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Date	November 2019	Marks available	1	Reference code	19N.2.sl.TZ0.5
Level	SL	Paper	2	Time zone	TZ0
Command term	Formulate and Apply	Question number	5	Adapted from	N/A

Question

Copper forms two chlorides, copper(I) chloride and copper(II) chloride.

An electrolysis cell was assembled using graphite electrodes and connected as shown.

State the electron configuration of the Cu⁺ ion.

[1]

a(i).

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.

[2]

a(ii).

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)}.

[2]

a(iii).

Explain how the catalyst increases the rate of the reaction.

[2]

a(iv).

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 $x$ .

[3]

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

Wires:

Electrolyte:

[2]

c(i).

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

[1]

c(ii).

Markscheme

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

a(i).

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.

a(ii).

E_{a (cat)} to the left of E_a ✔

peak lower AND E_{a (cat)} smaller ✔

a(iii).

«catalyst provides an» alternative pathway ✔

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

a(iv).

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