Chapter 12.1 - Displacement of Copper by Zinc

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.

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.

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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

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.

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• 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

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• 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

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In the next section, we will see a similar reaction. 

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Chapter 12 - Reactivity series and Electrochemistry

In the previous section, we completed a discussion on Equilibrium of chemical reactions. In this chapter we will discuss about Reactivity series and Electrochemistry.

A few metals are given below:
Sodium, magnesium, iron and copper.
Let us consider the reaction of each of the above metals with water:
First we take sodium.
1. Take a little water in a test tube and two drops of phenolphthalein to it
2. Pour equal volume of kerosene to this mixture. See fig.12.1 below:

Fig.12.1

 3. Put a small piece of sodium metal into it.
■ A reaction will occur. Note down the observations:
• Bubbles are formed on the surface of sodium
• When more bubbles are formed, the sodium piece goes up the kerosene and when all the bubbles break, it comes back to the water surface
• The water becomes pink in colour
■ Now we will see the explanation for each of the above observations:
1. Sodium and water reacts together according to the balanced equation given below:
2Na (s) + 2H₂O(l) ⟶ 2NaOH (aq) + H₂ (g)
• We can see that hydrogen gas is formed as one of the products. This hydrogen gas is the reason for the formation of bubbles. 
2. When more bubbles are formed, they rise to the top surface of kerosene. At the top surface of kerosene, the bubbles come into contact with air. Then the bubbles break and the sodium falls back into the top surface of water.
3. We can see that NaOH is also formed as a product. This makes the water in the test tube alkaline. We have already added phenolphthalein as the indicator. So the water turns pink in colour.
• Sodium does not react with kerosene. In the above experiment, kerosene was used so that we can see the formation of gas bubbles clearly. 

■ From the experiment, it is clear that sodium reacts with water even if the water is at ordinary temperature.

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Next we take magnesium.

1. Take a little cold water in one test tube 
2. Take a little hot water in another test tube 
3. Add one drop of phenolphthalein to each test tube 
4. Put some magnesium in each test tube 
Note down the observations:
• Bubbles come out of the test tube containing hot water. The water in this test tube becomes pink in colour
• There is no notable change in the test tube containing cold water
    ♦ But if we observe this test tube carefully for a long time, we will be able to see some bubbles come out slowly
■ Now we will see the explanation for the above observations:
1. Magnesium and water reacts according to the balanced equation given below:
Mg (s) + 2H₂O(l) ⟶ Mg(OH)₂ (aq) + H₂ (g)
2. But for this reaction, a higher temperature is required. So the magnesium reacts readily with hot water. It reacts with cold water only slowly.
• So we see quicker formation of hydrogen gas bubbles in the test tube containing hot water
3. We can see that magnesium hydroxide is also a product. This makes the water in the test tube alkaline. We have already added phenolphthalein as the indicator. So the water turns pink in colour. 
• Again note that, the pink colour for water is observed only in the test tube containing hot water.

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Next we take iron.

• Reaction of iron with water is some what familiar to us. We know that rust is formed as a result. But this reaction is slow. Will a 'higher temperature' increase the speed of this reaction?
• The answer is 'yes'. But we will need a very high temperature for this reaction. But at such high temperature, water will become steam. In fact, we need 'steam heated to high temperature'. It is called super heated steam. So we can say:
• Iron reacts with super heated steam. The balanced equation is:
3Fe (s) + 4H₂O(g) ⟶ Fe₃O₄ (s) + 4H₂ (g)
    ♦ Note that '(g)' is written after H₂O. This indicates that water is in the gaseous form.

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Finally we take copper from our list.

• But there is not much to write about the 'reaction between copper and water'.
• This is because copper will never react with water. Not even with super heated steam.

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■ So we see that some metals react quickly with water. While some others react slowly. And some others does not react at all. 

• If we are given a list of metals, we can arrange them in a 'decreasing order of reactivity' with water. Let us see an example:
• The given list is: Copper, Magnesium, Iron, Sodium 
• The arrangement in the order of decreasing reactivity is:
Sodium, Magnesium, Iron, Copper

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Now we will consider the reaction between metals and atmospheric air

First we take sodium. 
1. Cut a piece of sodium using a knife. 
2. Observe the freshly cut portion. 
• Within a short span of time, the shining lustre fades away. 
Let us see the reason:
• Sodium reacts with oxygen in the atmospheric air to form sodium oxide. The balanced equation is:
4Na (s) + O₂ (g) ⟶  2Na₂O (s)
• Sodium also reacts with water vapour in the atmospheric air and form sodium hydroxide. The balanced equation is:
2Na (s) + 2H₂O(l) ⟶ 2NaOH (s) + H₂ (g)
• The sodium hydroxide thus formed reacts with carbon dioxide and form sodium carbonate. The balanced equation is:
2NaOH (s) + CO₂ (g) ⟶ Na₂CO₃ (aq) + H₂ (g)
• So we see some new products: 2Na₂O,  2NaOH, Na₂CO₃
• These new products form a layer over the freshly cut portion and thus the shining lustre fades away.

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Next we take magnesium

• Magnesium also reacts with atmospheric air. 
• It also loses it's lustre. 
    ♦ But it will take a few days for this loss in lustre. 
• That means, this reaction is slow.

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Next we take aluminum.

• It too loses it's lustre. 

• But it takes even more time than magnesium.  

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• Next we take copper. 

• It too loses it's lustre. 

• But it takes even more time than aluminum. 

    ♦ In fact, it will takes months for the formation of verdigris on the surface of copper. 

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Finally we take gold. Gold does not react with atmospheric air at all. 

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• So it is clear that, like the reaction between 'metals and water', reaction between 'metals and atmospheric air' also varies. 

■ Some metals react quickly with atmospheric air. While some others react slowly. And some others does not react at all. 

• If we are given a list of metals, we can arrange them in a 'decreasing order of reactivity' with air. Let us see an example:
• The given list is: Magnesium, Copper, Gold, Sodium and Aluminum
• The arrangement in the order of decreasing reactivity is:
Sodium, Magnesium, Aluminum, Copper, Gold

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Now we will consider the reaction between metals and hydrochloric acid (HCl).

• We know that some metals react with acids like HCl and H₂SO₄ and produce hydrogen gas. We have seen the reaction between zinc and hydrochloric acid here. The balanced equation is:

Zn + 2HCl ⟶ ZnCl₂ + H₂ 

• In our present discussion we want to know the 'reactivity of various metals with HCl'. 

• We can get a general idea about the reactivity by observing the 'quantity of bubbles' formed. These bubbles are formed due to the production of hydrogen. 

    ♦ If more bubbles are formed, we can say that that metal has a greater reactivity with HCl. 

• The fig.12.2 below shows such an experiment.  In each test tube, a small metal piece is put in HCl. Different metals are put in different test tubes.  

Fig.12.2

• We can say that:

    ♦ The metal in test tube no.2 has a greater reactivity with HCl

    ♦ The metal in test tube no.4 has very low reactivity with HCl  

    ♦ The metal in test tube no.3 has no reactivity at all with HCl

• It is important to ensure that the following two conditions are satisfied while doing the experiment in fig.12.2:

    ♦ The 'quantity of bubbles' formed should be observed within the 'same time span' in all the test tubes

    ♦ Equal quantities of the reactants should be taken in all the test tubes
■ If the metals taken in the test tubes are:
Fe, Cu, Pb, Mg and zn, we will get the following result:
Mg> Zn> Fe> Pb> Cu

• That is., Mg will show the greatest reactivity with HCl and Cu will show the least reactivity

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In the next section, we will see Reactivity Series. 

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Chapter 11.3 - Factors which affect Equilibrium of a Chemical Reaction

In the previous section, we saw the equilibrium state of a chemical reaction. In this section we will learn about the factors which affect the Equilibrium.

Influence of concentration on Chemical equilibrium

■ Consider a system in equilibrium. If some quantities of a reactant is added to the system, what will happen? 
• Ans: The concentration of that reactant must decrease back to the previous level.  For that, the rate of forward reaction will increase, so as to convert the extra reactant to product.
Consider the example: N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
■ If we add some nitrogen to the system, what will happen?
Ans: The rate of forward reaction will increase so as to convert more nitrogen into ammonia. As a result, production of ammonia will increase.
■ If we add some ammonia to the system, what will happen?
Ans: The rate of backward reaction will increase so as to convert more ammonia back into nitrogen and hydrogen.
■ If we remove ammonia continuously from the system, what will happen?
Ans: The rate of forward reaction will increase so as to produce more ammonia

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• Effects of change in concentrations in a system producing ammonia can be written in a tabular form as follows:

Action	Change of Concentration	Change in rate
More hydrogen is added	Increases the concentration of the reactant	Rate of forward reaction increases
More ammonia is added	Increases the concentration of the product	Rate of backward reaction increases
Ammonia is removed	Decreases the concentration of the product	Rate of forward reaction increases
More nitrogen is added	Increases the concentration of the reactant	Rate of forward reaction increases

• The above discussion gives us an idea about the influence of concentrations on chemical equilibrium.

Influence of pressure on Chemical equilibrium

• Influence of pressure is felt in reactions involving gases. It can be analyzed using an example:
• Consider the reaction between nitrogen and hydrogen which gives ammonia as the product. The balanced chemical equation is: 
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
• One molecule of nitrogen reacts with 3 molecules of hydrogen to produce two molecules of ammonia. 
• So, if we take N_A (Avogadro number) of molecules of nitrogen, we must take 3N_A molecules of hydrogen. The result will be 2N_A molecules of ammonia.
• It is clear that, there are greater number of molecules on the reactant side than in the product side.
• In gaseous systems, larger number of molecules will produce greater pressure.

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Consider a state of equilibrium in the manufacture of ammonia.
■ If at this equilibrium we increase the pressure, what will happen?
Ans: The system will try to reduce that pressure. For that, the number of molecules must decrease.
• For a decrease in the number of molecules, more quantities of the product should be formed. Because, in this reaction, number of molecules on the product side is lower.
• So rate of the forward reaction will increase
■ If at this equilibrium we decrease the pressure, what will happen?
Ans: The system will try to increase the pressure. For that, the number of molecules must increase.
• For an increase in the number of molecules, more quantities of the reactants should be formed. Because, in this reaction, number of molecules on the reactants side is higher.
• So rate of the backward reaction will increase
■ In the manufacture of ammonia, why is a pressure of 200 - 900 atm used?
Ans: In the manufacture of ammonia, high pressure is used so that, the rate of the forward reaction will increase and more ammonia will be formed.

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Is an increase in pressure always helpful? Let us see another example:
Consider the balanced equation:
H₂ (g) + I₂ (g) ⇌ 2HI (g)
• One molecule of hydrogen reacts with 1 molecule of iodine to produce two molecules of hydrogen iodide. 
• So, if we take N_A (Avogadro number) of molecules of hydrogen, we must take N_A molecules of iodine. The result will be 2N_A molecules of hydrogen iodide.
• It is clear that, there are equal number of molecules on the reactant side and in the product side.
■ If there is no change in the number of reactant or product molecules as a result of forward and backward reactions, in gaseous reactions, pressure will not have any effect on the equilibrium state.

Influence of temperature on equilibrium

We have seen endothermic reactions and exothermic reactions in our earlier classes.
• Endothermic reactions are those in which heat is absorbed during the reaction
• Exothermic reactions are those in which heat is liberated during the reaction
Consider the equation of a reversible reaction:

1. In the forward reaction, N₂O₄ changes to NO₂. This forward reaction is endothermic. That is., if this forward reaction is to take place, we have to supply heat
2. In the backward reaction, NO₂ changes to N₂O₄. This backward reaction is exothermic. That is., when this backward reaction takes place, we will get heat.
3. Consider a system in which the above reaction is at equilibrium.
• If this system at equilibrium is placed in ice, the temperature will be lowered.
• The system will then try to increase the temperature by producing more heat.
• To produce heat, exothermic reaction must take place.
• So rate of the backward reaction increases. Thus So more NO₂ will be converted to N₂O₄.
4. Consider a system in which the above reaction is at equilibrium.
• If this system at equilibrium is placed in hot water, the temperature will be increased.
• The system will then try to decrease the temperature.
• To reduce temperature, heat must be absorbed. For heat to get absorbed, endothermic reaction must take place.
• So rate of the forward reaction increases. Thus So more N₂O₄ will be converted to NO₂.
■ Note that nitric acid (HNO₃) is a very corrosive acid and Nitrogen dioxide (NO₂) is a poisonous gas. All safety precautions should be taken while doing experiments.

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Let us now see the effect of temperature in the manufacture of ammonia.
1. We know that ammonia is manufactured by Haber process. It is a reversible reaction. It can be represented by the following balanced equation:
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
2. Here the forward reaction is exothermic in nature. That is., the forward reaction releases heat. So we can rewrite the equation as:
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g) + Heat
3. If we increase the temperature, the system will try to reduce it.
• For that, less heat must be produced. So the rate of backward reaction will increase. We will be getting more nitrogen and hydrogen instead of ammonia. So it is not advisable to increase the temperature.
4. If we cannot increase the temperature, can we decrease it?
• Reducing the temperature to low values has adverse effects. When the temperature is low, the rates of both forward and backward reactions become very low. (This is because, the molecules must have a minimum amount of energy for the collisions to take place)
• As a result, the system will take more time to reach equilibrium. Hence in the manufacture of ammonia, 450^o c is taken as optimum temperature.

Influence of catalyst on equilibrium

• We know that Positive catalysts are those substances which increase the rate of a reaction.
• In reversible reactions, there is forward as well as backward reactions taking place.
• A positive catalyst will increase the rate of both forward and backward reactions. So equilibrium is reached in a lesser time.
• Once equilibrium is reached, the rates of forward or backward reactions can be controlled by altering the other factors like temperature, pressure etc.,
• So, in many reversible reactions, positive catalyst is used at the beginning of the reaction.

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The French scientist Henry-Louis Le Chatelier put forward a principle regarding the effects of various factors on chemical equilibrium.
Le Chatelier's Principle:
When the concentration, pressure or temperature of a system at equilibrium is changed, the system will readjust itself so as to nullify the effect of that change. As a result of the readjustment, the system will attain a new state of equilibrium

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Now we will see a solved example:

Solved example 11.1

The manufacture of sulphuric acid by contact process has different stages. The chemical equation of one such stage is:

2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g) + heat

■ Identify how the factors given below influence this reaction

• Increase in the amount of oxygen

• Increase in pressure

• Maintains the temperature at optimum level

• Catalyst (V₂O₅) is added

• SO₃ is removed

Solution:

1. Increase in the amount of oxygen:

• Oxygen is one of the reactants.
• The system will try to nullify the effect of increase in concentration of a reactant by increasing the rate of the forward reaction.
• So more products will be formed
2. Increase in pressure:
• From the balanced equation, it is clear that, there are greater number of molecules on the reactant side than in the product side.
• In gaseous systems, larger number of molecules will produce greater pressure.
• The system will increase the rate of the forward reaction. So the number of molecules will become less and pressure will become low.
• Thus the effect of increase in pressure will be nullified
3. Maintains temperature at optimum level
• From the given balanced equation it is clear that, the forward reaction is exothermic
• If we increase the temperature, the system will try to nullify it by increasing the rate of the backward reaction. In that case we will get lesser quantities of the product.
• On the other hand, a decrease in temperature is also not advisable. Because, the molecules need a minimum amount of energy for the collisions to take place. 
• So, if we maintain an optimum temperature, we will get more products
4. Catalyst V₂O₅ is added
• The catalyst increases the rate of both forward and backward reactions. So it helps to achieve equilibrium in a lesser duration of time.
5. SO₃ is removed
• When SO₃ is removed, concentration of product decreases. To nullify the effect of this 'decrease in concentration', more SO₃ has to be produced. For that, the rate of forward reaction increases

Solved example 11.2
(a) In which of the following reactions does the change in pressure not influence equilibrium? What is the reason?
(i) H₂ (g) + I₂ (g) ⇌ 2HI (g)
(ii) N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
(b) What is the use of applying high pressure duriing the formation of ammonia from nitrogen and hydrogen?
Solution:
(a) In reaction (i), increase or decrease in pressure will not influence equilibrium. Because, the number of molecules in the reactants side and products side are equal
(b) If we increase the pressure, the system will try to reduce that pressure. For that, the number of molecules must decrease.
• For a decrease in the number of molecules, more quantities of the product should be formed. Because, in this reaction, number of molecules on the product side is lower.
• So rate of the forward reaction will increase and more ammonia will be formed.

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We have completed the present discussion on Chemical equilibrium. In the next chapter, we will see Reactivity Series and Electrochemistry. 

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