Determine the oxidation state of vanadium in each of the following species.

Identify, from the table, a non-vanadium species that can reduce VO²⁺(aq) to V³⁺(aq) but no further.

Identify, from the table, a non-vanadium species that could convert ${\text{VO}}_2^{+}\text{(aq)}$ to V²⁺(aq).

Formulate an equation for the reaction between VO²⁺(aq) and V²⁺(aq) in acidic solution to form V³⁺(aq).

Comment on the spontaneity of this reaction by calculating a value for $\mathrm{\Delta }{G}^{\theta }$ using the data given in (b) and in section 1 of the data booklet.

$V_2{O}_5:+5$
$V{O}^{2+}:+4$

Do not penalize incorrect notation twice.

[2 marks]

H₂SO₃(aq)
OR
Pb(s)

[1 mark]

Zn(s)

[1 mark]

$\text{V}{\text{O}}^{2+}(\text{aq)}+{\text{V}}^{2+}(\text{aq)}+\text{2}{\text{H}}^{+}(\text{aq)}\to \text{2}{\text{V}}^{3+}(\text{aq)}+{\text{H}}_{\text{2}}\text{O(l)}$

Accept equilibrium sign.

[1 mark]

$E^{\theta }\ll =+0.34\text{ V}-(-0.26\text{ V})\gg =+0.60\ll \text{V}\gg$

$\mathrm{\Delta }{G}^{\theta }=\ll -nF{E}^{\theta }=-9.65\times {10}^4\text{ C}\text{mo}{\text{l}}^{-1}\times 0.60\text{ J}{\text{C}}^{-1}=\gg -57900\ll \text{J}\text{mo}{\text{l}}^{-1}\gg /-57.9\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg$

spontaneous as $\mathrm{\Delta }{G}^{\theta }$ is negative

Do not award M3 as a stand-alone answer.

Accept “spontaneous” for M3 if answer given for M2 is negative.

Accept “spontaneous as $E^{\theta }$ is positive” for M3.

[3 marks]

Magnesium ions produce no emission or absorption lines in the visible region of the electromagnetic spectrum. Suggest why most magnesium compounds tested in a school laboratory show traces of yellow in the flame.

(i) Explain the convergence of lines in a hydrogen emission spectrum.

(ii) State what can be determined from the frequency of the convergence limit.

Magnesium chloride can be electrolysed.

(i) Deduce the half-equations for the reactions at each electrode when molten magnesium chloride is electrolysed, showing the state symbols of the products. The melting points of magnesium and magnesium chloride are 922K and 987K respectively.

(ii) Identify the type of reaction occurring at the cathode (negative electrode).

(iii) State the products when a very dilute aqueous solution of magnesium chloride is electrolysed.

Standard electrode potentials are measured relative to the standard hydrogen electrode. Describe a standard hydrogen electrode.

A magnesium half-cell, Mg(s)/Mg²⁺(aq), can be connected to a copper half-cell, Cu(s)/Cu²⁺(aq).

(i) Formulate an equation for the spontaneous reaction that occurs when the circuit is completed.

(ii) Determine the standard cell potential, in V, for the cell. Refer to section 24 of the data booklet.

(iii) Predict, giving a reason, the change in cell potential when the concentration of copper ions increases.

contamination with sodium/other «compounds»

i
energy levels are closer together at high energy / high frequency / short wavelength

ii
ionisation energy

i)

Anode (positive electrode):

2Cl^– → Cl₂ (g) + 2e^–

Cathode (negative electrode):

Mg²⁺ + 2e^– → Mg (l)

Penalize missing/incorrect state symbols at Cl₂ and Mg once only.

Award [1 max] if equations are at wrong electrodes.

Accept Mg (g).

ii)

reduction

iii)

Anode (positive electrode):
oxygen/O₂
OR
hydogen ion/proton/H⁺ AND oxygen/O₂
Cathode (negative electrode):
hydrogen/H₂
OR
hydroxide «ion»/OH^– AND hydrogen/H₂

Award [1 max] if correct products given at wrong electrodes.

Any two of:

«inert» Pt electrode
OR
platinum black conductor

1 mol dm^–3 H⁺ (aq)

H₂ (g) at 100 kPa

Accept 1 atm H₂ (g).
Accept 1 bar H₂ (g)
Accept a labelled diagram.
Ignore temperature if it is specified.

i

Mg(s) + Cu²⁺ (aq) → Mg²⁺ (aq) + Cu(s)

ii

«+0.34V – (–2.37V) = +»2.71 «V»

iii

cell potential increases

reaction «in Q4(k)(i)» moves to the right
OR
potential of the copper half-cell increases/becomes more positive

Accept correct answers based on the Nernst equation

Identify one organic functional group that can react with acidified K₂Cr₂O₇(aq).

Corrosion of iron is similar to the processes that occur in a voltaic cell. The initial steps involve the following half-equations:

Fe²⁺(aq) + 2e^– $\rightleftharpoons$ Fe(s)

$\frac{1}{2}$O₂(g) + H₂O(l) + 2e^– $\rightleftharpoons$ 2OH^–(aq)

Calculate E ^θ, in V, for the spontaneous reaction using section 24 of the data booklet.

Calculate the Gibbs free energy, ΔG ^θ, in kJ, which is released by the corrosion of 1 mole of iron. Use section 1 of the data booklet.

Explain why iron forms many different coloured complex ions.

Zinc is used to galvanize iron pipes, forming a protective coating. Outline how this process prevents corrosion of the iron pipes.

hydroxyl/OH
OR
aldehyde/CHO

Accept “hydroxy/alcohol” for “hydroxyl”.

Accept amino/amine/NH₂.

[1 mark]

«E ^θ =» +0.85 «V»

Accept 0.85 V.

[1 mark]

ΔG ^θ «= –nFE ^θ» = –2 «mol e^–» x 96500 «C mol^–1» x 0.85 «V»

«ΔG ^θ =» –164 «kJ»

Accept “«+»164 «kJ»” as question states energy released.

Award [1 max] for “+” or “–” 82 «kJ».

Do not accept answer in J.

[2 marks]

incompletely filled d-orbitals

colour depends upon the energy difference between the split d-orbitals

variable/multiple/different oxidation states

different «nature/identity of» ligands

different number of ligands

[3 marks]

Zn/zinc is a stronger reducing agent than Fe/iron
OR
Zn/zinc is oxidized instead of Fe/iron
OR
Zn/zinc is the sacrificial anode

Accept “Zn is more reactive than Fe”.

Accept “Zn oxide layer limits further corrosion”.

Do not accept “Zn layer limits further corrosion”.

[1 mark]

Outline how a catalyst increases the rate of reaction.

Explain why an increase in temperature increases the rate of reaction.

Discuss, referring to intermolecular forces present, the relative volatility of propanone and propan-2-ol.

The diagram shows an unlabelled voltaic cell for the reaction

${\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Ni}\left(\mathrm{s}\right)\to {\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Pb}\left(\mathrm{s}\right)$

Label the diagram with the species in the equation.

Calculate the standard cell potential, in $\mathrm{V}$, for the cell at $298 \mathrm{K}$. Use section 24 of the data booklet

Calculate the standard free energy change, $∆{\mathrm{G}}^{⦵}$, in $\mathbf{kJ}$, for the cell using sections 1 and 2 of the data booklet.

Suggest a metal that could replace nickel in a new half-cell and reverse the electron flow. Use section 25 of the data booklet.

Describe the bonding in metals.

Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the addition of another metal to nickel.

provides an alternative pathway/mechanism AND lower E_a ✔

Accept description of how catalyst lowers E_a (e.g. “reactants adsorb on surface «of catalyst»”, “reactant bonds weaken «when adsorbed»”).

more/greater proportion of molecules with E E_a ✔

greater frequency/probability/chance of collisions «between the molecules»
OR
more collision per unit of time/second ✔

hydrogen bonding/bonds «and dipole–dipole and London/dispersion forces are present in» propan-2-ol ✔

dipole–dipole «and London/dispersion are present in» propanone ✔

propan-2-ol less volatile AND hydrogen bonding/bonds stronger «than dipole–dipole »
OR
propan-2-ol less volatile AND «sum of all» intermolecular forces stronger ✔

$«-0.13 \mathrm{V}-(-0.26 \mathrm{V})=+»0.13«\mathrm{V}»$ ✔

$«{\mathrm{\Delta G}}^{\mathrm{\Theta }}=-{\mathrm{nFE}}^{\mathrm{\Theta }}=-2\times 96 500\times \frac{0.13}{1000}=»-25 «\mathrm{k} \mathrm{J}»$ ✔

$\mathrm{Bi}/\mathrm{Cu}/\mathrm{Ag}/\mathrm{Pd}/\mathrm{Hg}/\mathrm{Pt}/\mathrm{Au}$ ✔

Accept $Sb$ OR $As$.

electrostatic attraction ✔

between «a lattice of» metal/positive ions/cations AND «a sea of» delocalized electrons ✔

Accept “mobile/free electrons”.

Any of:

malleability/hardness
OR
«tensile» strength/ductility
OR
density
OR
thermal/electrical conductivity
OR
melting point
OR
thermal expansion ✔

Do not accept corrosion/reactivity or any chemical property.

Accept other specific physical properties.

Although fairly well done some candidates did not mention that providing an alternate pathway to the reaction was how the activation energy was lowered and hence did not gain the mark.

Almost all candidates earned at least 1 mark for the effect of temperature on rate. Some missed increase in collision frequency, others the idea that more particles reached the required activation energy.

The average mark was 1.9/3. Almost all candidates could recognize hydrogen bonding in alcohol but many missed the dipole-dipole attraction in propanone. There was also some confusion on the term volatility, with some thinking stronger IMF meant higher volatility.

A surprising number of No Response for a question where candidates simply had to label a diagram with the species in the equation. Some candidates had the idea but did not use the species for electrolytic cell, e.g., Pb(SO₄) instead of Pb²⁺(aq).

80% of candidates could correctly calculate a cell potential by using a reduction table and a balanced redox reaction. 

This was similar to 2f(ii) where many could apply the formula for Gibbs free energy change, ΔG^ө, correctly however some did not get the units correct.

80% could correctly pick a metal to reverse the electron flow, however some candidates thought a more reactive, rather than a less reactive metal than nickel would reverse the electron flow.

Most candidates were aware that metallic bonding involved a "sea of electrons", but were unsure about surrounding what and could not identify that it was electrostatic attraction holding the metal together.

Almost all candidates could correctly identify a physical property of a metal which might be altered when alloying.

Deduce a balanced equation for the overall reaction when the standard nickel and iodine half-cells are connected.

Predict, giving a reason, the direction of movement of electrons when the standard nickel and manganese half-cells are connected.

Calculate the cell potential, in V, when the standard iodine and manganese half-cells are connected.

Identify the best reducing agent in the table above.

State and explain the products of electrolysis of a concentrated aqueous solution of sodium chloride using inert electrodes. Your answer should include half-equations for the reaction at each electrode.

Ni (s) + I₂ (aq) → 2I^– (aq) + Ni²⁺ (aq)

electron movement «in the wire» from Mn(s) to Ni(s)

E^θ «for reduction» of Ni²⁺ is greater/less negative than E^θ «for reduction» of Mn²⁺

OR

Ni²⁺ is stronger oxidizing agent than Mn²⁺

OR

Mn is stronger reducing agent than Ni

«0.54 V – (–1.18 V) = +»1.72 «V»

Do not accept –1.72 V.

Mn «(s)»

Positive electrode (anode):
2Cl^– (aq) → Cl₂ (g) + 2e^–

Cl^– oxidized because higher concentration

OR

electrode potential/E depends on concentration

OR

electrode potential values «of H₂O and Cl^–» are close

Negative electrode (cathode):
2H₂O (l) + 2e^– → H₂ (g) + 2OH^– (aq)

OR

2H⁺ (aq) + 2e^– → H₂ (g)

H₂O/H⁺ reduced because Na⁺ is a weaker oxidizing agent

OR

Na⁺ not reduced to Na in water

OR

H⁺ easier to reduce than Na⁺
OR

H lower in activity series «than Na»

Accept $\rightleftharpoons$.

A student decides to build a voltaic cell consisting of an aluminium electrode, Al (s), a tin electrode, Sn (s), and solutions of aluminium nitrate, Al(NO₃)₃(aq) and tin(II) nitrate, Sn(NO₃)₂(aq).

Electron flow is represented on the diagram.

Label each line in the diagram using section 25 of the data booklet.

Write the equation for the expected overall chemical reaction in (a).

Calculate the cell potential using section 24 of the data booklet.

Calculate the Gibbs free energy change, ΔG^⦵, in kJ, for the cell, using section 1 of the data booklet.

Al/aluminium «electrode» AND aluminium nitrate/Al(NO₃)₃/Al³⁺ on left ✓

Sn/tin «electrode» AND tin«(II)» nitrate/Sn(NO₃)₂/Sn²⁺ on right ✓

salt bridge AND voltmeter/V/lightbulb ✓

Award [1] if M1 and M2 are reversed.

Award [1] for two correctly labelled solutions OR two correctly labelled electrodes for M1 and M2.

Accept a specific salt for “salt bridge”.

Accept other circuit components such as ammeter/A, fan, buzzer, resistor/heating element/R/Ω.

3Sn²⁺(aq) + 2Al (s) → 3Sn (s) + 2Al³⁺(aq)
OR
3Sn(NO₃)₂(aq) + 2Al (s) → 3Sn (s) + 2Al(NO₃)₃(aq) ✓

If half cells are reversed in part-question (a) then the equation must be reversed to award the mark.

Do not penalize equilibrium arrows.

«1.66 + (−0.14) = +»1.52 «V» ✓

Calculation must be consistent with equation given in 3 b.

«ΔG^⦵ = −nFE^⦵ = −6 × 9.65 × 104 × 1.52 =» −880080 «J mol⁻¹»
OR
6 «electrons» ✓

«$\frac{-880080}{1000}$=» −880 «kJ» ✓

Award [1] for “«+»880”.

Award [2] for correct final answer.

Deduce the ratio of Fe²⁺:Fe³⁺ in Fe₃O₄.

State the type of spectroscopy that could be used to determine their relative abundances.

State the number of protons, neutrons and electrons in each species.

Iron has a relatively small specific heat capacity; the temperature of a 50 g sample rises by 44.4°C when it absorbs 1 kJ of heat energy.

Determine the specific heat capacity of iron, in J g⁻¹K⁻¹. Use section 1 of the data booklet.

A voltaic cell is set up between the Fe²⁺(aq) | Fe (s) and Fe³⁺ (aq) | Fe²⁺ (aq) half-cells.

Deduce the equation and the cell potential of the spontaneous reaction. Use section 24 of the data booklet.

The figure shows an apparatus that could be used to electroplate iron with zinc. Label the figure with the required substances.

Outline why, unlike typical transition metals, zinc compounds are not coloured.

Transition metals like iron can form complex ions. Discuss the bonding between transition metals and their ligands in terms of acid-base theory.

1:2 ✔

Accept 2 Fe³⁺: 1 Fe²⁺
Do not accept 2:1 only

mass «spectroscopy»/MS ✔

Award [1 max] for 4 correct values.

specific heat capacity « = $\frac{q}{m\times ∆T}/\frac{1000 \mathrm{J}}{50 \mathrm{g}\times 44 \mathrm{K}}$» = 0.45 «J g⁻¹K⁻¹» ✔

Equation:
2Fe³⁺(aq) + Fe(s) → 3Fe²⁺(aq) ✔

Cell potential:
«+0.77 V − (−0.45 V) = +»1.22 «V» ✔

Do not accept reverse reaction or equilibrium arrow.

Do not accept negative value for M2.

left electrode/anode labelled zinc/Zn AND right electrode/cathode labelled iron/Fe ✔

electrolyte labelled as «aqueous» zinc salt/Zn²⁺ ✔

Accept an inert conductor for the anode.

Accept specific zinc salts such as ZnSO₄.

« Zn²⁺» has a full d-shell
OR
does not form « ions with» an incomplete d-shell ✔

Do not accept “Zn is not a transition metal”.

Do not accept zinc atoms for zinc ions.

ligands donate pairs of electrons to metal ions
OR
forms coordinate covalent/dative bond✔

ligands are Lewis bases
AND
metal «ions» are Lewis acids ✔

Identify the missing component of the cell and its function.

Deduce the half-equations for the reaction at each electrode when current flows.

Annotate the diagram with the location and direction of electron movement when current flows.

Calculate the cell potential, in V, using section 24 of the data booklet.

Determine the loss in mass of one electrode if the mass of the other electrode increases by 0.10 g.

salt bridge

movement of ions

OR

balance charge

Do not accept “to complete circuit” unless ion movement is mentioned for M2.

[2 marks]

Positive electrode (cathode):

Ag⁺(aq) + e^– → Ag(s)

Negative electrode (anode):

Mg(s) → Mg²⁺(aq) + 2e^–

Award [1 max] if correct equations given at wrong electrodes.

[2 marks]

in external wire from left to right

[1 mark]

«E = +0.80 V – (–2.37 V) = +» 3.17 «V»

[1 mark]

«moles of silver $\frac{0.10\text{ g}}{107.87\text{ g mo}{\text{l}}^{-1}}$»

moles of magnesium $=\frac{0.5\times 0.10\ll \text{g}\gg }{107.87\ll \text{g mo}{\text{l}}^{-1}\gg }$

«loss in mass of magnesium $=\frac{24.31\text{ g mol}\times 0.5\times 0.10\text{ g}}{107.87\text{ g mo}{\text{l}}^{-1}}=$» 0.011 «g»

Award [2] for correct final answer.

[2 marks]

In acidic solution, bromate ions, BrO₃⁻(aq), oxidize iodide ions, I⁻(aq).

BrO₃⁻(aq) + 6H⁺(aq) + 6e⁻ $\rightleftharpoons$ Br⁻(aq) + 3H₂O(l)

2I⁻(aq) $\rightleftharpoons$ I₂(s) + 2e⁻

Formulate the equation for the redox reaction.

The change in the free energy for the reaction under standard conditions, ΔG^Θ, is −514 kJ at 298 K.

Determine the value of E^Θ, in V, for the reaction using sections 1 and 2 of the data booklet.

Calculate the standard electrode potential, in V, for the BrO₃⁻/Br⁻ reduction half‑equation using section 24 of the data booklet.

BrO₃^–(aq) + 6H⁺(aq) + 6I^–(aq) $\rightleftharpoons$ Br^–(aq) + 3I₂(s) + 3H₂O(l)

Accept → for $\rightleftharpoons$.

[1 mark]

n = 6

«ΔG^Θ = –nFE^Θ»

«$E^{\mathrm{\Theta }}=-\frac{\mathrm{\Delta }{G}^{\mathrm{\Theta }}}{\text{n}F}=\frac{514\times {10}^3\text{ J mo}{\text{l}}^{-1}}{6\times 9.65\times {10}^4\text{ C mo}{\text{l}}^{-1}}=$» 0.888 «V»

[2 marks]

«E^Θ = E^Θ(BrO₃^–/Br^–) – E^Θ(I₂/I^–)»

«E^Θ(BrO₃^–/Br^–) = E^Θ + E^Θ(I₂/I^–) = 0.888 + 0.54 =» «+» 1.43 «V»

[1 mark]

Hydrogen spectral data give the frequency of 3.28 × 10¹⁵ s⁻¹ for its convergence limit.

Calculate the ionization energy, in J, for a single atom of hydrogen using sections 1 and 2 of the data booklet.

Calculate the wavelength, in m, for the electron transition corresponding to the frequency in (a)(iii) using section 1 of the data booklet.

Deduce any change in the colour of the electrolyte during electrolysis.

Deduce the gas formed at the anode (positive electrode) when graphite is used in place of copper.

Explain why transition metals exhibit variable oxidation states in contrast to alkali metals.

IE «= ΔE = hν = 6.63 × 10^–34 J s × 3.28 × 10¹⁵ s^–1» = 2.17 × 10^–18 «J»

[1 mark]

«$\lambda =\frac{C}{\text{v}}=\frac{3.00\times {10}^8\text{ m}{\text{s}}^{-1}}{3.28\times {10}^{15}{\text{s}}^{-1}}=$» 9.15 × 10^–8 «m»

[1 mark]

no change «in colour»

Do not accept “solution around cathode will become paler and solution around the anode will become darker”.

[1 mark]

oxygen/O₂

Accept “carbon dioxide/CO2”.

[1 mark]

Transition metals:

«contain» d and s orbitals «which are close in energy»

OR

«successive» ionization energies increase gradually

Alkali metals:

second electron removed from «much» lower energy level

OR

removal of second electron requires large increase in ionization energy

[2 marks]

Determine the limiting reactant showing your working.

The mass of copper obtained experimentally was 0.872 g. Calculate the percentage yield of copper.

The reaction was carried out in a calorimeter. The maximum temperature rise of the solution was 7.5 °C.

Calculate the enthalpy change, ΔH, of the reaction, in kJ, assuming that all the heat released was absorbed by the solution. Use sections 1 and 2 of the data booklet.

State another assumption you made in (b)(i).

The only significant uncertainty is in the temperature measurement.

Determine the absolute uncertainty in the calculated value of ΔH if the uncertainty in the temperature rise was ±0.2 °C.

Sketch a graph of the concentration of iron(II) sulfate, FeSO₄, against time as the reaction proceeds.

Outline how the initial rate of reaction can be determined from the graph in part (c)(i).

Explain, using the collision theory, why replacing the iron powder with a piece of iron of the same mass slows down the rate of the reaction.

A student electrolyzed aqueous iron(II) sulfate, FeSO₄ (aq), using platinum electrodes. State half-equations for the reactions at the electrodes, using section 24 of the data booklet.

n_CuSO4 «= 0.0800 dm³ × 0.200 mol dm^–3» = 0.0160 mol AND

n_Fe «$\frac{3.26\text{g}}{55.85\text{g}\text{mo}{\text{l}}^{-1}}$» = 0.0584 mol ✔

CuSO₄ is the limiting reactant ✔

Do not award M2 if mole calculation is not shown.

ALTERNATIVE 1:
«0.0160 mol × 63.55 g mol^–1 =» 1.02 «g»  ✔

«$\frac{0.872\text{g}}{1.02\text{g}}\times 100=$» 85.5 «%»  ✔

ALTERNATIVE 2:
«$\frac{0.872\text{g}}{63.55\text{g}\text{mo}{\text{l}}^{--1}}=$» 0.0137 «mol»  ✔

«$\frac{0.0137\text{mol}}{0.0160\text{mol}}\times 100=$» 85.6 «%»  ✔

Accept answers in the range 85–86 %.

Award [2] for correct final answer.

ALTERNATIVE 1:

q = «80.0 g × 4.18 J g^–1 K^–1 × 7.5 K =» 2.5 × 10³ «J»/2.5 «kJ» ✔

«per mol of CuSO₄ = $\frac{-2.5\text{kJ}}{0.0160\text{mol}}=-1.6\times {10}^2$ kJ mol^–1»

«for the reaction» ΔH = –1.6 × 10² «kJ» ✔

ALTERNATIVE 2:

q = «80.0 g × 4.18 J g^–1 K^–1 × 7.5 K =» 2.5 × 10³ «J»/2.5 «kJ» ✔

«n_Cu = $\frac{0.872}{63.55}$ = 0.0137 mol»

«per mol of CuSO₄ = $\frac{-2.5\text{kJ}}{0.0137\text{mol}}=-1.8\times {10}^2$ kJ mol^–1»

«for the reaction» ΔH = –1.8 × 10² «kJ» ✔

Award [2] for correct final answer.

density «of solution» is 1.00 g cm⁻³

OR

specific heat capacity «of solution» is 4.18 J g⁻¹ K⁻¹/that of «pure» water

OR

reaction goes to completion

OR

iron/CuSO₄ does not react with other substances ✔

The mark for “reaction goes to completion” can only be awarded if 0.0160 mol was used in part (b)(i).

Do not accept “heat loss”.

ALTERNATIVE 1:

«$0.2{}^{\circ }\text{C}\times \frac{100}{7.5{}^{\circ }\text{C}}=$» 3 %/0.03 ✔

«0.03 × 160 kJ» = «±» 5 «kJ» ✔

ALTERNATIVE 2:

«$0.2{}^{\circ }\text{C}\times \frac{100}{7.5{}^{\circ }\text{C}}=$» 3 %/0.03 ✔

«0.03 × 180 kJ» = «±» 5 «kJ» ✔

Accept values in the range 4.1–5.5 «kJ».

Award [2] for correct final answer.

initial concentration is zero AND concentration increases with time ✔

decreasing gradient as reaction proceeds ✔

«draw a» tangent to the curve at time = 0 ✔

«rate equals» gradient/slope «of the tangent» ✔

Accept suitable diagram.

piece has smaller surface area ✔

lower frequency of collisions

OR

fewer collisions per second/unit time ✔

Accept “chance/probability” instead of “frequency”.

Do not accept just “fewer collisions”.

Anode (positive electrode):

2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4e^– ✔

Cathode (negative electrode):

2H₂O (l) + 2e^– → H₂ (g) + 2OH^– (aq)
OR
2H⁺ (aq) + 2e^– → H₂ (g) ✔

Accept “4OH^– (aq) → O₂ (g) + 2H₂O (l)  + 4e^–” OR “Fe²⁺ (aq) → Fe³⁺ (aq) + e^–” for M1.

Accept “Fe²⁺ (aq) + 2e^– → Fe (s)” OR “SO₄^2- (aq) 4H⁺ (aq) + 2e^– → 2H₂SO₃(aq) + H₂O (l)”
for M2.

The standard electrode potential of zinc can be measured using a standard hydrogen electrode (SHE).

Draw and annotate the diagram to show the complete apparatus required to measure the standard electrode potential of zinc.

H₂(g) entering at «298 K and» 100 kPa ✔

platinum electrode on left ✔

voltmeter connecting electrodes AND salt bridge connecting electrolytes ✔

1 mol dm^–3 H⁺ on the left AND 1 mol dm^–3Zn²⁺ on the right ✔

Voltmeter and salt bridge need to be drawn but not necessarily annotated for M3.

Concentrations, but not state symbols, required for M4.

Suggest an experiment that shows that magnesium is more reactive than zinc, giving the observation that would confirm this.

Calculate the standard potential, in V, of a cell formed by magnesium and steel half-cells. Use section 24 of the data booklet and assume steel has the standard electrode potential of iron.

Calculate the free energy change, ΔG^⦵, in kJ, of the cell reaction. Use sections 1 and 2 of the data booklet.

This cell causes the electrolytic reduction of water on the steel. State the half-equation for this reduction.

Use the graph to deduce the dependence of the reaction rate on the amount of Mg.

The reaction is first order with respect to HCl. Calculate the time taken, in seconds (s), for half of the Mg to dissolve when [HCl] = 0.5 mol dm^–3.

Carbonates also react with HCl and the rate can be determined by graphing the mass loss. Suggest why this method is less suitable for the reaction of Mg with HCl.

Alternative 1

put Mg in Zn²⁺(aq) ✔

Zn/«black» layer forms «on surface of Mg» ✔

Award [1 max] for “no reaction when Zn placed in Mg²⁺(aq)”.

Alternative 2

place both metals in acid ✔

bubbles evolve more rapidly from Mg
OR
Mg dissolves faster ✔

Alternative 3

construct a cell with Mg and Zn electrodes ✔

Accept “electrons flow from Mg to Zn”.

Accept Mg is negative electrode/anode
OR
Zn is positive electrode/cathode

bulb lights up
OR
shows (+) voltage
OR
size/mass of Mg(s) decreases <<over time>>
OR
size/mass of Zn increases <<over time>>

Accept other correct methods.

Cell potential: «(–0.45 V – (–2.37 V)» = «+»1.92 «V» ✔

«ΔGº = -nFEº»
n = 2
OR
ΔGº = «-»2×96500×1.92 / «-»370,560 «J» ✔

-371 «kJ» ✔

For n = 1, award [1] for –185 «kJ».

Award [1 max] for (+)371 «kJ»

2 H₂O + 2 e^- → H₂ + 2 OH^- ✔

Accept equation with equilibrium arrows.

independent / not dependent ✔

Accept “zero order in Mg”.

«2×170 s» = 340 «s» ✔

Accept 320 – 360 «s».

Accept 400 – 450 «s» based on no more gas being produced after 400 to 450s.

«relative/percentage» decrease in mass is «too» small/«much» less ✔

Accept “«relative/percentage» uncertainty in mass loss «too» great”. OR “density/molar mass of H₂ is «much» less than CO₂”.

Mediocre performance; some experiments would not have worked such as adding magnesium to zinc salt without reference to aqueous environment, adding Zn to magnesium ions, or Mg combustion reaction being more exothermic. In the last one, an inference wad made instead of identifying an observation or measuring temperature using a thermometer or a temperature probe.

Good performance; instead of E° = 1.92 V, answer such as −1.92 V + or −2.82 V showed a lack of understanding of how to calculate E° cell.

Satisfactory performance; two major challenges in applying the equation ΔG° = −nFE° from the data booklet included:

Using n = 1, not 2, the number of electrons transferred in the redox reaction.

ΔG° unit from the equation is in J; some did not convert J to kJ as asked for.

Mediocre performance; some candidates had difficulty writing the reduction half-equation for water, the typical error included O₂(g) gas in the reactant or product, rather than H₂(g) in the product or including an equation with Fe(s) and H₂O(l) as reactants.

Candidates found this to be a tough question (see comments for parts (ii) and (iii)).

Mediocre performance in calculating time from the graph for the data provided. Some wrote the rate expression, which only contains [HCl] and not mass or amount in mol Mg (as a solid, [Mg] is constant). This presented a challenge in arriving at a reasonable answer.

Poorly done; many candidates did not grasp the question and answer it appropriately. Candidates generally did not realize that decrease in mass (due to H₂(g) as a product for the reaction of Mg with HCl) is «too» small/«much» less compared to that of CO₂(g) from the reaction of carbonates with HCl.

An aqueous solution of silver nitrate, AgNO₃ (aq), can be electrolysed using platinum electrodes.

Formulate the half-equations for the reaction at each electrode during electrolysis.

Cathode (negative electrode):

Anode (positive electrode):

Cathode (negative electrode):
Ag⁺ (aq) + e⁻ → Ag (s)    [✔]

Anode (positive electrode):
2H₂O(l) → O₂ (g) + 4H⁺ (aq) + 4e⁻    [✔]

Note: Accept 4OH⁻ (aq) → O₂ (g) + 2H₂O(l) + 4e⁻

Accept multiple or fractional coefficients in both half-equations.

Very few answers were correct, even for stronger candidates. Many failed to formulate the correct half equation for the reaction at the anode and used the nitrate ion instead of oxidation of H₂O. Some candidates lost one of the marks for using equilibrium arrows in an electrolysis equation.

Deduce the full electron configuration of Fe²⁺.

Explain why, when ligands bond to the iron ion causing the d-orbitals to split, the complex is coloured.

State the nuclear symbol notation, ${}_{\text{Z}}^{\text{A}}\text{X}$, for iron-54.

Mass spectrometry analysis of a sample of iron gave the following results:

Calculate the relative atomic mass, A_r, of this sample of iron to two decimal places.

An iron nail and a copper nail are inserted into a lemon.

Explain why a potential is detected when the nails are connected through a voltmeter.

Calculate the standard electrode potential, in V, when the Fe²⁺ (aq) | Fe (s) and Cu²⁺ (aq) | Cu (s) standard half-cells are connected at 298 K. Use section 24 of the data booklet.

Calculate ΔG^θ, in kJ, for the spontaneous reaction in (f)(i), using sections 1 and 2 of the data booklet.

Calculate a value for the equilibrium constant, K_c, at 298 K, giving your answer to two significant figures. Use your answer to (f)(ii) and section 1 of the data booklet. 

(If you did not obtain an answer to (f)(ii), use −140 kJ mol⁻¹, but this is not the correct value.)

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶   [✔]

«frequency/wavelength of visible» light absorbed by electrons moving between d levels/orbitals    [✔]

colour due to remaining frequencies
OR
complementary colour transmitted    [✔]

${}_{\text{26}}^{\text{54}}\text{Fe}$     [✔]

«A_r =» 54 × 0.0584 + 56 × 0.9168 + 57 × 0.0217 + 58 × 0.0031

OR

«A_r =» 55.9111    [✔]

«A_r =» 55.91    [✔]

Note: Award [2] for correct final answer

Do not accept data booklet value (55.85).

lemon juice is the electrolyte
OR
lemon juice allows flow of ions
OR
each nail/metal forms a half-cell with the lemon juice    [✔]

Any one of:
iron is higher than copper in the activity series
OR
each half-cell/metal has a different redox/electrode potential     [✔]

iron is oxidized
OR
Fe → Fe²⁺ + 2e⁻
OR
Fe → Fe³⁺ + 3e⁻
OR
iron is anode/negative electrode of cell   [✔]

copper is cathode/positive electrode of cell
OR
reduction occurs at the cathode
OR
2H⁺ + 2e⁻ → H₂   [✔]

electrons flow from iron to copper   [✔]

«E^θ = +0.34 V −(−0.45 V) = +»0.79 «V»   [✔]

«ΔG^θ = −nFE^θ = −2mol × 96 500 C mol⁻¹ × $\frac{0.79\text{ J }{\text{C}}^{-1}}{1000}$ =» −152 «kJ»    [✔]

Note: Accept range 150−153 kJ.

«lnK_c = $-\frac{\mathrm{\Delta }{G}^{\theta }}{RT}=-\frac{-152\times {10}^3\text{ Jmo}{\text{l}}^{-1}}{8.31\text{J}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1}\times 298\text{K}}$ =» 61.38    [✔]

K = 4.5 × 10²⁶    [✔]

Note: Accept answers in range 2.0 × 10²⁶ to 5.5 × 10²⁶.

Do not award M2 if answer not given to two significant figures.

If −140 kJmol⁻¹ used, answer is 3.6 × 10²⁴.

Done fairly well with common mistakes leaving in the 4s² electrons as part of Fe²⁺ electron configuration, or writing 4s¹ 3d⁵

This was poorly answered and showed a clear misconception and misunderstanding of the concepts. Most of the candidates failed to explain why the complex is coloured and based their answers on the emission of light energy when electrons fall back to ground state and not on light absorption by electrons moving between the split d-orbitals and complementary colour transmitted of certain frequencies.

Many candidates wrote the nuclear notation for iron as Z over A.

This question on average atomic mass was the best answered question on the exam. A few candidates did not write the answer to two decimal places as per instructions.

Very few candidates scored M1 regarding the lemon juice role as electrolyte. Some earned M2 but a lot of answers were too vague, such as ‘electrons move through the circuit’, etc.

Only 50 % of candidates earned this relatively easy mark on calculate EMF from 2 half-cell electrode potentials.

Average performance; typical errors were using the incorrect value for n, the number of electrons, or not using consistent units and making a factor of 1000 error in the final answer.

This question was left blank by quite a few candidates. Common errors included not using correct units, or more often, calculation error in converting ln K_c into K_c value.

Iron(II) is oxidized by bromine.

2Fe²⁺ (aq) + Br₂ (l) $\rightleftharpoons$ 2Fe³⁺ (aq) + 2Br⁻ (aq)

Calculate the E^⦵_cell, in V, for the reaction using section 24 of the data booklet.

Determine, giving a reason, if iodine will also oxidize iron(II). 

Molten zinc chloride undergoes electrolysis in an electrolytic cell at 450 °C.

Deduce the half-equations for the reaction at each electrode.

Deduce the overall cell reaction including state symbols. Use section 7 of the data booklet.

«${E^{⦵}}_{\text{cell}}$ = 1.09 – 0.77 =» 0.32 «V» ✔

«2Fe²⁺ (aq) + $\text{I}$₂(s) → 2Fe³⁺ (aq) + 2$\text{I}$^– (aq) »

no/non-spontaneous AND ${E^{\ddot{\text{O}}}}_{\text{cell}}$ «= 0.54 – 0.77 »= –0.23 «V»/ $E^{\ddot{\text{O}}}<0$
OR
no AND reduction potential of $\text{I}$₂ lower «than Fe3⁺ »/ 0.54 <0.77 ✔

Accept “standard electrode potential of $I$₂ lower /less positive than iron”.

Cathode (negative electrode):

Zn²⁺ + 2e⁻ → Zn (l) ✔

Anode (positive electrode):

2Cl⁻ → Cl₂ (g) + 2e⁻
OR
Cl⁻ → ½ Cl₂ (g) + e⁻ ✔

ZnCl₂ (l) → Zn (l) + Cl₂ (g)

balanced equation ✔

correct state symbols ✔

Accept ionic equation.

Only 50% got this straightforward calculation right, the most common error being to multiply both E₀ values by 2, reflecting a lack of practice with this type of exercises.

Only 10% were able to correctly justify the feasibility of the reaction with I₂; the MS showed the best answer using the E(v) values but also allowed simpler explanations referring to E₀ of iron; even then many candidates wrote Fe⁺² instead of Fe⁺³, understandably perhaps as this was mentioned in the question. However, it also revealed some difficulty in using and understanding data from the E₀ table in the data booklet.

3(bi)/(bii) Answers to both these questions revealed that many candidates struggle to conceptualize the equations that describe electrolysis. The question asked for products of the easiest case of electrolysis, a molten salt. However, many candidates proposed oxidation or reduction equations at both electrodes, or Zn and Cl₂ (with no charge) as the initial species rather than the product; the average mark was 1.2/2 as only 55% answered correctly.

3(bi)/(bii) Answers to both these questions revealed that many candidates struggle to conceptualize the equations that describe electrolysis. The question asked for products of the easiest case of electrolysis, a molten salt. However, many candidates proposed oxidation or reduction equations at both electrodes, or Zn and Cl₂ (with no charge) as the initial species rather than the product; the average mark was 1.2/2 as only 55% answered correctly.

The determination of the states proved to be even more difficult, with many stating the ions were aqueous in spite of the fact that the question is clearly about molten zinc chloride. Allowing ECF for the overall equation allowed marks for many candidates, but very few realised that both ionic species in ZnCl₂ were actually liquid (being a molten salt). As a result, correct answers were below 45% and the average mark was 0.9/2.

The stable isotope of rhenium contains 110 neutrons.

State the nuclear symbol notation ${}_{\text{Z}}^{\text{A}}\text{X}$ for this isotope.

Suggest the basis of these predictions.

A scientist wants to investigate the catalytic properties of a thin layer of rhenium metal on a graphite surface.

Describe an electrochemical process to produce a layer of rhenium on graphite.

Predict two other chemical properties you would expect rhenium to have, given its position in the periodic table.

Describe how the relative reactivity of rhenium, compared to silver, zinc, and copper, can be established using pieces of rhenium and solutions of these metal sulfates.

State the name of this compound, applying IUPAC rules.

Calculate the percentage, by mass, of rhenium in ReCl₃.

Suggest why the existence of salts containing an ion with this formula could be predicted. Refer to section 6 of the data booklet.

Deduce the coefficients required to complete the half-equation.

ReO₄⁻ (aq) + ____H⁺ (aq) + ____e⁻ ⇌ [Re(OH)₂]²⁺ (aq) + ____H₂O (l)        E^θ = +0.36 V

Predict, giving a reason, whether the reduction of ReO₄⁻ to [Re(OH)₂]²⁺ would oxidize Fe²⁺ to Fe³⁺ in aqueous solution. Use section 24 of the data booklet.

${}_{\text{75}}^{\text{185}}\text{Re}$    [✔]

gap in the periodic table
OR
element with atomic number «75» unknown
OR
break/irregularity in periodic trends     [✔]

«periodic table shows» regular/periodic trends «in properties»      [✔]

electrolyze «a solution of /molten» rhenium salt/Reⁿ⁺     [✔]

graphite as cathode/negative electrode
OR
rhenium forms at cathode/negative electrode     [✔]

Note: Accept “using rhenium anode” for M1.

Any two of:
variable oxidation states     [✔]

forms complex ions/compounds     [✔]

coloured compounds/ions     [✔]

«para»magnetic compounds/ions     [✔]

Note: Accept other valid responses related to its chemical metallic properties.

Do not accept “catalytic properties”.

place «pieces of» Re into each solution    [✔]

if Re reacts/is coated with metal, that metal is less reactive «than Re»    [✔]

Note: Accept other valid observations such as “colour of solution fades” or “solid/metal appears” for “reacts”.

rhenium(III) chloride
OR
rhenium trichloride    [✔]

«M_r ReCl₃ = 186.21 + (3 × 35.45) =» 292.56    [✔]
«100 × $\frac{186.21}{292.56}$ =» 63.648 «%»   [✔]

same group as Mn «which forms MnO₄^-»
OR
in group 7/has 7 valence electrons, so its «highest» oxidation state is +7    [✔]

ReO₄⁻ (aq) + 6H⁺ (aq) + 3e⁻ ⇌ [Re(OH)₂]²⁺ (aq) + 2H₂O (l)    [✔]

no AND ReO₄⁻ is a weaker oxidizing agent than Fe³⁺
OR
no AND Fe³⁺ is a stronger oxidizing agent than ReO₄⁻
OR
no AND Fe²⁺ is a weaker reducing agent than [Re(OH)₂]²⁺
OR
no AND [Re(OH)₂]²⁺ is a stronger reducing agent than Fe²⁺
OR
no AND cell emf would be negative/–0.41 V     [✔]

It was expected that this question would be answered correctly by all HL candidates. However, many confused the A-Z positions or calculated very unusual numbers for A, sometimes even with decimals.

This is a NOS question which required some reflection on the full meaning of the periodic table and the wealth of information contained in it. But very few candidates understood that they were being asked to explain periodicity and the concept behind the periodic table, which they actually apply all the time. Some were able to explain the “gap” idea and other based predictions on properties of nearby elements instead of thinking of periodic trends. A fair number of students listed properties of transition metals in general.

Generally well done; most described the cell identifying the two electrodes correctly and a few did mention the need for Re salt/ion electrolyte.

Generally well answered though some students suggested physical properties rather than chemical ones.

Many candidates chose to set up voltaic cells and in other cases failed to explain the actual experimental set up of Re being placed in solutions of other metal salts or the reaction they could expect to see.

Almost all candidates were able to name the compound according to IUPAC.

Most candidates were able to answer this stoichiometric question correctly.

This should have been a relatively easy question but many candidates sometimes failed to see the connection with Mn or the amount of electrons in its outer shell.

Surprisingly, a great number of students were unable to balance this simple half-equation that was given to them to avoid difficulties in recall of reactants/products.

Many students understood that the oxidation of Fe²⁺ was not viable but were unable to explain why in terms of oxidizing and reducing power; many students simply gave numerical values for E^Θ often failing to realise that the oxidation of Fe²⁺ would have the inverse sign to the reduction reaction.

State the electron configuration of the Cu⁺ ion.

Copper(II) chloride is used as a catalyst in the production of chlorine from hydrogen chloride.

4HCl (g) + O₂ (g) → 2Cl₂ (g) + 2H₂O (g)

Calculate the standard enthalpy change, ΔH^θ, in kJ, for this reaction, using section 12 of the data booklet.

The diagram shows the Maxwell–Boltzmann distribution and potential energy profile for the reaction without a catalyst.

Annotate both charts to show the activation energy for the catalysed reaction, using the label E_{a (cat)}.

Explain how the catalyst increases the rate of the reaction.

Solid copper(II) chloride absorbs moisture from the atmosphere to form a hydrate of formula CuCl₂•$x$H₂O.

A student heated a sample of hydrated copper(II) chloride, in order to determine the value of $x$. The following results were obtained:

Mass of crucible = 16.221 g
Initial mass of crucible and hydrated copper(II) chloride = 18.360 g
Final mass of crucible and anhydrous copper(II) chloride = 17.917 g

Determine the value of $x$.

State how current is conducted through the wires and through the electrolyte.

Wires: 

Electrolyte:

Write the half-equation for the formation of gas bubbles at electrode 1.

Bubbles of gas were also observed at another electrode. Identify the electrode and the gas.

Electrode number (on diagram):

Name of gas: 

Deduce the half-equation for the formation of the gas identified in (c)(iii).

Determine the enthalpy of solution of copper(II) chloride, using data from sections 18 and 20 of the data booklet.

The enthalpy of hydration of the copper(II) ion is −2161 kJ mol⁻¹.

Calculate the cell potential at 298 K for the disproportionation reaction, in V, using section 24 of the data booklet.

Comment on the spontaneity of the disproportionation reaction at 298 K.

Calculate the standard Gibbs free energy change, ΔG^θ, to two significant figures, for the disproportionation at 298 K. Use your answer from (e)(i) and sections 1 and 2 of the data booklet.

Suggest, giving a reason, whether the entropy of the system increases or decreases during the disproportionation.

Deduce, giving a reason, the sign of the standard enthalpy change, ΔH^θ, for the disproportionation reaction at 298 K.

Predict, giving a reason, the effect of increasing temperature on the stability of copper(I) chloride solution.

Describe how the blue colour is produced in the Cu(II) solution. Refer to section 17 of the data booklet.

Deduce why the Cu(I) solution is colourless.

When excess ammonia is added to copper(II) chloride solution, the dark blue complex ion, [Cu(NH₃)₄(H₂O)₂]²⁺, forms.

State the molecular geometry of this complex ion, and the bond angles within it.

Molecular geometry:

Bond angles: 

Examine the relationship between the Brønsted–Lowry and Lewis definitions of a base, referring to the ligands in the complex ion [CuCl₄]²⁻.

[Ar] 3d¹⁰
OR
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ ✔

ΔH^θ = ΣΔH^θ_f (products) − ΣΔH^θ_f (reactants) ✔
ΔH^θ = 2(−241.8 «kJ mol⁻¹») − 4(−92.3 «kJ mol⁻¹») = −114.4 «kJ» ✔

NOTE: Award [2] for correct final answer.

E_{a (cat)} to the left of E_a ✔                        

peak lower AND E_{a (cat)} smaller ✔

«catalyst provides an» alternative pathway ✔

«with» lower E_a
OR
higher proportion of/more particles with «kinetic» E ≥ E_a(cat) «than E_a» ✔

mass of H₂O = «18.360 g – 17.917 g =» 0.443 «g» AND mass of CuCl₂ = «17.917 g – 16.221 g =» 1.696 «g» ✔

moles of H₂O = «$\frac{0.443 \text{g}}{18.02 \text{g\hspace{0.05em}mo}{\text{l}}^{-1}}$=» 0.0246 «mol»
OR
moles of CuCl₂ =«$\frac{1.696 \text{g}}{134.45 \text{g\hspace{0.05em}mo}{\text{l}}^{-1}}$= » 0.0126 «mol» ✔

«water : copper(II) chloride = 1.95 : 1»

«$x$ =» 2 ✔

NOTE: Accept «$x$ =» 1.95.

NOTE: Award [3] for correct final answer.

Wires:
«delocalized» electrons «flow» ✔

Electrolyte:
«mobile» ions «flow» ✔

2Cl⁻ → Cl₂ (g) + 2e⁻
OR
Cl⁻ → $\frac{1}{2}$Cl₂ (g) + e⁻ ✔

NOTE: Accept e for e⁻.

«electrode» 3 AND oxygen/O₂ ✔

NOTE: Accept chlorine/Cl₂.

2H₂O (l) → 4H⁺ (aq) + O₂ (g) + 4e^– ✔

NOTE: Accept 2Cl^– (aq) → Cl₂ (g) + 2e^–.
Accept 4OH⁻ → 2H₂O + O₂ + 4e⁻

enthalpy of solution = lattice enthalpy + enthalpies of hydration «of Cu²⁺ and Cl⁻» ✔

«+2824 kJ mol^–1 − 2161 kJ mol^–1 − 2(359 kJ mol^–1) =» −55 «kJ mol^–1» ✔

NOTE: Accept enthalpy cycle.
Award [2] for correct final answer.

E^θ = «+0.52 – 0.15 = +» 0.37 «V» ✔

spontaneous AND E^θ positive ✔

ΔG^θ = «−nFE = −1 mol × 96 500 C Mol^–1 × 0.37 V=» −36 000 J/−36 kJ ✔

NOTE: Accept “−18 kJ mol^–1 «per mole of Cu⁺»”.

Do not accept values of n other than 1.

Apply SF in this question.

Accept J/kJ or J mol⁻¹/kJ mol⁻¹ for units.

2 mol (aq) → 1 mol (aq) AND decreases ✔

NOTE: Accept “solid formed from aqueous solution AND decreases”.
Do not accept 2 mol → 1 mol without (aq).

ΔG^θ < 0 AND ΔS^θ < 0 AND ΔH^θ < 0
OR
ΔG^θ + TΔS^θ < 0 AND ΔH^θ < 0 ✔

TΔS more negative «reducing spontaneity» AND stability increases ✔

NOTE: Accept calculation showing non-spontaneity at 433 K.

«ligands cause» d-orbitals «to» split ✔

light absorbed as electrons transit to higher energy level «in d–d transitions»
OR
light absorbed as electrons promoted ✔

energy gap corresponds to «orange» light in visible region of spectrum ✔

colour observed is complementary ✔

full «3»d sub-level/orbitals
OR
no d–d transition possible «and therefore no colour» ✔

octahedral AND 90° «180° for axial» ✔

NOTE: Accept square-based bi-pyramid.

Any two of:
ligand/chloride ion Lewis base AND donates e-pair ✔
not Brønsted–Lowry base AND does not accept proton/H⁺ ✔
Lewis definition extends/broader than Brønsted–Lowry definition ✔

The electron configuration of copper makes it a useful metal.

Determine the frequency of a photon that will cause the first ionization of copper. Use sections 1, 2 and 8 of the data booklet.

The electron configuration of copper makes it a useful metal.

Explain why a copper(II) solution is blue, using section 17 of the data booklet.

The electron configuration of copper makes it a useful metal.

Copper plating can be used to improve the conductivity of an object.

State, giving your reason, at which electrode the object being electroplated should be placed.

$«E=\frac{745 000 \mathrm{J} {\mathrm{mol}}^{-1}}{6.02\times {10}^{23} {\mathrm{mol}}^{-1}}=»1.24\times {10}^{-18} \mathrm{J}$ ✔

$«E=h\nu »$
$«1.24\times {10}^{-18} \mathrm{J}=6.63\times {10}^{-34} \mathrm{J} \mathrm{s}\times \mathrm{\nu }»$
$\mathrm{\nu }=1.87\times {10}^{15} «{\mathrm{s}}^{-1}/\mathrm{Hz}»$ ✔

Award [2] for correct final answer.
Award [1] for 1.12 × 10³⁹ «Hz».

orange light is absorbed «and the complementary colour is observed» ✔

Any TWO from:
partially filled d-orbitals ✔
«ligands/water cause» d-orbitals «to» split ✔
light is absorbed as electrons move to a higher energy orbital «in d–d transitions»
OR
light is absorbed as electrons are promoted ✔
energy gap corresponds to «orange» light in the visible region of the spectrum ✔

cathode/negative «electrode» AND ${\mathrm{Cu}}^{2+}$ reduced «at that electrode» ✔

Accept cathode/negative «electrode» AND copper forms «at that electrode».

Determining the frequency of a photon that will cause the first ionization of copper was the most challenging question on the exam. Many could not do it all, although some came up with the answer that came from using the result that would arise from the ionization energy in J/mole (and frequently kJ/mole) rather than J/atom.

Many students were able to fully explain why solutions containing Cu²⁺ appear blue, however the misconception between absorption and emission spectra is still quite evident.

Surprisingly not that well answered. Most students identified the cathode as the electrode where electroplating occurs but few could adequately justify why.

State the electron configuration of a bromine atom.

Sketch the orbital diagram of the valence shell of a bromine atom (ground state) on the energy axis provided. Use boxes to represent orbitals and arrows to represent electrons.

Draw two Lewis (electron dot) structures for BrO₃⁻.

Determine the preferred Lewis structure based on the formal charge on the bromine atom, giving your reasons.

Predict, using the VSEPR theory, the geometry of the BrO₃⁻ ion and the O−Br−O bond angles.

Bromate(V) ions act as oxidizing agents in acidic conditions to form bromide ions.

Deduce the half-equation for this reduction reaction.

Bromate(V) ions oxidize iron(II) ions, Fe²⁺, to iron(III) ions, Fe³⁺.

Deduce the equation for this redox reaction.

Calculate the standard Gibbs free energy change, ΔG^Θ, in J, of the redox reaction in (ii), using sections 1 and 24 of the data booklet.

E^Θ (BrO₃⁻ / Br⁻) = +1.44 V

State and explain the magnetic property of iron(II) and iron(III) ions.

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵

OR

[Ar] 4s² 3d¹⁰ 4p⁵ ✔

Accept 3d before 4s.

Accept double-headed arrows.

Structure I - follows octet rule:

Structure II - does not follow octet rule:

Accept dots, crosses or lines to represent electron pairs.

«structure I» formal charge on Br = +2

OR

«structure II» formal charge on Br = 0/+1 ✔

structure II is preferred AND it produces formal charge closer to 0 ✔

Ignore any reference to formal charge on oxygen.

Geometry:
trigonal/pyramidal ✔

Reason:
three bonds AND one lone pair
OR
four electron domains ✔

O−Br−O angle:
107° ✔

Accept “charge centres” for “electron domains”.

Accept answers in the range 104–109°.

BrO₃⁻ (aq) + 6e⁻ + 6H⁺ (aq) → Br⁻ (aq) + 3H₂O (l)

correct reactants and products ✔

balanced equation ✔

Accept reversible arrows.

BrO₃⁻ (aq) + 6Fe²⁺ (aq) + 6H⁺ (aq) → Br⁻ (aq) + 3H₂O (l) + 6Fe³⁺ (aq) ✔

E^Θ_reaction = «+1.44 V – 0.77 V =» 0.67 «V» ✔

ΔG^Θ = «–nFE^Θ_reaction = – 6 × 96500 C mol^–1 × 0.67 V =» –3.9 × 10⁵ «J» ✔

both are paramagnetic ✔

«both» contain unpaired electrons ✔

Accept orbital diagrams for both ions showing unpaired electrons.