Extraction of Metals

Lessons

Senior 6 · Chemistry

Extraction of Metals

4 lessons

Lesson 3 of 4

Electrolytic Extraction: Sodium and Aluminium

25 min

When carbon is not strong enough

Sodium and aluminium sit above carbon in the reactivity series. Heating their compounds with coke does nothing useful: on the Ellingham diagram of Lesson 1, the carbon line reaches the aluminium oxide line only above 2000 °C. The only practical way to force the electrons back onto these metal ions is to supply them from a direct-current source: electrolysis. You studied electrolysis and Faraday's laws in S5. Here you will see how industry uses them on a giant scale.

Definition

Electrolytic extraction: obtaining a metal by passing a direct current through a molten (or dissolved) compound of the metal. The metal ions gain electrons at the cathode: $\mathrm{M^{n+} + ne^- \rightarrow M}$ .

Faraday constant, $F = 96\,485\,\mathrm{C\,mol^{-1}}$ : the charge on one mole of electrons. The charge passed is $Q = I t$ , and the moles of metal formed are $\frac{Q}{nF}$ .

Sodium: the Downs cell

Sodium is made by electrolysing molten sodium chloride. An aqueous solution will not do: water is reduced at the cathode more easily than sodium ions, so hydrogen forms instead of sodium.

$\text{cathode (−): } \mathrm{Na^+(l) + e^- \rightarrow Na(l)} \qquad \text{anode (+): } \mathrm{2Cl^-(l) \rightarrow Cl_2(g) + 2e^-}$

$\text{overall: } \mathrm{2NaCl(l) \rightarrow 2Na(l) + Cl_2(g)}$

Pure sodium chloride melts at 801 °C. Adding calcium chloride lowers the melting point of the mixture to about 600 °C, which saves energy and keeps the sodium (boiling point 883 °C) from vaporising. The liquid sodium is less dense than the molten salt, so it floats up and is collected; the chlorine is a valuable by-product.

Figure 3.1: The Downs cell for sodium (simplified cross-section). The diaphragm keeps the sodium and chlorine apart so they cannot react again.

Aluminium, step 1: purifying bauxite (the Bayer process)

The ore, bauxite, is hydrated aluminium oxide mixed with iron(III) oxide and silica. The process uses the fact that aluminium oxide is amphoteric while iron(III) oxide is basic. Hot concentrated sodium hydroxide dissolves the aluminium oxide:

$\mathrm{Al_2O_3(s) + 2NaOH(aq) + 3H_2O(l) \rightarrow 2Na[Al(OH)_4](aq)}$

Iron(III) oxide does not dissolve and is filtered off as red mud. The solution is cooled and seeded with crystals of aluminium hydroxide, which precipitates:

$\mathrm{Na[Al(OH)_4](aq) \rightarrow Al(OH)_3(s) + NaOH(aq)}$

Heating the hydroxide strongly (at about 1100 °C) gives pure alumina:

$\mathrm{2Al(OH)_3(s) \rightarrow Al_2O_3(s) + 3H_2O(g)}$

Aluminium, step 2: electrolysis (the Hall–Héroult process)

Aluminium oxide melts at 2072 °C, too hot for any practical cell. Instead it is dissolved in molten cryolite, $\mathrm{Na_3AlF_6}$ , which gives a conducting electrolyte at about 950 °C.

$\text{cathode (−): } \mathrm{Al^{3+} + 3e^- \rightarrow Al(l)} \qquad \text{anode (+): } \mathrm{2O^{2-} \rightarrow O_2(g) + 4e^-}$

At 950 °C the oxygen attacks the carbon anodes, which burn away and must be replaced every few weeks:

$\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$

$\text{overall: } \mathrm{2Al_2O_3(l) + 3C(s) \rightarrow 4Al(l) + 3CO_2(g)}$

The molten aluminium is denser than the electrolyte, so it collects at the bottom of the cell and is siphoned off.

Figure 3.2: A Hall–Héroult cell for aluminium. A large modern cell carries several hundred thousand amperes.

Worked example

Problem. A Hall–Héroult cell runs at a current of $1.50 \times 10^{5}$ A. What mass of aluminium does it produce in 24.0 hours? (Al = 26.98, $F = 96\,485\,\mathrm{C\,mol^{-1}}$ )

Solution.

$Q = It = 1.50 \times 10^{5} \times 24.0 \times 3600 = 1.296 \times 10^{10}\,\mathrm{C}$

$n(e^-) = \frac{1.296 \times 10^{10}}{96\,485} = 1.343 \times 10^{5}\,\mathrm{mol} \qquad n(\mathrm{Al}) = \frac{1.343 \times 10^{5}}{3} = 4.477 \times 10^{4}\,\mathrm{mol}$

$m(\mathrm{Al}) = 4.477 \times 10^{4} \times 26.98 = 1.208 \times 10^{6}\,\mathrm{g}$

Answer: about 1.21 tonnes per day. Three moles of electrons are needed for every mole of aluminium, which is one reason aluminium costs so much electricity.

Worked example

Problem. A cell operates at 4.5 V. Calculate the electrical energy needed to make 1.00 kg of aluminium, in joules and in kilowatt-hours (1 kWh = $3.6 \times 10^{6}$ J).

Solution. 1.00 kg of Al is $\frac{1000}{26.98} = 37.06$ mol, which needs $3 \times 37.06 = 111.2$ mol of electrons.

$Q = 111.2 \times 96\,485 = 1.073 \times 10^{7}\,\mathrm{C} \qquad E = QV = 1.073 \times 10^{7} \times 4.5 = 4.83 \times 10^{7}\,\mathrm{J}$

$E = \frac{4.83 \times 10^{7}}{3.6 \times 10^{6}} = 13.4\,\mathrm{kWh}$

Answer: about $4.8 \times 10^{7}$ J, or 13 kWh, per kilogram, before any losses. This is why aluminium smelters are built where electricity is cheap, often beside large hydroelectric dams, and why recycling, which only melts the metal, saves most of this energy.

Table 3.1: The two electrolytic extractions compared

	Sodium (Downs cell)	Aluminium (Hall–Héroult cell)
Electrolyte	molten NaCl with $\mathrm{CaCl_2}$	$\mathrm{Al_2O_3}$ dissolved in molten cryolite
Temperature	about 600 °C	about 950 °C
Why the additive	lowers the melting point from 801 °C	dissolves alumina, which melts at 2072 °C
Anode and product	graphite; chlorine	carbon, which burns away; oxygen, then $\mathrm{CO_2}$
Cathode and product	steel; liquid sodium, which floats	carbon lining; liquid aluminium, which sinks
Electrons per metal atom	1	3

Warning

Hazards. Molten sodium reacts violently with water, and chlorine is toxic. Aluminium smelters release carbon dioxide from the anodes and small amounts of fluorine compounds from the cryolite, which must be scrubbed from the waste gas; the red mud from the Bayer process is strongly alkaline and has to be stored in sealed ponds.

Tip

In electrolysis the cathode is where metals are made: "cations go to the cathode". For mass calculations, always divide the moles of electrons by the charge on the ion: 1 for $\mathrm{Na^+}$ , 3 for $\mathrm{Al^{3+}}$ .

Cognito: Electrolysis to Extract Metals From Oxides (6 min). Watch for why cryolite is used and why the carbon anodes must be replaced.

✦

Practice

Explain why sodium is made from molten sodium chloride, not from a solution of sodium chloride in water.
Give two reasons for adding calcium chloride in the Downs cell.
Explain how the Bayer process separates aluminium oxide from iron(III) oxide. Which property of aluminium oxide does it use?
Write the half-equations at the anode and cathode in the Hall–Héroult cell, and explain why the anodes must be replaced.
What mass of aluminium is produced by a current of $1.00 \times 10^{5}$ A flowing for 8.00 hours?
Calculate the maximum mass of sodium, and of chlorine, obtainable from 1.00 tonne of sodium chloride. (Na = 22.99, Cl = 35.45)
Repeat the energy calculation in the second worked example for a cell running at 5.0 V. Suggest one reason why real cells use more energy than this.
"Aluminium cans should be recycled rather than thrown away." Evaluate this statement using the chemistry of this lesson.

Show answersHide answers

In aqueous solution, water is reduced at the cathode more easily than $\mathrm{Na^+}$ , so hydrogen forms, not sodium. And any sodium formed would react at once with the water.
It lowers the melting point of the electrolyte from 801 °C to about 600 °C, which saves energy, and it keeps the temperature well below sodium's boiling point (883 °C), so the sodium stays liquid.
Hot concentrated NaOH dissolves aluminium oxide as $\mathrm{Na[Al(OH)_4]}$ , because it is amphoteric; basic iron(III) oxide does not dissolve and is filtered off as red mud. The property used is amphoterism.
Cathode: $\mathrm{Al^{3+} + 3e^- \rightarrow Al}$ . Anode: $\mathrm{2O^{2-} \rightarrow O_2 + 4e^-}$ . The oxygen reacts with the hot carbon anodes, $\mathrm{C + O_2 \rightarrow CO_2}$ , so they burn away.
$Q = 1.00 \times 10^{5} \times 8.00 \times 3600 = 2.88 \times 10^{9}$ C; $n(e^-) = \frac{2.88 \times 10^{9}}{96\,485} = 2.985 \times 10^{4}$ mol; $n(\mathrm{Al}) = 9950$ mol; mass = $9950 \times 26.98 = 2.68 \times 10^{5}$ g = 268 kg.
$M(\mathrm{NaCl}) = 58.44$ ; $n = \frac{1.00 \times 10^{6}}{58.44} = 1.711 \times 10^{4}$ mol. Na: $1.711 \times 10^{4} \times 22.99 = 3.93 \times 10^{5}$ g = 393 kg. $\mathrm{Cl_2}$ : $\frac{1.711 \times 10^{4}}{2} \times 70.90 = 6.07 \times 10^{5}$ g = 607 kg.
$E = 1.073 \times 10^{7} \times 5.0 = 5.36 \times 10^{7}$ J = 14.9 kWh per kg. Real cells lose energy as heat (the electrolyte's resistance keeps it molten), some current is wasted in side reactions, and some aluminium is re-oxidised.
Strongly supported. Making new aluminium needs bauxite mining, the Bayer process (leaving red mud), and about 13–15 kWh of electricity per kg, with $\mathrm{CO_2}$ released from the anodes. Recycling only melts the metal, so it saves most of the energy, the ore and the waste. A limitation: cans must be collected and sorted, which needs organisation and some transport energy.

Sources and credits

Image: Pisolitic bauxite (Alcoa Bauxite Mine, Arkansas, USA) 1.jpg by James St. John, via Wikimedia Commons, licensed under CC BY 2.0.
Video: Electrolysis Part 2 of 3: Electrolysis to Extract Metals From Oxides, Cognito, embedded from YouTube.
The text, cell diagrams, table, worked examples and Practice items were written and drawn by Ganzaa.

1When carbon is not strong enough

2Sodium: the Downs cell

3Aluminium, step 1: purifying bauxite (the Bayer process)

4Aluminium, step 2: electrolysis (the Hall–Héroult process)

When carbon is not strong enough

Sodium: the Downs cell

Aluminium, step 1: purifying bauxite (the Bayer process)

Aluminium, step 2: electrolysis (the Hall–Héroult process)