Carbon Can Pull Oxygen Out of Iron Ore but Not Out of Aluminium Ore

Heat iron ore with coke in a blast furnace and molten iron runs out at the bottom. Heat bauxite with coke at any temperature a furnace can reach and no aluminium appears. Both ores are oxides, and both have oxygen that must be removed. The difference is how tightly each metal holds on to its oxygen.

Removing oxygen from a metal oxide is reduction. A reducing agent such as carbon takes the oxygen for itself. It can do that only if it attracts oxygen more strongly than the metal does.

- Iron holds its oxygen moderately, so carbon and carbon monoxide can take it away
- Aluminium holds its oxygen far more strongly than carbon can, so its oxide has to be broken apart by electricity instead

That contrast follows the activity series. Metals low in the series form oxides that are easy to reduce; metals high in the series form very stable oxides that only electrolysis can split.

But first, most ores are not oxides at all. Zinc blende is a sulphide and calamine a carbonate. Before reduction, ores are converted to oxides, by one of two heating processes:

- Roasting — heating in plenty of air, mainly for sulphide ores
- Calcination — heating with little or no air, mainly for carbonate and hydrated ores

So this part follows the concentrated ore from Part 1 through three stages:

- Conversion to oxide by roasting or calcination
- Reduction with hydrogen, carbon or carbon monoxide — or by electrolysis for the most reactive metals
- Refining of the impure metal by electrolysis

An everyday connection. India's large steel plants run blast furnaces around the clock, reducing iron oxide with coke and carbon monoxide by exactly the equations on this page. Aluminium plants, by contrast, run enormous electrolytic cells — a direct result of where the two metals sit in the activity series.

This page covers the second part of the ICSE Class 10 Chemistry chapter on metallurgy: roasting and calcination, reduction of metal oxides, why reactive metal oxides need electrolysis, and electrorefining.

Roasting heats an ore, usually a sulphide, strongly in excess air to form the oxide and sulphur dioxide; calcination heats an ore, usually a carbonate or hydrated oxide, in little or no air to drive off carbon dioxide or water.

Roasting. Heating a concentrated ore strongly in a plentiful supply of air, below its melting point.

- Used for: sulphide ores
- Chemical change: the sulphide is oxidised to the metal oxide and sulphur dioxide is released
- Also removes volatile impurities such as sulphur and arsenic as their oxides

$$2\text{ZnS}+3{\text{O}}_2\to 2\text{ZnO}+2{\text{SO}}_2$$
$$2\text{PbS}+3{\text{O}}_2\to 2\text{PbO}+2{\text{SO}}_2$$
$$4{\text{FeS}}_2+11{\text{O}}_2\to 2{\text{Fe}}_2{\text{O}}_3+8{\text{SO}}_2$$

Calcination. Heating a concentrated ore strongly in the absence or a limited supply of air, below its melting point.

- Used for: carbonate ores and hydrated ores
- Chemical change: decomposition — carbon dioxide or water of hydration is driven off
- Also makes the ore porous, which helps the reduction that follows

$${\text{ZnCO}}_3\to \text{ZnO}+{\text{CO}}_2$$
$${\text{PbCO}}_3\to \text{PbO}+{\text{CO}}_2$$
$${\text{Al}}_2{\text{O}}_3\cdot 2{\text{H}}_2\text{O}\to {\text{Al}}_2{\text{O}}_3+2{\text{H}}_2\text{O}$$
$$2{\text{Fe}}_2{\text{O}}_3\cdot 3{\text{H}}_2\text{O}\to 2{\text{Fe}}_2{\text{O}}_3+3{\text{H}}_2\text{O}$$

The differences in one list:

- Air: roasting — excess air; calcination — little or no air
- Ores: roasting — sulphides; calcination — carbonates and hydrated oxides
- Gas released: roasting — sulphur dioxide; calcination — carbon dioxide or water vapour
- Type of change: roasting — oxidation; calcination — decomposition

Worked check — balancing the pyrite equation.

- Iron: $4$ on the left; $2\times 2=4$ on the right
- Sulphur: $4\times 2=8$ on the left; $8$ on the right
- Oxygen: $11\times 2=22$ on the left; $2\times 3+8\times 2=6+16=22$ on the right

Balanced.

Why convert to an oxide at all. Carbon and carbon monoxide reduce metal oxides readily, but they do not reduce sulphides in the same way, and carbonates must lose carbon dioxide before reduction anyway. The oxide is the common starting point from which every reduction method works.

An everyday example of calcination. Traditional lime kilns heat limestone to make quicklime for whitewash and mortar: ${\text{CaCO}}_3\to \text{CaO}+{\text{CO}}_2$. That is calcination of a carbonate, exactly as for calamine, but the product wanted is the oxide itself rather than a metal.

The boundary case — air decides the name. The reason roasting needs air is that oxygen is a reactant: sulphur has to be converted to sulphur dioxide. Calcination is a decomposition that needs only heat, so air is unnecessary. The sulphur dioxide from roasting is a pollutant, which is why modern plants capture it and convert it into sulphuric acid instead of releasing it.

Each reducing agent removes oxygen from the heated metal oxide — hydrogen forms water, carbon forms carbon monoxide, and carbon monoxide forms carbon dioxide — leaving the metal behind.

Reduction here means removal of oxygen from the metal oxide; in electron terms, the metal ions gain electrons and become atoms. The reducing agent is itself oxidised.

1. Copper(II) oxide — reduced easily, since copper is low in the activity series.

$$\text{CuO}+{\text{H}}_2\to \text{Cu}+{\text{H}}_2\text{O}$$
$$\text{CuO}+\text{C}\to \text{Cu}+\text{CO}$$
$$\text{CuO}+\text{CO}\to \text{Cu}+{\text{CO}}_2$$

2. Lead(II) oxide.

$$\text{PbO}+{\text{H}}_2\to \text{Pb}+{\text{H}}_2\text{O}$$
$$\text{PbO}+\text{C}\to \text{Pb}+\text{CO}$$
$$\text{PbO}+\text{CO}\to \text{Pb}+{\text{CO}}_2$$

3. Iron(III) oxide.

$${\text{Fe}}_2{\text{O}}_3+3{\text{H}}_2\to 2\text{Fe}+3{\text{H}}_2\text{O}$$
$${\text{Fe}}_2{\text{O}}_3+3\text{C}\to 2\text{Fe}+3\text{CO}$$
$${\text{Fe}}_2{\text{O}}_3+3\text{CO}\to 2\text{Fe}+3{\text{CO}}_2$$

In a blast furnace, carbon monoxide made from coke is the main reducing agent for iron ore.

4. Zinc oxide — needs a higher temperature, since zinc is more reactive than lead and iron; reduced industrially with carbon.

$$\text{ZnO}+\text{C}\to \text{Zn}+\text{CO}$$
$$\text{ZnO}+\text{CO}\to \text{Zn}+{\text{CO}}_2$$

Identifying what is oxidised and reduced — in ${\text{Fe}}_2{\text{O}}_3+3\text{CO}\to 2\text{Fe}+3{\text{CO}}_2$:

- Iron(III) oxide is reduced — it loses oxygen, and each ${\text{Fe}}^{3+}$ gains $3$ electrons
- Carbon monoxide is oxidised — it gains oxygen to become carbon dioxide
- Carbon monoxide is the reducing agent; iron(III) oxide is the oxidising agent

Worked example — iron from iron(III) oxide. How much iron is produced from $160\text{ g}$ of iron(III) oxide by carbon monoxide, and what mass of carbon monoxide is used?

$$M({\text{Fe}}_2{\text{O}}_3)=160\text{moles}=\frac{160}{160}=1$$

- Iron: $2$ mol $=2\times 56=112\text{ g}$
- Carbon monoxide used: $3$ mol $=3\times 28=84\text{ g}$
- Carbon dioxide formed: $3$ mol $=3\times 44=132\text{ g}$, or $3\times 22.4=67.2\text{ L}$ at STP
- Check: $160+84=244$ and $112+132=244$

The ease of reduction follows the activity series.

- Copper and lead oxides reduce readily, even with hydrogen at moderate heat
- Iron oxide needs strong heating with carbon or carbon monoxide
- Zinc oxide needs a still higher temperature

A laboratory example. Black copper(II) oxide heated in a stream of dry hydrogen turns reddish-brown as copper forms, while droplets of water collect on the cooler part of the tube — the first equation above, seen directly.

The boundary case — hydrogen is not used for every oxide in industry. Hydrogen reduces copper and lead oxides easily in the laboratory, but for large-scale production of iron and zinc, carbon and carbon monoxide are used, because coke is cheap and also supplies the heat. The equations show what is chemically possible; the choice in a factory also weighs cost.

Aluminium and magnesium are so reactive that they hold oxygen more strongly than carbon, hydrogen or carbon monoxide can, so their oxides can be broken down only by supplying electrons directly through electrolysis of the molten compound.

The reasoning in steps:

- Aluminium and magnesium lie high in the activity series, just below calcium
- Metals high in the series have a very strong attraction for oxygen, so their oxides are extremely stable
- A reducing agent works only if it attracts oxygen more strongly than the metal does
- Carbon, hydrogen and carbon monoxide attract oxygen less strongly than aluminium and magnesium at the temperatures furnaces can reach
- So these reducing agents cannot remove the oxygen, and chemical reduction fails

Electrolysis gets round the problem by supplying electrons at the cathode, so no chemical reducing agent is needed:

$${\text{Al}}^{3+}+3e^{-}\to \text{Al}$$
$${\text{Mg}}^{2+}+2e^{-}\to \text{Mg}$$

Why the compound must be molten, not dissolved in water. In aqueous solution, hydrogen ions would be discharged in preference to aluminium or magnesium ions, as Part 2 of the electrolysis chapter showed. Only a molten electrolyte, containing no water, allows these metals to be deposited.

The extraction method follows the activity series:

- Potassium, sodium, calcium, magnesium, aluminium — electrolysis of molten compounds
- Zinc, iron, lead, copper — reduction of the oxide with carbon or carbon monoxide
- Very unreactive metals such as gold — often found uncombined in nature

Worked example — the charge each metal needs. Compare the moles of electrons needed per gram of metal for aluminium, atomic mass $27$, and magnesium, atomic mass $24$.

$$\text{Al}:\frac{3}{27}=\frac{1}{9}\text{ mol of electrons per gram}$$
$$\text{Mg}:\frac{2}{24}=\frac{1}{12}\text{ mol of electrons per gram}$$

Aluminium needs more electricity per gram than magnesium, because each aluminium ion takes three electrons — one reason aluminium smelting uses so much electrical energy.

The strongest evidence of aluminium's attraction for oxygen. Aluminium powder heated with iron(III) oxide takes the oxygen from the iron, in a fiercely hot reaction:

$${\text{Fe}}_2{\text{O}}_3+2\text{Al}\to {\text{Al}}_2{\text{O}}_3+2\text{Fe}$$

Aluminium is acting as the reducing agent for iron oxide. A metal that can strip oxygen from iron oxide is clearly holding its own oxygen far too tightly for carbon to remove.

An everyday example. The magnesium used in some alloys for lightweight parts, and the aluminium in cooking vessels and foil, both come from electrolytic plants rather than furnaces — the practical consequence of this section.

The boundary case. Electrolysis is not impossible for the less reactive metals — copper is purified by electrolysis, for example. It is simply unnecessary and more costly for them, because a cheap chemical reducing agent already works. For aluminium and magnesium, electrolysis is not a choice but the only practical route.

The impure metal is made the anode and a thin strip of pure metal the cathode, in a solution of a salt of that metal; metal dissolves from the anode and deposits pure on the cathode, while impurities stay in solution or fall as anode mud.

Where refining fits. Metal from reduction is rarely pure — iron from a blast furnace contains carbon, and copper or silver from reduction contains other metals. Refining is the final stage, and electrorefining is the method used when very high purity is needed.

The general arrangement for a metal M:

- Anode: a block of impure M
- Cathode: a thin strip of pure M
- Electrolyte: an aqueous solution of a soluble salt of M

The general electrode reactions, for a metal forming ions of charge $n+$:

$$\text{Anode:}\text{M}\to {\text{M}}^{n+}+n{e}^{-}$$
$$\text{Cathode:}{\text{M}}^{n+}+n{e}^{-}\to \text{M}$$

Example 1 — copper: impure copper anode, pure copper cathode, copper sulphate solution with dilute sulphuric acid.

$$\text{Cu}\to {\text{Cu}}^{2+}+2e^{-}{\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$$

Example 2 — silver: impure silver anode, pure silver cathode, silver nitrate solution with a little nitric acid.

$$\text{Ag}\to {\text{Ag}}^{+}+e^{-}{\text{Ag}}^{+}+e^{-}\to \text{Ag}$$

What happens to the impurities:

- More reactive impurities dissolve from the anode as ions but stay in solution, because their ions are harder to discharge than those of the metal being refined — copper impurities in silver behave this way
- Less reactive impurities do not dissolve and fall beneath the anode as anode mud — gold in impure silver or copper behaves this way

Worked example — comparing two refining cells. The same charge, $0.2$ mol of electrons, passes through a copper refining cell and a silver refining cell. Atomic masses Cu $63.5$, Ag $108$.

$$\text{Cu}:\frac{0.2}{2}=0.1\text{ mol}=0.1\times 63.5=6.35\text{ g}$$
$$\text{Ag}:\frac{0.2}{1}=0.2\text{ mol}=0.2\times 108=21.6\text{ g}$$

The silver cell deposits far more mass for the same electricity, because each silver ion needs one electron and each silver atom is heavier than a copper atom.

Why electrorefining gives such pure metal. At the cathode, only the ion easiest to discharge is deposited — and in a refining cell that is always the ion of the metal being purified, which is present in large concentration. Every impurity is excluded either by staying dissolved or by never dissolving.

An everyday example. Silver used for jewellery and coins, and copper for electrical wiring, are refined electrolytically. Precious metals recovered from the anode mud of copper refineries add a valuable by-product to the process.

The link across the chapter. Refining completes the sequence that began in Part 1: crushing, concentrating, converting to oxide, reducing and refining. Each step removes something unwanted, and electrolysis appears twice — as the only reduction method for the most reactive metals, and as the final purification for many others.