Extraction of Metals

Lessons

Senior 6 · Chemistry

Extraction of Metals

4 lessons

Lesson 1 of 4

How Reactivity Decides the Method of Extraction

25 min

Why was aluminium once a luxury?

Aluminium is the most abundant metal in the Earth's crust, and iron comes second. Yet people have smelted iron in charcoal furnaces for about three thousand years, while aluminium was a rare, costly metal until 1886, when Charles Martin Hall in the USA and Paul Héroult in France each found a cheap way to make it by electrolysis. Gold, meanwhile, has been picked up as shiny nuggets since prehistory. The difference is not how much of each metal there is, but how tightly each is bound in its compounds. That depends on the metal's reactivity, and it decides how the metal is extracted.

Definition

Mineral: a naturally occurring solid compound with a definite composition. Example: cassiterite, $\mathrm{SnO_2}$ .

Ore: a rock that contains enough of a mineral for the metal to be extracted at a profit. Example: haematite ore for iron, which contains $\mathrm{Fe_2O_3}$ .

Gangue: the unwanted rock (mostly sand and silicates) mixed with the mineral in an ore.

Figure 1.1: The general steps in extracting a metal. Step 4, the reduction, is where the method depends on the metal's reactivity.

Step 2: concentrating the ore

Gravity separation uses the difference in density. Cassiterite (about 7 g cm⁻³) and coltan are much denser than quartz sand (2.65 g cm⁻³), so a stream of water washes the light gangue away and leaves the heavy mineral behind.
Magnetic separation removes a mineral attracted by a magnet from one that is not. It is used to separate wolframite, which is weakly magnetic, from cassiterite.
Froth flotation is used for sulfide ores such as chalcopyrite, $\mathrm{CuFeS_2}$ . The powdered ore is stirred with water, an oil-like "collector" and air. The collector coats the sulfide particles and makes them water-repellent, so they stick to the air bubbles and float off in the froth, while the wet gangue sinks.

Step 3: roasting and calcination

Sulfide and carbonate ores are first turned into oxides, which are easier to reduce. Roasting heats a sulfide in air, for example $\mathrm{2ZnS(s) + 3O_2(g) \rightarrow 2ZnO(s) + 2SO_2(g)}$ . Calcination heats a carbonate with little or no air, for example $\mathrm{ZnCO_3(s) \rightarrow ZnO(s) + CO_2(g)}$ .

Industrial froth flotation of copper sulfide ore. The mineral particles are carried off in the yellowish froth at the top of the cell, while the gangue sinks. Photo: Andreslan, public domain, via Wikimedia Commons

Step 4: the reactivity series decides the reduction

Table 1.1: The reactivity series and the method of extraction

Metal	Position	Method of reduction	Example
K, Na, Ca, Mg, Al	above carbon: most reactive	electrolysis of a molten compound	Na from molten NaCl; Al from $\mathrm{Al_2O_3}$ dissolved in molten cryolite (Lesson 3)
carbon
Zn, Fe, Sn, Pb	below carbon	heat the oxide with carbon (coke) or carbon monoxide	Fe in the blast furnace; Sn from cassiterite (Lesson 2)
hydrogen
Cu	below hydrogen	roast the sulfide; the $\mathrm{Cu_2S}$ is converted to Cu with air	Cu from chalcopyrite, then electrolytic refining
Ag, Au, Pt	least reactive	often found uncombined (native); separated physically or dissolved and recovered	gold nuggets and gold in quartz veins

A few metals fit none of these routes. Titanium and tungsten react with carbon to form hard, brittle carbides, so titanium is made by reducing $\mathrm{TiCl_4}$ with magnesium, and tungsten by reducing $\mathrm{WO_3}$ with hydrogen (Lesson 4).

The thermodynamic reason: the Ellingham diagram

A reduction such as $\mathrm{ZnO + C \rightarrow Zn + CO}$ is feasible only if its Gibbs free energy change is negative. Recall from S5 that $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ . An Ellingham diagram plots $\Delta G^\circ$ for forming each oxide, per mole of $\mathrm{O_2}$ , against temperature (Figure 1.2). The lower a line, the more stable that oxide.

Figure 1.2: A simplified Ellingham diagram, calculated by Ganzaa from standard 298 K data (ΔG° = ΔH° − TΔS°), ignoring changes of state. Below its crossing point, a metal's oxide cannot be reduced by carbon.

Read the diagram with two rules.

A substance can reduce the oxide of any metal whose line lies above its own at that temperature. The two lines combine: for example $\mathrm{2ZnO + 2C \rightarrow 2Zn + 2CO}$ has $\Delta G^\circ = \Delta G^\circ(\mathrm{C \rightarrow CO}) - \Delta G^\circ(\mathrm{Zn \rightarrow ZnO})$ , which is negative only where the carbon line is lower.
The slope is $-\Delta S^\circ$ . Forming a metal oxide uses up a mole of gas, so $\Delta S^\circ$ is negative and the line rises. But $\mathrm{2C(s) + O_2(g) \rightarrow 2CO(g)}$ turns one mole of gas into two, so $\Delta S^\circ$ is positive and the line falls. Carbon therefore becomes a stronger reducing agent the hotter it gets.

The carbon line crosses the FeO line near 750 °C and the ZnO line near 990 °C, which is why iron and zinc can be smelted with coke. It crosses the $\mathrm{Al_2O_3}$ line only above 2000 °C, far too hot for a practical furnace, and at such temperatures aluminium reacts with carbon to form the carbide $\mathrm{Al_4C_3}$ . Aluminium must be made by electrolysis. The copper line lies above the carbon line at every temperature above about 80 °C, so copper(I) oxide is reduced by carbon very easily.

Worked example

Problem. Use the equations of the lines in Figure 1.2 to find the lowest temperature at which carbon can reduce zinc oxide under standard conditions.

$\mathrm{2C + O_2 \rightarrow 2CO}$ : $\Delta G^\circ = -221.0 - 0.1788\,T$ $\mathrm{2Zn + O_2 \rightarrow 2ZnO}$ : $\Delta G^\circ = -701.0 + 0.2011\,T$ (kJ, with T in K)

Solution. The reduction becomes feasible where the lines cross:

$-221.0 - 0.1788\,T = -701.0 + 0.2011\,T$

$480.0 = 0.3799\,T \qquad T = \frac{480.0}{0.3799} = 1263\,\mathrm{K}$

Answer: about 1263 K, or 990 °C. Zinc boils at 907 °C, so at this temperature the zinc forms as a vapour, which is distilled off and condensed (Lesson 2).

Worked example

Problem. In the blast furnace, iron(III) oxide is reduced by carbon monoxide: $\mathrm{Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)}$ . Use the data to calculate $\Delta G^\circ$ at 1000 K.

	$\mathrm{Fe_2O_3(s)}$	$\mathrm{CO(g)}$	$\mathrm{Fe(s)}$	$\mathrm{CO_2(g)}$
$\Delta_f H^\circ$ / kJ mol⁻¹	−824.2	−110.5	0	−393.5
$S^\circ$ / J K⁻¹ mol⁻¹	87.4	197.7	27.3	213.8

Solution.

$\Delta H^\circ = 3(-393.5) - [(-824.2) + 3(-110.5)]$

$\Delta H^\circ = -1180.5 + 1155.7 = -24.8\,\mathrm{kJ\,mol^{-1}}$

$\Delta S^\circ = [2(27.3) + 3(213.8)] - [87.4 + 3(197.7)]$

$\Delta S^\circ = 696.0 - 680.5 = +15.5\,\mathrm{J\,K^{-1}\,mol^{-1}}$

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -24.8 - 1000 \times 0.0155 = -40.3\,\mathrm{kJ\,mol^{-1}}$

Answer: −40.3 kJ mol⁻¹. $\Delta G^\circ$ is negative, so the reduction is feasible; with both $\Delta H^\circ$ negative and $\Delta S^\circ$ positive, it is feasible at every temperature. (Data are standard 298 K values, used here as an approximation at 1000 K.)

Warning

Feasible is not the same as fast or cheap. A negative $\Delta G^\circ$ says only that a reduction can happen. The rate, the cost of energy and reducing agent, and side reactions such as carbide formation also decide which method industry uses.

Tip

Remember the order with a rule of thumb: above carbon, electrolysis; below carbon, carbon; very unreactive, found native. On an Ellingham diagram, "lower line, stronger reducer".

Cognito: Separating Metals from Metal Oxides (5 min). Watch for how the position of a metal relative to carbon in the reactivity series decides whether carbon or electrolysis is used.

✦

Practice

Define ore and gangue. Why is not every rock that contains a metal an ore?
Suggest a method of extraction for (a) magnesium, (b) lead, (c) gold, (d) calcium. Give a reason for each.
Explain how froth flotation separates copper sulfide particles from the gangue.
Use the line equations for carbon ( $-221.0 - 0.1788\,T$ ) and iron(II) oxide ( $-544.0 + 0.1383\,T$ ) to find the lowest temperature, in °C, at which carbon reduces FeO under standard conditions.
Explain why the line for $\mathrm{2C + O_2 \rightarrow 2CO}$ slopes downwards on an Ellingham diagram while the lines for metal oxides slope upwards.
For the thermite reaction $\mathrm{2Al(s) + Fe_2O_3(s) \rightarrow Al_2O_3(s) + 2Fe(s)}$ , $\Delta H^\circ = -851.5$ kJ mol⁻¹ and $\Delta S^\circ = -38.5$ J K⁻¹ mol⁻¹. Calculate $\Delta G^\circ$ at 298 K. Suggest why this reaction is used to weld railway lines but not to produce iron on a large scale.
Titanium is below carbon in the reactivity series, yet it is not made by heating $\mathrm{TiO_2}$ with coke. Explain, and state how it is made instead.
A student writes: "Carbon can reduce every metal oxide if it is hot enough, so electrolysis is never really necessary." Evaluate this statement.

Show answersHide answers

Ore: a rock containing enough of a metal compound to extract the metal at a profit. Gangue: the unwanted rock mixed with the mineral. A rock with only a trace of metal would cost more to process than the metal is worth.
(a) Electrolysis of molten magnesium chloride: Mg is above carbon. (b) Heating the oxide with carbon: Pb is below carbon. (c) Found native; separated physically (panning) or dissolved and recovered: very unreactive. (d) Electrolysis of molten calcium chloride: very reactive.
The powdered ore is mixed with water, a collector and air. The collector makes the sulfide particles water-repellent, so they attach to air bubbles and float to the top in the froth, which is skimmed off. The gangue is wetted by the water and sinks.
$-221.0 - 0.1788\,T = -544.0 + 0.1383\,T$ ; $323.0 = 0.3171\,T$ ; $T = 1019$ K, which is 746 °C.
$\mathrm{2C(s) + O_2(g) \rightarrow 2CO(g)}$ makes 2 mol of gas from 1 mol, so $\Delta S^\circ > 0$ and $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ becomes more negative as T rises. Forming a solid metal oxide uses up gaseous oxygen, so $\Delta S^\circ < 0$ and $\Delta G^\circ$ becomes less negative as T rises.
$\Delta G^\circ = -851.5 - 298 \times (-0.0385) = -851.5 + 11.5 = -840.0$ kJ mol⁻¹: very feasible, and very exothermic, so it melts the iron for welding. But aluminium is itself made by expensive electrolysis, so using it to make cheap iron would waste money.
Carbon reacts with titanium to form titanium carbide, TiC, which makes the metal brittle. Titanium is made by converting $\mathrm{TiO_2}$ to $\mathrm{TiCl_4}$ and reducing that with magnesium: $\mathrm{TiCl_4 + 2Mg \rightarrow Ti + 2MgCl_2}$ (the Kroll process).
Partly true in theory: the carbon line falls with temperature and eventually crosses many oxide lines. In practice it is false: for aluminium and the reactive metals the crossing is above 2000 °C (or never reached), furnaces cannot run economically at such temperatures, and the metals react with carbon to form carbides. Electrolysis is necessary for K, Na, Ca, Mg and Al.

Sources and credits

Written by Ganzaa. The Ellingham lines were calculated by Ganzaa from standard enthalpies of formation and standard entropies at 298 K, as listed in the CRC Handbook of Chemistry and Physics.
Image: Froth flotation.JPG by Andreslan, via Wikimedia Commons, public domain.
Video: GCSE Chemistry - Separating Metals from Metal Oxides, Cognito, embedded from YouTube.
The flowchart, Ellingham diagram, table, worked examples and Practice items were written and drawn by Ganzaa.

1Why was aluminium once a luxury?

2Step 2: concentrating the ore

3Step 3: roasting and calcination

4Step 4: the reactivity series decides the reduction

5The thermodynamic reason: the Ellingham diagram

Why was aluminium once a luxury?

Step 2: concentrating the ore

Step 3: roasting and calcination

Step 4: the reactivity series decides the reduction

The thermodynamic reason: the Ellingham diagram