrogen 2H+ plus 2e- gives H2 sodium more reactive 1 mark -- anode oxygen 4OH- gives 2H2O plus O2 plus 4e- 1 mark -- sulfate too stable effectively electrolysis water

Question 2 -- Selective Discharge in Depth

Total: 12 marks

A research student in Chennai compares the electrolysis of three different aqueous solutions using inert platinum electrodes: concentrated NaCl, dilute NaCl, and dilute CuSO₄.

(a) [4]

(i) State the product at the cathode. Explain your reasoning using the reactivity series. [2]

(ii) State the product at the anode. Explain why concentration is the deciding factor here. [2]

Model Answer -- 2(a)

Cathode: hydrogen gas; sodium is more reactive than hydrogen (higher in reactivity series) [1]

So H⁺ ions are preferentially discharged because the less reactive cation is discharged [1]

Anode: chlorine gas [1]

Although OH⁻ would normally be discharged in preference, the very high concentration of Cl⁻ ions means they are discharged instead; concentration overrides the usual rule for anions [1]

⚠ If you missed marks here: Naming 'hydrogen' without the reasoning loses the explain mark – you must say sodium is MORE reactive than hydrogen, so the less reactive ion (H⁺) is discharged. At the anode, 'chlorine because there is chloride' is not enough either: the mark is for saying the HIGH CONCENTRATION of Cl⁻ overrides the usual rule that OH⁻ would be discharged.

1 mark -- cathode hydrogen sodium more reactive higher reactivity series 1 mark -- H+ preferentially discharged less reactive cation discharged 1 mark -- anode chlorine gas 1 mark -- high concentration Cl- discharged instead concentration overrides usual rule

(b) [3]

(i) State how the product at the anode differs from part (a). [1]

(ii) Explain why this change occurs when the solution is dilute. [1]

(iii) Write the half-equation for the reaction at the anode in dilute NaCl. [1]

Model Answer -- 2(b)

Oxygen is produced at the anode instead of chlorine [1]

In dilute solution, the concentration of Cl⁻ is low, so concentration no longer overrides the normal rule; OH⁻ ions are discharged preferentially [1]

4OH⁻ → 2H₂O + O₂ + 4e⁻ [1]

4OH− → 2H₂O + O₂ + 4e−

⚠ If you missed marks here: If you said chlorine again, remember dilution flips the anode product: with few Cl⁻ ions the normal rule takes over and OH⁻ is discharged, giving oxygen. The half-equation mark usually goes to unbalanced attempts like 4OH⁻ → O₂ + 2H₂O + 2e⁻ – four OH⁻ carry a 4− charge, so four electrons must be released.

1 mark -- oxygen produced anode instead chlorine 1 mark -- dilute low concentration Cl- hydroxide discharged preferentially 1 mark -- equation 4OH- gives 2H2O plus O2 plus 4e-

(i) State the product at the cathode and explain why it differs from the NaCl experiments. [2]

(ii) Explain why the blue colour of the solution gradually fades during prolonged electrolysis. [1]

Model Answer -- 2(c)

Copper metal is deposited at the cathode [1]

Copper is less reactive than hydrogen (below hydrogen in the reactivity series), so Cu²⁺ ions are discharged in preference to H⁺; in NaCl, sodium is above hydrogen so H⁺ is discharged instead [1]

Cu²⁺ ions (which cause the blue colour) are removed from solution and deposited as copper; with inert electrodes, no new Cu²⁺ ions are added to replace them [1]

⚠ If you missed marks here: Writing 'hydrogen' at the cathode is the big trap – copper sits BELOW hydrogen in the reactivity series, so Cu²⁺ is discharged instead of H⁺, the opposite of the NaCl cells. For the fading colour, 'the copper gets used up' only scores if you say Cu²⁺ ions (the source of the blue colour) leave the solution AND the inert platinum electrodes add nothing to replace them.

1 mark -- copper metal deposited cathode 1 mark -- copper less reactive than hydrogen below Cu2+ discharged preference H+ sodium above hydrogen 1 mark -- Cu2+ ions blue colour removed deposited inert electrodes no new Cu2+ added

(d) [2]

Complete the table summarising the products:

Electrolyte	Cathode product	Anode product
Conc. NaCl(aq)	Hydrogen	Chlorine
Dilute NaCl(aq)	Hydrogen	.......
Dilute CuSO₄(aq)	.......	Oxygen

Model Answer -- 2(d)

Dilute NaCl anode product: oxygen [1]

Dilute CuSO₄ cathode product: copper [1]

⚠ If you missed marks here: These marks are usually lost by carrying over the concentrated-solution answers: dilute NaCl gives OXYGEN at the anode (not chlorine, because the Cl⁻ concentration is too low), and dilute CuSO₄ gives COPPER at the cathode (not hydrogen, because copper is below hydrogen in the reactivity series).

1 mark -- dilute NaCl anode oxygen 1 mark -- dilute CuSO4 cathode copper

Question 3 -- Aluminium Extraction: Advanced Analysis

Total: 12 marks

An aluminium smelter in Iceland uses geothermal energy to power the electrolytic extraction of aluminium. The bauxite ore is imported from Australia.

(a) [4]

(i) Write the half-equation at the cathode. [1]

(ii) Write the half-equation at the anode. [1]

(iii) Using both half-equations, show that the ratio of moles of aluminium produced to moles of oxygen produced is 4:3. [2]

Hint: the number of electrons lost at the anode must equal the number gained at the cathode.

Model Answer -- 3(a)

Cathode: Al³⁺ + 3e⁻ → Al [1]

Al³+ + 3e− → Al

Anode: 2O²⁻ → O₂ + 4e⁻ [1]

2O²− → O₂ + 4e−

To balance electrons: multiply cathode by 4 (gives 12e⁻) and anode by 3 (gives 12e⁻) [1]

4Al³⁺ + 12e⁻ → 4Al and 6O²⁻ → 3O₂ + 12e⁻, giving 4Al : 3O₂ [1]

⚠ If you missed marks here: For the 4:3 proof, just quoting the overall equation 2Al₂O₃ → 4Al + 3O₂ scores nothing – the marks are for SHOWING the electron balance: the lowest common multiple of 3 and 4 is 12, so multiply the cathode equation by 4 and the anode by 3 so that 12e⁻ gained equals 12e⁻ lost. Also check you did not write 2 electrons for Al³⁺.

1 mark -- cathode Al3+ plus 3e- gives Al 1 mark -- anode 2O2- gives O2 plus 4e- 1 mark -- multiply cathode 4 anode 3 gives 12 electrons balanced 1 mark -- ratio 4Al to 3O2 demonstrated correctly

(b) [3]

(i) State the formula of cryolite. [1]

(ii) Explain why it is essential to lower the melting point of aluminium oxide. Include specific temperatures in your answer. [2]

Model Answer -- 3(b)

Na₃AlF₆ [1]

Aluminium oxide has a very high melting point of about 2072 °C; dissolving it in cryolite reduces the operating temperature to about 950 °C [1]

This lower temperature greatly reduces the energy (electricity) needed, making the process economically viable / prevents damage to the equipment [1]

⚠ If you missed marks here: 'Cryolite lowers the melting point' alone cannot earn both marks because the question asked for specific temperatures – quote about 2072 °C for pure Al₂O₃ falling to about 950 °C when dissolved in cryolite, then link it to less electricity/energy and lower cost. Also note the oxide DISSOLVES in molten cryolite; it does not react with it.

1 mark -- Na3AlF6 formula cryolite 1 mark -- aluminium oxide melting point 2072 cryolite reduces 950 1 mark -- lower temperature reduces energy electricity economically viable

(i) Write the equation for the reaction between carbon and oxygen at the anode. [1]

(ii) Explain why using inert anodes made of a non-carbon material (e.g. ceramic) would be advantageous, despite the higher initial cost. [2]

Model Answer -- 3(c)

C + O₂ → CO₂ [1]

Inert anodes would not need regular replacement, saving material costs and reducing downtime for the process [1]

No CO₂ would be produced at the anode, reducing greenhouse gas emissions and making the process more environmentally friendly [1]

⚠ If you missed marks here: For the two explain marks you need two DIFFERENT advantages: (1) inert anodes are not burned away by the reaction C + O₂ → CO₂, so no regular replacement or downtime, and (2) no CO₂ greenhouse gas is released. Giving the same idea twice ('cheaper' and 'saves money') only scores once.

1 mark -- equation C plus O2 gives CO2 1 mark -- inert anodes not need replacement saving costs reducing downtime 1 mark -- no CO2 produced reducing greenhouse gas emissions environmentally friendly

(d) [2]

(i) Explain why Iceland is a good location for an aluminium smelter, even though no bauxite is mined there. [1]

(ii) Suggest why recycling aluminium cans saves approximately 95% of the energy compared to extracting new aluminium from ore. [1]

Model Answer -- 3(d)

Iceland has abundant cheap geothermal and hydroelectric energy; the cost of electricity is the main expense in aluminium extraction, so cheap power outweighs the cost of importing bauxite [1]

Recycling only requires remelting the aluminium (relatively low energy), whereas extraction requires electrolysis at high temperatures with huge amounts of electricity, plus mining and purifying the bauxite ore [1]

⚠ If you missed marks here: 'Iceland has geothermal energy' needs the economic link to score: electricity is the BIGGEST cost of aluminium electrolysis, so cheap power outweighs the cost of shipping bauxite in. For recycling, the mark is for the comparison – remelting needs only a little heat energy while extraction needs huge electrical energy for electrolysis; 'recycling is better for the environment' does not answer why 95% of the energy is saved.

1 mark -- Iceland abundant cheap geothermal hydroelectric energy electricity main expense outweighs importing 1 mark -- recycling only remelting low energy extraction electrolysis high temperatures huge electricity mining purifying

Question 4 -- Copper Purification: Quantitative Analysis

Total: 12 marks

A copper refinery in Khetri, Rajasthan purifies blister copper. The anode slabs weigh 350 kg each and contain 99.0% copper, 0.5% silver, 0.3% gold, and 0.2% other impurities.

(a) [3]

(i) State what is used as the anode, the cathode, and the electrolyte in this process. [3]

Model Answer -- 4(a)

Anode: impure / blister copper [1]

Cathode: thin sheet of pure copper [1]

Electrolyte: copper(II) sulfate solution [1]

⚠ If you missed marks here: The classic error is swapping the electrodes: the IMPURE blister copper must be the anode (the anode dissolves away) and the thin PURE copper sheet is the cathode where pure metal builds up. Writing just 'copper' for the electrolyte also loses the mark – it must be copper(II) sulfate SOLUTION, which supplies the mobile Cu²⁺ ions.

1 mark -- anode impure blister copper 1 mark -- cathode thin sheet pure copper 1 mark -- electrolyte copper sulfate solution

(b) [3]

(i) Write the half-equation for oxidation at the anode. [1]

(ii) Write the half-equation for reduction at the cathode. [1]

(iii) Explain why the concentration of Cu²⁺ ions in the electrolyte remains approximately constant. [1]

Model Answer -- 4(b)

Anode (oxidation): Cu → Cu²⁺ + 2e⁻ [1]

Cu → Cu²+ + 2e−

Cathode (reduction): Cu²⁺ + 2e⁻ → Cu [1]

Cu²+ + 2e− → Cu

For every Cu²⁺ ion deposited at the cathode, another Cu²⁺ ion enters the solution from the dissolving anode, maintaining the concentration [1]

⚠ If you missed marks here: Check the direction of each half-equation: oxidation at the anode means copper LOSES electrons (Cu → Cu²⁺ + 2e⁻) – writing both electrodes the same way round is the usual slip. For the constant concentration you must link the two rates: one Cu²⁺ enters solution at the anode for every one removed at the cathode; 'the copper sulfate is not used up' with no reason scores zero.

1 mark -- anode oxidation Cu gives Cu2+ plus 2e- 1 mark -- cathode reduction Cu2+ plus 2e- gives Cu 1 mark -- every Cu2+ deposited cathode another enters from anode maintaining concentration

(i) Calculate the mass of copper in one 350 kg anode slab. [1]

(ii) Calculate the mass of silver in one anode slab. [1]

(iii) Explain what happens to the silver and gold during electrolysis, and why they behave differently from the copper. [2]

Model Answer -- 4(c)

Mass of copper = 99.0% × 350 = 346.5 kg [1]

Mass of silver = 0.5% × 350 = 1.75 kg [1]

Silver and gold do not dissolve at the anode because they are less reactive than copper / below copper in the reactivity series [1]

They fall to the bottom of the cell as anode sludge / anode mud, which can be collected and sold to offset the cost of refining [1]

⚠ If you missed marks here: If your copper mass came out as 34.65 kg you slipped a decimal place – 99.0% of 350 kg is 0.99 × 350 = 346.5 kg; likewise 0.5% means 0.005 × 350 = 1.75 kg (an answer of 175 kg means you multiplied by 0.5 instead of 0.005). For the silver and gold, 'they are impurities' is not enough: say they are LESS reactive than copper, so they do not dissolve at the anode and instead drop to the bottom as anode sludge.

1 mark -- mass copper 346.5 kg 1 mark -- mass silver 1.75 kg 1 mark -- silver gold not dissolve less reactive below copper reactivity series 1 mark -- fall bottom anode sludge mud collected sold offset cost

(d) [2]

Explain the difference in the products obtained when copper(II) sulfate solution is electrolysed using:

(i) copper electrodes [1]

(ii) platinum (inert) electrodes [1]

Model Answer -- 4(d)

With copper electrodes: copper dissolves from the anode and is deposited at the cathode; the electrolyte concentration remains constant; no gas is produced [1]

With platinum electrodes: copper is deposited at the cathode and oxygen is released at the anode; the Cu²⁺ concentration decreases, the solution becomes paler, and eventually becomes acidic (dilute H₂SO₄) [1]

⚠ If you missed marks here: The difference is all at the ANODE: copper electrodes dissolve (Cu → Cu²⁺ + 2e⁻, no gas, blue colour unchanged), but inert platinum cannot dissolve, so OH⁻ is discharged and OXYGEN gas is released while the blue colour fades. Answers that only describe the cathode lose marks because copper is deposited there in BOTH cases.

1 mark -- copper electrodes dissolves deposited concentration constant no gas 1 mark -- platinum inert copper deposited oxygen released concentration decreases paler acidic

Question 5 -- Electrochemical Cells: Predicting Voltages

Total: 10 marks

A student in Singapore sets up several electrochemical cells using different metal pairs in sodium chloride solution. She records the voltage and the direction of electron flow for each.

(a) [4]

Cell	Metals	Voltage / V	Negative electrode
A	Mg and Cu	2.71	Mg
B	Zn and Cu	1.10	Zn
C	Fe and Cu	0.78	Fe
D	Mg and Fe	.......	Mg

(i) State why the more reactive metal is always the negative electrode. [1]

(ii) Predict the approximate voltage for Cell D. Show your reasoning. [2]

(iii) Predict which metal pair from the table would give the lowest voltage if combined. [1]

Model Answer -- 5(a)

The more reactive metal loses electrons more readily / has a greater tendency to form ions, so it pushes electrons into the external circuit, making it the negative terminal [1]

Mg-Cu = 2.71 V and Fe-Cu = 0.78 V; so Mg-Fe should be approximately 2.71 − 0.78 = 1.93 V [1]

Reasoning: the voltage of Mg-Fe equals the difference between the Mg-Cu and Fe-Cu voltages (since Cu is the common reference) [1]

Zn and Fe (closest together in reactivity series); voltage would be approximately 1.10 − 0.78 = 0.32 V [1]

⚠ If you missed marks here: If you predicted about 3.49 V you ADDED 2.71 and 0.78 – because Cu is the common electrode, the Mg–Fe voltage is the DIFFERENCE: 2.71 − 0.78 = 1.93 V. For (i), 'magnesium is more reactive' alone is incomplete: say it loses electrons more readily and pushes them into the external circuit, which is what makes it the negative terminal. The lowest voltage comes from the pair CLOSEST together in reactivity, Zn and Fe (about 0.32 V).

1 mark -- more reactive loses electrons readily forms ions pushes electrons negative terminal 1 mark -- predicted voltage approximately 1.93 V 1 mark -- reasoning difference Mg-Cu minus Fe-Cu common reference 1 mark -- zinc iron lowest voltage closest reactivity series approximately 0.32

(b) [3]

For Cell A (Mg and Cu):

(i) Write the half-equation for the reaction at the magnesium electrode. State whether this is oxidation or reduction. [2]

(ii) Describe what happens at the copper electrode and explain why. [1]

Model Answer -- 5(b)

Mg → Mg²⁺ + 2e⁻ [1]

Mg → Mg²+ + 2e−

This is oxidation (loss of electrons) [1]

At the copper electrode, H⁺ ions from the electrolyte gain the electrons that have flowed through the external circuit, producing hydrogen gas bubbles on the copper surface; copper acts as an inert surface for the reduction reaction [1]

⚠ If you missed marks here: Remember OIL RIG: Mg → Mg²⁺ + 2e⁻ is OXIDATION because electrons are lost – putting the electrons on the left, or labelling it reduction, are the usual errors. At the copper electrode many answers wrongly say 'copper is deposited': nothing happens to the copper itself – H⁺ ions gain the incoming electrons and hydrogen bubbles form on its surface.

1 mark -- equation Mg gives Mg2+ plus 2e- 1 mark -- oxidation loss electrons 1 mark -- copper electrode H+ ions gain electrons hydrogen gas bubbles surface inert

(i) Explain why the voltage of Cell A decreases over time. [1]

(ii) Suggest why these simple cells are not suitable as replacements for commercial batteries. [1]

(iii) State one way commercial batteries differ from simple cells in their construction. [1]

Model Answer -- 5(c)

The magnesium electrode dissolves / is used up over time, and the concentration of the electrolyte changes / products build up, reducing the driving force for the reaction [1]

Simple cells produce a low voltage, have a short lifespan, and cannot deliver a constant voltage / current; they are impractical for most applications [1]

Commercial batteries use carefully chosen chemicals / electrolytes in sealed containers / may be rechargeable / use paste electrolytes instead of liquid / are designed to maintain constant voltage [1]

⚠ If you missed marks here: 'The cell runs out' is too vague for (i) – name the cause: the magnesium electrode dissolves away and the electrolyte changes as products build up, so the driving force falls. For (ii) and (iii) you need a concrete limitation (low voltage, short lifespan, no constant output) and a concrete construction difference (sealed container, paste electrolyte, rechargeable design) – 'commercial batteries are just better' scores nothing.

1 mark -- magnesium dissolves used up concentration changes products build up reducing 1 mark -- simple cells low voltage short lifespan cannot constant impractical 1 mark -- commercial batteries chosen chemicals sealed rechargeable paste electrolytes constant voltage

Question 6 -- Electroplating: Advanced Applications

Total: 12 marks

A manufacturer of surgical instruments in Sheffield uses electrolysis for two different purposes: gold-plating connectors and anodising aluminium handles.

(a) [4]

The manufacturer gold-plates copper connectors using gold(III) chloride solution as the electrolyte.

(i) State what the cathode and anode should be. [2]

(ii) Write the half-equation at the cathode. [1]

(iii) Explain why gold plating is used on electrical connectors despite gold being expensive. [1]

Model Answer -- 6(a)

Cathode: the copper connector (object to be plated) [1]

Anode: pure gold [1]

Cathode: Au³⁺ + 3e⁻ → Au [1]

Au³+ + 3e− → Au

Gold does not corrode / tarnish / oxidise, ensuring reliable electrical contact over time; gold is an excellent conductor [1]

⚠ If you missed marks here: Electroplating rule: the object being plated is ALWAYS the cathode (here the copper connector) and the plating metal (pure gold) is the anode – reversing them is the most common error. For the half-equation check the charge: gold(III) means Au³⁺ + 3e⁻ → Au, not 2e⁻. And 'gold looks nice' scores nothing – the marking point is that gold never corrodes or tarnishes, so the electrical contact stays reliable.

1 mark -- cathode copper connector object plated 1 mark -- anode pure gold 1 mark -- cathode Au3+ plus 3e- gives Au 1 mark -- gold does not corrode tarnish oxidise reliable electrical contact conductor

(b) [4]

Another manufacturer plates steel cutlery with nickel using nickel(II) sulfate solution.

(i) Write the half-equation at the cathode. [1]

(ii) Write the half-equation at the anode (nickel anode). [1]

(iii) Explain why a nickel anode is used rather than a platinum anode. [1]

(iv) A student accidentally connects the steel cutlery as the anode. Predict what would happen. [1]

Model Answer -- 6(b)

Cathode: Ni²⁺ + 2e⁻ → Ni [1]

Ni²+ + 2e− → Ni

Anode: Ni → Ni²⁺ + 2e⁻ [1]

Ni → Ni²+ + 2e−

A nickel anode replaces the Ni²⁺ ions being removed from the electrolyte, keeping the concentration constant; a platinum anode would not dissolve, so the electrolyte would gradually be depleted [1]

The steel cutlery would dissolve / iron from the steel would enter the solution as Fe²⁺ ions; nickel would be deposited on the nickel electrode instead [1]

⚠ If you missed marks here: The two half-equations are mirror images – Ni²⁺ + 2e⁻ → Ni at the cathode, Ni → Ni²⁺ + 2e⁻ at the anode; writing both as reduction is the classic slip. For (iii), 'nickel is cheaper than platinum' misses the chemistry: the nickel anode dissolves to REPLACE the Ni²⁺ ions being deposited, keeping the electrolyte concentration constant. For (iv), with the connections swapped the steel becomes the anode and dissolves instead of being plated.

1 mark -- cathode Ni2+ plus 2e- gives Ni 1 mark -- anode Ni gives Ni2+ plus 2e- 1 mark -- nickel anode replaces Ni2+ ions concentration constant platinum not dissolve depleted 1 mark -- steel dissolve iron Fe2+ ions enter solution nickel deposited wrong electrode

(i) Explain how increasing the current affects the rate of metal deposition and the quality of the coating. [2]

(ii) Explain why the temperature of the electrolyte can affect the quality of the coating. [1]

(iii) Suggest why the surface must be degreased and cleaned before plating. [1]

Model Answer -- 6(c)

Increasing current increases the rate of deposition (more ions discharged per second) [1]

But too high a current produces a rough, powdery, poorly adherent coating because the metal atoms do not have time to arrange into a smooth layer [1]

Higher temperature increases ion mobility / conductivity of the electrolyte, but excessive temperature may cause uneven deposition or decomposition of the electrolyte [1]

Grease and dirt act as insulators / prevent the plating metal from bonding to the surface, causing an uneven or peeling coating [1]

⚠ If you missed marks here: Part (i) carries 2 marks, so 'more current = faster plating' only earns one – you also need the quality trade-off: too high a current gives a rough, powdery, poorly adherent coating because the atoms have no time to settle into a smooth layer. For grease, say it acts as an insulating barrier that stops the metal bonding to the surface – 'it makes it dirty' is not an explanation.

1 mark -- increasing current increases rate deposition more ions discharged per second 1 mark -- too high current rough powdery poorly adherent atoms not time smooth 1 mark -- higher temperature increases ion mobility conductivity excessive uneven decomposition 1 mark -- grease dirt insulators prevent bonding uneven peeling coating

Question 7 -- Fuel Cells: Comparative Analysis

Total: 10 marks

The Indian Space Research Organisation (ISRO) evaluates hydrogen fuel cells for powering remote communication equipment. A fuel cell engineer presents data comparing different power sources.

(a) [3]

In an alkaline hydrogen fuel cell:

(i) Write the overall equation for the reaction. [1]

(ii) Explain why this reaction produces electricity rather than just heat, as in combustion. [1]

(iii) State the type of energy conversion that occurs in a fuel cell. [1]

Model Answer -- 7(a)

2H₂ + O₂ → 2H₂O [1]

In a fuel cell, the hydrogen and oxygen react at separate electrodes; electrons are transferred through an external circuit rather than directly, converting chemical energy into electrical energy; in combustion, the reaction occurs directly and energy is released as heat and light [1]

Chemical energy to electrical energy [1]

⚠ If you missed marks here: Check the balancing: it is 2H₂ + O₂ → 2H₂O – the unbalanced H₂ + O₂ → H₂O loses the mark. For (ii), 'it makes electricity' just restates the question; the marking point is that the reactions happen at SEPARATE electrodes, so the electrons must travel through the external circuit, unlike combustion where transfer is direct and the energy leaves as heat. In (iii) name BOTH energy types: chemical → electrical.

1 mark -- equation 2H2 plus O2 gives 2H2O 1 mark -- separate electrodes electrons external circuit chemical electrical combustion direct heat light 1 mark -- chemical energy electrical energy

(b) [4]

Compare hydrogen fuel cells with each of the following. For each, give one advantage of the fuel cell and one disadvantage.

(i) Petrol generators [2]

(ii) Lithium-ion batteries [2]

Model Answer -- 7(b)

vs Petrol generators -- Advantage: fuel cells produce no pollutant gases (only water), whereas petrol generators produce CO₂, CO, NO_x, and particulates [1]

vs Petrol generators -- Disadvantage: hydrogen is more difficult and expensive to store and transport than petrol / hydrogen refuelling infrastructure is limited [1]

vs Lithium-ion batteries -- Advantage: fuel cells operate continuously as long as hydrogen is supplied, whereas batteries need recharging / fuel cells can be refuelled quickly [1]

vs Lithium-ion batteries -- Disadvantage: fuel cells require a supply of hydrogen gas which must be stored safely / fuel cell systems are currently more expensive / heavier [1]

⚠ If you missed marks here: Marks slip away when the points are not truly comparative – each one must say why the fuel cell beats or loses to THAT power source. 'Fuel cells are clean' needs the contrast: only water is produced, versus CO₂, CO and particulates from petrol. Versus batteries, the advantage is continuous running or fast refuelling rather than waiting to recharge, and the disadvantage must mention hydrogen storage, cost or weight – not just 'batteries are better'.

1 mark -- vs petrol advantage no pollutant gases only water CO2 particulates 1 mark -- vs petrol disadvantage hydrogen difficult expensive store transport infrastructure limited 1 mark -- vs battery advantage continuously supplied recharging refuelled quickly 1 mark -- vs battery disadvantage hydrogen stored safely expensive heavier

Scientists talk about a future "hydrogen economy" where hydrogen replaces fossil fuels.

(i) Describe two ways hydrogen gas can be produced. For each, state whether the method is carbon-neutral. [2]

(ii) Explain why some scientists argue that hydrogen fuel cells are not truly "zero emission" when the full production cycle is considered. [1]

Model Answer -- 7(c)

Method 1: Electrolysis of water using renewable energy (solar, wind, hydroelectric); this is carbon-neutral as no fossil fuels are burned [1]

Method 2: Steam reforming of natural gas (CH₄ + H₂O → CO + 3H₂); this is NOT carbon-neutral because CO₂ is produced as a by-product [1]

Most hydrogen is currently produced from fossil fuels (steam reforming), which generates CO₂; even electrolysis requires electricity, which often comes from fossil fuel power stations; the manufacturing and transport of hydrogen also have carbon footprints [1]

⚠ If you missed marks here: Each production method needs its carbon-neutral verdict attached: electrolysis of water is only carbon-neutral IF the electricity is renewable, while steam reforming of methane (CH₄ + H₂O → CO + 3H₂) is NOT because CO₂ is produced – naming two methods without judging each loses marks. For (ii), the key idea is that the fuel cell itself emits only water, but MAKING the hydrogen releases CO₂.

1 mark -- electrolysis water renewable energy carbon neutral no fossil fuels 1 mark -- steam reforming natural gas not carbon neutral CO2 by-product 1 mark -- most hydrogen fossil fuels steam reforming CO2 electricity fossil fuel power stations manufacturing transport carbon footprints

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