In the previous section, we saw the following three cases:
• Metal + water
• Metal + atmospheric air
• Metal + hydrochloric acid
Based on that discussion we can form a chart:
• In the above chart, as we move from top to bottom in each column, the reactivity decreases.
• It is clear from the chart that metals differ in their reactivity.
■ So scientists have prepared a series in which metals are arranged in their decreasing order of reactivity. It is known as the reactivity series. A portion of that series is shown below:
Note that the series shown above shows only some of the metals in the actual series. This is done to keep the list 'short' for our present discussion.
• The arrangement is in the decreasing order of reactivity. That is., as we move down the list, reactivity decreases.
♦ Take any metal from the series. It will have a lower reactivity than any metal above it.
■ Note that hydrogen is also shown in the list. This is done to mark a 'border'.
• All the metals above hydrogen will produce hydrogen while reacting with dilute acids.
♦ In other words: All the metals above hydrogen will displace hydrogen while reacting with dil. acids
Metals below hydrogen will not produce hydrogen while reacting with acids
♦ In other words: The metals below hydrogen will not displace hydrogen while reacting with dil. acids
• Why use the word 'displace'? We will see the answer soon.
• Some metals donate electrons more readily than others
• Reactivity of a metal is it's capacity to donate electrons.
• If a metal shows greater reactivity, it indicates that that metal has a greater capacity to donate electrons. That is why it enters into reactions more vigorously. And that metal will occupy a position near the top of the reactivity series.
■ Now we will learn about displacement reactions in a step by step manner:
1. Consider 'salt solutions' of metals. Some examples are:
• Sodium chloride (NaCl) is a salt of the metal sodium. So a solution of NaCl is a salt solution of the metal sodium
• Copper sulphate (CuSO4) is a salt of the metal copper. So a solution of CuSO4 is a salt solution of the metal copper
2. In a salt solution of a metal, that metal will be present as ions. Since metals are ready donors of electrons, the ions will be positive ions. So we can write:
• In a solution of NaCl, sodium will be present as Na+
• In a solution of , copper will be present as Cu2+
3. Suppose we have such a solution of metal A.
• Let us introduce another metal B in solid form into that solution.
• This new metal B has a greater reactivity than the metal A.
♦ That is., the new metal occupies a higher position in the reactivity series.
4. We want to know the result of the reaction between the two:
• Solution of the salt of metal A
• Metal B in solid form
5. Let us see an example:
• In a solution of CuSO4,we introduce a zinc rod.
• We know that zinc occupies a higher position than copper in the reactivity series. Then what will be the result?
Ans: The zinc has a greater tendency to give off electrons. So it will become Zn2+ ions.
• The copper, which is present as Cu2+ ions will receive these electrons and will become Cu atoms. Thus we will get pure copper.
• This is a displacement reaction. Because, copper has been displaced from it's solution. So we can write the definition for displacement reaction:
■ Displacement reaction is a chemical reaction in which a more reactive element displaces a less reactive element from it's compound.
Note down the observations:
1. Initially the zinc rod has a shining appearance. See image here
• But after the experiment, it's colour changes to reddish brown
2. Initially, the solution is blue in colour.
• But after the experiment, it's colour changes to pale blue
■ Let us see the explanation for the above two observations:
• The blue colour of the solution is due to the presence of Cu2+ ions.
• When the zinc rod is dipped in the solution, a reaction takes place. The equation for this reaction is:
Zn (s) + CuSO4 (aq) ⟶ ZnSO4 (aq) + Cu (s)
• We can see that pure copper is formed as one of the products.
• This is because, the Zn atoms, which is more reactive than Cu, donated electrons and became Zn2+ ions. These electrons were received by the Cu2+ ions, and they became Cu atoms.
• The newly formed Cu atoms stick to the surface of zinc rod. So it's colour become reddish brown. This is shown in the enlarged view inside the yellow circle in fig.12.3 above. The enlarged view shows what is actually happening in a small circular portion on the surface of the zinc rod
• As more and more Cu2+ ions becomes Cu atoms, the number of Cu2+ ions will obviously decrease. So the blue colour fades and become pale blue
■ If instead of zinc, we place a silver rod, the displacement reaction will not take place. Because silver is less reactive than copper.
♦ oxidation number of Zn increased from 0 to +2
♦ oxidation number of Cu decreased from +2 to 0
• So we can write:
♦ Zn is oxidized
★ In other words, Zn which was in it's pure form, is oxidized to a positive ion
♦ Cu is reduced
★ In other words, positive Cu ion is reduced to pure copper
■ In this reaction, oxidation and reduction take place simultaneously. It is a redox reaction.
■ Note the following points:
• Zn is more reactive than copper. Zn is situated above copper in the reactivity series
• As a result of the reaction:
♦ Zn donated electrons and got oxidized
♦ Cu received electrons and got reduced
In the next section, we will see a similar reaction.
• Metal + water
• Metal + atmospheric air
• Metal + hydrochloric acid
Based on that discussion we can form a chart:
• It is clear from the chart that metals differ in their reactivity.
■ So scientists have prepared a series in which metals are arranged in their decreasing order of reactivity. It is known as the reactivity series. A portion of that series is shown below:
• The arrangement is in the decreasing order of reactivity. That is., as we move down the list, reactivity decreases.
♦ Take any metal from the series. It will have a lower reactivity than any metal above it.
■ Note that hydrogen is also shown in the list. This is done to mark a 'border'.
• All the metals above hydrogen will produce hydrogen while reacting with dilute acids.
♦ In other words: All the metals above hydrogen will displace hydrogen while reacting with dil. acids
Metals below hydrogen will not produce hydrogen while reacting with acids
♦ In other words: The metals below hydrogen will not displace hydrogen while reacting with dil. acids
• Why use the word 'displace'? We will see the answer soon.
Reactivity series and displacement reactions
We know that metals in general have a tendency to donate electrons. We saw details when we learned about periodic trends.• Some metals donate electrons more readily than others
• Reactivity of a metal is it's capacity to donate electrons.
• If a metal shows greater reactivity, it indicates that that metal has a greater capacity to donate electrons. That is why it enters into reactions more vigorously. And that metal will occupy a position near the top of the reactivity series.
■ Now we will learn about displacement reactions in a step by step manner:
1. Consider 'salt solutions' of metals. Some examples are:
• Sodium chloride (NaCl) is a salt of the metal sodium. So a solution of NaCl is a salt solution of the metal sodium
• Copper sulphate (CuSO4) is a salt of the metal copper. So a solution of CuSO4 is a salt solution of the metal copper
2. In a salt solution of a metal, that metal will be present as ions. Since metals are ready donors of electrons, the ions will be positive ions. So we can write:
• In a solution of NaCl, sodium will be present as Na+
• In a solution of , copper will be present as Cu2+
3. Suppose we have such a solution of metal A.
• Let us introduce another metal B in solid form into that solution.
• This new metal B has a greater reactivity than the metal A.
♦ That is., the new metal occupies a higher position in the reactivity series.
4. We want to know the result of the reaction between the two:
• Solution of the salt of metal A
• Metal B in solid form
5. Let us see an example:
• In a solution of CuSO4,we introduce a zinc rod.
• We know that zinc occupies a higher position than copper in the reactivity series. Then what will be the result?
Ans: The zinc has a greater tendency to give off electrons. So it will become Zn2+ ions.
• The copper, which is present as Cu2+ ions will receive these electrons and will become Cu atoms. Thus we will get pure copper.
• This is a displacement reaction. Because, copper has been displaced from it's solution. So we can write the definition for displacement reaction:
■ Displacement reaction is a chemical reaction in which a more reactive element displaces a less reactive element from it's compound.
Now we will see an experiment to demonstrate the above findings.
• Prepare some CuSO4 solution in a beaker. Dip a Zn rod in it. See fig.12.3 below: |
| Fig.12.3 |
1. Initially the zinc rod has a shining appearance. See image here
• But after the experiment, it's colour changes to reddish brown
2. Initially, the solution is blue in colour.
• But after the experiment, it's colour changes to pale blue
■ Let us see the explanation for the above two observations:
• The blue colour of the solution is due to the presence of Cu2+ ions.
• When the zinc rod is dipped in the solution, a reaction takes place. The equation for this reaction is:
Zn (s) + CuSO4 (aq) ⟶ ZnSO4 (aq) + Cu (s)
• We can see that pure copper is formed as one of the products.
• This is because, the Zn atoms, which is more reactive than Cu, donated electrons and became Zn2+ ions. These electrons were received by the Cu2+ ions, and they became Cu atoms.
• The newly formed Cu atoms stick to the surface of zinc rod. So it's colour become reddish brown. This is shown in the enlarged view inside the yellow circle in fig.12.3 above. The enlarged view shows what is actually happening in a small circular portion on the surface of the zinc rod
• As more and more Cu2+ ions becomes Cu atoms, the number of Cu2+ ions will obviously decrease. So the blue colour fades and become pale blue
■ If instead of zinc, we place a silver rod, the displacement reaction will not take place. Because silver is less reactive than copper.
• We must analyse the above reaction in terms of Oxidation and Reduction
• We have learned about oxidation and reduction in a previous lesson. Details here.
• Consider the equation again:
Zn (s) + CuSO4 (aq) ⟶ ZnSO4 (aq) + Cu (s)
We will write the oxidation state of each component on the reactants side as well as the products side.
Reactants:
1. Zn is an atom of the element. So it's oxidation number will be obviously zero. We can write: Zn0
2. CuSO4 is an ionic compound. Let us write it as a combination: (Cu)(SO4)
• The sulphate ion (SO4)-2 has a known oxidation number of -2
• So the oxidation number of Cu in CuSO4 must be +2
• Thus we can write: Cu+2SO4-2
3. ZnSO4 is similar to CuSO4
• ZnSO4 is an ionic compound. Let us write it as a combination: (Zn)(SO4)
• The sulphate ion (SO4)-2 has a known oxidation number of -2
• So the oxidation number of Zn in must be +2
• Thus we can write: Zn+2SO4-2
4. Cu is an atom of the element. So it's oxidation number will be obviously zero. We can write: Cu0
• So the equation can be written as:
Zn0 (s) + Cu+2SO4-2 ⟶ Zn+2SO4-2 + Cu0
• In the above equation, we can see that:♦ oxidation number of Zn increased from 0 to +2
♦ oxidation number of Cu decreased from +2 to 0
• So we can write:
♦ Zn is oxidized
★ In other words, Zn which was in it's pure form, is oxidized to a positive ion
♦ Cu is reduced
★ In other words, positive Cu ion is reduced to pure copper
■ In this reaction, oxidation and reduction take place simultaneously. It is a redox reaction.
■ Note the following points:
• Zn is more reactive than copper. Zn is situated above copper in the reactivity series
• As a result of the reaction:
♦ Zn donated electrons and got oxidized
♦ Cu received electrons and got reduced
In the next section, we will see a similar reaction.
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