Extraction of Metals Worked ExamplesIGCSE Chemistry 0620

(a) Metals at the very bottom of the series are so unreactive that they do not combine with oxygen or other elements, so they are found uncombined. Gold.

(b) Q is below carbon in the series, so carbon is more reactive than Q. Heating the oxide of Q with carbon (coke) reduces it: carbon removes the oxygen and the metal is left.

(c) Magnesium is above carbon — it is more reactive than carbon, so carbon cannot remove oxygen from magnesium oxide. Metals above carbon are extracted by electrolysis of a molten compound.

The general rule: the more reactive the metal, the more stable its compounds and the harder (and more energy-consuming) it is to obtain the metal from its ore.

(a) The solid charge is hematite (iron ore, mainly iron(III) oxide), coke (carbon) and limestone (calcium carbonate).

(b) Hot air is blown in at B. Its oxygen burns the coke, releasing the heat that keeps the furnace hot.

(c) Two liquid layers collect at the bottom. The upper, less dense layer is molten slag (calcium silicate), tapped off at C. The lower, denser layer is molten iron, tapped off at D.

(d) The limestone undergoes thermal decomposition to calcium oxide and carbon dioxide. Calcium oxide then reacts with silicon(IV) oxide to form slag (calcium silicate), which is removed as a liquid.

(a) Coke burns in the hot air blast — a strongly exothermic combustion.

(b) Higher up, the carbon dioxide meets more hot coke and is reduced to carbon monoxide.

(c) Carbon monoxide removes oxygen from iron(III) oxide. The iron forms as a liquid because the furnace is so hot. Balance: 2 Fe on each side; 3 O in $\text{Fe}_2\text{O}_3$ needs 3 CO, making 3 $\text{CO}_2$.

(d) Limestone (calcium carbonate) decomposes in the heat.

(e) Calcium oxide (a basic oxide) reacts with silicon(IV) oxide (an acidic oxide) to make calcium silicate, the slag.

(f) The reducing agent is carbon monoxide. It is itself oxidised because it gains oxygen, becoming carbon dioxide; iron(III) oxide loses oxygen, so it is reduced.

Carbon gains oxygen (C → CO), so carbon is oxidised. Carbon dioxide loses oxygen ($\text{CO}_2$ → CO), so carbon dioxide is reduced. Answer: B.

A is impossible: in a redox reaction one substance is oxidised while another is reduced. D has the roles reversed — carbon is the reducing agent here, because it removes oxygen from carbon dioxide.

C is the tempting distractor. This reaction is actually endothermic; the furnace is heated by the combustion of coke, $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$.

(a) P: the carbon (graphite) anodes — connected to the positive terminal. Q: the electrolyte, aluminium oxide dissolved in molten cryolite. R: molten aluminium. S: the carbon lining, which is the cathode (negative electrode).

(b) Aluminium oxide has a very high melting point. Dissolving it in molten cryolite lowers the operating temperature, which saves energy (and therefore cost), and gives a liquid in which the ions can move.

(c) At the negative electrode (cathode), aluminium ions gain electrons — reduction:

At the positive electrode (anode), oxide ions lose electrons — oxidation:

(d) The oxygen formed at the anodes reacts with the hot carbon to form carbon dioxide, so the anodes gradually burn away and must be replaced.

Statement 1 is correct: $\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}$ is a gain of electrons — reduction at the cathode.

Statement 2 is wrong: oxide ions lose electrons at the anode ($2\text{O}^{2-} \rightarrow \text{O}_2 + 4\text{e}^-$). Negative ions are oxidised.

Statement 3 is correct. Statement 4 is wrong: cryolite lowers the operating temperature. So the answer is A.

B and D are chosen by candidates who remember "gain of electrons" but attach it to the wrong electrode. C and D are chosen by those who know that cryolite is involved but not what it does.

(a) Take $64.0\%$ of the ore.

(b) First write the equation: $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$. $M_r(\text{Fe}_2\text{O}_3) = (2 \times 56) + (3 \times 16) = 160$. Convert to grams and find the moles.

The equation gives 2 mol Fe for every 1 mol $\text{Fe}_2\text{O}_3$.

(c) Percentage yield compares actual with maximum (theoretical) yield.

(d) 3 mol CO are needed per mol $\text{Fe}_2\text{O}_3$; each mole of gas occupies $24 \text{ dm}^3$ at r.t.p.

(a) Balance oxygen first. $\text{V}_2\text{O}_5$ has 5 O and $\text{Al}_2\text{O}_3$ has 3 O; the lowest common multiple is 15, so use $3\,\text{V}_2\text{O}_5$ and $5\,\text{Al}_2\text{O}_3$.

Then 3 × 2 = 6 V atoms and 5 × 2 = 10 Al atoms.

Check: V 6 = 6; O 15 = 15; Al 10 = 10. ✓

(b) Aluminium is more reactive than vanadium, so it removes the oxygen from vanadium(V) oxide.

(c) Vanadium(V) oxide loses oxygen (it is reduced), so it is the oxidising agent — it oxidises the aluminium to aluminium oxide.

(d) Moles of vanadium, then use the 10 : 6 ratio of Al : V.

Mass of aluminium (keep the unrounded mole value):

Prepare the ore: crush the malachite to a fine powder with a pestle and mortar, to increase the surface area.

Method 1 — reduction with carbon. Heat the powder in a crucible: copper(II) carbonate decomposes (green → black copper(II) oxide; the gas turns limewater milky). Mix the cooled solid with excess powdered carbon, put the lid on and heat strongly.

Allow to cool with the lid on (so hot copper is not re-oxidised by air), then tip into water: the light carbon powder can be poured off with the water, leaving red-brown (pink-brown) copper.

Method 2 — acid then displacement. Add the powder to excess dilute sulfuric acid: effervescence and a blue solution of copper(II) sulfate. Filter off the rock. Add excess iron filings to the filtrate: a red-brown solid forms and the solution turns pale green.

Filter off the copper, add more dilute acid to remove the excess iron (the copper does not react), then wash with distilled water and dry. Either method scores; the marks are for the crushing, the reducing agent or more reactive metal, strong heating or filtering, and the observations.

(a) The less reactive the metal, the earlier it was obtained. Gold (least reactive) was known first; the most reactive metals, sodium and aluminium, were obtained last.

(b) Copper and iron are below carbon in the reactivity series, so their ores can be reduced by heating with carbon (charcoal) — possible with a simple fire. Sodium and aluminium are above carbon, so carbon cannot reduce their compounds.

They need electrolysis, which only became possible once a supply of electricity (batteries) was available around 1800.

(c) Potassium is more reactive than aluminium, so it can displace aluminium from its compounds. But potassium itself had to be made first — by electrolysis — before it could be used.

(a) The electrons must cancel. Multiply the cathode equation by 4 and the anode equation by 3 so that both involve 12 electrons.

Add and cancel the electrons:

(b) Each $\text{Al}^{3+}$ ion needs 3 electrons.

(c) Each $\text{O}_2$ molecule releases 4 electrons.

(d) $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$: 1 mol C reacts with 1 mol $\text{O}_2$.

(a) Mass of silicon(IV) oxide $= 0.060 \times 1000 = 60$ tonnes. $M_r(\text{SiO}_2) = 28 + 32 = 60$.

Link the equations: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$, then $\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3$. So 1 mol $\text{CaCO}_3$ removes 1 mol $\text{SiO}_2$. $M_r(\text{CaCO}_3) = 40 + 12 + 48 = 100$.

(b) $M_r(\text{CaSiO}_3) = 40 + 28 + 48 = 116$, and 1 mol slag forms per mol $\text{SiO}_2$.

(c) 1 mol $\text{CO}_2$ per mol $\text{CaCO}_3$.

(d) Calcium oxide is a basic oxide (a metal oxide) and silicon(IV) oxide is an acidic oxide (a non-metal oxide), so they react together in a neutralisation to form a salt, calcium silicate.