Important Questions for CBSE Class 11 Chemistry Chapter 8 – Redox Reactions
Important Questions Class 11 Chemistry Chapter 8 – Redox Reactions
Chemistry is concerned with different kinds of matter and discusses how one type of matter can change into another. Through various reactions, matter can change from one type to another.
Chapter 8 of Class 11 Chemistry is about ‘Redox Reactions’. Redox reactions are a significant component of a variety of reactions that change matter from one type to another. This chapter is designed to help students develop a better understanding of the redox reactions that find extensive use in pharmaceutical, biological, industrial, metallurgical, and agricultural areas.
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Students can use our question bank of Chapter 8 Class 11 Chemistry Important Questions that have been compiled from a variety of sources, including NCERT textbooks, reference books, and past years’ exams. The solutions to these questions are prepared by expert Science teachers and strictly adhere to the CBSE exam guidelines. So students can confidently rely on our solutions for their exam preparation.
Extramarks Chemistry Class 11 Chapter 8 Important Questions contain distinct sections of questions which includes multiple choice questions, short answer questions, medium answer questions, and long answer questions. The questions cover all topics from the chapter so it would also help students to revise all concepts and gain a solid base of the theoretical concepts of redox reactions and their utilisation in various processes.
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|CBSE Class 11 Chemistry Important Questions|
|Sr No||Chapters||Chapter Name|
|1||Chapter 1||Some Basic Concepts of Chemistry|
|2||Chapter 2||Structure of Atom|
|3||Chapter 3||Classification of Elements and Periodicity in Properties|
|4||Chapter 4||Chemical Bonding and Molecular Structure|
|5||Chapter 5||States of Matter|
|8||Chapter 8||Redox Reactions|
|10||Chapter 10||The S-Block Elements|
|11||Chapter 11||The P-Block Elements|
|12||Chapter 12||Organic Chemistry Some Basic Principles and Techniques|
|14||Chapter 14||Enviornmental Chemistry|
Redox Reaction Class 11 Important Questions with Answers
Extramarks team has given step-by-step and comprehensive answers to all the questions that are covered in our Important questions Class 11 Chemistry Chapter 8. By using this information effectively, students will be able to comprehend redox reactions and other electrolytic reactions in an enhanced manner.
Below is a list of a few of the questions from our question bank of Important Questions Class 11 Chemistry Chapter 8
Question 1: Which of the following statement(s) is/are not true about the following decomposition reaction?
2KClO3 → 2KCl + 3O2
(i) Potassium is undergoing oxidation
(ii) Chlorine is undergoing oxidation
(iii) Oxygen is reduced
(iv) None of the species are undergoing oxidation or reduction
Answer 1: (i) and (iv)
Explanation: The following statements are incorrect because it is clear from the reaction that oxygen is being oxidised while potassium remains in the same oxidation state.
Question 2: Identify disproportionation reaction
(i) CH4 + 2O2 → CO2 + 2H2O
(ii) CH4 + 4Cl2 → CCl4 + 4HCl
(iii) 2F2 + 2OH– → 2F– + OF2 + H2O
(iv) 2NO2 + 2OH– → NO2– + NO3– + H2O
Answer 2: (iv) 2NO2 + 2OH– → NO2– + NO3– + H2O
Explanation: Oxidation number of nitrogen decreases by 1 from NO2 to NO2– and increase by +1 from NO2 to NO3–.
Question 3: Which of the following arrangements represents an increasing oxidation number of the central atom?
(i) CrO2– , ClO3– , CrO2-4 , MnO-4
(ii) ClO3– , CrO2-4 , MnO-4 , CrO2–
(iii) CrO2– , ClO3– , MnO-4 , CrO2-4
(iv) CrO2-4 , MnO-4 , CrO2– , ClO3– ,
Answer 3: (i) CrO2– , ClO3– , CrO2-4 , MnO-4
Explanation: As the central element’s oxidation number grows as +3, +5, +6, and +7, respectively, the above arrangement depicts the rising oxidation number of the central atom.
Question 4: The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
(i) The oxidation number of hydrogen is always +1.
(ii) The algebraic sum of all the oxidation numbers in a compound is zero.
(iii) An element in the free or uncombined state bears oxidation number zero.
(iv) In all its compounds, the oxidation number of fluorine is -1.
Answer 4: (i)
Explanation: As hydrogen picks up a negative charge when it is with its companion, ionic hydrides have hydrogen in the -1 oxidation state.
Question 5: Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
E values: Fe3+ / Fe2+ = 0.77; I2 / I– = + 0.54;
Cu2+ / Cu = 0.34; Ag+ /Ag= + 0.80 V
(i) Fe3+ and I–
(ii) Ag+ and Cu
(iii) Fe3+ and Cu
(iv) Ag and Fe3+
Answer 5: (iv) Ag and Fe3+
Explanation: To make the reaction feasible, E0the cell needs to be positive for the pair Ag and Fe3+, but it is negative. Hence the reaction is not feasible.
Question 6: The more positive the value of E, the greater the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent.
E values: Fe3+ / Fe2+ = + 0.77; I2(s)/ I– = +0.54;
Cu2+ /Cu = + 0.34; Ag+ /Ag = + 0.80 V
Answer 6: (iv) Ag+
Explanation: The reduction potential of Ag+ /Ag is the highest.
Question 7: What is an oxidation reaction?
Answer 7: Oxidation reactions are those which involve either the addition of oxygen or the removal of hydrogen.
Question 8: What is the reduction reaction?
Answer 8: Reduction reactions are those in which oxygen is either removed, or hydrogen is added.
Question 9: What are the most essential conditions that must be satisfied in a redox reaction?
Answer 9: It shouldn’t interfere with the theory of electron conservation. The total amount of electrons lost must balance the number of electrons gained by the oxidising agent.
Question 10: What happens to the oxidation number of an element in oxidation?
Answer 10: During oxidation, the oxidation number of the element increases. It is oxidised if the oxidation number increases from 0 to +1.
Question 11: Name the different types of redox reactions.
Answer 11: The different types of redox reactions are:
- Combination reactions
- Decomposition reactions
- Displacement reactions
- Disproportionate reactions
Question 12: All decomposition reactions are not redox reactions. Give justification. .
Answer 12: Calcium carbonate decomposition is not a redox reaction. This is true because there must be at least one elemental substance present during the decomposition of calcium carbonate on the product side.
Question 13: Define half-cell.
Answer 13: A half-cell is made up of an electrode structure and conducting electrolyte that is divided by a Helmholtz double layer.
Question 14: What is the role of a salt bridge in an electrochemical cell?
Answer 14: The role of a salt bridge in an electrochemical cell is that it provides electrical neutrality and prevents the mixing of the electrolytes.
Question 15: Define a redox couple.
Answer 15: A substance that participates in an oxidation and reduction half-reaction is said to be in a redox couple when its reduced and oxidised forms are present together.
Question 16: Explain why
3Fe3O4(s) + 8Al(s) → 9Fe(s) + 4Al2O3(g)
is an oxidation reaction?
Answer 16: It is an oxidation reaction because aluminium is getting oxidised. It forms Al2O3 in the product indicating that the addition of oxygen has taken place.
Question 17: The reaction
Cl2(g) + 2OH– (aq) → CIO– (aq) + Cl– (aq) + H2O(l)
represents the process of bleaching. Identify and name the species that bleaches the substances due to their oxidising action.
Here, Cl2 is converted into ClO- and Cl-, respectively, through oxidation and reduction. Because Cl- is not an oxidising agent (O.A.). As a result of the hypochlorite ClO- ion’s oxidising effect, Cl2 bleaches many materials.
Question 18: Fluorine reacts with ice and results in the change:
H2O(s) + F2(g) → HF(g)+ HOF(g)
Justify that this reaction is a redox reaction:
The oxidation number of each atom involved in the process should be written above its symbol as follows:
The oxidation number of F goes from 0 in F2 to +1 in HOF, as seen above. Additionally, the number of oxidations decreases from 0 in F2 to -1 in HF. F is thus both oxidised and reduced in the given reaction. As a result, the given reaction is a redox reaction.
Question 19: MnO42- undergoes a disproportionation reaction in an acidic medium but MnO4– does not. Give a reason.
A disproportionation reaction is a redox reaction in which one component with an intermediate oxidation state results in the formation of two molecules with greater and lower oxidation states.
The oxidation states of manganese in their different compounds range from +2 to +7. Disproportionation is not possible with MnO4– because of its maximum oxidation state of +7; however, MnO42- has a +6 oxidation state and can be both oxidised and reduced.
Question 20: Write the formula for the following compounds:
Answer 20:The chemical formula of the given compounds are as follows:
(a)Mercury (II) chloride: HgCl2
(b)Nickel (II) sulphate: NiSO4
(c)Tin (IV) oxide: SnO2
(d)Thallium (I) sulphate: TI2SO4
(e)Iron (III) sulphate: Fe2(SO4)3
(f)Chromium (III) oxide: Cr2O3
Question 21: PbO and PbO₂ react with HCl according to the following chemical equations:
2PbO + 4HCl → 2PbCl2 + 2H2O
PbO2 + 4HCI → PbCl2 + Cl2 + 2H2O
Why do these compounds differ in their reactivity?
Answer 21: In the first reaction, none of the atoms’ O.N. changes. Hence, it cannot be categorised as a redox reaction. It is an acid-base interaction because PbO is a basic oxide that reacts with HCl acid.
In the second reaction, which is a redox reaction, PbO2 is reduced and acts as an oxidising agent.
Question 22: Nitric acid is an oxidising agent and reacts with PbO, but it does not react with PbO₂. Explain why.
PbO, a basic oxide, reacts with HNO3 in a conventional acid-base manner. On the other hand, lead cannot be further oxidised in PbO2 because it is in the +4 oxidation state. As a result, no reaction takes place. Consequently, PbO2 is inactive, and only PbO interacts with HNO3.
2PbO + 4HNO3 → 2Pb(NO3)2 + 2H2O
Question 23: The compound AgF2 is an unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?
Answer 23: Ag in AgF2 has an oxidation state of +2, which is an unstable oxidation state of Ag. Silver hence readily accepts an electron to generate Ag+ whenever AgF2 is formed. This brings the oxidation state of Ag from +2 to +1, which is more stable. Hence, AgF2 acts as a very strong oxidising agent.
Question 24: Calculate the oxidation number of phosphorus in the following species.
(a) HPO32- and
(a). Let the oxidation number of phosphorus in HPO32- be x.
H + P + 3O2-
⇒ +1 + x + (-2)×3 = -2
⇒ +1 + x – 6 = -2
⇒ x – 5 = -2
⇒ x = – 2 + 5
⇒ x = +3
Thus, the oxidation number of phosphorus in HPO32- is +3.
(b). Let the oxidation number of phosphorus in PO43-
⇒ x + 4 ×(-2) = -3
⇒ x = -3 + 8 = +5
⇒ x = +5
Thus, the oxidation number of phosphorus in PO43- is +5.
Question 25: What sorts of information can you draw from the following reaction?
(CN)2(g) + 2OH–(aq) → CN–(aq) + CNO–(aq) + H2O(l)
The carbon in (CN)2, CN− and CNO− have oxidation numbers +3, +2 and +4, respectively. These are obtained as follows:
Let the oxidation number of C be x.
2(x – 3) = 0
⇒x = 3
CN− x – 3 = –1
⇒x = 2
CNO–x – 3 – 2 = –1
⇒x = 4
The oxidation number of carbon in the various species is:
It is clear from the equation that the same compound is being reduced and oxidised at the same time. Disproportionation reactions are those in which the same compound is reduced and oxidised. As a result, it is possible to say that the alkaline decomposition of cyanogen is an illustration of a disproportionation reaction.
Question 26: Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.
(i) 3 HCl(aq) + HNO3(aq) → Cl2(g) + NOCl (g) +2H2O(l)
(ii) HgCl2(aq) + 2KI (aq) → HgI2(s) + 2KCl (aq)
(iii) Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
(iv) PCl3(I) + 3H2O(l) → 3HCl(aq) + H3PO4 (aq)
(v) 4NH3 + 3O2(g) → 2N2(g) + 6H2O(g)
(i) Putting the oxidation number of each atom, we get:
3HCl + HNO3 → Cl2 + NOCl + 2H2O
The oxidation number of Cl goes from -1 (in HCl) to 0 (in Cl2). HCl is used as a reducing agent because of the oxidation of Cl–.
HNO3 functions as an oxidising agent because the oxidation number of N reduces from +5 (in HNO3) to +3 (in NOCl).
Thus, reaction (i) is a redox reaction.
(ii) HgCl2 + 2Kl → Hgl + 2KCl
There is no change in the oxidation number. Hence this reaction doesn’t qualify as a redox reaction.
(iii) Fe2O3 + 3CO → 2Fe +3CO2
Fe2O3 works as an oxidising agent because the oxidation number of Fe changes from +3 (in Fe2O3) to 0 (in Fe).
CO works as a reducing agent as the oxidation number of C increases from +2 (in CO) to +4 (in CO2). As a result, it is a redox reaction.
(iv) PCl3 + 3H2O → 3HCl + H2PO3
There is no change in the oxidation number of any of the atoms. Hence it is not a redox reaction.
(v) 4NH3 + 3O2 → 2N2 + 6H2O
Because the oxidation number of N in N2 grows from -3 to 0, NH3 functions as a reducing agent.
Furthermore, O2 functions as an oxidising agent because the oxidation number of O drops from 0 in O2 to -2 in H2O.
Hence, this is a redox reaction.
Question 27: Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non-metals that can show a disproportionation reaction.
(b) Select three metals that can show a disproportionation reaction.
One of the reacting compounds in disproportionation reactions always contains an element that can exist in at least three oxidation states,
(a) As the non-metals P, Cl, and S can exist in three or more oxidation states, they can exhibit disproportionation reactions.
(b) Mn, Cu, and Ga can also exhibit such reactions as these metals can exist in three or more oxidation states.
Question 28: Explain redox reactions on the basis of electron transfer. Give suitable examples.
A chemical reaction in which electrons are transferred between two substances is known as an oxidation-reduction (redox) reaction. An oxidation-reduction reaction is any chemical process in which a molecule, atom, or ion’s oxidation number changes as a result of gaining or losing an electron.
As we know, the reactions,
2Na(s) + Cl2(g) → 2NaCl(s)
4Na(s) + O2(g) → 2Na2(s)
are redox reactions since they all entail the addition of oxygen or a more electronegative element to sodium.
While sodium, an electropositive element, has been added, chlorine and oxygen are simultaneously being depleted. We also realise that sodium chloride and sodium oxide are ionic compounds, and may be better stated as Na+Cl–(s) and Na2+O2, respectively, based on our knowledge of chemical bonding (s).
The reactions can be represented as follows:
One phase of each of the processes above involves electron loss, and the other phase involves electron gain. We can elaborate on one of them, like the production of sodium chloride.
2Na(s) → 2Na+(g) + 2e−
Cl2 + 2e− → 2Cl−(g)
Since the involvement of electrons is evident, each preceding step is referred to as a half-reaction. The sum of the half-reactions determines the total reaction:
2Na(s) + Cl2(g) → 2Na + Cl−(s) or 2NaCl(s)
According to the reactions above, oxidation reactions account for 50% of all reactions involving electron loss. Similar to this, electron gain-related half-reactions are referred to as reduction reactions.
Redox reactions are the fundamental processes of life, including photosynthesis, respiration, combustion, and corrosion or rusting.
Question 29: Arrange the given metals in the order in which they displace each other from the solution of their salts.
In a salt solution, a metal with a higher reducing power pushes out metal with lower reducing power.
The following list of metals is in increasing order of their reducing power:
Cu < Fe < Zn < Al < Mg
Thus, Mg can displace Al from its salt solution, but Al cannot displace Mg. Therefore, we can conclude that the order in which the given metals displace each other from the solution of their salts is as given below: Mg >Al>Zn> Fe >Cu
Question 30: Why does fluorine not show a disproportionation reaction?
A disproportionation reaction is a redox reaction in which one component with an intermediate oxidation state results in the formation of two molecules with greater and lower oxidation states.
2F2(g) + 2OH–(aq) → 2F–(aq) + OF2(g) + H2O(I)
Such a redox reaction cannot take place unless the element is in at least three oxidation states. As a result, during the disproportionation reaction, that element is in the intermediate state and is capable of switching between higher and lower oxidation levels.
Of all the halogens, fluorine is the most electronegative and oxidising element. It is also the smallest.
The disproportionation reaction does not occur, and it only has one positive oxidation state.
Question 31: Which method can be used to find out the strength of the reductant/oxidant in a solution? Explain with an example.
The relative electrode potential can be determined when a reductant (reducing agent) or oxidant (oxidising agent) is connected in a solution using a cell. The element under discussion can be used as an electrode in a typical cell with a known electrode potential. If the electrode of the given species is positive, it acts as a reductant; if it is negative, it acts as an oxidant.
Let us take the example of Fe3+/Fe with a Standard Hydrogen electrode (SHE). For Fe and H, the half-cell reaction is as follows:
H+ + e– → H2Eo = 0.0V
Fe3+ + e– → Fe2+Eo = 0.77
Any element that requires evaluation can be employed in SHE as an electrode. The amount of emf an element produces in a cell is referred to as its potential.
Eocell = Eocathode – Eoanode
Eocell= 0 – Eoanode
Eocell = 0 – 0.77
The Fe3+ ion is more likely to go through reduction than hydrogen is. The strength of Fe as a reductant can then be determined by altering the previously believed Fe anode arrangement. As a result, the strength of the oxidant can be identified.
Benefits of Solving Class 11 Chemistry Chapter 8 Important Questions
Important questions Class 11 Chemistry Chapter 8 from Extramarks will help students identify redox reactions as a class of reactions in which oxidation and reduction reactions occur simultaneously; use the concept of oxidation number to identify oxidant and reductant in a reaction, and learn the concept of redox reactions in terms of electrode processes. These questions not only offer a thorough list of questions to consider, but they also facilitate understanding of the chapter.
The following are the benefits of solving questions from our important questions of redox reaction class 11:
- Our expert Science faculty members have prepared the answers with step-by-step solutions in easy to understand language. It makes it easy for students to understand the complex concepts and also retain knowledge for a longer period of time.
- By practising the exam specific questions given in our question bank
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Q.1 What will happen when chlorine is passed through an aqueous solution of potassium bromide? Write the chemical reaction also.
When chlorine is passed through an aqueous solution of potassium bromide, the solution will acquire an orange colour due to the vapours of bromine that are evolved. Chlorine is a stronger oxidising agent than bromine and oxidises Br– ions (in KBr) to Br2 and itself is reduced to Cl– ions.
Q.2 Why does H2S acts only as a reducing agent while SO2 can act an oxidising as well as reducing agent.
Oxidation state of S in H2S is 2.
Oxidation state of S in SO2 +4.
In H2S, S is in its lowest oxidation state i.e., 2 and it can loss electrons to attain the maximum oxidation state of +6 but it cannot gain electron. Thus, H2S acts only as a reducing agent. On the other hand, the oxidation state of SO2 is +4. So, it can lose its 2 more electrons to attain +6 oxidation state and it can gain electrons to attain 2 oxidation state. Due to the capability of S in SO2 to gain and lose electron, it acts as an oxidising as well as reducing agent.
Q.3 Why does fluorine not show disproportionation reaction?
In disproportionation reaction, an element in one oxidation state simultaneously undergoes oxidation and reduction. So, such type of redox reaction takes place when an element exists in at least 3 oxidation states. An element present in intermediate state can change to both higher and lower oxidation state during disproportionation reaction. Fluorine is the most electronegative element and a strong oxidising agent. It does not show positive oxidation state and does not undergo disproportionation reaction.
Q.4 What is the oxidation number of C in CH2O?
Q.5 Pick the one which is not an example of redox couple.
FAQs (Frequently Asked Questions)
1. Where can students find a comprehensive list of Class 11 Chemistry Chapter 8 Important Questions?
Extramarks provides a comprehensive list of Important questions Class 11 Chemistry Chapter 8 for students to revise easily and do well in their school tests and also competitive exams.
2. What is the importance of studying redox reactions in Class 11?
The importance of studying redox reactions in Class 11 is that students learn about electrolysis, which is based on redox reactions and is used to manufacture many of the chemicals we require on a daily basis, such as caustic soda, chlorine, fluorine, etc. Redox titrations are a powerful tool for quantitative analysis.
3. What is the economic impact of redox reactions?
The economic impact of redox reactions is that rust on commercial objects is formed due to redox processes. Numerous metal objects are destroyed by rust. We can safeguard them by giving them coverings. Food oxidation, which results in bad food odour, is also fairly prevalent and should be stopped.