The method used to extract copper from its ores depends on the nature of the ore. Sulfide ores such as chalcopyrite (\(CuFeS_2\)) are converted to copper by a different method from silicate, carbonate or sulfate ores. Chalcopyrite (also known as copper pyrites) and similar sulfide ores are the commonest ores of copper. The ores typically contain low percentages of copper and have to be concentrated before refining (e.g., via froth flotation).
The Process
The concentrated ore is heated strongly with silicon dioxide (silica) and air or oxygen in a furnace or series of furnaces.
- The copper(II) ions in the chalcopyrite are reduced to copper(I) sulfide (which is reduced further to copper metal in the final stage).
- The iron in the chalcopyrite ends up converted into an iron(II) silicate slag which is removed.
- Most of the sulfur in the chalcopyrite turns into sulfur dioxide gas. This is used to make sulfuric acid via the Contact Process.
An overall equation for this series of steps is:
\[2CuFeS_2 + 2SiO_2 +4O_2 \rightarrow Cu_2S + 2FeSiO_3 + 3SO_2 \label{1} \]
The copper(I) sulfide produced is converted to copper with a final blast of air.
\[ Cu_2S + O_2 \rightarrow 2Cu + SO_2 \label{2} \]
The end product of this is called blister copper - a porous brittle form of copper, about 98 - 99.5% pure.
Exploring the redox processes in this reaction
It is worthwhile spending some time sorting out what the reducing agent is in these reactions, because at first sight there does not appear to be one! Or, if you look superficially, it seems as if it might be oxygen! But that's silly! We'll start by looking at the second reaction because it is much easier to see what is happening.
\[ Cu_2S + O_2 \rightarrow 2Cu + SO_2 \label{3} \]
Let's look at the oxidation states of everything.
- In the copper(I) sulfide, the copper is +1 and the sulfur -2.
- The oxidation states of the elements oxygen (in the gas) and copper (in the metal) are 0.
- In sulfur dioxide, the oxygen has an oxidation state of -2 and the sulfur +4.
That means that both the copper and the oxygen have been reduced (decrease in oxidation state). The sulfur has been oxidized (increase in oxidation state). The reducing agent is therefore the sulfide ion in the copper(I) sulfide.
The other reaction is more difficult to deal with, because you can't work out all of the oxidation states by following the simple rules - there are too many variables in some of the substances. You have to use some chemical knowledge as well.
\[ 2CuFeS_2 + 2SiO_2 + 4O_2 \rightarrow Cu_2S + 2FeSiO_3 + 3SO_2 \label{4} \]
In the CuFeS2, you would have to know that the copper and iron are both in oxidation state +2, for example. You would also have to know that the oxidation state of the silicon remains unchanged at +4. So use that information to work out what has been oxidized and what reduced in this case!
You should find that copper has been reduced from +2 to +1; oxygen (in the gas) has been reduced from 0 to -2 (oxygen in the SiO2 is unchanged); and three of the four sulfurs on the left-hand side have been oxidized from -2 to +4 (the other is unchanged). Once again, the sulfide ions are acting as the reducing agent.