Extracting copper from its ores

Extracting copper from its ores

The method used to extract copper from its ores depends on the nature of the ore. Sulphide ores such as chalcopyrite are converted to copper by a different method from silicate, carbonate or sulphate ores.


Getting copper from chalcopyrite, CuFeS2

Chalcopyrite (also known as copper pyrites) and similar sulphide ores are the commonest ores of copper. The ores typically contain low percentages of copper and have to be concentrated by, for example, froth flotation before refining.

he 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) sulphide (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 sulphur in the chalcopyrite turns into sulphur dioxide gas. This is used to make sulphuric acid via the Contact Process.

An overall equation for this series of steps is:

The copper(I) sulphide produced is converted to copper with a final blast of air.



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 doesn't 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.

Let's look at the oxidation states of everything.

• In the copper(I) sulphide, the copper is +1 and the sulphur -2.
• The oxidation states of the elements oxygen (in the gas) and copper (in the metal) are 0.
• In sulphur dioxide, the oxygen has an oxidation state of -2 and the sulphur +4.

That means that both the copper and the oxygen have been reduced (decrease in oxidation state). The sulphur has been oxidised (increase in oxidation state).
The reducing agent is therefore the sulphide ion in the copper(I) sulphide.

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.

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 oxidised 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 sulphurs on the left-hand side have been oxidised from -2 to +4 (the other is unchanged).
Once again, the sulphide ions are acting as the reducing agent.

Extracting copper from other ores
Copper can be extracted from non-sulphide ores by a different process involving three separate stages:

1. Reaction of the ore (over quite a long time and on a huge scale) with a dilute acid such as dilute sulphuric acid to produce a very dilute copper(II) sulphate solution.
2. Concentration of the copper(II) sulphate solution by solvent extraction.
   The very dilute solution is brought into contact with a relatively small amount of an organic solvent containing something which will bind with copper(II) ions so that they are removed from the dilute solution. The solvent mustn't mix with the water.
   The copper(II) ions are removed again from the organic solvent by reaction with fresh sulphuric acid, producing a much more concentrated copper(II) sulphate solution than before.
3. Electrolysis of the new solution. Copper(II) ions are deposited as copper on the cathode (for the electrode equation, see under the purification of copper below).
   The anodes for this process were traditionally lead-based alloys, but newer methods use titanium or stainless steel.
   The cathode is either a strip of very pure copper which the new copper plates on to, or stainless steel which it has to be removed from later.

Purification of copperWhen copper is made from sulphide ores by the first method above, it is impure. The blister copper is first treated to remove any remaining sulphur (trapped as bubbles of sulphur dioxide in the copper - hence "blister copper") and then cast into anodes for refining using electrolysis.

Electrolytic refining
The purification uses an electrolyte of copper(II) sulphate solution, impure copper anodes, and strips of high purity copper for the cathodes.
The diagram shows a very simplified view of a cell.

At the cathode, copper(II) ions are deposited as copper.

At the anode, copper goes into solution as copper(II) ions.

For every copper ion that is deposited at the cathode, in principle another one goes into solution at the anode. The concentration of the solution should stay the same.
All that happens is that there is a transfer of copper from the anode to the cathode. The cathode gets bigger as more and more pure copper is deposited; the anode gradually disappears.
In practice, it isn't quite as simple as that because of the impurities involved.

What happens to the impurities?
Any metal in the impure anode which is below copper in the electrochemical series (reactivity series) doesn't go into solution as ions. It stays as a metal and falls to the bottom of the cell as an "anode sludge" together with any unreactive material left over from the ore. The anode sludge will contain valuable metals such as silver and gold.
Metals above copper in the electrochemical series (like zinc) willform ions at the anode and go into solution. However, they won't get discharged at the cathode provided their concentration doesn't get too high.
The concentration of ions like zinc will increase with time, and the concentration of the copper(II) ions in the solution will fall. For every zinc ion going into solution there will obviously be one fewer copper ion formed. (See the next note if you aren't sure about this.)
The copper(II) sulphate solution has to be continuously purified to make up for this.