Consider the following situations of daily life and think what happens when -

• milk is left at room temperature during summers.
• an iron tawa/pan/nail is left exposed to humid atmosphere.
• grapes get fermented.
• food is cooked.
• food gets digested in our body.
• we respire.

In all the above situations, the nature and the identity of the initial substance have somewhat changed. We have already learnt about physical and chemical changes of matter in our previous classes. Whenever a chemical change occurs, we can say that a chemical reaction has taken place.

You may perhaps be wondering as to what is actually meant by a chemical reaction. How do we come to know that a chemical reaction has taken place? Let us perform some activities to find the answer to these questions.

Activity 1.1

CAUTION: This Activity needs the teacher’s assistance. It would be better if students wear suitable eyeglasses.

• Clean a magnesium ribbon about 3-4 cm long by rubbing it with sandpaper.
• Hold it with a pair of tongs. Burn it using a spirit lamp or burner and collect the ash so formed in a watch-glass as shown in Fig. 1.1. Burn the magnesium ribbon keeping it away as far as possible from your eyes.
• What do you observe?

Figure 1.1

Burning of a magnesium ribbon in air and collection of magnesium oxide in a watch-glass

Figure 1.2 Formation of hydrogen gas by the action of dilute sulphuric acid on zinc

You must have observed that magnesium ribbon burns with a dazzling white flame and changes into a white powder. This powder is magnesium oxide. It is formed due to the reaction between magnesium and oxygen present in the air.

Activity 1.2

• Take lead nitrate solution in a test tube.
• Add potassium iodide solution to this.
• What do you observe?

Activity 1.3

• Take a few zinc granules in a conical flask or a test tube.
• Add dilute hydrochloric acid or sulphuric acid to this (Fig. 1.2).

CAUTION: Handle the acid with care.

• Do you observe anything happening around the zinc granules?
• Touch the conical flask or test tube. Is there any change in its temperature?

From the above three activities, we can say that any of the following observations helps us to determine whether a chemical reaction has taken place –

• change in state
• change in colour
• evolution of a gas
• change in temperature.

As we observe the changes around us, we can see that there is a large variety of chemical reactions taking place around us. We will study about the various types of chemical reactions and their symbolic representation in this Chapter.

1.1 CHEMICAL EQUATIONS

Activity 1.1 can be described as – when a magnesium ribbon is burnt in oxygen, it gets converted to magnesium oxide. This description of a chemical reaction in a sentence form is quite long. It can be written in a shorter form. The simplest way to do this is to write it in the form of a word-equation. The word-equation for the above reaction would be –

$ \underset{(\text{Reactants})}{\mathrm{Magnesium}+\mathrm{Oxygen}} \rightarrow \underset{(\text{Product})}{\mathrm{Magnesium Oxide}} $

The substances that undergo chemical change in the reaction (1.1), magnesium and oxygen, are the reactants. The new substance is magnesium oxide, formed during the reaction, as a product.

A word-equation shows change of reactants to products through an arrow placed between them. The reactants are written on the left-hand side (LHS) with a plus sign (+) between them. Similarly, products are written on the right-hand side (RHS) with a plus sign (+) between them. The arrowhead points towards the products, and shows the direction of the reaction

1.1.1 Writing a Chemical Equation

Is there any other shorter way for representing chemical equations? Chemical equations can be made more concise and useful if we use chemical formulae instead of words. A chemical equation represents a chemical reaction. If you recall formulae of magnesium, oxygen and magnesium oxide, the above word-equation can be written as -

$$ \begin{equation*} \mathrm{Mg}+\mathrm{O} _{2} \rightarrow \mathrm{MgO} \tag{1.2} \end{equation*} $$

Count and compare the number of atoms of each element on the LHS and RHS of the arrow. Is the number of atoms of each element the same on both the sides? If yes, then the equation is balanced. If not, then the equation is unbalanced because the mass is not the same on both sides of the equation. Such a chemical equation is a skeletal chemical equation for a reaction. Equation (1.2) is a skeletal chemical equation for the burning of magnesium in air.

1.1.2 Balanced Chemical Equations

Recall the law of conservation of mass that you studied in Class IX; mass can neither be created nor destroyed in a chemical reaction. That is, the total mass of the elements present in the products of a chemical reaction has to be equal to the total mass of the elements present in the reactants.

In other words, the number of atoms of each element remains the same, before and after a chemical reaction. Hence, we need to balance a skeletal chemical equation. Is the chemical Eq. (1.2) balanced? Let us learn about balancing a chemical equation step by step.

The word-equation for Activity 1.3 may be represented as -

$\mathrm{Zinc + Sulphuric \quad acid}\rightarrow \mathrm{Zinc \quad sulphate + Hydrogen}$

The above word-equation may be represented by the following chemical equation -

$$\mathrm{Zn}+\mathrm{H} _{2} \mathrm{SO} _{4} \rightarrow \mathrm{ZnSO} _{4}+\mathrm{H} _{2} \tag{1.3}$$

Let us examine the number of atoms of different elements on both sides of the arrow.

As the number of atoms of each element is the same on both sides of the arrow, Eq. (1.3) is a balanced chemical equation.

Let us try to balance the following chemical equation -

$$ \begin{equation*} \mathrm{Fe}+\mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{Fe} _{3} \mathrm{O} _{4}+\mathrm{H} _{2} \tag{1.4} \end{equation*} $$

Step I: To balance a chemical equation, first draw boxes around each formula. Do not change anything inside the boxes while balancing the equation.

$$ \begin{equation*} \mathrm{Fe}+\mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{Fe} _{3} \mathrm{O} _{4}+\mathrm{H} _{2} \tag{1.5} \end{equation*} $$

Step II: List the number of atoms of different elements present in the unbalanced equation (1.5).

Step III: It is often convenient to start balancing with the compound that contains the maximum number of atoms. It may be a reactant or a product. In that compound, select the element which has the maximum number of atoms. Using these criteria, we select $Fe_3 O_4$ and the element oxygen in it. There are four oxygen atoms on the RHS and only one on the LHS.

To balance the oxygen atoms -

To equalise the number of atoms, it must be remembered that we cannot alter the formulae of the compounds or elements involved in the reactions. For example, to balance oxygen atoms we can put coefficient ‘4’ as $4 H_2 O$ and not $H_2 O_4$ or $(H_2 O)_4$. Now the partly balanced equation becomes -

$$ \begin{equation*}\boxed{\mathrm{Fe}}+4\boxed{\mathrm{H_2O}}\rightarrow \boxed{\mathrm{Fe_3O_4}}+\boxed{\mathrm{H_2}} \hspace{123 px} \tag {1.6} \end{equation*} $$

(partly balanced equation)

Step IV: Fe and $H$ atoms are still not balanced. Pick any of these elements to proceed further. Let us balance hydrogen atoms in the partly balanced equation.

To equalise the number of $H$ atoms, make the number of molecules of hydrogen as four on the RHS.

The equation would be -

$$ \begin{equation*} \mathrm{Fe}+4 \mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{Fe} _{3} \mathrm{O} _{4}+4 \mathrm{H} _{2} \tag{1.7} \end{equation*} $$

(partly balanced equation)

Step V: Examine the above equation and pick up the third element which is not balanced. You find that only one element is left to be balanced, that is, iron.

To equalise $Fe$, we take three atoms of $Fe$ on the LHS.

$$ \begin{equation*} 3 \boxed{\mathrm{Fe}}+4 \mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{Fe} _{3} \mathrm{O} _{4}+4 \mathrm{H} _{2} \tag{1.8} \end{equation*} $$

Step VI: Finally, to check the correctness of the balanced equation, we count atoms of each element on both sides of the equation.

$$ \begin{equation*} 3 \mathrm{Fe}+4 \mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{Fe} _{3} \mathrm{O} _{4}+4 \mathrm{H} _{2} \tag{1.9} \end{equation*} $$

(balanced equation)

The numbers of atoms of elements on both sides of Eq. (1.9) are equal. This equation is now balanced. This method of balancing chemical equations is called hit-and-trial method as we make trials to balance the equation by using the smallest whole number coefficient.

Step VII: Writing Symbols of Physical States Carefully examine the above balanced Eq. (1.9). Does this equation tell us anything about the physical state of each reactant and product? No information has been given in this equation about their physical states.

To make a chemical equation more informative, the physical states of the reactants and products are mentioned along with their chemical formulae. The gaseous, liquid, aqueous and solid states of reactants and products are represented by the notations (g), (l), (aq) and (s), respectively. The word aqueous (aq) is written if the reactant or product is present as a solution in water.

The balanced Eq. (1.9) becomes

$$ \begin{equation*} 3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H} _{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe} _{3} \mathrm{O} _{4}(\mathrm{~s})+4 \mathrm{H} _{2}(\mathrm{~g}) \tag{1.10} \end{equation*} $$

Note that the symbol $(g)$ is used with $H_2 O$ to indicate that in this reaction water is used in the form of steam.

Usually physical states are not included in a chemical equation unless it is necessary to specify them.

Sometimes the reaction conditions, such as temperature, pressure, catalyst, etc., for the reaction are indicated above and/or below the arrow in the equation. For example -

$$\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_2(\mathrm{~g}) \xrightarrow{340 \mathrm{~atm}} \mathrm{CH}_3 \mathrm{OH}(\mathrm{l}) \quad \tag{1.11}$$

$6 \mathrm{CO_2} (aq) +12 \mathrm{H_2 O(l)} \xrightarrow[\text{ Chlorophyll }]{\text{ Sunlight }} \underset{\text{ (Glucose) }}{\mathrm{C_6 H_{12} O_6} \text{ (aq) }}+6 \mathrm{O_2} (aq) +6 \mathrm{H_2 O}(l) \tag {1.12}$

Using these steps, can you balance Eq. (1.2) given in the text earlier?

QUESTIONS

1. Why should a magnesium ribbon be cleaned before burning in air?

2. Write the balanced equation for the following chemical reactions.

(i) Hydrogen + Chlorine → Hydrogen chloride

(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride

(iii) Sodium + Water → Sodium hydroxide + Hydrogen

3. Write a balanced chemical equation with state symbols for the following reactions.

(i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.

(ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.

1.2 TYPES OF CHEMICAL REACTIONS

We have learnt in Class IX that during a chemical reaction atoms of one element do not change into those of another element. Nor do atoms disappear from the mixture or appear from elsewhere. Actually, chemical reactions involve the breaking and making of bonds between atoms to produce new substances. You will study about types of bonds formed between atoms in Chapters 3 and 4.

1.2.1 Combination Reaction

Activity 1.4

• Take a small amount of calcium oxide or quick lime in a beaker.
• Slowly add water to this.
• Touch the beaker as shown in Fig. 1.3.
• Do you feel any change in temperature?

Figure 1.3 Formation of slaked lime by the reaction of calcium oxide with water

Calcium oxide reacts vigorously with water to produce slaked lime (calcium hydroxide) releasing a large amount of heat.

$$ \underset{\text{ (Quick lime) }}{\mathrm{CaO(s)}}+\quad \mathrm{H_2 O(l)} \to \underset{\text{ (Slaked lime) }}{\mathrm{Ca(OH)_2(aq)}}+\text{ Heat } \tag {1.13} $$

In this reaction, calcium oxide and water combine to form a single product, calcium hydroxide. Such a reaction in which a single product is formed from two or more reactants is known as a combination reaction.

Do you Know?

A solution of slaked lime produced by the reaction 1.13 is used for whitewashing walls. Calcium hydroxide reacts slowly with the carbon dioxide in air to form a thin layer of calcium carbonate on the walls. Calcium carbonate is formed after two to three days of whitewashing and gives a shiny finish to the walls. It is interesting to note that the chemical formula for marble is also CaCO3.

$\underset{(\text{Calcium hydroxide})}{\mathrm{Ca(OH)_2}}+\mathrm{CO_2(g)} \to \underset{(\text{Calcium carbonate})}{\mathrm{CaCO_3(s)}} + \mathrm{H_2O(l)} \tag {1.14}$

Let us discuss some more examples of combination reactions.

(i) Burning of coal

$$ \begin{equation*} \mathrm{C}(\mathrm{s})+\mathrm{O} _{2}(\mathrm{~g}) \rightarrow \mathrm{CO} _{2}(\mathrm{~g}) \tag{1.15} \end{equation*} $$

(ii) Formation of water from $H_2(g)$ and $O_2(g)$

$$ \begin{equation*} 2 \mathrm{H} _{2}(\mathrm{~g})+\mathrm{O} _{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H} _{2} \mathrm{O}(\mathrm{l}) \tag{1.16} \end{equation*} $$

In simple language we can say that when two or more substances (elements or compounds) combine to form a single product, the reactions are called combination reactions.

In Activity 1.4, we also observed that a large amount of heat is evolved. This makes the reaction mixture warm. Reactions in which heat is released along with the formation of products are called exothermic chemical reactions.

Other examples of exothermic reactions are -

(i) Burning of natural gas

$$ \begin{equation*} \mathrm{CH_4(g)+2 O_2(g) \to CO_2(g)+2 H_2 O(g)} \tag{1.17} \end{equation*} $$

(ii) Do you know that respiration is an exothermic process?

We all know that we need energy to stay alive. We get this energy from the food we eat. During digestion, food is broken down into simpler substances. For example, rice, potatoes and bread contain carbohydrates. These carbohydrates are broken down to form glucose. This glucose combines with oxygen in the cells of our body and provides energy. The special name of this reaction is respiration, the process of which you will study in Chapter 6.

$ \begin{equation*} \underset{\text{ (Glucose) }}{C_6 H_{12} O_6(aq)}+6 O_2(aq) \to 6 CO_2(aq)+6 H_2 O(l)+\text{ energy } \tag{1.18} \end{equation*} $

(iii) The decomposition of vegetable matter into compost is also an example of an exothermic reaction.

Identify the type of the reaction taking place in Activity 1.1, where heat is given out along with the formation of a single product.

1.2.2 Decomposition Reaction

Activity 1.5

• Take about $2 g$ ferrous sulphate crystals in a dry boiling tube.
• Note the colour of the ferrous sulphate crystals.
• Heat the boiling tube over the flame of a burner or spirit lamp as shown in Fig. 1.4.
• Observe the colour of the crystals after heating.

Figure 1.4 Correct way of heating the boiling tube containing crystals offerrous sulphate and of smelling the odour

Have you noticed that the green colour of the ferrous sulphate crystals has changed? You can also smell the characteristic odour of burning sulphur.

$ \underset{\text{(Ferrous sulphate)}}{\mathrm{2 FeSO_4(s)}} \xrightarrow[\text{ (Ferric oxide) }]{\text{ Heat }} \underset{\text{ Fe }}{\mathrm{Fe_2 O_3(s)+SO_2(g)+SO_3(g)}} \quad {1.19} $

In this reaction you can observe that a single reactant breaks down to give simpler products. This is a decomposition reaction. Ferrous sulphate crystals $(FeSO_4, 7 H_2 O)$ lose water when heated and the colour of the crystals changes. It then decomposes to ferric oxide $(Fe_2 O_3)$, sulphur dioxide $(SO_2)$ and sulphur trioxide $(SO_3)$. Ferric oxide is a solid, while $SO_2$ and $SO_3$ are gases.

Decomposition of calcium carbonate to calcium oxide and carbon dioxide on heating is an important decomposition reaction used in various industries. Calcium oxide is called lime or quick lime. It has many uses - one is in the manufacture of cement. When a decomposition reaction is carried out by heating, it is called thermal decomposition.

$\underset{\text{ (Limestone) }}{CaCO_3(s)} \stackrel{\text{ Heat }}{\longrightarrow} \underset{\text{ (Quick lime) }}{CaO(s)}+\underset{2}{CO_2(g)} \quad (1.20)$

Another example of a thermal decomposition reaction is given in Activity 1.6.

Activity 1.6

• Take about $2 g$ lead nitrate powder in a boiling tube.
• Hold the boiling tube with a pair of tongs and heat it over a flame, as shown in Fig. 1.5.
• What do you observe? Note down the change, if any.

You will observe the emission of brown fumes. These fumes are of nitrogen dioxide $(NO_2)$. The reaction that takes place is -

Figure 1.5 Heating of lead nitrate and emission of nitrogen dioxide

$\underset{\text { (Lead nitrate) }}{2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2(\mathrm{~s})} \xrightarrow{\text { Heat }} \underset{\text { (Lead oxide) }}{2 \mathrm{PbO}(\mathrm{s})}+\underset{\begin{array}{l}\text { (Nitrogen } \ \text { dioxide) }\end{array}}{4 \mathrm{NO}_2(\mathrm{~g})}+\underset{\text { (Oxygen) }}{\mathrm{O}_2(\mathrm{~g})} \quad (1.21)$

Let us perform some more decomposition reactions as given in Activities 1.7 and 1.8.

Activity 1.7

• Take a plastic mug. Drill two holes at its base and fit rubber stoppers in these holes. Insert carbon electrodes in these rubber stoppers as shown in Fig. 1.6.
• Connect these electrodes to a 6 volt battery.
• Fill the mug with water such that the electrodes are immersed. Add a few drops of dilute sulphuric acid to the water.
• Take two test tubes filled with water and invert them over the two carbon electrodes.
• Switch on the current and leave the apparatus undisturbed for some time.
• You will observe the formation of bubbles at both the electrodes. These bubbles displace water in the test tubes.
• Is the volume of the gas collected the same in both the test tubes?
• Once the test tubes are filled with the respective gases, remove them carefully.
• Test these gases one by one by bringing a burning candle close to the mouth of the test tubes.

CAUTION: This step must be performed carefully by the teacher.

• What happens in each case?
• Which gas is present in each test tube?

Figure 1.6 Electrolysis of water

Activity 1.8

• Take about $2 g$ silver chloride in a china dish.
• What is its colour?
• Place this china dish in sunlight for some time (Fig. 1.7).
• Observe the colour of the silver chloride after some time.

Figure 1.7 Silver chloride turns grey in sunlight to form silver metal light.

You will see that white silver chloride turns grey in sunlight. This is due to the decomposition of silver chloride into silver and chlorine by light.

$ \mathrm{2 AgCl(s) \xrightarrow{\text{ Sunlight }} 2 Ag(s)+Cl_2(g)} \quad (1.22) $

Silver bromide also behaves in the same way.

$\mathrm{2 AgBr(s) \xrightarrow{\text{ Sunlight }} 2 Ag(s)+Br_2(g)} \quad (1.23)$

The above reactions are used in black and white photography. What form of energy is causing these decomposition reactions?

We have seen that the decomposition reactions require energy either in the form of heat, light or electricity for breaking down the reactants. Reactions in which energy is absorbed are known as endothermic reactions.

Carry out the following Activity

Take about $2 g$ barium hydroxide in a test tube. Add $1 g$ of ammonium chloride and mix with the help of a glass rod. Touch the bottom of the test tube with your palm. What do you feel? Is this an exothermic or endothermic reaction?

QUESTIONS

1. A solution of a substance ’ $X$ ’ is used for whitewashing.

(i) Name the substance ’ $X$ ’ and write its formula.

(ii) Write the reaction of the substance ’ $X$ ’ named in (i) above with water.

2. Why is the amount of gas collected in one of the test tubes in Activity 1.7 double of the amount collected in the other? Name this gas.

1.2.3 Displacement Reaction

Activity 1.9

• Take three iron nails and clean them by rubbing with sand paper.
• Take two test tubes marked as (A) and (B). In each test tube, take about $10 mL$ copper sulphate solution.
• Tie two iron nails with a thread and immerse them carefully in the copper sulphate solution in test tube B for about 20 minutes [Fig. 1.8 (a)]. Keep one iron nail aside for comparison.
• After 20 minutes, take out the iron nails from the copper sulphate solution.
• Compare the intensity of the blue colour of copper sulphate solutions in test tubes (A) and (B) [Fig. 1.8 (b)].
• Also, compare the colour of the iron nails dipped in the copper sulphate solution with the one kept aside [Fig. 1.8 (b)].

Figure 1.8 (a) Iron nails dipped in copper sulphate solution

Figure 1.8 (b) Iron nails and copper sulphate solutions compared before and after the experiment

Why does the iron nail become brownish in colour and the blue colour of copper sulphate solution fades?

The following chemical reaction takes place in this Activity-

$\mathrm{Fe(s)}+\underset{\text{(Copper sulphate)}}{\mathrm{CuSO_4(aq)}} \rightarrow \underset{\text{Iron sulphate}}{\mathrm{FeSO_4(aq)}}+ \mathrm{Cu(s)} \quad (1.24)$

In this reaction, iron has displaced or removed another element, copper, from copper sulphate solution. This reaction is known as displacement reaction.

Other examples of displacement reactions are

$\mathrm{Zn(s)}+\underset{\text{(Copper sulphate)}}{\mathrm{CuSO_4(aq)}} \rightarrow \underset{\text{Zinc sulphate}}{\mathrm{PbCl_2(aq)}}+ \mathrm{Cu(s)}\quad (1.25)$

$\mathrm{Pb(s)}+\underset{\text{(Copper chloride)}}{\mathrm{CuSO_4(aq)}} \rightarrow \underset{\text{Lead sulphate}}{\mathrm{FeSO_4(aq)}}+ \mathrm{Cu(s)}\quad (1.26)$

Zinc and lead are more reactive elements than copper. They displace copper from its compounds.

1.2.4 Double Displacement Reaction

Activity 1.10

• Take about $3 mL$ of sodium sulphate solution in a test tube.
• In another test tube, take about $3 mL$ of barium chloride solution.
• Mix the two solutions (Fig. 1.9).
• What do you observe?

You will observe that a white substance, which is insoluble in water, is formed. This insoluble substance formed is known as a precipitate. Any reaction that produces a precipitate can be called a precipitation reaction.

Figure 1.9 Formation of barium sulphate and sodium chloride

$ \underset{\text{(Sodium sulphate)}}{\mathrm{Na_2SO_4(aq)}}+\underset{\text{(Barium chloride)}}{\mathrm{BaCl_2(aq)}} \rightarrow \underset{\text{(Barium sulphate)}}{\mathrm{BaSO_4(s)}}+ \underset{\text{(Sodium chloride)}}{\mathrm{2NaCl(aq)}} \quad (1.27)$

What causes this? The white precipitate of $BaSO_4$ is formed by the reaction of $SO_4^{2-}$ and $Ba^{2+}$. The other product formed is sodium chloride which remains in the solution. Such reactions in which there is an exchange of ions between the reactants are called double displacement reactions.

Recall Activity 1.2, where you have mixed the solutions of lead(II) nitrate and potassium iodide.

(i) What was the colour of the precipitate formed? Can you name the compound precipitated?

(ii) Write the balanced chemical equation for this reaction.

(iii) Is this also a double displacement reaction?

1.2.5 Oxidation and Reduction

Activity 1.11

• Heat a china dish containing about $1 g$ copper powder (Fig. 1.10).
• What do you observe?
• The surface of copper powder becomes coated with black copper(II) oxide. Why has this black substance formed?
• This is because oxygen is added to copper and copper oxide is formed.

$$ \begin{equation*} 2 \mathrm{Cu}+\mathrm{O} _{2} \xrightarrow{\text { Heat }} 2 \mathrm{CuO} \tag{1.28} \end{equation*} $$

Figure 1.10 Oxidation of copper to copper oxide

If hydrogen gas is passed over this heated material $(CuO)$, the black coating on the surface turns brown as the reverse reaction takes place and copper is obtained.

$$ \begin{equation*} \mathrm{CuO}+\mathrm{H} _{2} \xrightarrow{\text { Heat }} \mathrm{Cu}+\mathrm{H} _{2} \mathrm{O} \tag{1.29} \end{equation*} $$

If a substance gains oxygen during a reaction, it is said to be oxidised. If a substance loses oxygen during a reaction, it is said to be reduced.

During this reaction (1.29), the copper(II) oxide is losing oxygen and is being reduced. The hydrogen is gaining oxygen and is being oxidised. In other words, one reactant gets oxidised while the other gets reduced during a reaction. Such reactions are called oxidation-reduction reactions or redox reactions.

Some other examples of redox reactions are:

$$ \mathrm{Zno + C \to Zn + CO} \quad (1.31)$$

$$ \mathrm{MnO_2+4HCl \to MnCl_2 + 2H_2O + Cl_2} \quad (1.32)$$

In reaction (1.31) carbon is oxidised to $CO$ and $ZnO$ is reduced to $Zn$. In reaction (1.32) $HCl$ is oxidised to $Cl_2$ whereas $MnO_2$ is reduced to $MnCl_2$.

From the above examples we can say that if a substance gains oxygen or loses hydrogen during a reaction, it is oxidised. If a substance loses oxygen or gains hydrogen during a reaction, it is reduced.

Recall Activity 1.1, where a magnesium ribbon burns with a dazzling flame in air (oxygen) and changes into a white substance, magnesium oxide. Is magnesium being oxidised or reduced in this reaction?

1.3 HAVE YOU OBSERVED THE EFFECTS OF OXIDATION REACTIONS IN EVERYDAY LIFE?

1.3.1 Corrosion

You must have observed that iron articles are shiny when new, but get coated with a reddish brown powder when left for some time. This process is commonly known as rusting of iron. Some other metals also get tarnished in this manner. Have you noticed the colour of the coating formed on copper and silver? When a metal is attacked by substances around it such as moisture, acids, etc., it is said to corrode and this process is called corrosion. The black coating on silver and the green coating on copper are other examples of corrosion.

Corrosion causes damage to car bodies, bridges, iron railings, ships and to all objects made of metals, specially those of iron. Corrosion of iron is a serious problem. Every year an enormous amount of money is spent to replace damaged iron. You will learn more about corrosion in Chapter 3.

1.3.2 Rancidity

Have you ever tasted or smelt the fat/oil containing food materials left for a long time?

When fats and oils are oxidised, they become rancid and their smell and taste change. Usually substances which prevent oxidation (antioxidants) are added to foods containing fats and oil. Keeping food in air tight containers helps to slow down oxidation. Do you know that chips manufacturers usually flush bags of chips with gas such as nitrogen to prevent the chips from getting oxidised?

Questions

1. Why does the colour of copper sulphate solution change when an iron nail is dipped in it?

2. Give an example of a double displacement reaction other than the one given in Activity 1.10.

3. Identify the substances that are oxidised and the substances that are reduced in the following reactions.

(i) $\mathrm{4Na(s)+O_2(g) \to 2 Na_2 O(s)}$

(ii) $\mathrm{CuO(s)+H_2(g) \to Cu(s)+H_2 O(l)}$

What you have learnt

• A complete chemical equation represents the reactants, products and their physical states symbolically.
• A chemical equation is balanced so that the numbers of atoms of each type involved in a chemical reaction are the same on the reactant and product sides of the equation. Equations must always be balanced.
• In a combination reaction two or more substances combine to form a new single substance.
• Decomposition reactions are opposite to combination reactions. In a decomposition reaction, a single substance decomposes to give two or more substances.
• Reactions in which heat is given out along with the products are called exothermic reactions.
• Reactions in which energy is absorbed are known as endothermic reactions.
• When an element displaces another element from its compound, a displacement reaction occurs.
• Two different atoms or groups of atoms (ions) are exchanged in double displacement reactions.
• Precipitation reactions produce insoluble salts.
• Reactions also involve the gain or loss of oxygen or hydrogen by substances. Oxidation is the gain of oxygen or loss of hydrogen. Reduction is the loss of oxygen or gain of hydrogen.

Group Activity

Perform the following activity.

• Take four beakers and label them as A, B, C and D.
• Put $25 mL$ of water in $A, B$ and $C$ beakers and copper sulphate solution in beaker $D$.
• Measure and record the temperature of each liquid contained in the beakers above.
• Add two spatulas of potassium sulphate, ammonium nitrate, anhydrous copper sulphate and fine iron fillings to beakers A, B, C and D respectively and stir.
• Finally measure and record the temperature of each of the mixture above.

Find out which reactions are exothermic and which ones are endothermic in nature.