Class 11 Chemistry NCERT Solutions 2021 for Chapter 8 Redox Reaction

NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reaction: NCERT which stands for the National Council Of Educational Research and Training is responsible for designing and publishing textbooks for all the classes and subjects. NCERT textbooks covered all the topics and apply to the Central Board Of Secondary Education (CBSE) and various state boards.

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• NCERT Solutions For 11th Chemistry All Chapters

The NCERT Solutions offer an insight into the exam pattern, latest syllabus, and complete marking scheme. The solutions are available in the pdf format which can be easily downloaded and saved for future reference as well. We are providing you free PDF of NCERT solutions for class 11 chemistry chapter 8 Redox reactions for download. 

NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reaction

The chemistry Class 11 Ncert Solutions Chapter 8 Redox Reaction deals with the permutations and combinations between the oxidation and reduction reactions. There are a number of phenomena, biological as well as physical that is concerned with the redox reactions.

Also, reactions find their use in biological, pharmaceutical, metallurgical, industrial, and agricultural areas. Therefore, to guide you in your preparation for NCERT Class 11, we have offered the downloadable PDF for NCERT solutions for class 11 chemistry chapter 8. 

This chapter is solved by our subject matter expert and prepared as per the NCERT guidelines. These PDF consists of the exercise questions along with solutions that can help you revise the complete syllabus and thereby help you score higher marks.

You can download CBSE NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reaction from below.

NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reaction

 

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What will you learn in CBSE Class 11 Chemistry Chapter 8 Redox Reaction?

After studying our 11th Class Chemistry Chapter 8 Solutions pdf students will be able to identify the redox reactions as a reaction in which reduction and oxidation reactions are occurring simultaneously. Also, students will be defining the terms reduction, oxidation, oxidant, and reductant. Additionally, students will explain the mechanism of redox reactions using the electron transfer process.

Along with that, you will use the concept of oxidation number in identifying reductant and oxidation in a reaction. So, you learn to classify the redox reaction into combination, displacement, decomposition, and disproportionation. Adding to that, you will suggest a comparative order among the various oxidants and reductants. You will also learn about balancing the chemical equations using the half-reaction method and oxidation number.

Whether you are unable to solve the difficult problems in the NCERT textbook or you are facing difficulty in understanding the complex theories of Science, Download class 11 chemistry chapter 8 NCERT solutions pdf to resolve all your doubts.

NCERT Solutions are a wholesome source of study material which provides the students access to solved queries and information. If you practice the solutions on a regular basis, you will give yourself a better chance of scoring higher points in your exam. With its help, students will be able to learn easy methods and ways to understand difficult concepts of redox reactions and oxidation reactions.

Furthermore, they will also get to study competitive electron transfer reaction, electron transfer reach, oxidation number, types of redox reactions, and balancing the redox reactions. The chemistry chapter 8 redox reactions are a part of unit 8. In the final exam, unit 8, 9, 10, and 11 holds the total weight of 16 marks.

Below are the subtopics gave in this chapter.

• Ex 8.1 – Classical Idea of Redox Reactions – Oxidation and Reduction Reactions
• Ex 8.2 – Redox Reactions in terms of Electron Transfer Reactions
• Ex 8.2.1 – Competitive Electron Transfer Reactions
• Ex 8.3 – Oxidation Number
• Ex 8.3.1 – Types of Redox Reactions
• Ex 8.3.2 – Balancing of Redox Reactions
• Ex 8.3.3 – Redox Reactions as the Basis for Titrations
• Ex 8.3.4 – Limitations of Concept of Oxidation Number
• Ex 8.4 – Redox Reaction and Electrode Process

Advantages of NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reaction

Chemistry is one of the challenging subjects for most of the students. It is equally important because in Class 11th students prepare a foundation that will be of immense help in Class 12th. With the right study material and proper exam preparation, students can score very good marks on their Boards Exam.

• NCERT Solution for Class 11 chemistry chapter 8 covers all the important topics given in the chapter.
• It will prescribe you an idea about the marking scheme, pattern, and questions and prepare you for your exam.
• Solutions will help you build your performance before the actual exam.
• Practicing solutions will help you increase your pace and manage time during real exams.
• NCERT solutions are framed by extracting content from the NCERT textbook.
• Diagrams used are neatly labelled and self-explanatory
• Conceptual based learning promoted
• Answering methodologies as per the expected pattern
• Solutions developed by subject experts

Access NCERT Solutions For Class 11 Chemistry Chapter 8

Question 1. What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30°C?

Answer: P₁ = 1 bar,P₂ = ?       V₁= 500 dm³ ,V₂=200 dm³
As temperature remains constant at 30°C,
P₁V₁=P₂V₂
1 bar x 500 dm³ = P₂ x 200 dm³ or P₂=500/200 bar=2.5 bar

Question 2. A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure?

Answer:  V₁= 120 mL, P₁=1.2 bar,
V₂ = 180 mL, P₂ = ?
As temperature remains constant, P₁V₁ = P₂V₂
(1.2 bar) (120 mL) = P2 (180mL)

Question 3. Using the equation of state PV = nRT, show that at a given temperature, density of a gas is proportional to the gas pressure P.

Answer: According to ideal gas equation
PV = nRT or PV=nRT/V

Question 4.  At 0°C, the density of a gaseous oxide at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Answer: Using the expression, d =MP/RT , at the same temperature and for same density,
M₁P₁ = M₂P₂ (as R is constant)
(Gaseous oxide) (N₂)
or
M₁ x 2 = 28 x 5(Molecular mass of N₂ = 28 u)
or M₁ = 70u

Question 5. Pressure of l g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses.

Answer: Suppose molecular masses of A and B are M_A and M_B respectively. Then their number of moles will be

Question 6. The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminium reacts?

Answer: The chemical equation for the reaction is
2 Al + 2 NaOH + H₂0 -> 2 NaAl0₂ + 3H₂ (3 x 22400 mL At N.T.P)
2 x 27 = 54 g.
54 g of Al at N.T.P release
H₂ gas = 3 x 22400 0.15 g of Al at N.T.P release

Question 7. What will be the pressure exerted by a mixture of 3.2g of methane and 4.4g of carbon dioxide contained in a 9 dm³ flask at 27 °C?

Answer:

Question 8. What will be the pressure of the gas mixture when 0.5 L of H₂ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in all vessel at 27 °C?

Answer: Calculation of partial pressure of H₂ in 1L vessel P₁= 0.8 bar,
P₂= ? V₁= 0.5 L , V₂ = 1.0 L
As temperature remains constant, P₁V₁ = P₂V₂
(0.8 bar) (0.5 L) = P₂ (1.0 L) or P₂ = 0.40 bar, i.e., PH₂ = 0.40 bar
Calculation of partial pressure of 02 in 1 L vessel
P₁‘ V₁ = P₂‘V₂‘
(0.7 bar) (2.0 L) = P₂ (1L) or P₂‘ = 1.4 bar, i.e.,Po₂= 1.4 bar
Total pressure =P_Hz + P_Q2 = 0.4 bar + 1.4 bar = 1.8 bar

Question 9. Density of a gas is found to be 5.46 g/dm³ at 27 °C and at 2 bar pressure. What will be its density at STP?

Answer:

Question 10. 34.05 mL of phosphorus vapor weighs 0.0625 g at 546°C and 1.0 bar pressure. What is the molar mass of phosphorus?

Answer:

Question 11. A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead, he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer, he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out?

Answer:

Question 12.Calculate the temperature of 4.0 moles of a gas occupying 5 dm³ at 3.32 bar (R = 0.083 bar dm³ K^-1 mol^-1)
Answer:

Question 13. Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Answer: Molecular mass of N₂ = 28g
28 g of N₂ has No. of molecules = 6.022 x 10²³ 1.4 g of
N₂ has No. of molecules = 6.022 x 10²³ x 1.4 g/28 g
= 3.011 x 10²² molecules.
Atomic No. of Nitrogen (N) = 7
1 molecule of N₂ has electrons = 7 x 2 = 14
3.011 x 10²² molecules of N₂ have electrons
= 14 x 3.011 x 10²²
= 4.215 x 10²³ electrons.

Question 14. How much time would it take to distribute one Avogadro number of wheat grains if 10¹⁰ grains are distributed each second ?

Answer:

Question 15. Calculate the total pressure in a mixture of 8g of oxygen and 4g of hydrogen confined in a vessel of l dm³ at 27°C. R = 0.083 bar dm³ K^-1 mol^-1.

Answer:

Question 16. Pay load is defined as the difference between the mass of the displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C (Density of air = 1.2 kg m^-3 and R = 0.083 bar dm³ K^-1 mol^-1).

Answer:

Question 17. Calculate the volume occupied by 8.8 g of CO₂ at 31.1 °C and 1 bar pressure. R = 0.083 bar LK^-1 mol^-1

Answer:

Question 18. 2.9 g of a gas at 95°C occupied the same volume as 0.184 g of hydrogen at 17°C at the same pressure. What is the molar mass of the gas ?

Answer:

Question 19. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Answer: As the mixture H₂ and O₂ contains 20% by weight of dihydrogen, therefore, if H₂ = 20g, then O₂ = 80g

Question 20. What would be the SI unit for the quantity PV²T²/n?

Answer:

Question 21. In terms of Charles’ law explain why -273°C is the lowest possible temperature.

Answer: At -273°C, volume of the gas becomes equal to zero, i.e., the gas ceases to exist.

Question 22. Critical temperature for CO₂ and CH₄ are 31.1°C and -81.9°C respectively. Which of these has stronger intermolecular forces and why?

Answer:  Higher the critical temperature, more easily the gas can be liquefied, i.e., greater are the intermolecular forces of attraction. Hence, Co₂ has stronger intermolecular forces than CH₄.

Question 23. Explain the physical significance of vander Waals parameters.

Answer: ‘a’ is a pleasure of the magnitude of the intermolecular forces of attraction, while b is a measure of the effective size of the gas molecules.

MORE QUESTIONS SOLVED

I. Very Short Answer Type Questions

Question 1. What is the value of the gas constant in SI units?

Answer: 8.314 JK^-1 mol^-1.

Question 2. Define boiling point of a liquid.

Answer: The temperature at which the vapor pressure of a liquid is equal to external pressure is called boiling point of liquid.

Question 3. What is SI unit of (i) Viscosity (ii) Surface tension?

Answer: (i) Unit of viscosity is Nsm^-2
(ii) Unit of surface tension is Nm^-1

Question 4. What is the effect of temperature on (i) surface tension and (ii) Viscosity?

Answer: (i) Surface tension decreases with increase of temperature.
(ii) Viscosity decreases with increase of temperature.

Question 5. What is the unit of coefficient of viscosity?

Ans. Poise.

Question 6. What do you understand by laminar flow of a liquid?

Answer: The type of flow in which there is regular gradation of velocity in passing from one layer to the next is called laminar flow.

Question 7. What do you mean by compressibility factor?

Answer:  The deviation from ideal behaviour can be measured in terms of compressibility factor Z.
Z=PV/nRT

Question 8. What is Boyle Temperature?

Answer:  The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure, is called Boyle temperature or Boyle point.

Question 9. What is meant by elastic collision ?

Answer: Collision in which there is no loss of kinetic energy but there is transfer of energy, is called elastic collision.

Question 10. Define critical temperature of gas.

Answer: The temperature above which a gas cannot be liquefied.

Question 11. What are real gases ?

Answer:  A gas which can deviate from ideal gas behaviour at higher pressure and lower temperature, is called a real gas.

Question 12. Define an ideal gas.

Answer: A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly, is called an ideal gas.

Question 13. Name four properties of gases.

Answer:

• Gases, have no definite shape and no definite volume.
• There is no force of attraction existing between the molecules of gases.
• Gases are highly compressible.
• Gases can mix evenly and can spread in whole space.

Question 14. State Dalton’s law of partial pressure.

Answer:  Daltons’ Law states that, total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

Question 15. What do you mean by aqueous tension?

Answer: Pressure exerted by saturated water vapor is called aqueous tension.

Question 16. Give mathematical expression for ideal gas equation.

Answer:  PV = nRT
Where R is called Gas constant.

Question 17. Write van der Waals equation for n moles of a gas.

Answer:

Where ‘a’ and ‘V are van der walls constants.

Question 18. How is compressibility factor expressed in terms of molar volume of the real gas and that of the ideal gas?

Answer:

Question 19. Why liquids diffuse slowly as compared to gases?

Answer: In liquids, the molecules are more compact in comparison to gases.

Question 20. What is the effect of temperatures on the vapor pressure of a liquid?

Answer: Vapour pressure increases with rise in temperature.

Question 21. Why falling liquid drops are spherical?

Answer:  Because of the property of surface tension, liquid tends to minimise its area.

II. Short Answer Type Questions

Question 1. A weather balloon has a volume of 175 dm³ when filled with hydrogen gas at a pressure of 1.0 bar. Calculate the volume of the balloon when it rises to a height where the atmospheric pressure is 0.8 bar. Assume that temperature is constant.

Answer:

Question 2. A certain amount of a gas at 27°C and 1 bar pressure occupies a volume of 25 m³. If the pressure is kept constant and the temperature is raised to 77°C, what will be the volume of the gas?

Answer:  From the available data: V₁ = 25 m³, T₁= 27 + 273 = 300 K
V₂ = ?                                                                T₂ = 77 + 273 = 350 K

Question 3. A flask was heated from 27°C to 227°C at constant pressure. Calculate the volume of the flask if 0.1 dm³ of air measured at 227°C was expelled from the flask.

Answer:  Let the volume of the flask = V dm³ (after expelling the air)
V₁ = V dm³, T₁ = 27 + 273 = 300K
VT₂ = (V + 0.1) dm³, T₂ = 227 + 273 = 500 K

Question 4. A gas occupying a volume of 100 liters is at 20°C under a pressure of 2 bar. What temperature will it have when it is placed in an evacuated chamber of volume 175 liters? The pressure of the gas in the chamber is one-third of its initial pressure.

Answer: 

Question 5. At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 mL?

Answer:

Question 6. A 34.0 dm³ cylinder contains 212 g of oxygen gas at 21°C. What mass of oxygen must be released to reduce the pressure in the cylinder to 1.24 bar?

Answer:

Question 7. The values of the van der Waal’s constants for a gas are a = 4.10 dm⁶ bar mol^-2 and b = 0.035 dm³ bar mol^-1. Calculate the values of the critical temperature and critical pressure for the gas.

Answer:

Question 8. The pressure of a mixture of H₂ and N₂ in a container is 1200 torr. The partial pressure of nitrogen in the mixture is 300 torr. What is the ratio of H₂ and N₂ molecules in the mixture?

Answer:

Question 9. (a) What do you mean by Surface Tension of a liquid?
(b) Explain the factors which can affect the surface tension of a liquid.

Answer:  (a) Surface tension: It is defined as the force acting per unit length perpendicular to the line drawn on the surface. It’s unit is Nnr1.
(b) Surface tension of a liquid depends upon following factors.
(i) Temperature: Surface tension decreases with rise in temperature. As the temperature of the liquid increases, the average kinetic energy of the molecules increases. Thus, there is a decrease in intermolecular force of attraction which decreases the surface tension.
(ii) Nature of the liquid: Greater the magnitude of intermolecular forces of attraction in the liquid, greater will be the value of surface tension.

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