Question
Metals are extracted from their ores by various means.
Aluminium is produced by the electrolysis of alumina (aluminium oxide) dissolved in cryolite.
a. Discuss why different methods of reduction are needed to extract metals.
b(i)Determine the percentage of ionic bonding in alumina using sections 8 and 29 of the data booklet.
b(ii)Write half-equations for the electrolysis of molten alumina using graphite electrodes, deducing the state symbols of the products.
Cathode (negative electrode):
▶️Answer/Explanation
Markscheme
a. ions of more reactive metals are harder to reduce
OR
more reactive metals have more negative electrode potentials
electrolysis is needed/used for most reactive metals
OR
carbon is used to reduce metal oxides of intermediate reactivity/less reactive than carbon OR
heating ore is sufficient for less reactive metals
NOTE: Award [1 max] for “«ease of reduction/extraction” depends on reactivity”.
$57 \ll \% »$
NOTE: Accept any value in the range 52-65\%.
Award [2] for correct final answer.
b(ii)Anode (positive electrode):
$$
2 \mathrm{O}^{2-} \rightarrow 4 \mathrm{e}^{-}+\mathrm{O}_2(\mathrm{~g})
$$
OR
$$
2 \mathrm{O}^{2-}+\mathrm{C} \rightarrow 4 \mathrm{e}^{-}+\mathrm{CO}_2(\mathrm{~g})
$$
NOTE: Award [1 max] for M1 and M2 if correct half-equations are given at the wrong electrodes OR if incorrect reversed half-equations are given at the correct electrodes.
Cathode (negative electrode):
$\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{I})$
$\mathrm{O}_2$ gas $\boldsymbol{A N D}$ Al liquid
NOTE: Only state symbols of products required, which might be written as (g) and (l) in half-equations. Ignore any incorrect or missing state symbols for reactants.
Question
Lithium has many uses.
The emission spectra obtained by ICP-OES for a mixture containing the isotope ${ }^6 \mathrm{Li}(\mathrm{Li}-6)$ and naturally occurring lithium (Li (N)) is shown .
a(i)Identify the type of bonding in lithium hydride, using sections 8 and 29 of the data booklet.
a(iiExplain why lithium is paramagnetic while lithium hydride is diamagnetic by referring to electron configurations.
b(i)Suggest why ICP-OES does not give good quantitative results for distinguishing ${ }^6$ Li from naturally occurring lithium.
b(ii)uggest a better method.
c. Lithium is obtained by electrolysis of molten lithium chloride. Calculate the time, in seconds, taken to deposit $0.694 \mathrm{~g} \mathrm{Li}$ using a current of $2.00 \mathrm{~A}$.
$$
Q(\text { charge })=I(\text { current }) \times t \text { (time })
$$
▶️Answer/Explanation
Markscheme
a(i)ionic [ $\boldsymbol{\swarrow}]$
a(ii)ithium has an unpaired electron $[\boldsymbol{V}]$
all electrons in lithium hydride are paired [ $\boldsymbol{J}$
Note: Award [1 max] for correct electron configurations of Li AND Li’ AND H’ without discussion of pairing.
b(i)emission spectra of both « 6 Li and natural Li» give same colour/produce same «range of» wavelengths
OR
they have same electron transitions/same nuclear charge $[\boldsymbol{U}]$
Note: Accept “the spectra are almost identical”.
b(ii)CP-MS [ $\sim$
Note: Accept “MS/mass spectrometry”.
c. $\mathrm{n} \ll=\frac{\mathrm{m}}{\mathrm{M}_{\mathrm{r}}}=\frac{0.694}{6.94} »=0.100 \ll \mathrm{mol}$ [ $]$
Note: Accept “4820” OR “4825 «S»”.
Award [2] for correct final answer.
Question
The presence of very small amounts of lead in calcium-based antacids can be determined using inductively coupled plasma-mass spectroscopy (ICPMS).
An unknown antacid sample has a lead ion concentration of $0.50 \mu \mathrm{g} \mathrm{dm}^{-3}$.
Chelating agents can be used to treat heavy metal poisoning.
a. State the type of particle present in the plasma formed.
b.i. Calculate the concentration of lead ions in the sample in $\mathrm{mol} \mathrm{dm}^{-3}$.
b.ii Lead ions are toxic and can be precipitated using hydroxide ions.
$$
\mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{Pb}(\mathrm{OH})_2(\mathrm{~s})
$$
Sufficient sodium hydroxide solid is added to the antacid sample to produce a $1.0 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3}$ hydroxide ion solution at $298 \mathrm{~K}$.
Deduce if a precipitate will be formed, using section 32 of the data booklet.
If you did not calculate the concentration of lead ions in (b)(i), use the value of $2.4 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}$, but this is not the correct value.
c. Electrolysis is used to obtain lead from $\mathrm{Pb}^{2+}(\mathrm{aq})$ solution.
Determine the time, in hours, required to produce $0.0500 \mathrm{~mol}$ lead using a current $(I)$ of $1.34 \mathrm{~A}$. Use section 2 of the data booklet and the equation, charge $(Q)=$ current $(I) \times$ time $(t$, in seconds).
d.i. State one feature of a chelating agent.
d.iiAn aqueous lead(II) ion reacts with three ethane-1,2-diamine molecules to form an octahedral chelate ion.
Outline why the chelate ion is more stable than the reactants.
▶️Answer/Explanation
Markscheme
a. positive ions/cations/ $\mathrm{Pb}^{2+}$
OR
free electrons
Accept “ions” OR “charged species/particle”.
b.i. $\left[\mathrm{Pb}^{2+}\right]=0.50 \times 10^{-6} / 5.0 \times 10^{-7} « \mathrm{~g} \mathrm{dm}^{-3} »$
$\left[\mathrm{Pb}^{2+}\right] «=\frac{0.50 \times 10^{-6} \mathrm{~g} \mathrm{dm}^{-3}}{207.20 \mathrm{~g} \mathrm{~mol}^{-1}} »=2.4 \times 10^{-9} « \mathrm{~mol} \mathrm{dm}{ }^{-3} »$
Award [2] for correct final answer.
b.ii.« $K_{\mathrm{sp}}=1.43 \times 10^{-20} »$
ALTERNATIVE 1:
$$
« Q=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2=2.4 \times 10^{-9} \times\left(1.0 \times 10^{-2}\right)^2 »=2.4 \times 10^{-13}
$$
$Q>K_{\mathrm{sp}} \boldsymbol{A N D}$ precipitate will form
OR
$2.4 \times 10^{-13}>1.43 \times 10^{-20}$ AND precipitate will form
ALTERNATIVE 2:
critical $\left[\mathrm{Pb}^{2+}\right]$ for hydroxide solution $«=\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{OH}^{-}\right]^2}=\frac{1.43 \times 10^{-20}}{\left(1.0 \times 10^{-2}\right)^2} »=1.4 \times 10^{-16}$
initial concentration > critical concentration $A N D$ precipitate will form
OR
$2.4 \times 10^{-9}>1.4 \times 10^{-16}$ AND precipitate will form $\boldsymbol{V}$
If value given is used:
ALTERNATIVE 3:
$$
« Q=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2=2.4 \times 10^{-4} \times\left(1.0 \times 10^{-2}\right)^2 »=2.4 \times 10^{-8}
$$
$Q>K_{\mathrm{sp}} A N D$ precipitate will form
OR
$2.4 \times 10^{-8}>1.43 \times 10^{-20} \boldsymbol{A N D}$ precipitate will form
c. «Faraday’s constant, $F=9.65 \times 10^4 \mathrm{C} \mathrm{mol}^{-1}$ and $1 \mathrm{~A}=1 \mathrm{C} \mathrm{s}^{-1}$ »
$\mathrm{Q} \ll=0.0500 \mathrm{~mol} \times 2 \times 96500 \mathrm{C} \mathrm{mol}^{-1} »=9650 « \mathrm{C} »$
$\mathrm{t} «=\frac{\mathrm{Q}}{\mathrm{I}}=\frac{9650 \mathrm{C}}{1.34 \mathrm{C} \mathrm{s}^{-1}} \approx 7200 \mathrm{~s}$ so $\frac{7200 \mathrm{~s}}{60 \times 60 \mathrm{sh}^{-1}} »=2.00$ «hours»
Award [2] for correct final answer.
d.i. Any one of:
two «or more» lone/non-bonding pairs on different atoms
OR
two «or more» atoms/centres that act as Lewis bases
form «at least» two coordination/coordinate bonds
OR
«at least» two atoms can form coordination/coordinate bonds
Reference to “on DIFFERENT atoms” required.
Accept “dative “covalent» bond” for “coordination/coordinate bond”.
d.ii.increase in entropy
OR
$\Delta S>0 / \Delta S$ positive
Accept ” $\Delta G<0$ ” but not ” $\Delta H<0$ “.
Question
Alloys are important substances in industries that use metals.
Describe an alloy.
Explain how alloying can modify the structure and properties of metals.
▶️Answer/Explanation
Markscheme
homogeneous mixture of metals/a metal and non-metal;
alloying element(s) disrupts regular/repeating (metal) lattice;
difficult for one layer to slide over another / atoms smaller than the metal cations can fit into the (holes of) metal lattice disrupting bonding;
can make metal harder/stronger/more corrosion resistant/brittle;
Examiners report
Many candidates did not gain the mark as they omitted the required word homogeneous.
Many candidates did not gain the mark as they omitted the required word homogeneous (a), but they could explain how alloying can modify the structure (b).
Question
Aluminium is extracted by the electrolysis of a molten mixture containing alumina, Al2O3, using graphite electrodes.
Explain why the molten electrolyte also contains cryolite.
State a half-equation for the reaction at the negative electrode (cathode).
Oxygen is produced at the positive electrode (anode). State the name of another gas produced at this electrode.
State two properties of aluminium that make it suitable for use as an overhead electric cable.
Alloys of aluminium with nickel are used to make engine parts. Explain, by referring to the structure of these alloys, why they are less malleable than pure aluminium.
▶️Answer/Explanation
Markscheme
it lowers the operating temperature/melting point (of alumina) / it saves heat/energy / improves conductivity / acts as a solvent;
Do not accept lowers melting point of aluminium.
Do not accept “lowers boiling point”.
\({\text{Al(l}}{{\text{)}}^{3 + }} + {\text{3}}{{\text{e}}^ – } \to {\text{Al (l)}}\);
Ignore state symbols.
Accept e instead of e–.
carbon dioxide / carbon monoxide / fluorine / tetrafluoromethane;
Do not accept formulas since the name is asked for specifically.
high/good (electrical) conductivity and low density;
Do not accept lighter.
Accept malleable/ductile/resistant to (further) corrosion as one property.
Reference to high/good conductivity or low density needed.
in alloy different sized/Ni atoms/ions/particles disrupt regular structure;
stops layers from slipping/sliding / OWTTE;
Do not accept “stop layers moving”.
Accept diagrams if explanation clear.
Examiners report
This was reasonably well answered. However a surprising number of candidates were unable to write the half equation for reaction at the negative electrode in (a) (ii), and few candidates were able to give an explanation of the reduced malleability of the alloys in terms of their structure in b (ii).
This was reasonably well answered. However a surprising number of candidates were unable to write the half equation for reaction at the negative electrode in (a) (ii), and few candidates were able to give an explanation of the reduced malleability of the alloys in terms of their structure in b (ii).
This was reasonably well answered. However a surprising number of candidates were unable to write the half equation for reaction at the negative electrode in (a) (ii), and few candidates were able to give an explanation of the reduced malleability of the alloys in terms of their structure in b (ii).
This was reasonably well answered. However a surprising number of candidates were unable to write the half equation for reaction at the negative electrode in (a) (ii), and few candidates were able to give an explanation of the reduced malleability of the alloys in terms of their structure in b (ii).
This was reasonably well answered. However a surprising number of candidates were unable to write the half equation for reaction at the negative electrode in (a) (ii), and few candidates were able to give an explanation of the reduced malleability of the alloys in terms of their structure in b (ii).
Question
Aluminium and its alloys are widely used in industry.
Aluminium metal is obtained by the electrolysis of alumina dissolved in molten cryolite.
Explain the function of the molten cryolite.
State the half-equations for the reactions that take place at each electrode.
Positive electrode (anode):
Negative electrode (cathode):
Outline two different ways that carbon dioxide may be produced during the production of aluminium.
▶️Answer/Explanation
Markscheme
melting point of the cryolite solution is much lower than the melting point of alumina/Al2O3 / it lowers the melting point of the mixture / cell operates at lower temperature;
Allow lowers melting point or lowers melting point of aluminium oxide.
Do not allow lowers melting point of aluminium.
Positive electrode:
\({\text{2}}{{\text{O}}^{2 – }} \to {{\text{O}}_2} + {\text{4}}{{\text{e}}^ – }/{{\text{O}}^{2 – }} \to \frac{1}{2}{{\text{O}}_2} + {\text{2}}{{\text{e}}^ – }\);
Negative electrode:
\({\text{A}}{{\text{l}}^{3 + }} + {\text{3}}{{\text{e}}^ – } \to {\text{Al}}\);
Award [1] for correct equations but wrong electrodes.
Allow e instead of \({e^ – }\).
use of fossil fuels (to provide energy);
oxidation of the (graphite) positive electrode/anode;
Examiners report
This was either answered very well or very poorly.
Only the best candidates could state half-equations for the electrolysis of aluminium.
Many candidates scored one mark out of two for outlining how carbon dioxide may be produced during aluminium production in (b). C.1.10 states that candidates should know the environmental impacts of aluminium production.
Question
The Industrial Revolution was the result of large-scale extraction of iron from its ore and had significant impact worldwide.
In a blast furnace, a large volume of air is introduced under pressure near the bottom while a mixture of limestone, coke and iron(III) oxide is introduced at the top.
State the equation for the reaction of coke with air in the blast furnace.
The product formed in part (i) reacts with coke to produce carbon monoxide. Explain, giving an equation, why this reaction is important in the extraction of iron.
▶️Answer/Explanation
Markscheme
\({\text{C(s)}} + {{\text{O}}_2}{\text{(g)}} \to {\text{C}}{{\text{O}}_2}{\text{(g)}}\);
Ignore state symbols.
CO acts as a reducing agent / reaction is endothermic/cools (this part of) furnace;
\({\text{C}}{{\text{O}}_2}{\text{(g)}} + {\text{C(s)}} \to {\text{2CO(g)}}/{\text{F}}{{\text{e}}_2}{{\text{O}}_3}{\text{(s)}} + {\text{3CO(g)}} \to {\text{2Fe(l)}} + {\text{3C}}{{\text{O}}_2}{\text{(g)}}\);
Ignore state symbols.
Examiners report
Generally the equation for the reaction of coke with air was given correctly.
The correct equation for the reaction of \({\text{C}}{{\text{O}}_2}\) with coke was sometimes missing
Question
Aluminium is the most abundant metal on Earth and its alloys are widely used.
Describe what is meant by the term alloy.
State the main improvement made to the properties of aluminium when it is alloyed.
▶️Answer/Explanation
Markscheme
homogeneous mixture of metals/metal(s) and non-metal(s);
increase in aluminium’s strength;
Examiners report
Most candidates did not gain the mark for alloys in part (b) because they did not state that the mixture was homogeneous which was needed for the mark.
The majority of candidates knew that alloying aluminium increased its strength in part (c).
Question
The main ore used to produce aluminium by electrolysis is bauxite. Bauxite is mainly aluminium hydroxide, and contains iron(III) oxide and titanium(IV) oxide as impurities.
Explain how pure aluminium oxide is obtained from bauxite.
Explain why sodium hexafluoroaluminate, \({\text{N}}{{\text{a}}_{\text{3}}}{\text{Al}}{{\text{F}}_{\text{6}}}\), (cryolite) is added to the aluminium oxide before electrolysis takes place to produce aluminium.
State the half-equations for the reactions taking place at the positive and negative electrodes during the production of aluminium by electrolysis.
Positive electrode (anode):
Negative electrode (cathode):
Before the introduction of the electrolytic method by Hall and Héroult in the 1880s it was very difficult to obtain aluminium metal from its ores. Suggest one way in which it was achieved.
The worldwide production of aluminium by electrolysis makes a significant impact on global warming. Suggest two different ways in which the process increases the amount of carbon dioxide in the atmosphere.
▶️Answer/Explanation
Markscheme
(bauxite) is reacted with (concentrated) sodium hydroxide/NaOH (solution at high temperature);
forms sodium aluminate / \({\text{Al(OH}}{{\text{)}}_3} + {\text{O}}{{\text{H}}^ – } \to {\text{Al(OH)}}_4^ – \);
Accept both ionic and non-ionic equations and different, correct representations of the aluminate ion (Al(OH )4–, AlO2–).
solution is filtered / insoluble impurities removed (by filtration);
reaction reversed by cooling / diluting solution / adding water;
Accept passing CO2 through the solution.
mixture seeded with alumina crystals;
pure hydroxide precipitated / \({\text{Al(OH)}}_4^ – \to {\text{Al (OH}}{{\text{)}}_3} + {\text{O}}{{\text{H}}^ – }\);
Accept both ionic and non-ionic equations and different, correct representations of the aluminate ion (Al(OH)4–, AlO2–).
(pure) \({\text{Al(OH}}{{\text{)}}_3}\) heated / \({\text{2Al(OH}}{{\text{)}}_3} \to {\text{A}}{{\text{l}}_2}{{\text{O}}_3} + {\text{3}}{{\text{H}}_2}{\text{O}}\);
Award [1 max] for “Alumina is soluble in alkali, but impurities are not” / OWTTE.
Ignore state symbols.
melting point of the cryolite solution is much lower than the melting point of alumina/Al2O3 / it lowers the melting point (of the mixture);
Do not allow lowers melting point of aluminium.
Do not allow lowers required/operating temperature.
Accept improves conductivity of the electrolyte/aluminium oxide.
Positive electrode (anode):
\({\text{2}}{{\text{O}}^{2 – }} \to {{\text{O}}_2} + {\text{4}}{{\text{e}}^ – }/{{\text{O}}^{2 – }} \to \frac{1}{2}{{\text{O}}_2} + {\text{2}}{{\text{e}}^ – }/{\text{C}} + {\text{2}}{{\text{O}}^{2 – }} \to {\text{C}}{{\text{O}}_2} + {\text{4}}{{\text{e}}^ – }\);
Negative electrode (cathode):
\({\text{A}}{{\text{l}}^{3 + }} + {\text{3}}{{\text{e}}^ – } \to {\text{Al}}\);
Allow e instead of e–.
Accept multiples of the correct equations, such as 2Al3+ + 6e– \( \to \) 2Al .
Award [1 max] if correct equations but at wrong electrodes.
Ignore state symbols.
by reduction with a more reactive metal/metal above Al in electrochemical
series/ECS/reactivity series / OWTTE;
Accept equations for displacement reactions of Al2O3 with more reactive metals.
graphite/carbon electrodes converted/oxidized (into CO2);
the fossil fuels used to provide energy/transport (produce CO2);
Examiners report
This question was probably the worst answered question on the whole paper. In the first section many candidates confused the purification process with the electrolytic extraction and answers that scored any marks were rare. Many candidates knew the reasons for the addition of cryolite, but it was unusual to find both electrode equations correct and balanced. Hardly any had the lateral thinking skills to suggest displacement by a more reactive metal as a possible way of obtaining aluminium, but most students knew of at least one way in which aluminium production resulted in the emission of carbon dioxide.
This question was probably the worst answered question on the whole paper. In the first section many candidates confused the purification process with the electrolytic extraction and answers that scored any marks were rare. Many candidates knew the reasons for the addition of cryolite, but it was unusual to find both electrode equations correct and balanced. Hardly any had the lateral thinking skills to suggest displacement by a more reactive metal as a possible way of obtaining aluminium, but most students knew of at least one way in which aluminium production resulted in the emission of carbon dioxide.
This question was probably the worst answered question on the whole paper. In the first section many candidates confused the purification process with the electrolytic extraction and answers that scored any marks were rare. Many candidates knew the reasons for the addition of cryolite, but it was unusual to find both electrode equations correct and balanced. Hardly any had the lateral thinking skills to suggest displacement by a more reactive metal as a possible way of obtaining aluminium, but most students knew of at least one way in which aluminium production resulted in the emission of carbon dioxide.
This question was probably the worst answered question on the whole paper. In the first section many candidates confused the purification process with the electrolytic extraction and answers that scored any marks were rare. Many candidates knew the reasons for the addition of cryolite, but it was unusual to find both electrode equations correct and balanced. Hardly any had the lateral thinking skills to suggest displacement by a more reactive metal as a possible way of obtaining aluminium, but most students knew of at least one way in which aluminium production resulted in the emission of carbon dioxide.
This question was probably the worst answered question on the whole paper. In the first section many candidates confused the purification process with the electrolytic extraction and answers that scored any marks were rare. Many candidates knew the reasons for the addition of cryolite, but it was unusual to find both electrode equations correct and balanced. Hardly any had the lateral thinking skills to suggest displacement by a more reactive metal as a possible way of obtaining aluminium, but most students knew of at least one way in which aluminium production resulted in the emission of carbon dioxide.
Question
Iron ore can be reduced in a blast furnace.
The properties of a metal can be altered by alloying or heat treatment.
Explain why alloying can modify the structure and properties of a metal.
▶️Answer/Explanation
Markscheme
(alloying element(s)) atoms/ions have different size;
Allow suitable diagram.
disrupts regular/repeating (metal) lattice;
difficult for one layer to slide over another / added atoms/ions smaller than metal atoms/ions can fit into the (holes of) metal lattice disrupting bonding;
If “particles” is penalised in M1, allow “particles” in M3.
Do not award mark for different or unique properties of alloys.
Examiners report
The common errors in (a) were to give a formula rather than the name of an ore and to add it at R. Whilst part (ii) was generally answered correctly it was slightly alarming to see oxygen given from time to time. Few seemed to know the equation of the reaction primarily responsible for the temperature of 1900 °C. Slag was usually correctly identified in (c) but the equations given seldom started from the raw material, \({\text{CaC}}{{\text{O}}_{\text{3}}}\). Part (d) was often answered successfully whilst candidates were generally unable to earn both marks in (e) (i). Most understood the effect of tempering on steel.
Question
Aluminium is an important metal to modern society.
Aluminium is often used to produce lightweight alloys for use in the aerospace industry.
(i) Describe the production of aluminium from its purified ore. Explain the role of cryolite and deduce the equations for the reactions occurring at the two electrodes.
Production of aluminium:
Role of cryolite:
Negative electrode (cathode):
Positive electrode (anode):
(ii) Outline why aluminium was not available in large quantities before 1900.
(i) State one advantage of using an alloy rather than the pure metal.
(ii) Outline why the range of metals alloyed with aluminium for this use is very limited.
Suggest one possible environmental impact that can result from the large-scale production of aluminium.
▶️Answer/Explanation
Markscheme
(i) Production of aluminium:
electrolysis of molten alumina/aluminium oxide/\({\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\);
Role of cryolite:
(molten) cryolite (saves money due to) lower operating temperature / solvent with a lower melting point (than aluminium oxide);
Accept lowers the melting point of aluminium oxide.
Negative electrode (cathode):
\({\text{A}}{{\text{l}}^{3 + }}{\text{(l)}} + {\text{3}}{{\text{e}}^ – } \to {\text{Al(l)}}\);
Positive electrode (anode):
\({\text{2}}{{\text{O}}^{2 – }}{\text{(l)}} \to {{\text{O}}_2}{\text{(g)}} + {\text{4}}{{\text{e}}^ – }\);
Allow e for e–.
Ignore state symbols.
Penalize use of equilibrium sign once only.
Award [1 max] for M3 and M4 if correct equations are given but at the wrong electrodes.
(ii) no electricity / electricity not widely available before 1900;
(i) to control/improve properties / alloys are stronger/more durable/less reactive/less malleable/less ductile than pure metals;
(ii) only a small number of metals have low densities / many low density metals are too reactive / alloys need presence of other metallic atoms of slightly different size (few metals like this);
(purification of ore produces) waste \({\text{F}}{{\text{e}}_2}{{\text{O}}_3}\)/iron(III) oxide/red mud;
carbon dioxide/ \({\text{C}}{{\text{O}}_2}\) from burning electrodes;
environmental impacts of power generation;
aluminium production a significant contributor to global warming;
mining the ore damages the landscape/local ecology;
generation of fluorides/polyfluorinated carbons/fluorine containing waste products;
Examiners report
(i) Many candidates understood the principles behind the production of aluminium and the role of the cryolite, although not always managing to state the principles accurately to be able to score the first two marks, but many candidates showed weakness in writing the correct half-equations at the correct electrodes.
(ii) About half of the candidates stated the there was no electricity available at that time.
(i) More than half of the candidates were able to give an advantage of using an alloy over using the pure metal. A few candidates stated the misconception that alloys were less brittle than pure metals.
(ii) Some candidates offered sensible suggestions mainly focusing on the importance of finding a low density metal.
Quite well answered. Not many candidates discussed the impact of the purification of the bauxite or its mining. Candidates focused on the impact of the generation of large amounts of electricity and global warming.
Question
Modern society is very dependent on electrical power for portable devices.
Two common rechargeable batteries are lead-acid and nickel-cadmium (NiCad) batteries.
(i) State equations for the reactions that occur at each electrode in a lead-acid battery when it delivers a current.
Positive electrode (cathode):
Negative electrode (anode):
(ii) State equations for the reactions that occur at each electrode in a nickel-cadmium (NiCad) battery when it delivers a current.
Positive electrode (cathode):
Negative electrode (anode):
Another source of power for portable devices is the fuel cell. Compare fuel cells with lead-acid rechargeable batteries, stating one similarity and two differences.
Similarity:
Differences:
▶️Answer/Explanation
Markscheme
(i) Lead-acid:
Positive electrode (cathode):
\({\text{Pb}}{{\text{O}}_2}{\text{(s)}} + {\text{SO}}_4^{2 – }{\text{(aq)}} + {\text{4}}{{\text{H}}^ + }{\text{(aq)}} + {\text{2}}{{\text{e}}^ – } \to {\text{PbS}}{{\text{O}}_4}{\text{(s)}} + {\text{2}}{{\text{H}}_2}{\text{O(l)}}\) /
\({\text{Pb}}{{\text{O}}_2}{\text{(s)}} + {\text{HSO}}_4^ – {\text{(aq)}} + {\text{3}}{{\text{H}}^ + }{\text{(aq)}} + {\text{2}}{{\text{e}}^ – } \to {\text{PbS}}{{\text{O}}_4}{\text{(s)}} + {\text{2}}{{\text{H}}_2}{\text{O(l)}}\);
Negative electrode (anode):
\({\text{Pb(s)}} + {\text{SO}}_4^{2 – }{\text{(aq)}} \to {\text{PbS}}{{\text{O}}_4}{\text{(s)}} + {\text{2}}{{\text{e}}^ – }\) /
\({\text{Pb(s)}} + {\text{HSO}}_4^ – {\text{(aq)}} \to {\text{PbS}}{{\text{O}}_4}{\text{(s)}} + {\text{2}}{{\text{e}}^ – } + {{\text{H}}^ + }{\text{(aq)}}\);
Allow e for e– throughout.
Ignore state symbols.
Award [1 max] if correct equations are given but at the wrong electrodes.
(ii) NiCad:
Positive electrode (cathode):
\({\text{NiO(OH)(s)}} + {{\text{H}}_2}{\text{O(l)}} + {{\text{e}}^ – } \to {\text{Ni(OH}}{{\text{)}}_2}{\text{(s)}} + {\text{O}}{{\text{H}}^ – }{\text{(aq)}}\);
Negative electrode (anode):
\({\text{Cd(s)}} + {\text{2O}}{{\text{H}}^ – }{\text{(aq)}} \to {\text{Cd(OH}}{{\text{)}}_2}{\text{(s)}} + {\text{2}}{{\text{e}}^ – }\);
Allow e for e– throughout.
Ignore state symbols.
Award [1 max] if correct equations are given but at the wrong electrodes.
Similarity:
(both) convert chemical energy to electrical energy / (both are) voltaic cells;
Differences:
Award [2 max] for any two.
rechargeable batteries employ reversible reactions while fuel cells have irreversible reactions;
fuel cells work non-stop while rechargeable batteries take time to recharge;
fuel cells need a constant supply of reactants/fuel while rechargeable batteries do not need any other substances;
fuel cells convert energy and rechargeable batteries store energy;
fuel cell products must be constantly removed but not for rechargeable batteries;
fuel cells are less polluting/more expensive/weigh less/last longer (than lead-acid rechargeable batteries);
fuel cells have inert/Pt electrodes/components while lead-acid rechargeable batteries have active/non-inert/Pb and \({\text{Pb}}{{\text{O}}_{\text{2}}}\) electrodes;
fuel cells run at higher temperatures than rechargeable batteries;
fuel cells are less portable than rechargeable batteries / fuel cells require pumps/cooling systems while rechargeable batteries do not;
Award [2 max] if three valid points (one similarity and two differences) are given without comparison and [1 max] if two valid points are given without comparison.
Examiners report
(i) Poorly answered even by strong candidates. Very few candidates gave the correct equations for the reactions occurring at the electrodes of a lead-acid battery.
(ii) This part-question was also poorly answered with very few candidates scoring one out of the two possible marks.
Many candidates scored at least one mark, but many answers only reflected a shallow understanding of fuel cells and lead-acid batteries. Quite a few candidates are still neglecting to satisfy the demands of the “compare” command term that requires reference to both items in every point of comparison.
Question
Crude oil (petroleum) is initially separated into its components by fractional distillation, but subsequent cracking of the heavier fractions is usually required.
Ethene can be polymerized to form poly(ethene) and, depending on the conditions used, either high-density poly(ethene) (HDPE) or low-density poly(ethene) (LDPE) is formed.
State a balanced equation for the thermal cracking of \({{\text{C}}_{{\text{20}}}}{{\text{H}}_{{\text{42}}}}\) in which octane and ethene are products.
(i) Other than density, state two differences in the physical properties of HDPE and LDPE.
(ii) Outline how the differences in (b)(i) relate to differences in their chemical structure.
It has been said that bitumen and heavy fuel oils are too valuable a resource to use for road surfacing and electricity generation. Comment on this statement.
▶️Answer/Explanation
Markscheme
\({{\text{C}}_{20}}{{\text{H}}_{42}} \to {{\text{C}}_8}{{\text{H}}_{18}} + {{\text{C}}_2}{{\text{H}}_4} + {{\text{C}}_{10}}{{\text{H}}_{20}}/{{\text{C}}_{20}}{{\text{H}}_{42}} \to {{\text{C}}_{\text{8}}}{{\text{H}}_{{\text{18}}}} + {\text{6}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\)
Accept any correctly balanced equation that includes octane and at least one ethene molecule as products.
correct reactants and products;
balanced equation;
M2 can only be scored if M1 is correct.
(i) Award [1] for any two.
HDPE has higher mp;
HDPE is more rigid / less flexible;
HDPE is stronger;
Accept opposite statements for LDPE.
(ii) HDPE has straight chain and LDPE has branched chain / LDPE has more branched chains;
more valuable for (cracking to provide) chemical precursors/petrochemicals / may be cracked to produce same substances now obtained from lighter fractions / OWTTE;
Examiners report
Only about a third of the candidates were able to score both marks by giving a balanced equation producing octane and ethene. Others scored one mark as they failed to balance the equation.
(i) Less than half the candidates gave two physical properties that differed between LDPE and HDPE.
(ii) A small number of candidates attributed the difference to the branching in the chains.
The answers were mostly unsatisfactory as they failed to recognize the value of cracking products for the petrochemical industry.
Question
Aluminium is chemically reactive so it has to be extracted by the electrolysis of aluminium oxide dissolved in molten cryolite.
Deduce an equation for the discharge of the ions at each electrode.
Positive electrode (anode):
Negative electrode (cathode):
(i) Outline why aluminium is alloyed with copper and magnesium when used to construct aircraft bodies.
(ii) State two properties of aluminium that make it suitable for use in overhead power cables.
▶️Answer/Explanation
Markscheme
Positive electrode (anode):
\({\text{2}}{{\text{O}}^{2 – }} \to {{\text{O}}_2}{\text{(g)}} + {\text{4}}{{\text{e}}^ – }/{{\text{O}}^{2 – }} \to \frac{1}{2}{{\text{O}}_2}{\text{(g)}} + {\text{2}}{{\text{e}}^ – }/{\text{2}}{{\text{O}}^{2 – }} – {\text{4}}{{\text{e}}^ – } \to {{\text{O}}_2}{\text{(g)}}/\)
\({{\text{O}}^{2 – }} – {\text{2}}{{\text{e}}^ – } \to \frac{1}{2}{{\text{O}}_2}{\text{(g)}}\);
Allow C(s) + 2O2– \( \to \) CO2(g) + 4e–.
Negative electrode (cathode):
\({\text{A}}{{\text{l}}^{3 + }} + {\text{3}}{{\text{e}}^ – } \to {\text{Al(l)}}\);
Accept e instead of e–.
Ignore state symbols.
If correct equations shown at wrong electrodes, award [1 max].
(i) harder/stronger (than pure aluminium);
(ii) Award [1] for any two of:
good conductor of electricity;
resists corrosion;
Do not allow rusting.
low density;
Do not allow lighter/light mass/light weight.
ductile;
Do not allow malleable.
Examiners report
In (a), most were able to write the correct half-equation for the cathode though incorrect states were commonly seen, e.g. (aq). The anode half-equation was not well known.
Both parts of (b) were well done. In (ii), incorrect answers included malleability and light mass.
Question
The large-scale production of iron is important for the industrial development of many countries.
Magnetite, Fe3O4, is a common ore of iron. Calculate the average oxidation state of iron in the compound and comment on your answer.
State the equation for the reduction of this ore to iron with carbon monoxide.
Outline why iron is obtained from its ores using chemical reducing agents but aluminium is obtained using electrolysis.
▶️Answer/Explanation
Markscheme
\( + \frac{8}{3}/ + 2\frac{2}{3}\) ;
Accept +2.7 but not +3.
+2 and +3 / contains two (or more) iron ions with different oxidation states / contains \({\text{F}}{{\text{e}}^{{\text{2}} + }}\) and \({\text{F}}{{\text{e}}^{{\text{3}} + }}\);
Accept II and III oxidation number notation for oxidation states but not 2+ and 3+ unless ions are referred to explicitly.
Accept “contains different iron compounds/FeO and Fe2O3”.
\({\text{F}}{{\text{e}}_3}{{\text{O}}_4}{\text{(s)}} + {\text{4CO(g)}} \to {\text{3Fe(l)}} + {\text{4C}}{{\text{O}}_2}{\text{(g)}}\);
Accept “Fe3O4(s) + CO(g) \( \to \) 3FeO(s) + CO2(g) and
FeO(s) + CO(g) \( \to \) Fe(l ) + CO2(g)”.
Ignore state symbols.
Al is more reactive than Fe / Al is higher than Fe in the reactivity series / Al is a stronger reducing agent / it is harder to reduce aluminium ores compared to iron ores / \({\text{F}}{{\text{e}}^{3 + }}/{\text{F}}{{\text{e}}^{2 + }}\) is a stronger oxidizing agent / \({\text{A}}{{\text{l}}^{3 + }}\) has a very negative E° value;
Examiners report
This was a challenging question but some candidates managed to calculate the average oxidation state of iron as +8/3, and very few were able to deduce that both +2 and +3 oxidation states were present in the ore.
It was disappointing that most candidates could not give a correct equation for the reduction, considering that both reactants and one product are already given in the question.
Some candidates gave the higher reactivity of aluminium as the reason why it could not be extracted by the use of reducing agents.
Question
Iron is extracted from its ore by reduction in a blast furnace.
State an equation for the reaction by which iron (III) oxide, Fe2O3, is reduced to iron in the blast furnace.
▶️Answer/Explanation
Markscheme
\({\text{F}}{{\text{e}}_2}{{\text{O}}_3}{\text{(s)}} + {\text{3CO(g)}} \to {\text{2Fe(l)}} + {\text{3C}}{{\text{O}}_2}{\text{(g)}}/{\text{F}}{{\text{e}}_2}{{\text{O}}_3}{\text{(s)}} + {\text{3}}{{\text{H}}_2}{\text{(g)}} \to {\text{2Fe(l)}} + {\text{3}}{{\text{H}}_2}{\text{O(g)}}\) /
\({\text{2F}}{{\text{e}}_2}{{\text{O}}_3}{\text{(s)}} + {\text{3C(s)}} \to {\text{4Fe(l)}} + {\text{3C}}{{\text{O}}_2}{\text{(g)}}/{\text{F}}{{\text{e}}_2}{{\text{O}}_3}{\text{(s)}} + {\text{3C(s)}} \to {\text{2Fe(l)}} + {\text{3CO(g)}}\);
Ignore state symbols.
Examiners report
It was pleasing to see a good number of correctly balanced equations here. The precise explanation of how alloying of steel affects physical properties was rarely seen. The correct description of how quenched steel is tempered was rarely seen with the most common answer being heated and rapidly cooled.
Question
Iron may be extracted from an ore containing Fe2O3 in a blast furnace by reaction with coke, limestone and air. Aluminium is obtained by electrolysis of an ore containing Al2O3.
State the overall redox equation when carbon monoxide reduces Fe2O3 to Fe.
Predict the magnetic properties of Fe2O3 and Al2O3 in terms of the electron structure of the metal ion, giving your reasons.
Fe2O3:
Al2O3:
Molten alumina, Al2O3(l), was electrolysed by passing 2.00×106 C through the cell. Calculate the mass of aluminium produced, using sections 2 and 6 of the data booklet.
▶️Answer/Explanation
Markscheme
Fe2O3 (s) + 3CO (g) → 2Fe (l) + 3CO2 (g)
Fe2O3:
paramagnetic
AND
unpaired electrons present «so magnetic moments do not cancel out»
Al2O3:
diamagnetic
AND
no unpaired electrons/all electrons are paired «so magnetic moments cancel out»
Award [1 max] for “Fe2O3 paramagnetic AND Al2O3 diamagnetic”.
Award [1 max] for “Fe2O3 unpaired electrons present AND Al2O3 no unpaired electrons/all electrons are paired”.
Award [1 max] for “Magnetic moments do not cancel out in Fe2O3 but do in Al2O3”.
Unpaired and paired electrons may also be conveyed by orbital diagrams for the respective ions.
\(n\left( {\rm{e}} \right) = \frac{{2.00 \times {{10}^6}}}{{96500}}/20.7 \ll {\rm{mol}} \gg \)
OR
n(Al)=\(\frac{1}{3}\)n(e)/6.91«mol»
m(Al)=«6.91×26.98=»186«g»
Award [2] for correct final answer for any value within the range 186–189 «g».
Question
A student wanted to determine the formula of indium sulfate. She applied an electrical current of 0.300A to an aqueous solution of indium sulfate for 9.00 × 103 s and found that 1.07 g of indium metal deposited on the cathode.
Calculate the charge, in coulombs, passed during the electrolysis.
\(\left( {{\text{current }}I = \frac{{{\text{charge }}Q\,}}{{{\text{time }}t}}} \right)\)
Calculate the amount, in mol, of electrons passed using section 2 of the data booklet.
Calculate the mass of indium deposited by one mole of electrons.
Calculate the number of moles of electrons required to deposit one mole of indium. Relative atomic mass of indium, Ar=114.82.
Deduce the charge on the indium ion and the formula of indium sulfate.
▶️Answer/Explanation
Markscheme
«0.300A × 9.00 × 103 s =» 2.70 × 103 «C»
«mol e− = \(\frac{{2700\,{\text{C}}}}{{96\,500\,{\text{C}}\,{\text{mo}}{{\text{l}}^{ – 1}}}}\) =» 2.80 × 10−2 «mol»
«\(\frac{{1.07{\rm{g}}}}{{0.0280{\rm{mol}}}}\)=» 38.2 «g»
«\(\frac{{114.82\,{\text{g}}}}{{38.2\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ – 1}}}}\) e− =» 3.01/3.00 «mol e−»
In3+ /3+ AND In2(SO4)3
Do not accept “+3/3”
Question
Lanthanum metal may be produced by the electrolysis of molten LaBr3.
State why lanthanum cannot be produced by reducing its oxide with carbon.
Calculate the current (I), in A, required to produce 1.00 kg of lanthanum metal per hour. Use the formula \(Q(C) = I(A) \times t(s)\) and sections 2 and 6 of the data booklet.
▶️Answer/Explanation
Markscheme
too high/higher than carbon in the reactivity series
OR
carbon/C is a weaker reducing agent than lanthanum/La
Accept “lanthanum is more reactive than carbon”.
Accept “lanthanum is a weaker oxidizing agent than carbon”.
Accept converse arguments.
[1 mark]
amount of La «\( = \frac{{1000{\text{ g}}}}{{138.91{\text{ g}}\,{\text{mo}}{{\text{l}}^{ – 1}}}}\)» = 7.20 «mol»
Q «= 7.20 mol x 3 x 96\(\,\)500 C\(\,\)mol–1» = 2.08 x 106 «C»
I «\(\frac{{2.08 \times {{10}^6}{\text{ C}}}}{{60 \times 60{\text{ s}}}}\)» = 579 «A»
Award [3] for “578 «A»” (from premature rounding) or “579 «A»”.
[3 marks]
Question
Antimony oxide is widely used as a homogeneous catalyst in the reaction of benzene-1,4-dicarboxylic acid with ethane-1,2-diol in the production of polyethylene terephthalate (PETE) shown below.
Catalysts reduce the activation energy. Outline how homogeneous catalysts are involved in the reaction mechanism.
Suggest why it is important to know how catalysts function.
Antimony and its compounds are toxic, so it is important to check that the catalyst is removed from the final product. One technique to detect antimony is Inductively Coupled Plasma Mass Spectroscopy (ICP-MS).
Outline the nature of the plasma state and how it is produced in ICP-MS.
▶️Answer/Explanation
Markscheme
combine with reactants to form a «temporary» activated complex/intermediate
OR
consumed in one reaction/step AND regenerated in a later reaction/step
[1 mark]
can modify/improve the catalyst/reaction «by making logical predictions»
OR
science relies on models to understand physical reality
Accept other reasonable, relevant answers.
Accept “to predict/select
[1 mark]
electrons AND positive ions «in gaseous state»
high frequency/alternating current passed through argon
OR
«oscillating» electromagnetic/magnetic field
OR
high frequency radio waves
Accept “gas” instead of “argon”.
[2 marks]
Question
Rhodium and palladium are often used together in catalytic converters. Rhodium is a good reduction catalyst whereas palladium is a good oxidation catalyst.
In a catalytic converter, carbon monoxide is converted to carbon dioxide. Outline the process for this conversion referring to the metal used.
Nickel is also used as a catalyst. It is processed from an ore until nickel(II) chloride solution is obtained. Identify one metal, using sections 24 and 25 of the data booklet, which will not react with water and can be used to extract nickel from the solution.
Deduce the redox equation for the reaction of nickel(II) chloride solution with the metal identified in (b)(i).
Another method of obtaining nickel is by electrolysis of a nickel(II) chloride solution. Calculate the mass of nickel, in g, obtained by passing a current of 2.50 A through the solution for exactly 1 hour. Charge (Q) = current (I) × time (t).
▶️Answer/Explanation
Markscheme
carbon monoxide/CO adsorbs onto palladium/Pd
bonds stretched/weakened/broken
OR
«new» bonds formed
OR
activation energy/Ea «barrier» lowered «in both forward and reverse reactions»
products/CO2 desorb «from catalyst surface»
[3 marks]
Fe/iron
OR
Zn/zinc
OR
Co/cobalt
OR
Cd/cadmium
OR
Cr/chromium
Accept “Mn/manganese”.
[1 mark]
Ni2+(aq) + Fe(s) → Ni(s) + Fe2+(aq)
OR
Ni2+(aq) + Zn(s) → Ni(s) + Zn2+(aq)
OR
Ni2+(aq) + Co(s) → Ni(s) + Co2+(aq)
OR
Ni2+(aq) + Cd(s) → Ni(s) + Cd2+(aq)
OR
Ni2+(aq) + Cr(s) → Ni(s) + Cr2+(aq)
Accept “3Ni2+(aq) + 2Cr(s) → 3Ni(s) + 2Cr3+(aq)”.
Do not penalize similar equations involving formation of Fe3+(aq), Mn2+(aq) OR Co3+(aq).
Ignore Cl− ions.
Accept correctly balanced non-ionic equations eg, “NiCl2(aq) + Zn(s) → Ni(s) + ZnCl2(aq)” etc.
Do not allow ECF from (b)(i).
[2 mark]
\(n{\text{(}}{{\text{e}}^ – }{\text{)}}\) «\( = \frac{{2.50{\text{ A}} \times 3600{\text{ s}}}}{{96500{\text{ C}}\,{\text{mo}}{{\text{l}}^{ – 1}}}}\)» = 0.09326 «mol»
OR
\(n{\text{(Ni)}}\) «\( = \frac{{0.09326{\text{ mol}}}}{2}\)» = 0.04663 «mol»
\(m{\text{(Ni)}}\) «= 0.04663 mol x 58.69 g\(\,\)mol–1» = 2.74 «g»
Award [2] for correct final answer.
[2 marks]
Question
Liquid Crystal on Silicon, LCoS, uses liquid crystals to control pixel brightness. The degree of rotation of plane polarized light is controlled by the voltage received from the silicon chip.
Two important properties of a liquid crystal molecule are being a polar molecule and having a long alkyl chain. Explain why these are essential components of a liquid crystal molecule.
Metal impurities during the production of LCoS can be analysed using ICP-MS. Each metal has a detection limit below which the uncertainty of data is too high to be valid. Suggest one factor which might influence a detection limit in ICP-MS/ICP-OES.
▶️Answer/Explanation
Markscheme
Polar molecule:
«orientation of molecule» influenced by electric field/«applied» voltage/«applied» potential «difference»/«applied» current
OR
can be switched on and off
Long alkyl chain:
prevent close packing of molecules
OR
molecules can align
OR
reduces the melting point of the liquid crystal/LC «phase making liquid at room temperature»
Accept “makes molecule rod-shaped” for M2.
[2 marks]
inability to replicate calibrations below certain levels
OR
variation in methodology
OR
variation between machines calibrated with the same samples
OR
variation in plasma torches
OR
different detection limits for MS AND OES
OR
interference from solvents/other chemicals
OR
inability to produce pure standards
OR
chance that low signal AND blank are same
[1 mark]
Question
It is wise to fill dental cavities before irreversible tooth decay sets in. An amalgam (alloy of mercury, silver, and other metals) is often used although many prefer a white composite material.
Outline the composition of an alloy and a composite.
Outline why an alloy is usually harder than its components by referring to its structure.
At present, composite fillings are more expensive than amalgam fillings.
Suggest why a patient might choose a composite filling.
Explain how Inductively Coupled Plasma (ICP) Spectroscopy could be used to determine the concentration of mercury in a sample of dental filling.
▶️Answer/Explanation
Markscheme
Alloy:
mixture of metal with other metals/non-metals
OR
mixture of elements that retains the properties of a metal
Composite:
reinforcing phase embedded in matrix phase
Award [1 max] for implying “composites only have heterogeneous/nonhomogeneous compositions”.
[Max 2 Marks]
difference in ionic/atomic radius prevents layers sliding over each other
Accept “difference in diameter/packing of cations prevents layers sliding over each other”.
concern about Hg poisoning
OR
«composite» is white «so looks more like tooth»
OR
galvanic response potential exists
OR
local allergic potential
OR
less damage/destruction of healthy tooth tissue
OR
long term corrosion requires replacement
OR
gradual darkening of tooth
Accept other correct responses.
Any three of:
sample injected into argon «plasma»
atoms «of sample» are excited/ionised
OR
electrons are promoted
electrons drop back/recombine with ions AND emit photons of characteristic energies/wavelengths/frequencies
total number of photons is proportional to concentration of element
actual concentration found from calibration/standard curve
Accept “graph/plot” for “curve”.
[Max 3 Marks]
Question
Both HDPE (high density polyethene) and LDPE (low density polyethene) are produced by the polymerization of ethene.
Both of these are thermoplastic polymers. Outline what this term means.
Compare and contrast the structures of HDPE and LDPE.
State one way in which a physical property of HDPE, other than density, differs from that of LDPE as a result of this structural difference.
The production of HDPE involves the use of homogeneous catalysts. Outline how homogeneous catalysts reduce the activation energy of reactions.
Trace amounts of metal from the catalysts used in the production of HDPE sometimes remain in the product. State a technique that could be used to measure the concentration of the metal.
Suggest two of the major obstacles, other than collection and economic factors, which have to be overcome in plastic recycling.
Suggest why there are so many different ways in which plastics can be classified. HDPE can, for example, be categorized thermoplastic, an addition polymer, having Resin Identification Code (RIC) 2, etc.
▶️Answer/Explanation
Markscheme
soften/melt when heated
OR
can be melted and moulded
Accept “low melting point” OR “can be moulded when heated”.
[1 mark]
both have «long» hydrocarbon chains
OR
both have chains comprising CH2 units
HDPE has little/no branching AND LDPE has «more» branching
Accept “CH2–CH2 units”.
Accept “HDPE more crystalline”.
[2 marks]
HDPE is more rigid/less flexible
OR
HDPE has a higher melting point
OR
HDPE has greater «tensile» strength
Accept “HDPE has lower ductility”.
[1 mark]
form «temporary» activated complexes/reaction intermediates
Accept “consumed in one reaction/step AND regenerated in a later reaction/step”.
Accept “provides alternative mechanism”.
[1 mark]
inductively coupled plasma/ICP spectroscopy using mass spectroscopy/mass spectrometry/MS/ICP-MS
OR
inductively coupled plasma/ICP spectroscopy using optical emission spectroscopy/OES/ICP-OES
Accept “atomic absorption/aa spectroscopy” or “MS/massspectroscopy/mass spectrometry”.
[1 mark]
Any two of:
many types «of plastics» exist
OR
«plastics» require sorting «by type»
«plastics» need to be separated from non-plastic materials
OR
«often» composites/moulded on/bound to non-plastic/other components
Accept other valid factors such as thermal decomposition of some plastics, production of toxic fumes, etc.
[2 marks]
«different classifications are appropriate for» different properties/applications/purposes
[1 mark]
Question
Aluminium is produced by the electrolysis of a molten electrolyte containing bauxite.
Determine the mass, in g, of aluminium produced by the passage of a charge of 1.296 × 1013 C. Use sections 2 and 6 of the data booklet.
▶️Answer/Explanation
Markscheme
ratio of electrons : aluminium ions = 3 : 1
amount Al «\(\frac{{1.296 \times {{10}^{13}}{\text{ C}}}}{{96500{\text{ C mo}}{{\text{l}}^{ – 1}} \times 3}}\) » = 4.48 × 107 «mol»
mass Al «= 4.48 × 107 mol × 26.98 g mol–1» = 1.21 × 109 «g»
Award [3] for correct final answer.
[3 marks]
Examiners report
Question
Inductively Coupled Plasma (ICP) used with Mass Spectrometry (MS) or Optical Emission Spectrometry (OES) can be used to identify and quantify elements in a sample.
The following graphs represent data collected by ICP-OES on trace amounts of vanadium in oil.
Graph 1: Calibration graph and signal for 10 μg kg−1 of vanadium in oil
Graph 2: Calibration of vanadium in μg kg−1
[Source: © Agilent Technologies, Inc.1998. Reproduced with Permission, Courtesy of Agilent Technologies, Inc.]
ICP-OES/MS can be used to analyse alloys and composites. Distinguish between alloys and composites.
ICP-MS is a reference mode for analysis. The following correlation graphs between ICP-OES and ICP-MS were produced for yttrium and nickel.
Each y-axis shows concentrations calculated by ICP-OES; each x-axis shows concentrations for the same sample as found by ICP-MS.
The line in each graph is y = x.
Discuss the effectiveness of ICP-OES for yttrium and nickel.
Identify the purpose of each graph.
Calculate, to four significant figures, the concentration, in μg kg−1, of vanadium in oil giving a signal intensity of 14 950.
Vanadium(V) oxide is used as the catalyst in the conversion of sulfur dioxide to sulfur trioxide.
SO2(g) + V2O5(s) → SO3(g) + 2VO2(s)
\(\frac{1}{2}\)O2(g) + 2VO2(s) → V2O5(s)
Outline how vanadium(V) oxide acts as a catalyst.
▶️Answer/Explanation
Markscheme
Alloy:
mixture of metal with other metals/non-metals
OR
mixture of elements that retains the properties of a metal
Composite:
reinforcing phase embedded in matrix phase
Award [1 max] for implying “composites only have heterogeneous/nonhomogeneous compositions”.
[2 marks]
effective for yttrium «but less/not for nickel»
points on nickel graph do not lie on «y = x» line
OR
cannot be used for low concentrations of nickel
OR
concentration of nickel is lower than recorded value
Accept “ICP-OES is more accurate for lower yttrium concentrations than higher concentrations” for M1.
Accept [Ni] and [Y] for concentrations of nickel and yttrium.
Accept “detection limit for yttrium is lower than for nickel” for M2.
Award [1 max] for “more accurate for yttrium at lower concentrations AND nickel at higher concentrations”.
[2 marks]
Graph 1: determines wavelength of maximum absorption/maximum intensity «for vanadium»
Graph 2: determines absorption of known concentrations «at that wavelength»
OR
estimates [V]/concentration in a sample using «the signal» intensity
Do not accept just “determines maximum wavelength/λmax” for M1.
Do not accept “calibration curve” for M2.
[2 marks]
«14 950 = 392.19x + 147.62»
x = 37.74 «μg kg–1»
Answer must be given to four significant figures.
Do not accept values obtained directly from the graph.
[1 mark]
vanadium reduced in first reaction AND oxidized in second reaction
OR
V2O5 oxidizes SO2 in first reaction AND VO2 reduces O2 in second reaction
OR
vanadium returns to original oxidation state «after reaction»
provides an alternative reaction pathway/mechanism «with a lower activation energy» ✔
Do not accept “reactants adsorb onto surface AND products desorb”.
Accept “oxidation number” for “oxidation state”.
[2 marks]