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Class 10th Chapters
1. Chemical Reactions And Equations 2. Acids, Bases And Salts 3. Metals And Non-Metals
4. Carbon And Its Compounds 5. Life Processes 6. Control And Coordination
7. How Do Organisms Reproduce? 8. Heredity 9. Light – Reflection And Refraction
10. The Human Eye And The Colourful World 11. Electricity 12. Magnetic Effects Of Electric Current
13. Our Environment



Chapter 1 Chemical Reactions And Equations



Introduction to Chemical Reactions

In our daily lives, we see many changes happening around us.

For instance, milk turns sour when left out in the summer. An iron pan or nail develops rust in humid air. Grapes get fermented, food is cooked, and our bodies digest the food we eat.

Even the process of respiration is a change. In all these examples, the original substance changes its nature and identity.

We have learned about physical and chemical changes before. Whenever a chemical change happens, we can say that a chemical reaction has taken place.

But what exactly is a chemical reaction? And how can we tell if one has occurred?

To find the answers to these questions, let us perform some simple activities.


Determining a Chemical Reaction

There are several key observations that help us determine if a chemical reaction has happened.

Activity 1: Burning of Magnesium Ribbon

First, a magnesium ribbon (about 3-4 cm long) is cleaned by rubbing it with sandpaper.

Using a pair of tongs, the ribbon is then burnt over a spirit lamp or burner. The ash that forms is collected in a watch-glass.

You will observe that the magnesium ribbon burns with a dazzling white flame.

It also changes into a white powder. This new substance is magnesium oxide.

It is formed because of the reaction between magnesium and the oxygen present in the air.

A diagram showing a pair of tongs holding a burning magnesium ribbon over a watch-glass which is collecting the white powder of magnesium oxide.

Observation: This activity clearly shows a change in state (from a solid metal to a powder) and a change in temperature (heat and light are produced).


Activity 2: Reaction between Lead Nitrate and Potassium Iodide

In another activity, we take lead nitrate solution in a test tube.

To this, we add potassium iodide solution. A distinct change is observed immediately.

Observation: A yellow, insoluble substance is formed. This is known as a precipitate.

This result demonstrates a change in colour and a change in state, as two clear liquids react to form a yellow solid.


Activity 3: Reaction between Zinc and Dilute Acid

For this activity, a few zinc granules are placed in a conical flask or a test tube.

Dilute hydrochloric acid or sulphuric acid is then carefully added.

You will notice something happening around the zinc granules. If you touch the flask, you will also feel a change.

A conical flask containing zinc granules and dilute sulphuric acid. A delivery tube is shown coming out, indicating gas formation.

Observations: Bubbles start to form around the zinc. This indicates the evolution of a gas, which is hydrogen gas.

The flask also becomes warm to the touch. This shows a change in temperature, as the reaction releases heat.


Activity 4: Reaction of Quicklime with Water

Take a small amount of calcium oxide, also known as quicklime, in a beaker.

Slowly and carefully add water to the beaker.

Touch the outside of the beaker to feel for any change.

Observations: You will hear a hissing sound as the water is added.

The beaker will become very hot, indicating a significant change in temperature.

The solid quicklime reacts with water to form a milky suspension, showing a change in the substance's appearance.


Activity 5: Reaction of Baking Soda with Vinegar

Take a spoonful of baking soda (sodium bicarbonate) and place it in a glass or a beaker.

Pour some vinegar (acetic acid) onto the baking soda.

Observe the reaction that takes place immediately.

Observations: The mixture will fizz and bubble vigorously. This is a clear sign of the evolution of a gas (carbon dioxide).

If you touch the beaker, you may notice it feels slightly cooler, indicating a change in temperature (this reaction absorbs a small amount of heat).



Writing Chemical Equations

Describing a chemical reaction in a sentence is often long and cumbersome. To represent chemical reactions in a more concise, informative, and efficient way, we use chemical equations.

Word Equations

The simplest method for describing a chemical reaction is to write it as a word equation.

In a word equation, the names of the substances that react are written on the left-hand side (LHS). These substances are called the reactants.

The names of the new substances formed are written on the right-hand side (RHS). These are called the products.

An arrow ($\rightarrow$) is placed between the reactants and products. The arrowhead points towards the products and indicates the direction of the reaction.

If there is more than one reactant or product, they are linked with a plus sign (+).

For the burning of magnesium ribbon (Activity 1), the word equation is:

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

Similarly, for the reaction of zinc with acid (Activity 3), the word equation would be:

Zinc + Sulphuric acid $\rightarrow$ Zinc sulphate + Hydrogen


Skeletal Chemical Equations

While word equations are a step up from sentences, they can be made even more concise and useful. This is done by using chemical formulae instead of words.

A chemical equation that uses symbols and formulae to represent a chemical reaction is the standard method used in chemistry.

Using formulae, the word equation for burning magnesium becomes:

$Mg + O_2 \rightarrow MgO$

This type of initial, unbalanced equation is known as a skeletal chemical equation.

It is called 'skeletal' or 'unbalanced' because the mass is not the same on both sides of the equation. If you count the atoms of each element, you will find they are not equal on the LHS and RHS.

For example, in the equation $Mg + O_2 \rightarrow MgO$:

Since the number of oxygen atoms is not the same on both sides, the equation is unbalanced. Such an equation violates the Law of Conservation of Mass, which is a fundamental principle in chemistry.



Balanced Chemical Equations

Recall the Law of Conservation of Mass that you studied in Class IX. It states that mass can neither be created nor destroyed in a chemical reaction.

This means that the total mass of the elements present in the products of a chemical reaction must be equal to the total mass of the elements present in the reactants.

For this to be true, the number of atoms of each element must remain the same before and after a chemical reaction. Hence, we need to balance skeletal chemical equations to make them consistent with this law. A balanced equation is a true representation of a chemical reaction.


Example of an Already Balanced Equation

Let's consider the reaction from Activity 1.3:

Word Equation: Zinc + Sulphuric acid $\rightarrow$ Zinc sulphate + Hydrogen

The skeletal equation is: $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$

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

Element Number of atoms in reactants (LHS) Number of atoms in products (RHS)
Zn 1 1
H 2 2
S 1 1
O 4 4

As the number of atoms of each element is the same on both sides of the arrow, this is a balanced chemical equation.


Steps to Balance a Chemical Equation (Hit-and-Trial Method)

Let us learn to balance an unbalanced equation step-by-step. We will use the reaction of iron with steam as an example:

$Fe + H_2O \rightarrow Fe_3O_4 + H_2$

Step I: Draw boxes around each formula.

First, draw boxes around each formula. It is important not to change anything inside the boxes while balancing. Changing the formula (e.g., changing $H_2O$ to $H_2O_2$) would mean changing the substance itself.

$\boxed{Fe} + \boxed{H_2O} \rightarrow \boxed{Fe_3O_4} + \boxed{H_2}$

Step II: List the number of atoms of different elements.

List the number of atoms for each element present in the unbalanced equation.

Element Number of atoms in reactants (LHS) Number of atoms in products (RHS)
Fe 1 3
H 2 2
O 1 4

Step III: Start balancing with the most complex compound.

It is often convenient to start balancing with the compound that contains the maximum number of atoms. In that compound, select the element which has the maximum number of atoms.

In our example, we select $Fe_3O_4$. The element oxygen has 4 atoms on the RHS and only 1 on the LHS. To balance oxygen atoms, we place a coefficient '4' in front of $H_2O$ on the LHS.

The partly balanced equation now becomes:

$\boxed{Fe} + 4\boxed{H_2O} \rightarrow \boxed{Fe_3O_4} + \boxed{H_2}$

Step IV: Balance the other atoms.

Now, Fe and H atoms are still not balanced. Let us balance hydrogen atoms. The number of H atoms on the LHS is now $4 \times 2 = 8$. On the RHS, there are only 2. To balance, we place a coefficient '4' in front of $H_2$ on the RHS.

The equation becomes:

$\boxed{Fe} + 4\boxed{H_2O} \rightarrow \boxed{Fe_3O_4} + 4\boxed{H_2}$

Step V: Balance the last remaining element.

Examine the equation again and pick the element which is still unbalanced. Here, it is iron (Fe). There is 1 Fe atom on the LHS and 3 on the RHS. To balance, we place a coefficient '3' in front of $Fe$ on the LHS.

$3\boxed{Fe} + 4\boxed{H_2O} \rightarrow \boxed{Fe_3O_4} + 4\boxed{H_2}$

Step VI: Finally, check the correctness of the balanced equation.

We count the atoms of each element on both sides of the final equation to verify the balancing.

$3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2$

The numbers of atoms are equal on both sides. The equation is now balanced.


More Examples of Balancing Equations

Example 1. Balance the equation for the combustion of methane.

Skeletal Equation: $CH_4 + O_2 \rightarrow CO_2 + H_2O$

Answer:

1. Balance Carbon (C): There is 1 C on the LHS and 1 C on the RHS. Carbon is already balanced.

2. Balance Hydrogen (H): There are 4 H on the LHS and 2 H on the RHS. Place a coefficient '2' in front of $H_2O$.

$CH_4 + O_2 \rightarrow CO_2 + 2H_2O$

3. Balance Oxygen (O): Now, there are 2 O on the LHS. On the RHS, there are 2 O in $CO_2$ and $2 \times 1 = 2$ O in $2H_2O$, making a total of 4 O. Place a coefficient '2' in front of $O_2$.

Balanced Equation: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

Example 2. Balance the equation for the reaction between barium chloride and aluminium sulphate.

Skeletal Equation: $BaCl_2 + Al_2(SO_4)_3 \rightarrow BaSO_4 + AlCl_3$

Answer:

1. Balance Polyatomic Ion (SO₄): Treat the sulphate ion ($SO_4$) as a single unit. There are 3 units on the LHS and 1 on the RHS. Place a '3' in front of $BaSO_4$.

$BaCl_2 + Al_2(SO_4)_3 \rightarrow 3BaSO_4 + AlCl_3$

2. Balance Barium (Ba): Now there are 3 Ba on the RHS and 1 on the LHS. Place a '3' in front of $BaCl_2$.

$3BaCl_2 + Al_2(SO_4)_3 \rightarrow 3BaSO_4 + AlCl_3$

3. Balance Aluminium (Al) and Chlorine (Cl): There are 2 Al on the LHS and 1 on the RHS. Place a '2' in front of $AlCl_3$. Now check Chlorine: LHS has $3 \times 2 = 6$ Cl atoms, and RHS has $2 \times 3 = 6$ Cl atoms. They are balanced.

Balanced Equation: $3BaCl_2 + Al_2(SO_4)_3 \rightarrow 3BaSO_4 + 2AlCl_3$


Step VII: Writing Symbols of Physical States and Reaction Conditions

To make a chemical equation fully informative, we add physical states and reaction conditions.

The physical states are: (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous solution.

The balanced equation for iron and steam becomes:

$3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)$

Reaction conditions like temperature, pressure, or a catalyst are indicated above or below the arrow.

Examples of Reaction Conditions

1. Photosynthesis:

This process requires sunlight and the catalyst chlorophyll.

$6CO_2(aq) + 6H_2O(l) \xrightarrow[Chlorophyll]{Sunlight} C_6H_{12}O_6(aq) + 6O_2(g)$

2. Thermal Decomposition of Calcium Carbonate:

This reaction requires a high temperature to proceed. The delta symbol ($\Delta$) signifies heat.

$CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)$

3. The Haber Process for Ammonia Production:

This industrial process requires specific conditions of temperature, pressure, and a catalyst.

$N_2(g) + 3H_2(g) \xrightarrow[200 \ atm, \ Fe \ catalyst]{450^\circ C} 2NH_3(g)$

4. Decomposition of Potassium Chlorate:

This reaction is sped up by a catalyst ($MnO_2$) and requires heat.

$2KClO_3(s) \xrightarrow[MnO_2]{\Delta} 2KCl(s) + 3O_2(g)$

5. Electrolysis of Water:

This reaction requires electrical energy to break down water.

$2H_2O(l) \xrightarrow{Electricity} 2H_2(g) + O_2(g)$



Types of Chemical Reactions

In Class IX, we learned that during a chemical reaction, atoms of one element do not change into those of another. 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. Reactions are classified into different types based on the nature of this chemical change.


1. Combination Reaction

A reaction in which a single product is formed from two or more reactants is known as a combination reaction. The reactants can be elements or compounds.

The general form of a combination reaction is: $A + B \rightarrow AB$

Activity 4: Reaction of Calcium Oxide with Water

Take a small amount of calcium oxide (quick lime) in a beaker. Slowly add water to this. When you touch the beaker, you will feel that it has become hot.

A diagram of a beaker containing water and calcium oxide, indicating that the reaction produces heat.

In this reaction, calcium oxide and water combine to form a single product, calcium hydroxide (slaked lime), releasing a large amount of heat.

$$\underset{\text{(Quick lime)}}{CaO(s)} + \underset{\text{(Water)}}{H_2O(l)} \rightarrow \underset{\text{(Slaked lime)}}{Ca(OH)_2(aq)} + Heat$$

Reactions in which heat is released along with the formation of products are called exothermic chemical reactions.

More Examples of Combination Reactions:

(i) Burning of coal: Carbon (coal) combines with oxygen to form a single product, carbon dioxide.

$C(s) + O_2(g) \rightarrow CO_2(g)$

(ii) Formation of water: Hydrogen gas combines with oxygen gas to form a single product, water.

$2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$

More Examples of Exothermic Reactions:

(i) Burning of natural gas (methane):

$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g) + Heat$

(ii) Respiration: During digestion, carbohydrates (from rice, potatoes, bread) are broken down into simpler substances like glucose. This glucose combines with oxygen in our cells and provides energy. This special reaction is called respiration.

$$\underset{\text{(Glucose)}}{C_6H_{12}O_6(aq)} + 6O_2(aq) \rightarrow 6CO_2(aq) + 6H_2O(l) + Energy$$

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


2. Decomposition Reaction

A reaction in which a single reactant breaks down to give simpler products is known as a decomposition reaction. It is the opposite of a combination reaction.

The general form is: $AB \rightarrow A + B$

Decomposition reactions require energy in the form of heat, light, or electricity for breaking down the reactants. Reactions that absorb energy are known as endothermic reactions.

Thermal Decomposition (using heat)

When a decomposition reaction is carried out by heating, it is called thermal decomposition.

(i) Decomposition of Ferrous Sulphate (Activity 1.5): When green crystals of ferrous sulphate ($FeSO_4 \cdot 7H_2O$) are heated, they first lose water. Upon further heating, the anhydrous ferrous sulphate decomposes to form ferric oxide (a brown solid), sulphur dioxide, and sulphur trioxide gas, which have a characteristic odour of burning sulphur.

$$\underset{\text{(Ferrous sulphate)}}{2FeSO_4(s)} \xrightarrow{Heat} \underset{\text{(Ferric oxide)}}{Fe_2O_3(s)} + SO_2(g) + SO_3(g)$$

(ii) Decomposition of Lead Nitrate (Activity 1.6): When lead nitrate powder is heated in a boiling tube, you will observe the emission of brown fumes. These fumes are of nitrogen dioxide ($NO_2$).

$$\underset{\text{(Lead nitrate)}}{2Pb(NO_3)_2(s)} \xrightarrow{Heat} \underset{\text{(Lead oxide)}}{2PbO(s)} + \underset{\text{(Nitrogen dioxide)}}{4NO_2(g)} + \underset{\text{(Oxygen)}}{O_2(g)}$$

Electrolytic Decomposition (using electricity)

Electrolysis of Water (Activity 1.7): When an electric current is passed through acidulated water, it decomposes into hydrogen and oxygen gas. Hydrogen gas is collected at the negative electrode (cathode) and oxygen gas is collected at the positive electrode (anode). The volume of hydrogen produced is double the volume of oxygen.

An apparatus showing the electrolysis of water, with two test tubes inverted over carbon electrodes in a mug of water, collecting hydrogen and oxygen gas.

$$2H_2O(l) \xrightarrow{Electricity} 2H_2(g) + O_2(g)$$

Photolytic Decomposition (using light)

Decomposition of Silver Chloride (Activity 1.8): Take some white silver chloride in a china dish and place it in sunlight. You will see that the white silver chloride turns grey. This is because light causes it to decompose into silver (grey) and chlorine gas.

A china dish containing silver chloride placed in sunlight.

$$2AgCl(s) \xrightarrow{Sunlight} 2Ag(s) + Cl_2(g)$$

Silver bromide also behaves in the same way. These reactions are used in black and white photography.

$$2AgBr(s) \xrightarrow{Sunlight} 2Ag(s) + Br_2(g)$$


3. Displacement Reaction

A reaction in which a more reactive element displaces or removes another, less reactive element from its salt solution. This is based on the reactivity series of metals.

The general form is: $A + BC \rightarrow AC + B$ (where A is more reactive than B)

Activity 1.9: Reaction of Iron with Copper Sulphate

When an iron nail is dipped in a blue copper sulphate solution, the following changes are observed after about 20 minutes: the iron nail becomes brownish in colour, and the blue colour of the copper sulphate solution fades.

Two test tubes with copper sulphate solution. One has iron nails dipped in it, showing a faded blue color. The nails removed from it are brownish.

This happens because iron is more reactive than copper. It displaces copper from the copper sulphate solution. The brown coating on the nail is copper metal, and the solution turns pale green due to the formation of iron sulphate.

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

More Examples of Displacement Reactions:

Zinc and lead are also more reactive than copper and will displace it from its compounds.

$$\underset{\text{(Zinc)}}{Zn(s)} + \underset{\text{(Copper sulphate)}}{CuSO_4(aq)} \rightarrow \underset{\text{(Zinc sulphate)}}{ZnSO_4(aq)} + \underset{\text{(Copper)}}{Cu(s)}$$

$$\underset{\text{(Lead)}}{Pb(s)} + \underset{\text{(Copper chloride)}}{CuCl_2(aq)} \rightarrow \underset{\text{(Lead chloride)}}{PbCl_2(aq)} + \underset{\text{(Copper)}}{Cu(s)}$$


4. Double Displacement Reaction

A reaction in which there is an exchange of ions between the reactants to form new compounds is called a double displacement reaction.

The general form is: $AB + CD \rightarrow AD + CB$

Activity 1.10: Reaction of Sodium Sulphate with Barium Chloride

When a solution of sodium sulphate is mixed with a solution of barium chloride, a white, insoluble substance is formed immediately. This insoluble substance is known as a precipitate. Any reaction that produces a precipitate can be called a precipitation reaction.

A diagram showing the mixing of two solutions from two test tubes, resulting in the formation of a precipitate.

The white precipitate of barium sulphate ($BaSO_4$) is formed by the reaction of $Ba^{2+}$ ions and $SO_4^{2-}$ ions. The other product, sodium chloride, remains dissolved in the solution.

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

Another Example of a Double Displacement Reaction:

When lead(II) nitrate solution is mixed with potassium iodide solution, a yellow precipitate of lead(II) iodide is formed.

$$\underset{\text{(Lead(II) nitrate)}}{Pb(NO_3)_2(aq)} + \underset{\text{(Potassium iodide)}}{2KI(aq)} \rightarrow \underset{\text{(Lead(II) iodide)}}{PbI_2(s)} + \underset{\text{(Potassium nitrate)}}{2KNO_3(aq)}$$



Oxidation-Reduction and its Effects in Daily Life

Oxidation and Reduction (Redox Reactions)

Oxidation and reduction are fundamental chemical processes that are often linked. They can be defined in several ways, but a simple definition involves the transfer of oxygen or hydrogen.

Activity 1.11: Oxidation of Copper

If you heat a china dish containing about 1 g of copper powder, you will observe that the surface of the copper powder becomes coated with a black substance. Why has this black substance formed?

A china dish containing copper powder is being heated over a tripod stand, leading to the oxidation of copper.

This is because oxygen from the air is added to copper, and copper(II) oxide is formed. In this reaction, copper has gained oxygen, so it is said to be oxidised.

$$\underset{\text{(Copper)}}{2Cu} + \underset{\text{(Oxygen)}}{O_2} \xrightarrow{Heat} \underset{\text{(Copper(II) oxide)}}{2CuO}$$

Now, if hydrogen gas is passed over this heated black material (CuO), the black coating on the surface turns brown. This is because the reverse reaction takes place, and copper is obtained again.

$$\underset{\text{(Copper(II) oxide)}}{CuO} + \underset{\text{(Hydrogen)}}{H_2} \xrightarrow{Heat} \underset{\text{(Copper)}}{Cu} + \underset{\text{(Water)}}{H_2O}$$

In this reaction (1.29), two things are happening simultaneously:

In other words, one reactant gets oxidised while the other gets reduced during a reaction. Such reactions are called oxidation-reduction reactions or, more commonly, redox reactions.

More Examples of Redox Reactions:

(i) $$\underset{\text{(Zinc oxide)}}{ZnO} + \underset{\text{(Carbon)}}{C} \rightarrow \underset{\text{(Zinc)}}{Zn} + \underset{\text{(Carbon monoxide)}}{CO}$$

In this reaction, Carbon (C) is oxidised to Carbon monoxide (CO), and Zinc oxide (ZnO) is reduced to Zinc (Zn).

(ii) $$\underset{\text{(Manganese dioxide)}}{MnO_2} + \underset{\text{(Hydrochloric acid)}}{4HCl} \rightarrow \underset{\text{(Manganese chloride)}}{MnCl_2} + \underset{\text{(Water)}}{2H_2O} + \underset{\text{(Chlorine)}}{Cl_2}$$

In this reaction, HCl is oxidised to $Cl_2$ (as it loses hydrogen), whereas $MnO_2$ is reduced to $MnCl_2$ (as it loses oxygen).


Effects of Oxidation in Everyday Life

Oxidation reactions have a significant impact on our daily lives, and not all of them are beneficial. Two common damaging effects are corrosion and rancidity.

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 the rusting of iron.

Corrosion is the process in which a metal is attacked by substances around it such as moisture, air, acids, etc. The metal is said to to corrode, and it slowly deteriorates.

Corrosion causes damage to car bodies, bridges, iron railings, ships, and all objects made of metals, especially those of iron. Corrosion of iron is a serious problem. Every year, an enormous amount of money is spent to replace damaged iron structures.

2. Rancidity

Have you ever tasted or smelt food containing fat or oil that has been left for a long time? When fats and oils are oxidised, they become rancid, and their smell and taste change, making them unsuitable for consumption.

This process of oxidation of fats and oils is called rancidity. There are several ways to slow down or prevent this process:



Intext Questions



Page No. 6

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

Answer:

A magnesium ribbon must be cleaned before burning in air to remove the protective layer of magnesium oxide from its surface.

Elaboration:

Magnesium ($Mg$) is a reactive metal. When it is stored and exposed to air, it slowly reacts with the oxygen ($O_2$) present in the atmosphere. This reaction forms a thin, dull, and stable layer of magnesium oxide ($MgO$) on the surface of the ribbon.

The chemical reaction for this slow oxidation is:

$2Mg(s) + O_2(g) \rightarrow 2MgO(s)$

This magnesium oxide layer is non-combustible and acts as a protective barrier. It prevents the underlying pure magnesium from coming into direct contact with the oxygen needed for burning. If you try to burn the ribbon without cleaning it, the ignition will be difficult, slow, or may not happen at all.

By cleaning the ribbon with sandpaper, this unreactive oxide layer is physically scraped off, exposing the fresh, shiny, and highly reactive magnesium metal. This allows the magnesium to react rapidly with oxygen when heated, resulting in the characteristic dazzling white flame.

Question 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

Answer:

(i) Hydrogen + Chlorine → Hydrogen chloride

1. Skeletal Equation: Hydrogen gas ($H_2$) reacts with Chlorine gas ($Cl_2$) to form Hydrogen chloride ($HCl$).

$H_2 + Cl_2 \rightarrow HCl$

2. Balancing: On the LHS, there are 2 atoms of Hydrogen and 2 atoms of Chlorine. On the RHS, there is 1 atom of each. To balance, we place a coefficient of 2 in front of HCl.

Balanced Equation: $H_2 + Cl_2 \rightarrow 2HCl$

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

1. Skeletal Equation: Barium chloride ($BaCl_2$) reacts with Aluminium sulphate ($Al_2(SO_4)_3$) to form Barium sulphate ($BaSO_4$) and Aluminium chloride ($AlCl_3$).

$BaCl_2 + Al_2(SO_4)_3 \rightarrow BaSO_4 + AlCl_3$

2. Balancing:

  • Balance Al: There are 2 Al atoms on the LHS, so we put a 2 in front of $AlCl_3$ on the RHS.
  • $BaCl_2 + Al_2(SO_4)_3 \rightarrow BaSO_4 + 2AlCl_3$

  • Balance Cl: Now there are $2 \times 3 = 6$ Cl atoms on the RHS. So, we put a 3 in front of $BaCl_2$ on the LHS.
  • $3BaCl_2 + Al_2(SO_4)_3 \rightarrow BaSO_4 + 2AlCl_3$

  • Balance Ba and SO₄: Now there are 3 Ba atoms on the LHS. So, we put a 3 in front of $BaSO_4$ on the RHS. This also balances the sulphate ions (3 on each side).

Balanced Equation: $3BaCl_2 + Al_2(SO_4)_3 \rightarrow 3BaSO_4 + 2AlCl_3$

(iii) Sodium + Water → Sodium hydroxide + Hydrogen

1. Skeletal Equation: Sodium ($Na$) reacts with Water ($H_2O$) to form Sodium hydroxide ($NaOH$) and Hydrogen gas ($H_2$).

$Na + H_2O \rightarrow NaOH + H_2$

2. Balancing:

  • Atom Count: LHS: 1 Na, 2 H, 1 O. RHS: 1 Na, 3 H, 1 O.
  • Balance H: The hydrogen count is odd on the RHS. To make it even, we place a 2 in front of $NaOH$.
  • $Na + H_2O \rightarrow 2NaOH + H_2$

  • Re-count & Balance: Now RHS has 2 Na, 4 H, and 2 O. We balance Na by putting a 2 in front of $Na$ on the LHS. Then we balance O and H by putting a 2 in front of $H_2O$ on the LHS.

Balanced Equation: $2Na + 2H_2O \rightarrow 2NaOH + H_2$

Question 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.

Answer:

(i) Reaction of Barium Chloride and Sodium Sulphate

1. Translate to Formulae with States:

  • "Solution of barium chloride": $BaCl_2(aq)$
  • "Solution of sodium sulphate": $Na_2SO_4(aq)$
  • "insoluble barium sulphate": This is a solid precipitate, so $BaSO_4(s)$
  • "solution of sodium chloride": $NaCl(aq)$

2. Skeletal Equation:

$BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + NaCl(aq)$

3. Balancing: On the LHS, there are 2 Na and 2 Cl atoms. On the RHS, there is only 1 of each. We place a coefficient of 2 in front of NaCl to balance both.

Balanced Equation with State Symbols:

$BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)$

(ii) Reaction of Sodium Hydroxide and Hydrochloric Acid

1. Translate to Formulae with States:

  • "Sodium hydroxide solution": $NaOH(aq)$
  • "hydrochloric acid solution": $HCl(aq)$
  • "sodium chloride solution": $NaCl(aq)$
  • "water": This is a pure liquid product, so $H_2O(l)$

2. Skeletal Equation:

$NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)$

3. Balancing: We count the atoms on both sides.

  • LHS: 1 Na, 1 O, 2 H, 1 Cl
  • RHS: 1 Na, 1 O, 2 H, 1 Cl

The equation is already balanced as written.

Balanced Equation with State Symbols:

$NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)$



Page No. 10

Question 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.

Answer:

(i) Name and Formula of Substance ‘X’

The substance ‘X’, a solution of which is used for whitewashing, is Calcium Oxide, also commonly known as Quick Lime.

The chemical formula for substance ‘X’ is CaO.

(ii) Reaction of Substance ‘X’ with Water

When substance ‘X’ (Calcium Oxide) is mixed with water, it reacts vigorously to produce Calcium Hydroxide, also known as Slaked Lime. This reaction releases a large amount of heat, making it an exothermic reaction. The solution of slaked lime is what is used for whitewashing.

The balanced chemical equation for the reaction is:

$$\underset{\text{(Quick Lime)}}{CaO(s)} + \underset{\text{(Water)}}{H_2O(l)} \rightarrow \underset{\text{(Slaked Lime)}}{Ca(OH)_2(aq)} + Heat$$

Question 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.

Answer:

Activity 1.7 describes the electrolysis of water, where water ($H_2O$) decomposes into hydrogen gas ($H_2$) and oxygen gas ($O_2$) when an electric current is passed through it.

The balanced chemical equation for this reaction is:

$2H_2O(l) \xrightarrow{Electricity} 2H_2(g) + O_2(g)$

Elaboration:

From the balanced equation, we can see that the decomposition of 2 molecules of water produces 2 molecules of hydrogen gas and only 1 molecule of oxygen gas. This means the molar ratio of hydrogen gas produced to oxygen gas produced is 2:1.

According to Avogadro's Law, at the same temperature and pressure, the volume of a gas is directly proportional to the number of moles (or molecules). Therefore, the volume of the gas collected will also be in the same 2:1 ratio.

This is why the amount (volume) of gas collected in one test tube is double the amount collected in the other.

The gas collected in double the amount is Hydrogen ($H_2$), which collects at the negative electrode (cathode). The other gas, collected in a single volume, is Oxygen ($O_2$), which collects at the positive electrode (anode).



Page No. 13

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

Answer:

The colour of the copper sulphate solution changes when an iron nail is dipped in it because a displacement reaction takes place.

Elaboration:

Iron (Fe) is more reactive than copper (Cu). According to the reactivity series, a more reactive metal can displace a less reactive metal from its salt solution. In this case, iron displaces copper from the copper sulphate solution.

The chemical reaction that occurs is:

$$\underset{\text{(Iron)}}{Fe(s)} + \underset{\text{(Copper Sulphate - Blue)}}{CuSO_4(aq)} \rightarrow \underset{\text{(Iron Sulphate - Pale Green)}}{FeSO_4(aq)} + \underset{\text{(Copper - Brown)}}{Cu(s)}$$

Initially, the copper sulphate solution ($CuSO_4$) is blue in colour. As the reaction proceeds, it is consumed and a new substance, iron sulphate ($FeSO_4$), is formed. Iron sulphate solution is pale green in colour.

Therefore, the blue colour of the solution gradually fades and turns into a pale green colour. Additionally, a reddish-brown layer of copper metal gets deposited on the surface of the iron nail.

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

Answer:

An example of a double displacement reaction, other than the reaction between sodium sulphate and barium chloride, is the reaction between silver nitrate and sodium chloride.

Elaboration:

When an aqueous solution of silver nitrate ($AgNO_3$) is mixed with an aqueous solution of sodium chloride ($NaCl$), an exchange of ions ($Ag^+$ and $Na^+$, $NO_3^-$ and $Cl^-$) takes place. This results in the formation of a white, insoluble precipitate of silver chloride ($AgCl$) and a solution of sodium nitrate ($NaNO_3$).

The balanced chemical equation for this reaction is:

$$\underset{\text{(Silver Nitrate)}}{AgNO_3(aq)} + \underset{\text{(Sodium Chloride)}}{NaCl(aq)} \rightarrow \underset{\text{(Silver Chloride - White Ppt.)}}{AgCl(s)} + \underset{\text{(Sodium Nitrate)}}{NaNO_3(aq)}$$

This is also an example of a precipitation reaction.

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

(i) $4Na(s) + O_2(g) \rightarrow 2Na_2O(s)$

(ii) $CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(l)$

Answer:

(i) $4Na(s) + O_2(g) \rightarrow 2Na_2O(s)$

  • Substance Oxidised: Sodium (Na).

    Reason: Sodium is gaining oxygen to form sodium oxide ($Na_2O$). The gain of oxygen is oxidation.

  • Substance Reduced: Oxygen ($O_2$).

    Reason: Oxygen is combining with a metal to form an oxide. In a redox reaction, if one substance is oxidised, the other must be reduced. Here, oxygen acts as the oxidising agent and is itself reduced.

(ii) $CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(l)$

  • Substance Oxidised: Hydrogen ($H_2$).

    Reason: Hydrogen is gaining oxygen to form water ($H_2O$). The gain of oxygen is oxidation.

  • Substance Reduced: Copper(II) oxide (CuO).

    Reason: Copper(II) oxide is losing its oxygen to become copper (Cu). The loss of oxygen is reduction.



Exercises



Question 1. Which of the statements about the reaction below are incorrect?

$2PbO(s) + C(s) \rightarrow 2Pb(s) + CO_2(g)$

(a) Lead is getting reduced.

(b) Carbon dioxide is getting oxidised.

(c) Carbon is getting oxidised.

(d) Lead oxide is getting reduced.

(i) (a) and (b)

(ii) (a) and (c)

(iii) (a), (b) and (c)

(iv) all

Answer:

(i) (a) and (b)

Explanation:

  • In the reaction, Lead oxide (PbO) is losing oxygen to become Lead (Pb). The loss of oxygen is reduction. So, statement (d) is correct.
  • Carbon (C) is gaining oxygen to become Carbon dioxide (CO₂). The gain of oxygen is oxidation. So, statement (c) is correct.
  • Statement (a) is incorrect because Lead oxide (a reactant) is getting reduced, not the element Lead (which is a product).
  • Statement (b) is incorrect because Carbon dioxide (a product) is not getting oxidised; it is the product of the oxidation of Carbon.

Therefore, the incorrect statements are (a) and (b).

Question 2. $Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe$

The above reaction is an example of a

(a) combination reaction.

(b) double displacement reaction.

(c) decomposition reaction.

(d) displacement reaction.

Answer:

(d) displacement reaction.

Explanation:

In this reaction, Aluminium (Al), being more reactive than Iron (Fe), is displacing Iron from its compound, Iron(III) oxide ($Fe_2O_3$). A more reactive element displacing a less reactive element from its compound is a characteristic of a displacement reaction.

Question 3. What happens when dilute hydrochloric acid is added to iron fillings? Tick the correct answer.

(a) Hydrogen gas and iron chloride are produced.

(b) Chlorine gas and iron hydroxide are produced.

(c) No reaction takes place.

(d) Iron salt and water are produced.

Answer:

(a) Hydrogen gas and iron chloride are produced.

Explanation:

Iron (Fe) is a metal that is above hydrogen in the reactivity series. When a reactive metal reacts with a dilute acid, it displaces hydrogen from the acid to form a metal salt and hydrogen gas.

The reaction is: $Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)$

Question 4. What is a balanced chemical equation? Why should chemical equations be balanced?

Answer:

Balanced Chemical Equation:

A balanced chemical equation is a chemical equation in which the number of atoms of each element on the reactant side is equal to the number of atoms of that element on the product side. It represents a chemical reaction in its true quantitative sense.

Reason for Balancing:

Chemical equations should be balanced to satisfy the Law of Conservation of Mass. This fundamental law of chemistry states that mass can neither be created nor destroyed in a chemical reaction. For the mass to remain conserved, the total number of atoms of each element participating in the reaction must be the same before and after the reaction. Balancing the equation ensures that this condition is met.

Question 5. Translate the following statements into chemical equations and then balance them.

(a) Hydrogen gas combines with nitrogen to form ammonia.

(b) Hydrogen sulphide gas burns in air to give water and sulpur dioxide.

(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.

(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.

Answer:

(a) $3H_2(g) + N_2(g) \rightarrow 2NH_3(g)$

(b) $2H_2S(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2SO_2(g)$

(c) $3BaCl_2(aq) + Al_2(SO_4)_3(aq) \rightarrow 2AlCl_3(aq) + 3BaSO_4(s)$

(d) $2K(s) + 2H_2O(l) \rightarrow 2KOH(aq) + H_2(g)$

Question 6. Balance the following chemical equations.

(a) $HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + H_2O$

(b) $NaOH + H_2SO_4 \rightarrow Na_2SO_4 + H_2O$

(c) $NaCl + AgNO_3 \rightarrow AgCl + NaNO_3$

(d) $BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + HCl$

Answer:

(a) $2HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2H_2O$

(b) $2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$

(c) $NaCl + AgNO_3 \rightarrow AgCl + NaNO_3$ (This equation is already balanced.)

(d) $BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2HCl$

Question 7. Write the balanced chemical equations for the following reactions.

(a) Calcium hydroxide + Carbon dioxide $\rightarrow$ Calcium carbonate + Water

(b) Zinc + Silver nitrate $\rightarrow$ Zinc nitrate + Silver

(c) Aluminium + Copper chloride $\rightarrow$ Aluminium chloride + Copper

(d) Barium chloride + Potassium sulphate $\rightarrow$ Barium sulphate + Potassium chloride

Answer:

(a) $Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O$

(b) $Zn + 2AgNO_3 \rightarrow Zn(NO_3)_2 + 2Ag$

(c) $2Al + 3CuCl_2 \rightarrow 2AlCl_3 + 3Cu$

(d) $BaCl_2 + K_2SO_4 \rightarrow BaSO_4 + 2KCl$

Question 8. Write the balanced chemical equation for the following and identify the type of reaction in each case.

(a) Potassium bromide(aq) + Barium iodide(aq) $\rightarrow$ Potassium iodide(aq) + Barium bromide(s)

(b) Zinc carbonate(s) $\rightarrow$ Zinc oxide(s) + Carbon dioxide(g)

(c) Hydrogen(g) + Chlorine(g) $\rightarrow$ Hydrogen chloride(g)

(d) Magnesium(s) + Hydrochloric acid(aq) $\rightarrow$ Magnesium chloride(aq) + Hydrogen(g)

Answer:

(a) $2KBr(aq) + BaI_2(aq) \rightarrow 2KI(aq) + BaBr_2(s)$

Type of Reaction: Double Displacement Reaction (and Precipitation Reaction).

(b) $ZnCO_3(s) \rightarrow ZnO(s) + CO_2(g)$

Type of Reaction: Decomposition Reaction (specifically, thermal decomposition).

(c) $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$

Type of Reaction: Combination Reaction.

(d) $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$

Type of Reaction: Displacement Reaction.

Question 9. What does one mean by exothermic and endothermic reactions? Give examples.

Answer:

Exothermic Reaction:

A chemical reaction in which energy is released, usually in the form of heat and/or light, is called an exothermic reaction. In these reactions, the products have less energy than the reactants, and the excess energy is given off to the surroundings, causing the temperature of the surroundings to rise.

Example: The burning of natural gas (methane).

$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) + Heat$

Endothermic Reaction:

A chemical reaction in which energy is absorbed from the surroundings is called an endothermic reaction. In these reactions, the products have more energy than the reactants. This energy must be supplied from the surroundings, causing the temperature of the surroundings to fall.

Example: The decomposition of calcium carbonate.

$CaCO_3(s) + Heat \rightarrow CaO(s) + CO_2(g)$

Question 10. Why is respiration considered an exothermic reaction? Explain.

Answer:

Respiration is considered an exothermic reaction because it releases energy.

Explanation:

To stay alive, our bodies need a constant supply of energy to carry out various life processes like muscle contraction, nerve impulse transmission, etc. We obtain this energy from the food we eat.

During digestion, complex carbohydrates from food (like rice, potatoes, bread) are broken down into simpler sugars, mainly glucose ($C_6H_{12}O_6$). This glucose is then transported by the blood to the body's cells.

Inside the cells, glucose combines with oxygen (which we inhale) in a series of reactions. This process, known as cellular respiration, breaks down the glucose molecule, releasing a significant amount of energy. This energy is then used by the body. Carbon dioxide and water are produced as byproducts.

The overall simplified equation for respiration is:

$C_6H_{12}O_6(aq) + 6O_2(aq) \rightarrow 6CO_2(aq) + 6H_2O(l) + Energy$

Since energy is released during this process, respiration is classified as an exothermic reaction.

Question 11. Why are decomposition reactions called the opposite of combination reactions? Write equations for these reactions.

Answer:

Decomposition reactions are called the opposite of combination reactions because of the fundamental difference in how substances are transformed.

  • In a combination reaction, two or more simple substances (reactants) combine to form a single, more complex substance (product). The general form is $A + B \rightarrow AB$.
  • In a decomposition reaction, a single complex substance (reactant) breaks down into two or more simpler substances (products). The general form is $AB \rightarrow A + B$.

Thus, one process involves synthesis (building up), while the other involves breakdown, making them opposites.

Example Equations:

Combination Reaction: Formation of water from hydrogen and oxygen.

$2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$

Decomposition Reaction: Decomposition of water into hydrogen and oxygen (electrolysis).

$2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$

Question 12. Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.

Answer:

Decomposition reactions require energy to break the bonds of the reactant. Here are examples for each type of energy input:

(i) Heat (Thermal Decomposition):

Heating limestone (calcium carbonate) causes it to decompose into quicklime (calcium oxide) and carbon dioxide.

$CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)$

(ii) Light (Photolytic Decomposition):

Exposing silver bromide to sunlight causes it to decompose into silver metal and bromine gas.

$2AgBr(s) \xrightarrow{Sunlight} 2Ag(s) + Br_2(g)$

(iii) Electricity (Electrolytic Decomposition):

Passing an electric current through water causes it to decompose into hydrogen gas and oxygen gas.

$2H_2O(l) \xrightarrow{Electricity} 2H_2(g) + O_2(g)$

Question 13. What is the difference between displacement and double displacement reactions? Write equations for these reactions.

Answer:

The main difference between displacement and double displacement reactions lies in the number and nature of the reactants and the mechanism of the reaction.

Basis of Difference Displacement Reaction Double Displacement Reaction
Reactants Involves one element and one compound. Involves two compounds (usually in aqueous solution).
Mechanism A more reactive element displaces a less reactive element from its compound. There is a mutual exchange of ions between the two reacting compounds.
Nature It is typically a redox reaction. It is typically a non-redox reaction, often resulting in a precipitate.
General Form $A + BC \rightarrow AC + B$ $AB + CD \rightarrow AD + CB$

Example Equation for Displacement Reaction:

Zinc displaces copper from copper sulphate solution.

$Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)$

Example Equation for Double Displacement Reaction:

Sodium sulphate reacts with barium chloride to form a precipitate of barium sulphate.

$Na_2SO_4(aq) + BaCl_2(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)$

Question 14. In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved.

Answer:

The recovery of silver from a silver nitrate solution using copper metal is a displacement reaction. Copper is more reactive than silver, so it can displace silver from its salt solution.

When a piece of copper metal is placed in a silver nitrate solution, the copper goes into the solution as copper(II) nitrate, and solid silver metal is deposited.

The balanced chemical equation for the reaction is:

$$\underset{\text{(Copper)}}{Cu(s)} + 2\underset{\text{(Silver Nitrate)}}{AgNO_3(aq)} \rightarrow \underset{\text{(Copper(II) Nitrate)}}{Cu(NO_3)_2(aq)} + 2\underset{\text{(Silver)}}{Ag(s)}$$

Question 15. What do you mean by a precipitation reaction? Explain by giving examples.

Answer:

A precipitation reaction is a type of chemical reaction in which two soluble compounds in aqueous solution react to form at least one insoluble product. This insoluble solid product is called a precipitate, which separates from the solution.

These reactions are often a type of double displacement reaction, where the cations and anions of the two reactants switch partners, and one of the new combinations is insoluble in water.

Example 1: Formation of Barium Sulphate

When a solution of sodium sulphate is mixed with a solution of barium chloride, a white precipitate of barium sulphate is formed immediately.

$Na_2SO_4(aq) + BaCl_2(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)$

Here, $BaSO_4$ is the insoluble white precipitate.

Example 2: Formation of Lead Iodide

When a solution of lead(II) nitrate is mixed with a solution of potassium iodide, a bright yellow precipitate of lead(II) iodide is formed.

$Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s) + 2KNO_3(aq)$

Here, $PbI_2$ is the insoluble yellow precipitate.

Question 16. Explain the following in terms of gain or loss of oxygen with two examples each.

(a) Oxidation

(b) Reduction

Answer:

(a) Oxidation

Oxidation is a chemical process that involves the gain of oxygen by a substance.

Example 1: When magnesium burns in air, it gains oxygen to form magnesium oxide.

$2Mg + O_2 \rightarrow 2MgO$

Example 2: When copper is heated in air, it gains oxygen to form black copper(II) oxide.

$2Cu + O_2 \rightarrow 2CuO$

(b) Reduction

Reduction is a chemical process that involves the loss of oxygen from a substance.

Example 1: When hydrogen gas is passed over heated copper(II) oxide, the CuO loses oxygen to form copper metal.

$CuO + H_2 \rightarrow Cu + H_2O$

Example 2: When zinc oxide is heated with carbon, the ZnO loses oxygen to form zinc metal.

$ZnO + C \rightarrow Zn + CO$

Question 17. A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.

Answer:

The shiny brown coloured element ‘X’ is Copper (Cu).

When copper is heated in air, it reacts with oxygen (undergoes oxidation) to form a new compound, which is black in colour.

The black coloured compound formed is Copper(II) Oxide (CuO).

The reaction is: $2Cu(s) + O_2(g) \xrightarrow{Heat} 2CuO(s)$

Question 18. Why do we apply paint on iron articles?

Answer:

We apply paint on iron articles primarily to prevent rusting.

Explanation:

Rusting is the corrosion of iron, a chemical process where iron reacts with oxygen and moisture (water) from the atmosphere to form hydrated iron(III) oxide, commonly known as rust. Rust weakens the iron structure.

Applying a coat of paint creates a protective, waterproof barrier on the surface of the iron article. This barrier cuts off the contact between the iron and the essential elements for rusting—air and moisture. Without contact with both oxygen and water, the chemical reaction of rusting cannot occur, thus protecting the iron article from corrosion.

Question 19. Oil and fat containing food items are flushed with nitrogen. Why?

Answer:

Oil and fat containing food items are flushed with nitrogen to prevent rancidity.

Explanation:

Rancidity is the process where fats and oils in food get oxidised when exposed to air (oxygen). This oxidation process changes the smell and taste of the food, making it unpleasant and stale.

Nitrogen ($N_2$) is a very unreactive and inert gas. By flushing the food packages (like bags of potato chips) with nitrogen, the oxygen inside the package is displaced and removed. This creates an inert atmosphere around the food item.

Without oxygen, the oxidation of fats and oils is significantly slowed down, preventing the food from becoming rancid. This increases the shelf-life of the food product and keeps it fresh for a longer period.

Question 20. Explain the following terms with one example each.

(a) Corrosion

(b) Rancidity

Answer:

(a) Corrosion

Corrosion is the gradual deterioration of a material, especially a metal, by a chemical or electrochemical reaction with its environment. It is essentially the process of a metal returning to its more stable, natural form (like an oxide).

Example: The most common example of corrosion is the rusting of iron. When iron objects are exposed to air and moisture, they get coated with a reddish-brown, flaky substance called rust (hydrated iron(III) oxide). This weakens the iron over time. Another example is the formation of a green layer on the surface of copper objects.

(b) Rancidity

Rancidity is the process of aerial oxidation of fats and oils present in food materials, resulting in an unpleasant change in their taste and smell. This makes the food unpalatable and unfit for consumption.

Example: If a packet of potato chips is left open to the air for a few days, the oil in the chips reacts with oxygen. The chips then develop a stale, foul smell and taste. This change is due to rancidity.