In the previous section, we completed a discussion on 'concentration of ore'. In this section, we will see the next stage.
The second stage after concentration of ore is:
■ Extraction of metal from concentrated ore
There are two stages in it:
(i) Conversion of the concentrated ore into it's oxide
(ii) Reduction of the oxide
• We will now see them in detail:
For the first stage, any one of the following two methods can be used.
(a) Calcination
(b) Roasting
The choice depends on the type of the ore
(a) Calcination:
1. The concentrated ore is heated to a high temperature in the absence of air
• But the temperature should not exceed the melting point of the ore
2. So the impurities which have lower melting points will melt and then evaporate away.
• The most common impurities which are removed in this way are water and organic matter
3. When the impurities are removed, the useful ore remains.
• These are the carbonates and hydroxides of the metal.
4. Due to the heat, these carbonates and hydroxides decompose to form the oxides of the metal.
• When oxides of the metal are formed, they are ready to be passed on to the next stage
5. Since there is no air, oxygen will not take part in this decomposition reaction.
Example: Zinc carbonate is converted into Zinc oxide
(b) Roasting:
1. The concentrated ore is heated to a high temperature in a current of air
• But the temperature should not exceed the melting point of the ore
• Because of the current of air, oxygen will also take part in the reaction
2. The impurities which have lower melting points will melt and then evaporate away.
• The most common impurities which are removed in this way are water and organic matter
3. When the impurities are removed, the useful ore remains.
• These remaining ore is converted into oxides because of the presence of oxygen
• When oxides of the metal are formed, they are ready to be passed on to the next stage
■ At this stage we will draw a flow chart to show various steps that we have seen so far:
• As we learn more details, we will fill up the missing spaces in the above chart.
• Now we move to the next step in 'extraction of metals', which is:
• We have seen some redox reactions in the previous chapter on 'Reactivity' also. Details here.
• When we 'reduce' the oxide of a metal, we will get the original metal.
• Suitable reducing agents can be used for this purpose.
Before we proceed further, let us discuss the reason for saying:
'reduce' the oxide of a metal.
The following steps will help us to understand the reason:
1. Let 'X' be a metal. So we can denote it's oxide as 'XO', where 'O' stands for oxygen.
2. Thus 'XO' is a compound. The individual elements in that compound are:
• The metal 'X'
• The oxygen 'O'
3. Now, X is a metal. Metals are electron donors.
• So in the compound XO, the metal X will be having a positive oxidation state. Let it be +m
• And O will be having a negative oxidation state. Let it be -n
4. So we can write: X+m O-n.
• Mg+2 O-2 is an example
5. Now we allow this X+m O-n to react with 'another compound'.
This 'another compound' must be able to donate electrons
6. The electrons thus donated will be received by X+m .
• Then X+m will become X0 or simply X
• The oxidation number is reduced from +m to zero
7. This X will no longer attach itself to O
• It becomes independent and we get the pure metal X
8. So we see that some 'reduction' is indeed involved
• The oxidation number decreases from +m to zero
• That is why we say:
'reduce' the oxide of a metal to get the pure metal
9. The 'another compound' that we saw in step (5) above is the reducing agent.
10. Electricity is the strongest reducing agent. It has an unlimited supply of electrons. Also, it's energy level can be increased by increasing the voltage.
Now we continue our original discussion:
Some examples of the application of reducing agents are:
• Carbon monoxide is used as the reducing agent to extract iron from haematite
■ So we completed the second stage also. The next stage is:
• Such impurities include small quantities of other metals and may be some non metals.
• We must remove them also to get the pure metal.
■ Following are some of the methods that can be used. The choice of the method will depend on the nature of the metals and the impurities present in them.
(a) Liquation
(b) Distillation
(c) Electrolytic refining
• Before seeing each of them in detail, we will complete the chart.
• This will give us a better understanding about our exact position in the whole discussion.
• The completed chart is shown in fig.13.5 below:
• Now we can continue the discussion. We were discussing the last step, which is 'Refining of the metal'.
• So we are in the bottom most green box. We can use any one of the three. The choice depends on the type of metal and the impurities.
(a) Liquation
(b) Distillation
(c) Electrolytic refining
♦ The pure metal to be extracted, has a low melting point
♦ The impurities are metals having high melting points
• Consider fig.13.6 below. We can write the steps:
1. The impure metal is supplied from the top of a specially shaped funnel
• This funnel is kept in an inclined position
2. The funnel is heated from it's bottom side
3. The pure metal which has low melting point will melt and will flow down along the inclined surface.
• It are collected at the bottom
4. The impurities having high melting points will be left behind
• Tin and lead, which have low melting points can be refined using this method
5. Note that, the structure in which heat is produced, is made of bricks
• But they are not ordinary bricks. They are 'refractory bricks'. They can withstand very high temperatures.
♦ The pure metal to be extracted, has a low boiling point
♦ The impurities are metals having high boiling points
• We can write the steps:
1. The impure metal is heated in a retort.
2. The pure metal alone will boil and turn into vapours
3. These vapours are collected and condensed to get the pure metal.
4. The impurities will be left behind
• Zinc, cadmium and mercury have low boiling points. So they can be refined by this method
• In this method, we use electrolysis to obtain the metal.
• We have seen the details about electrolysis in the previous chapter. (Details here).
• For electrolysis, we need an electrolyte. For our present case, we use a solution of the 'salt of the metal' as the electrolyte.
■ A small piece of the pure metal is used as the negative electrode.
■ A piece of the impure metal is used as the positive electrode.
• We want to know why pure metal is used as negative and the impure metal is used as positive electrodes.
• We will see the reason when we analyse the reaction that takes place when we turn on the switch.
We will write it in steps:
1. We want to refine the impure piece.
• For that, the pure metal should separate from the impurities and dissolve in the solution.
2. How does such a dissolution occur?
Ans: The positive metal ions get separated from the impure piece and goes into the solution.
3. Now we want these positive ions to be converted into pure metal.
• For that, the positive ions move to the negative electrode.
• There they receive electrons and will be converted into pure metal atoms.
4. So the formation of the 'new pure metal' occurs at the negative electrode.
• This newly formed pure metal should stick to the already available pure metal. Then only we can use it.
■ So the already available pure metal piece is used as the negative electrode.
5. The reaction taking place at the negative electrode is reduction.
• So the negative electrode is the cathode.
• The impure piece is the positive electrode and it is the anode
Also they are arranged in four groups.
Let us write the details:
1. The five members of the top most group are very reactive.
• We will need electricity to extract them from their compounds. These compounds are first heated and melted into the liquid state.
Then electrolysis is carried out
2. The five members of the second group are extracted by reducing their oxides.
• Depending on the metal, carbon or carbon monoxide can be used as the reducing agent.
3. There is only one member in the third group. It is copper.
• It has a lower reactivity than the first and second groups. It can be extracted more easily than the metals above it.
• First it's sulphide is converted to oxide and then the oxide is reduced.
4. The fourth group has two members.
• Their reactivity is very low. So they do not form any compounds.
• Thus they are found in free state in nature.
We can write a summary:
1. Metals have a tendency to lose electrons.
• So they are seen as positive ions in their compounds
2. The positive ions must be converted to ordinary atoms to obtain the pure metal
• So we must supply electrons to the positive ions
3. The positive ions receive these electrons and get reduced to the original atoms
■ So in short, if we are asked to choose between oxidation and reduction:
We must choose reduction to extract metal from it's compounds.
So we have seen the basics of metallurgy and the various stages involved in the extraction of metals. In the next section, we will see how these stages are applied in the case of iron.
The second stage after concentration of ore is:
■ Extraction of metal from concentrated ore
There are two stages in it:
(i) Conversion of the concentrated ore into it's oxide
(ii) Reduction of the oxide
• We will now see them in detail:
For the first stage, any one of the following two methods can be used.
(a) Calcination
(b) Roasting
The choice depends on the type of the ore
(a) Calcination:
1. The concentrated ore is heated to a high temperature in the absence of air
• But the temperature should not exceed the melting point of the ore
2. So the impurities which have lower melting points will melt and then evaporate away.
• The most common impurities which are removed in this way are water and organic matter
3. When the impurities are removed, the useful ore remains.
• These are the carbonates and hydroxides of the metal.
4. Due to the heat, these carbonates and hydroxides decompose to form the oxides of the metal.
• When oxides of the metal are formed, they are ready to be passed on to the next stage
5. Since there is no air, oxygen will not take part in this decomposition reaction.
Example: Zinc carbonate is converted into Zinc oxide
(b) Roasting:
1. The concentrated ore is heated to a high temperature in a current of air
• But the temperature should not exceed the melting point of the ore
• Because of the current of air, oxygen will also take part in the reaction
2. The impurities which have lower melting points will melt and then evaporate away.
• The most common impurities which are removed in this way are water and organic matter
3. When the impurities are removed, the useful ore remains.
• These remaining ore is converted into oxides because of the presence of oxygen
• When oxides of the metal are formed, they are ready to be passed on to the next stage
■ At this stage we will draw a flow chart to show various steps that we have seen so far:
 |
| Fig.13.4 |
• Now we move to the next step in 'extraction of metals', which is:
Reduction of the oxide
• We have seen the details about reducing agents and reduction reactions here.• We have seen some redox reactions in the previous chapter on 'Reactivity' also. Details here.
• When we 'reduce' the oxide of a metal, we will get the original metal.
• Suitable reducing agents can be used for this purpose.
Before we proceed further, let us discuss the reason for saying:
'reduce' the oxide of a metal.
The following steps will help us to understand the reason:
1. Let 'X' be a metal. So we can denote it's oxide as 'XO', where 'O' stands for oxygen.
2. Thus 'XO' is a compound. The individual elements in that compound are:
• The metal 'X'
• The oxygen 'O'
3. Now, X is a metal. Metals are electron donors.
• So in the compound XO, the metal X will be having a positive oxidation state. Let it be +m
• And O will be having a negative oxidation state. Let it be -n
4. So we can write: X+m O-n.
• Mg+2 O-2 is an example
5. Now we allow this X+m O-n to react with 'another compound'.
This 'another compound' must be able to donate electrons
6. The electrons thus donated will be received by X+m .
• Then X+m will become X0 or simply X
• The oxidation number is reduced from +m to zero
7. This X will no longer attach itself to O
• It becomes independent and we get the pure metal X
8. So we see that some 'reduction' is indeed involved
• The oxidation number decreases from +m to zero
• That is why we say:
'reduce' the oxide of a metal to get the pure metal
9. The 'another compound' that we saw in step (5) above is the reducing agent.
10. Electricity is the strongest reducing agent. It has an unlimited supply of electrons. Also, it's energy level can be increased by increasing the voltage.
Now we continue our original discussion:
Some examples of the application of reducing agents are:
• Carbon monoxide is used as the reducing agent to extract iron from haematite
♦ The equation is: Fe2O3 (s) + 3CO (s) → 2Fe (s) + 3CO2 (g)
♦ In Fe2O3, The Fe exists as positive ion. It receive electrons from the reducing agent CO.
♦ The Fe ion thus becomes Fe atom
• Carbon is used as reducing agent to extract zinc from zinc oxide.
♦ The equation is: ZnO (s) + C (s) → Zn (s) + CO (g)
♦ In ZnO, The Zn exists as positive ion. It receive electrons from the reducing agent C.
♦ The Zn ion thus becomes Zn atom
• Electricity is the strongest reducing agent. This electricity is required to extract highly reactive metals like sodium, potassium and calcium from their oxides.■ So we completed the second stage also. The next stage is:
Refining of metals
• The metal obtained by the reduction process that we discussed above, will still contain some impurities.• Such impurities include small quantities of other metals and may be some non metals.
• We must remove them also to get the pure metal.
■ Following are some of the methods that can be used. The choice of the method will depend on the nature of the metals and the impurities present in them.
(a) Liquation
(b) Distillation
(c) Electrolytic refining
• Before seeing each of them in detail, we will complete the chart.
• This will give us a better understanding about our exact position in the whole discussion.
• The completed chart is shown in fig.13.5 below:
 |
| Fig.13.5 |
• So we are in the bottom most green box. We can use any one of the three. The choice depends on the type of metal and the impurities.
(a) Liquation
(b) Distillation
(c) Electrolytic refining
Liquation
• This method is used when both the following condition are satisfied:♦ The pure metal to be extracted, has a low melting point
♦ The impurities are metals having high melting points
• Consider fig.13.6 below. We can write the steps:
 |
| Fig.13.6 |
• This funnel is kept in an inclined position
2. The funnel is heated from it's bottom side
3. The pure metal which has low melting point will melt and will flow down along the inclined surface.
• It are collected at the bottom
4. The impurities having high melting points will be left behind
• Tin and lead, which have low melting points can be refined using this method
5. Note that, the structure in which heat is produced, is made of bricks
• But they are not ordinary bricks. They are 'refractory bricks'. They can withstand very high temperatures.
Distillation
• This method is used when both the following condition are satified:♦ The pure metal to be extracted, has a low boiling point
♦ The impurities are metals having high boiling points
• We can write the steps:
1. The impure metal is heated in a retort.
2. The pure metal alone will boil and turn into vapours
3. These vapours are collected and condensed to get the pure metal.
4. The impurities will be left behind
• Zinc, cadmium and mercury have low boiling points. So they can be refined by this method
Electrolytic refining
• In this method, we use electrolysis to obtain the metal.
• We have seen the details about electrolysis in the previous chapter. (Details here).
• For electrolysis, we need an electrolyte. For our present case, we use a solution of the 'salt of the metal' as the electrolyte.
■ A small piece of the pure metal is used as the negative electrode.
■ A piece of the impure metal is used as the positive electrode.
• We want to know why pure metal is used as negative and the impure metal is used as positive electrodes.
• We will see the reason when we analyse the reaction that takes place when we turn on the switch.
We will write it in steps:
1. We want to refine the impure piece.
• For that, the pure metal should separate from the impurities and dissolve in the solution.
2. How does such a dissolution occur?
Ans: The positive metal ions get separated from the impure piece and goes into the solution.
3. Now we want these positive ions to be converted into pure metal.
• For that, the positive ions move to the negative electrode.
• There they receive electrons and will be converted into pure metal atoms.
4. So the formation of the 'new pure metal' occurs at the negative electrode.
• This newly formed pure metal should stick to the already available pure metal. Then only we can use it.
■ So the already available pure metal piece is used as the negative electrode.
5. The reaction taking place at the negative electrode is reduction.
• So the negative electrode is the cathode.
• The impure piece is the positive electrode and it is the anode
The chemistry of metallurgy
Consider the table given below. The metals are arranged in the decreasing order of their reactivity.Also they are arranged in four groups.
Let us write the details:
1. The five members of the top most group are very reactive.
• We will need electricity to extract them from their compounds. These compounds are first heated and melted into the liquid state.
Then electrolysis is carried out
2. The five members of the second group are extracted by reducing their oxides.
• Depending on the metal, carbon or carbon monoxide can be used as the reducing agent.
3. There is only one member in the third group. It is copper.
• It has a lower reactivity than the first and second groups. It can be extracted more easily than the metals above it.
• First it's sulphide is converted to oxide and then the oxide is reduced.
4. The fourth group has two members.
• Their reactivity is very low. So they do not form any compounds.
• Thus they are found in free state in nature.
We can write a summary:
1. Metals have a tendency to lose electrons.
• So they are seen as positive ions in their compounds
2. The positive ions must be converted to ordinary atoms to obtain the pure metal
• So we must supply electrons to the positive ions
3. The positive ions receive these electrons and get reduced to the original atoms
■ So in short, if we are asked to choose between oxidation and reduction:
We must choose reduction to extract metal from it's compounds.
So we have seen the basics of metallurgy and the various stages involved in the extraction of metals. In the next section, we will see how these stages are applied in the case of iron.
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