DP Chemistry (last assessment 2024)

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Question 19N.3.sl.TZ0.5

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Date November 2019 Marks available [Maximum mark: 7] Reference code 19N.3.sl.TZ0.5
Level sl Paper 3 Time zone TZ0
Command term Deduce, Determine, Discuss, Formulate Question number 5 Adapted from N/A
5.
[Maximum mark: 7]
19N.3.sl.TZ0.5

Metals are extracted from their ores by various means.

(a)

Discuss why different methods of reduction are needed to extract metals.

[2]

Markscheme

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”.

Aluminium is produced by the electrolysis of alumina (aluminium oxide) dissolved in cryolite.

(b(i))

Determine the percentage of ionic bonding in alumina using sections 8 and 29 of the data booklet.

[2]

Markscheme

electronegativity difference = 1.8 «and average electronegativity = 2.5» ✔
57 «%» ✔

NOTE: Accept any value in the range 52−65 %.
Award [2] for correct final answer.

(b(ii))

Write half-equations for the electrolysis of molten alumina using graphite electrodes, deducing the state symbols of the products.

Anode (positive electrode):

Cathode (negative electrode):

[3]

Markscheme

Anode (positive electrode):
2O2− → 4e− + O2(g)
OR
2O2− + C → 4e− + CO2 (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):
Al3+ + 3e− → Al (l) ✔
O2 gas AND 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.

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