Kerala Syllabus Class 9 Chemistry Chapter 4 Redox Reactions Notes Solutions

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Kerala SCERT Class 9 Chemistry Chapter 4 Solutions Redox Reactions

Kerala Syllabus Std 9 Chemistry Chapter 4 Redox Reactions Notes Solutions Questions and Answers

Class 9 Chemistry Chapter 4 Let Us Assess Answers Redox Reactions

Question 1.
The unbalanced chemical equation regarding the formation of ammonia from nitrogen and hydrogen is given below.
N₂ + H₂ → NH₃
a) Balance the chemical equation.
b) Find out the total number of atoms of the same type in both the reactants and the products.
c) If 28 g of nitrogen combines with 6 g of hydrogen, find out the mass of ammonia formed. (Hint: Atomic mass H = 1 u N = 14 u)
Answer:
a) N₂ + 3H₂ → NH₃

b)

Elements	Nitrogen	Hydrogen
Reactants	2	6
Products	2	6

c) The balanced chemical equation for the formation of Ammonia
N₂ + 3H₂ → 2NH₃
28g + 6g → 2 × 17 (34g)
Mass of ammonia formed = 34g

Question 2.
C + 4HNO₃ → 2H₂O + CO₂ + 4NO₂
a) Find out and mark the oxidation number of carbon in this reaction.
b) What happens to the oxidation number of carbon in this reaction?
c) What happens to carbon – oxidation or reduction?
d) What are the oxidising and reducing agents in this reaction?
Answer:
a) C + 4HNO₃ → 2H₂O + CO₂ + 4NO₂
In reactants carbon is in the elemental state, So the oxidation number is 0.
In products, let the oxidation number of carbon be ‘x’
x + (2x – 2) = 0 1
x + – 4 = 0
x = +4

b) Increases from zero to +4.

c) Oxidation

d) Oxidising agent – HNO₃
Reducing agent – C

Question 3.
Find out the oxidation number of sulphur in the following compounds.
(Hint: Oxidation number H = +1, O = -2)
a) SO₂
b) SO₃
c) H₂SO₃
d) H₂SO₄
Answer:
a) SO₂
S + (2 × -2) = 0
S + -4 = 0
S = +4

b) SO₃
S + (-2 × 3) = 0
S + -6 = 0
S = + 6

c) H₂SO₃
(+1 × 2) + S + (-2 × 3) = 0
+ 2 + S + -6 = 0
S + -4 = 0
S = + 4

d) H₂SO₄
(+1 × 2) + S + (-2 × 4) = 0
+ 2 + S + -8 = 0
S + -6 = 0
S = +6

Question 4.
Certain statements are given below. Write whether they are true or false.
a) The process involving an increase in oxidation number is oxidation.
b) The process involving a decrease ini oxidation number is oxidation.
c) In a chemical reaction, oxidising agent undergoes reduction.
d) In a chemical reaction, oxidising agent undergoes oxidation.
Answer:
a) True
b) False
c) True
d) False

Question 5.
Balance the chemical equations given below.
a) SO₂ + O₂ → SO₃
b) H₂O₂ → H₂O + O₂
c) CH₄ + O₂ → H₂O + CO₂
d) Fe + HCl → FeCl₂ + H₂
Answer:
a) 2SO₂ + O₂ → 2SO₃
b) 2H₂O₂ → 2H₂O + O₂
c) CH₄ + 2O₂ → 2H₂O + CO₂
d) Fe + 2HCl → FeCl₂ + H₂

Question 6.
Two chemical reactions are given below. Find out the oxidation number of atoms and check whether these reactions are redox reactions.
a) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
b) Zn + 2HCl → ZnCl₂ + H₂
Answer:
a)
Not a redox reaction. Oxidation and reduction do not take place.

b)
It is a redox reaction, because oxidation and reduction take place simultaneously-.

Question 7.
The gaseous fuel carbon monoxide burns in oxygen to form carbon dioxide.
a) Write the balanced equation of this chemical reaction.
b) Is this a redox reaction? Why?
c) What is the oxidising agent in this reaction? What is the reducing agent?
Answer:
a) 2CO + O₂ → 2CO₂

b)
Yes, this is a redox reaction and reduction takes place simultaneously.

c) Oxidising agent – Oxygen (O)
Reducing agent – Carbon monoxide (CO)

Question 8.
Analyse the chemical equation given below.
Zn + 2HCl → ZnCl₂ + H₂
a) Mark the oxidation number of atoms before and after the chemical reaction.
b) Which atom undergoes oxidation?
c) Which atom undergoes reduction?
d) What are the oxidising and reducing agents?
Answer:
a)
b) Zn
c) H
d) Oxidising agent – HCl
Reducing agent – Zn

Question 9.
Analyse the chemical equations given below and find out whether they are redox reactions.
a) NaOH + HCl → NaCl + H₂O
b) H₂S + Cl₂ → 2HCl + S
Answer:
a)
It is not a redox reaction. No oxidation and reduction reactions.

b)
It is a redox reaction, because oxidation and reduction reactions takes place simultaneously.

Question 10.
A chemical reaction is given in the concept map below. Find out the oxidation number of each atom. On the basis of this, fill up the blanks.
(Hint: Valency S = 2 Fe = 2)

Answer:

Extended Activities

Question 1.
Mix iron powder and sulphur in the mass ratio 7 : 4 in a china dish. Heat the mixture well. After some time cool the china dish. Check whether iron can be separated using magnet.
Examine whether the product dissolves in carbon disulphide.
What is your inference?
Write down the equation of the chemical reaction. Check whether it is a redox reaction.
Answer:
Iron and sulphur on heating combines to gether to form Iron sulphide (ferrous sulphide). Thereafter iron cannot be separated using a magnet. The product obtained does not dissolve in carbon disulphide.

The constituents of a compound never retain their fundamental properties. (Iron is attracted by a magnet, sulphur dissolves in carbon disulphide).

This is a redox reaction because oxidation and reduction take place simultaneously

Question 2.
Take some sand in a tray. Place calcium carbide (CaC₂) on it. Place some more sand on top of it. Place some ice cubes on the sand. Ignite the ice cubes carefully. What do you see?
Calcium carbide reacts with water and forms acetylene (C₂H₂) gas. Acetylene is an inflammable gas.
Write the chemical equation of the combustion.
Check whether it is a redox reaction.
Answer:

It is a redox reaction.
Because oxidation and reduction take place simultaneously.

Question 3.
Make a mixture of aluminium powder and powdered iodine crystals in the mass ratio 1 : 2. Make a heap of it in a china dish. Make a small hole at the top of the heap. Add one or two drops of water into the hole. What do you see?
Here aluminium and iodine combine to form aluminium triiodide.
The valency of Al= 3 I = 1
a) Write the equation of the chemical reaction.
b) Find out the oxidation number of each atom. Check whether it is a redox reaction.
Answer:
a) 2Al + 3I₂ → 2AlI₃
b)
It is a redox reaction.
Because oxidation and reduction take place simultaneously.

Question 4.
Conduct a study tour to understand the importance of redox reactions in industry.
Answer:
You can visit one among the following to understand the importance of Redox reactions:

• Visit a metal refinery (e.g., aluminum or copper). Observe the smelting process, where a reducing agent (like carbon) extracts the desired metal from its oxide ore through a redox reaction. Learn about the role of electrolysis in purifying metals, another application of redox reactions.
• Visit a factory producing a metal product (e.g., car parts or appliances). Discuss how corrosion protection techniques, like galvanization (coating with zinc), utilize redox reactions to prevent rust.
• Tour a power plant (e.g., coal or natural gas). Understand the combustion process, a redox reaction releasing energy to generate electricity. Discuss the environmental implications of these reactions and potential cleaner alternatives.
• Visit a battery manufacturing facility. Learn about the different types of batteries (e.g., lithium – ion) and how they rely on redox reactions to store and release electrical energy. Explore advancements in battery technology.
• Tour a chemical plant producing a product that utilizes redox reactions (e.g., chlorine bleach or fertilizers). Understand the specific redox chemistry involved in the manufacturing process. Discuss safety protocols for handling these chemicals.
• Visit a water treatment facility. Leam about water disinfection methods like ozonation, which uses redox reactions to eliminate bacteria and contaminants. Explore wastewater treatment processes that employ redox reactions to break down pollutants.

Tips and Tricks for Your Redox Reactions Study Tour:
Preparation:

• Pre-tour Research: For each industry visit, delve deeper. Research the specific redox reactions . involved and the role of different chemicals. This will allow you to ask insightful questions during the tours.
• Focus and Note Taking: With so much information, prioritize! Identify key aspects of each visit that align with your learning goals. Take clear notes with diagrams or sketches for better recall.
• Bring Curiosity. Don’t be afraid to ask questions. The best way to learn is through active engagement with industry professionals.

During the Tour:

• Observe Keenly: Pay close attention to the processes and equipment used. Look for evidence of redox reactions happening (e.g., color changes, gas evolution).
• Engage with the Guides: Ask clarifying questions when something.is unclear. Don’t hesitate to request additional information or demonstrations.
• Network with Professionals: Introduce yourself and exchange contact information with industry experts. This can be valuable for future research or career opportunities.

Post-Tour:

• Consolidate Your Learnings: Review your notes and revisit the research you conducted. Organize the information to identify patterns and connections across different industries,
• Group Discussions: Share your experiences and insights with fellow participants. Discuss the broader implications of redox reactions in industry and brainstorm potential future applications.
• Project or Presentation: Consider creating a project or presentation summarizing your learnings. This will help solidify your understanding and allow you to share your knowledge with others.

Redox Reactions Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
What changes are generally observed during chemical reactions?
Answer:

• Different types of energy are absorbed or released.
• Occur change in the number of substances
• Occur colour change.
• Occur change in physical state.
• New substances are formed.

Question 2.
Let us do an experiment.
Take a trough and fill three-fourths of it with water. Add two drops of phenolphthalein to it and stir well. Cut a small piece of sodium and put it into the trough carefully.

a) What changes can be observed?
Answer:

• Sodium reacts with water vigorously.
• Sodium moves faster on the surface of the water and catches fire.
• A gas is evolved.
• Water in the trough turns pink.

b) What is the reason? Analyse the chemical equation and find out.
2Na + 2H₂O → 2NaOH + H₂↑
Answer:
Reason – Sodium reacts with water, forming sodium hydroxide alkali and hydrogen. In the presence of alkali, phenolphthalein turns pink. During this reaction, heat is liberated. Because of this heat, hydrogen gas bums.

Question 3.
Is there any change in the total mass of the substances during chemical reactions?
Answer:
No, there are no changes happening in the total mass of the substances during chemical reactions.

Question 4.
Take 20 mL barium chloride (BaCl₂) solution in a beaker. Take 20 mL sodium sulphate (Na₂SO₄) solution in another beaker. Place both the beakers together on an electronic balance and note the reading.
Now, pour the solution from one beaker to the other.

a) What do you observe?
Answer:
A chemical reaction takes place, and a white precipitate is formed.

b) After some time, note the reading of the electronic balance again. Compare this with the previous reading. What is your inference?

Answer:
We can see that the reading is not having any changes from the previous reading. Both are the same. From this, we can infer that the total mass of the reactants and the total mass of the products in the chemical reaction is constant.

c) Is there any change in the total mass as a result of this chemical reaction?
Answer:
No

d) Let us write the equation of this chemical reaction.
Answer:
BaCl₂ + Na₂SO₄ → BaSO₄ ↓ + 2NaCl
In this chemical reaction, barium chloride reacts with sodium sulphate to form barium sulphate and sodium chloride.

Question 5.
Take 20 mL of dilute hydrochloric acid (HCl) in a conical flask. Drop some zinc (Zn) granules in a balloon. Fix the balloon firmly to the mouth of the conical flask, as shown in the figure. Place the conical flask on an electronic balance and note the mass. Then, carefully raise the balloon and drop the zinc granules into the acid in the flask.

a) What do you see?
Answer:
Zinc reacts vigorously with hydrochloric acid. A gas evolves. The balloon gets inflated as the gas is collected in it.

b) Note the reading of the electronic balance. Compare this reading with the previous one. What do you understand?
Answer:
There is no change in both readings.

c) Which gas is collected in the balloon?
Answer:
Hydrogen gas (H₂)

d) Let us write the equation of the chemical reaction,
Answer:
Zn + 2HCl → ZnCl₂ + H₂ ↑

Question 6.
What can be inferred from these experiments?
Answer:
The total mass of the reactants and products does not change due to chemical reactions.

Question 7.
Does the total mass change during chemical reactions?
Answer:
No.
During chemical reactions, there is no change in the total mass of substances.
During the combustion of fuels and the burning paper, the total mass of the reactants and the products is constant. The major products formed during these reactions are carbon dioxide and water vapour. They are lost in the atmosphere. But if we collect these products without any loss and weighed, we can see that There will be no change in total mass.

Based on experiments and observations, the French scientist Antoine Lavoisier stated the law of conservation of mass,

LAW OF CONSERVATION OF MASS In a chemical reaction, the total mass of the reactants will be equal to the total mass of the products.

Question 8.
Carbon and oxygen combine to form carbon dioxide. Analyse the symbolic representation of this chemical reaction.

a) Complete table given below.

Total mass of the reactants	……………………….
Total mass of the products	……………………….

Answer:

Total mass of the reactants	12 u + 32 u = 44 u
Total mass of the products	16u + 12u + 16u = 44u

b) Record your inference.
Answer:
The total mass of the reactants is equal to the total mass of the products in a chemical reaction.

Question 9.
Methane (CH₄) burns in air to form carbon dioxide and water vapour. The symbolic representation of this chemical reaction is given below.

a) Write down the equation of the chemical reaction.
Answer:
CH₄ + 2O₂ → CO₂ + 2H₂O

b) Check whether this chemical reaction obeys the law of conservation of mass.
(Hint: H = 1 u, C = 12 u, O = 16 u.)
Answer:
Total mass of the reactants = (1 × 4) + 12 + (16 × 4)
= 4 + 12 + 64
= 80
Total mass of the products = (16 × 2) + 12 + (18 × 2)
= 32 + 12 + 36
= 80
The total mass of the reactants is equal to the total mass of the products. So, this chemical reaction obeys the law of conservation of mass.

Question 10.
You know that oxygen and hydrogen are diatomic molecules.
a) How are these molecules represented using symbols?
Oxygen ………………….., Hydrogen …………………
Answer:
Oxygen – O₂
Hydrogen – H₂

b) What is the total number of atoms in water (H₂O) molecule?
Answer:
3 (Three)

c) Calculate the number of molecules and the total number of atoms present in 5H₂O.
Total number of molecules …………….. Total number of atoms ………………..
Answer:
Total number of molecules – 5
Total number of atoms – 5 × 3 = 15
Balancing the chemical equation of the formation of water from hydrogen and oxygen:
Step 1 :

Number of atoms in the reactants	Hydrogen = 2	Oxygen = 2
Number of atoms in the products	Hydrogen = 2	Oxygen = 1

According to the law of conservation of mass, the total number of atoms should be the same.
So, the number of oxygen atoms in the products and reactants should be the same. We must make the number of oxygen atom must be 2. For this make the number of water molecules 2.

Step 2: H₂ + O₂ → 2H₂O
Number of atoms in the reactants
Number of atoms in the products

Number of atoms in the reactants	Hydrogen = 2	Oxygen = 2
Number of atoms in the products	Hydrogen = 4	Oxygen = 2

Now, the number of Hydrogen atoms is not the same. So we have to make it as four.

Step 3: 2H₂ + O₂ → 2H₂O

Number of atoms in the reactants	Hydrogen = 4	Oxygen = 2
Number of atoms in the products	Hydrogen = 4	Oxygen = 2

Now the number of atoms in reactants and the products are the same. Now we can say that the reaction is balanced.
The balanced chemical equation for this reaction is:
2H₂ + O₂ → 2H₂O

Balancing a chemical equation is the method of equalising the number of the same type of atoms in both the reactants and the products. The equation thus obtained is known as a balanced chemical equation.

More examples:
• Magnesium + Oxygen → Magnesium Oxide
Step 1: Mg + O₂ → MgO
Reactants
Mg – 1
O – 2
Products:
Mg – 1
O – 1
Number of oxygen atoms is not the same. Making it equal.

Step 2: Mg + O₂ → 2MgO
Reactants
Mg – 1
O – 2
Products:
Mg – 2
O – 2
Number of Magnesium atoms is not equal. Making it equal.

Step 3: 2Mg + O₂ → 2MgO
Reactants
Mg – 2
O – 2
Products:
Mg – 1
O – 1
Now, the number of both atoms is equal on both sides.

Balanced Chemical Equation: 2Mg + O₂ → 2MgO

• Hydrogen + Chlorine → Hydrogen chloride
Step 1: H₂ + Cl₂ → HCl
Reactants
H – 2
Cl – 2
Products:
H – 1
Cl – 1
Number of both the atoms are not same. Making it equal.

Step 2: H₂ + Cl₂ → 2HCl
Reactants
H – 2
Cl – 2
Products:
H – 2
Cl – 2
Now, the number of both atoms is equal on both sides. The equation is now balanced.

Balanced Chemical Equation: H₂ + Cl₂ → 2HCl
• Aluminium + Oxygen → Aluminium oxide
Step 1: Al + O₂ → Al₂O₃
Reactants
Al – 1
O – 2
Products:
Al – 2
O – 3
(Number of oxygen atoms is not the equal. Making it equal)

Step 2: Al + 3O₂ → Al₂O₃
Reactants
Al – 1
O – 6
Products:
Al – 2
O – 3
(Again, number of oxygen atoms is not the same. Making it equal))

Step 3: Al + 3O₂ → 2Al₂O₃
Reactants
Al – 1
O – 6
Products:
Al – 4
O – 6
Now, the number of oxygen atoms is the same. But the number of Aluminium atoms is not the same. Making it equal.

Step 4: 4Al + 3O₂ → 2Al₂O₃
Reactants
Al – 4
O – 6
Products:
Al- 4
O – 6
Now, the number of both atoms is equal on the reactant and products. The equation is balanced.

Balanced Chemical Equation: 4Al + 3O₂ → 2Al₂O₃
• Nitrogen + Hydrogen → Ammonia
Step 1: N₂ + H₂ → NH₃
Reactants
N – 2
H – 2
Products:
N – 1
H – 3
Making the number of Nitrogen atoms equal.

Step 2: N₂ + H₂ → 2NH₃
Reactants
N – 2
H – 2
Products:
N – 2
H – 6
Now, the number of hydrogen atoms is not the same. Making it equal.

Step 3: N₂ + 3H₂ → 2NH₃
Reactants
N – 2
H – 6
Products:
N – 2
H – 6
Now, the number of both atoms is equal on the reactant and products. The equation is balanced.
Balanced Chemical Equation:N₂ + 3H₂ → 2NH₃

Question 11.
Balance the chemical equations given below and record them in science diary.
a) H₂ + I₂ → HI
b) Na + H₂O → NaOH + H₂
c) Mg + HCl → MgCl₂ + H₂
Answer:
a) H₂ + I₂ → HI
Step 1: H₂ + I₂ → HI
Step 2: H₂ + I₂ → 2HI
Balanced Chemical Equation: H₂ + I₂ → 2HI

b) Na + H₂O → NaOH + H₂
Step 1: Na + 2H₂O → NaOH + H₂
Step 2: Na + 2H₂O → 2NaOH + H₂
Step 3: 2Na + 2H₂O → 2NaOH + H₂
Balanced Chemical Equation: 2Na + 2H₂O → 2NaOH + H₂

c) Mg + HCl → MgCl₂ + H₂
Step 1: Mg + HCl → MgCl₂ + H₂
Step 2: Mg + 2HCl → MgCl₂ + H₂
Balanced Chemical Equation: Mg + 2HCl → MgCl₂ + H₂

Question 12.
Look at the chemical equation regarding the formation of sodium chloride.
2Na + Cl₂ → 2NaCl
a) Which atom undergoes oxidation?
Answer:
Sodium (Loses electron)

b) Which atom supports oxidation? (sodium/chlorine)
Answer:
Chlorine (gains electron)

c) Which atom gets reduced? olojDceadm nilauxnxBac&jcm and)Qo
Answer:
Chlorine (gains electron)

d) Which atom helps reduction?
Answer:
Sodium (loses electron)

The species that helps oxidation in a chemical reaction is the Oxidising agent. The oxidising agent gets reduced in a chemical reaction.
The species that helps reduction is the reducing agent. The reducing agent gets oxidised in a chemical reaction.

Question 13.
Analyse the chemical reactions given below and complete the table.
1) Mg + F₂ → MgF₂
2) Ca + Cl₂ → CaCl₂
3) 4Fe + 3O₂ → 2Fe₂O₃

Answer:

Question 14.
The oxidation number of magnesium is +2 and that of oxygen is -2 in magnesium oxide (MgO). What do you understand from this?
Answer:
MgO (Magnesium oxide) is composed of Magnesium ion (Mg²⁺) and Oxide ion (O^2-).
When an electron is lost, a positive ion is formed and when an electron is gained, a negative ion is formed. Magnesium loses two electrons and become a positively charged ion, oxide gains two electrons and becomes negatively charged electron. The positively charged Magnesium ion (Mg²⁺) and the negatively charged oxide ion (O^2-) combine together to form Magnesium oxide (MgO). Covalent compounds are formed by the sharing of electrons. In such compounds, the oxidation number is assigned assuming that the shared electrons are shifted to the more electronegative element.

Example – In the covalent compound HF, it is considered that the more electronegative fluorine (F) attracts the electron pair and attains -1 oxidation number. Hydrogen is assumed to lose one electron and it attains +1 oxidation number.

The sum of the oxidation numbers of all atoms in a compound is zero.
In element molecules, electrons are equally shared by the atoms. So, at elemental state the oxidation number is considered to be zero.

Question 15.
Find out the oxidation number of nitrogen in HNO₂ and NO₂.
Answer:
HNO₂
+ 1 + N + (-2 × 2) = 0
+ 1 + N + – 4 = 0
N + -3 = 0
N = +3
Oxidation number of nitrogen in HNO₃ is +3.

NO₂
N + (-2 × 2) = 0
N + -4 = 0
N = +4
Oxidation number of nitrogen in NO₂ is +4.

Finding out the oxidation number of chromium (Cr) in potassium dichromate (K₂Cr₂O₇).
If the oxidation number of chromium is considered to be ‘x’.

Question 16.
Find out the oxidation number of chromium in Cr₂O₃.
Answer:
2Cr + (-2 × 3) = 0
2Cr + -6 = 0
2Cr = +6
Cr = \(\frac{+6}{2}\) = +3
Oxidation number of Chromium in Cr₂O₃ is + 3.

Question 17.
Find out the oxidation number of manganese (Mn) in the following compounds and record it in your science diary. (Hint: Oxidation number of O = -2, K = +1.)
a) MnO₂
b) Mn₂O₇
c) KMnO₄
Answer:
a) MnO₂
Mn + (-2 × 2) = 0
Mn + -4 = 0
Mn = +4
Oxidation number of Manganese in MnO₂ is +4.

b) Mn₂O₇
2Mn + (-2 × 2) = 0
2Mn + -14 = 0
2Mn = +14
Mn = \(\frac{+14}{2}\) = +7
Oxidation number of Manganese in Mn₂O₇ is +7.

c) KMnO₄
+ 1 + Mn + (-2 × 4) = 0
+ 1 + Mn + -8 = 0
Mn + -7 -0
Mn = +7
Oxidation number of manganese in KMnO₄ is +7.

Question 18.
Analysing the chemical equation of the formation of sodium chloride (NaCl).
2Na + Cl₂ → 2NaCl
a) What is the oxidation number of sodium and chlorine in their elemental state?
Answer:
Zero

b) Write the the chemical equation including their oxidation numbers.
Answer:

c) What happened to the oxidation number of sodium as a result of this reaction (increased/decreased)?
Answer:
Increased (Increased from 0 to +1)

d) What happened to the oxidation number of chlorine?
Answer:
Decreased (Decreased from 0 to -1)
Oxidation number increases during oxidation reactions. Reduction reactions involve a decrease in oxidation number.

e) Which atom undergoes oxidation during the formation of sodium chloride?
Answer:
Sodium

f) What is the oxidising agent in this reaction? Why?
Answer:
Chlorine. Because Chlorine helps oxidation by gaining electron.

g) Which atom undergoes reduction in this reaction? Why?
Answer:
Chlorine. Because the oxidation number of chlorine decreased from zero to -1.

g) What is the reducing agent in this case?
Answer:
Sodium. Because podium loses electron.

Question 19.
Analyse the chemical equation given below. Find out the oxidation number of atoms and complete the table given below.
H₂ + Cl₂ → 2HCl

Answer:

Question 20.
Let us analyse another chemical equation.
Mg + 2HCl → MgCl₂ + H₂

a) The oxidation number of magnesium changes from ………………. to ……………….
Answer:
Zero to +2

b) The change that happened to magnesium, (oxidation/ reduction).
Answer:
Oxidation

c) What is the oxidising agent in this case? ………………… (Mg/HCl)
Answer:
HCl

d) What is the reducing agent? (Mg/HCl)
Answer:
Mg

Question 21.
Analyse the chemical reaction given below and complete the table given below.

Answer:

Question 22.
The equation for the chemical reaction between hydrogen and chlorine to form hydrogen chloride is given below.

a) Which atom has undergone oxidation in this reaction?
Answer:
Hydrogen (H₂)

b) Which atom has undergone reduction?
Answer:
Chlorine (Cl)

A chemical reaction in which oxidation and reduction take place simultaneously is a redox reaction. In a redox reaction oxidising agent gets reduced and the reducing agent gets oxidised.

Familiar Redox reactions:

• Glucose molecules decompose and release energy during cellular respiration.
• Formation of oxide coating on the surface of metals.
• Combustion of fuels.
• Decomposition of organic substances in the presence of oxygen.
• Production of electricity in electrochemical cells.

Question 23.
Analyse the above redox reactions and present a seminar on the importance of redox reactions in daily life.
Answer:
Hints that can be used for seminar preparation

Redox Reactions

• Briefly explain chemical reactions as the rearranging of atoms to form new substances.
• Introduce the concept of redox reactions (reduction-oxidation) involving electron transfer.
• Mention that one element loses electrons (oxidation) while another gains them (reduction).

Redox Reactions in Action
• Energy for Life:

• Explain cellular respiration, where glucose (food) breaks down using oxygen, releasing energy to power our cells.
• Highlight that this is a redox reaction with glucose losing electrons (oxidized) and oxygen gaining them (reduced).
• Briefly mention ATP as the energy currency in cells produced during this process.

• Fire and Fuel:

• Explain combustion of fuels like wood or gasoline as a redox reaction.
• Show how the fuel loses electrons (oxidized) while reacting with oxygen (reduced), releasing heat and light energy.
• Briefly mention the importance of combustion for cooking, heating, and transportation.

• Metal Mysteries:

• Explain rusting of iron as a redox reaction.
• Show how iron atoms lose electrons (oxidized) to oxygen and water Vapour in the environment, forming the reddish-brown rust.
• Briefly mention that painting or using galvanized steel can prevent rust (oxidation).

Amazing Applications
• Briefly discuss how redox reactions are used in:

• Batteries: Electrochemical cells where reactions convert chemical energy to electrical energy (e.g., AA batteries).
• Food Preservation: Adding antioxidants (reducing agents) to food slows down spoilage by preventing oxidation.

Real-world Examples and Activities

• Show a simple demonstration (depending on safety guidelines):
  Observe the browning of an apple slice (oxidation) or the burning of a steel wool pad – (oxidation).
• Encourage class mates to brainstorm other examples of redox reactions in daily life (e.g., food
  browning, bleaching clothes).

Conclusion

• Summarize how redox reactions are crucial for our energy needs, material protection, and technological advancements.
• Briefly discuss the negative aspects of some redox reactions, like corrosion, and ways to mitigate them.