<\/span><\/h2>\nAfter 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.<\/p>\n
Along with that, you will use the concept of oxidation number in identifying reductants 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.<\/p>\n
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.<\/p>\n
NCERT Solutions is a wholesome source of study material that provides the student with 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.<\/p>\n
Furthermore, they will also get to study competitive electron transfer reactions, electron transfer reach, oxidation numbers, 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 a total weight of 16 marks.<\/p>\n
Below are the subtopics given in this chapter.<\/p>\n
\n- Ex 8.1 \u2013 Classical Idea of Redox Reactions \u2013 Oxidation and Reduction Reactions<\/li>\n
- Ex 8.2 \u2013 Redox Reactions in terms of Electron Transfer Reactions<\/li>\n
- Ex 8.2.1 \u2013 Competitive Electron Transfer Reactions<\/li>\n
- Ex 8.3 \u2013 Oxidation Number<\/li>\n
- Ex 8.3.1 \u2013 Types of Redox Reactions<\/li>\n
- Ex 8.3.2 \u2013 Balancing of Redox Reactions<\/li>\n
- Ex 8.3.3 \u2013 Redox Reactions as the Basis for Titrations<\/li>\n
- Ex 8.3.4 \u2013 Limitations of Concept of Oxidation Number<\/li>\n
- Ex 8.4 \u2013 Redox Reaction and Electrode Process<\/li>\n<\/ul>\n
<\/span>Advantages of NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reaction<\/span><\/h2>\nChemistry 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 board exams.<\/p>\n
\n- NCERT Solution for Class 11 Chemistry Chapter 8 covers all the important topics given in the chapter.<\/li>\n
- It will give you an idea about the marking scheme, pattern, and questions and prepare you for your exam.<\/li>\n
- Solutions will help you build your performance before the actual exam.<\/li>\n
- Practicing solutions will help you increase your pace and manage time during real exams.<\/li>\n
- NCERT solutions are framed by extracting content from the NCERT textbook.<\/li>\n
- The diagrams used are neatly labeled and self-explanatory<\/li>\n
- Conceptual-based learning promoted<\/li>\n
- Answering methodologies as per the expected pattern<\/li>\n
- Solutions developed by subject experts<\/li>\n<\/ul>\n
<\/span>Access NCERT Solutions For Class 11 Chemistry Chapter 8<\/span><\/h2>\n<\/span>Question 1. What will be the minimum pressure required to compress 500 dm3<\/sup> of air at 1 bar to 200 dm3<\/sup> at 30\u00b0C?<\/strong><\/span><\/h3>\nAnswer: <\/strong>P1<\/sub> = 1 bar,P2<\/sub> = ? V1<\/sub>= 500 dm3<\/sup> ,V2<\/sub>=200 dm3<\/sup>
As temperature remains constant at 30\u00b0C,
P1<\/sub>V1<\/sub>=P2<\/sub>V2<\/sub>
1 bar x 500 dm3<\/sup> = P2<\/sub> x 200 dm3<\/sup> or P2<\/sub>=500\/200 bar=2.5 bar<\/p>\n<\/span>Question 2. A vessel of 120 mL capacity contains a certain amount of gas at 35\u00b0C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35\u00b0C. What would be its pressure?<\/strong><\/span><\/h3>\nAnswer: <\/strong>V1<\/sub>= 120 mL, P1<\/sub>=1.2 bar,
V2<\/sub> = 180 mL, P2<\/sub> =?
As temperature remains constant, P1<\/sub>V1<\/sub> = P2<\/sub>V2<\/sub>
(1.2 bar) (120 mL) = P2 (180mL)<\/p>\n<\/span>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.<\/strong><\/span><\/h3>\nAnswer:<\/strong> According to the ideal gas equation
PV = nRT or PV=nRT\/V
![\"NCERT](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2020\/11\/NCERT-Solutions-for-Class-11-Chemistry-Chapter-5-States-of-Matter-Q3.png\")
<\/p>\n<\/span>Question 4. At 0\u00b0C, 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?<\/strong><\/span><\/h3>\nAnswer:<\/strong> Using the expression, d =MP\/RT , at the same temperature and for same density,
M1<\/sub>P1<\/sub> = M2<\/sub>P2<\/sub> (as R is constant)
(Gaseous oxide) (N2<\/sub>)
or
M1<\/sub> x 2 = 28 x 5(Molecular mass of N2<\/sub> = 28 u)
or M1<\/sub> = 70u<\/p>\n<\/span>Question 5. The pressure of l g of an ideal gas A at 27\u00b0C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses.<\/strong><\/span><\/h3>\nAnswer:<\/strong> Suppose molecular masses of A and B are MA<\/sub> and MB<\/sub> respectively. Then the number of moles will be
![\"NCERT](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2020\/11\/NCERT-Solutions-for-Class-11-Chemistry-Chapter-5-States-of-Matter-Q5.png\")
<\/p>\n<\/span>Question 6. The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 \u00b0C and one bar will be released when 0.15g of aluminum reacts?<\/strong><\/span><\/h3>\nAnswer:<\/strong> The chemical equation for the reaction is
2 Al + 2 NaOH + H2<\/sub>0 -> 2 NaAl02<\/sub> + 3H2<\/sub> (3 x 22400 mL At N.T.P)
2 x 27 = 54 g.
54 g of Al at N.T.P release
H2<\/sub> gas = 3 x 22400 0.15 g of Al at N.T.P release
![\"NCERT](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2020\/11\/NCERT-Solutions-for-Class-11-Chemistry-Chapter-5-States-of-Matter-Q6.png\")
<\/p>\n![\"NCERT](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2020\/11\/NCERT-Solutions-for-Class-11-Chemistry-Chapter-5-States-of-Matter-Q6.1.png\")
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<\/span>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<\/sup> flask at 27 \u00b0C?<\/strong><\/span><\/h3>\nAnswer:<\/strong>
![\"NCERT](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2020\/11\/NCERT-Solutions-for-Class-11-Chemistry-Chapter-5-States-of-Matter-Q7.png\")
<\/p>\n<\/span>Question 8. What will be the pressure of the gas mixture when 0.5 L of H2<\/sub> at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in all vessel at 27 \u00b0C?<\/strong><\/span><\/h3>\nAnswer: <\/strong>Calculation of partial pressure of H
![\"NCERT](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/09\/NCERT-Solutions-For-Class-11-Chemistry-Chapter-8-Redox-Reactions-1.jpg\")
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