VMC VIQ Scholarship Test
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NCERT exemplar Class 11 Chemistry solutions chapter 8 is the quintessential workbook for students who have an inclination towards expanding their practical understanding regarding a theoretical concept. NCERT exemplar Class 11 Chemistry chapter 8 solutions commit towards focussing upon the practical implication of the principles and rules of a topic by forming them in challenging question format for the interested students. This chapter of NCERT Class 11 Chemistry Solutions introduces the students to redox reactions and guides them around their different types and how to deal with them by understanding their specific rules and properties.
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Scholarship Test: Vidyamandir Intellect Quest (VIQ)
Question:1
Which of the following is not an example of redox reaction?
Answer:
The answer is the option
Explanation: The correct answer is the option (iv) because only its oxidation no. changes while that of all elements remains unchanged.
Question:2
Answer:
The answer is the option
Explanation: Reduction potential of is the highest.
Question:3
values of some redox couples are given below. On the basis of these values choose the correct option.
values:
(i) Cu will reduce
(ii) Cu will reduce Ag
(iii) Cu will reduce
(iv) Cu will reduce
Answer:
The answer is the option (iv) Cu will reduce Br2
Explanation: Compared to Cu, Br2 is a better reducing agent.
Question:4
The answer is the option
Explanation: To make the reaction feasible, cell need to be positive for the pair Ag and , but it is negative. Hence the reaction is not feasible.
Question:5
Thiosulphate reacts differently with iodine and bromine in the reactions given below:
Which of the following statements justifies the above dual behaviour of thiosulphate?
(i) Bromine is a stronger oxidant than iodine.
(ii) Bromine is a weaker oxidant than iodine.
(iii) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.
(iv) Bromine undergoes oxidation and iodine undergoes reduction in these reactions.
Answer:
Ans. The answer is the option (i) Bromine is a stronger oxidant than iodine.
Explanation: To be a better oxidant, the standard reduction potential of the particular element should be higher than the other, but, here, the standard reduction potential of bromine is higher than iodine. Hence, bromine is a stronger oxidant than iodin
Question:6
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 the uncombined state bears oxidation number zero.
(iv) In all its compounds, the oxidation number of fluorine is -1.
Answer:
The answer is the option (i) The oxidation no. of hydrogen is always +1.
Explanation: In ionic hydrides, hydrogen exists in -1 oxidation state because the hydrogen acquires negative charge in the presence of its companion.
Question:7
In which of the following compounds, an element exhibits two different oxidation states.
Answer:
The answer is the option
Explanation: Here, Nitrogen exists in two different oxidation states, i.e., +5 and -3.
Question:8
Which of the following arrangements represent increasing oxidation number of the central atom?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (i)
Explanation: the above arrangement represents increasing oxidation no. of the central atom because here the oxidation no. of the central element increases as +3, +5, +6, +7, respectively.
Question:9
The answer is the option
Explanation: the above arrangement will exhibit the largest oxidation number since total no. of electrons present in d and s subshell = 7.
Question:10
Identify disproportionation reaction
Answer:
The answer is the option
Explanation: Oxidation no. of nitrogen decreases by 1 from to and increase by +1 from to
Question:11
Which of the following elements does not show disproportionation tendency?
(i) Cl
(ii) Br
(iii) F
(iv) I
Answer:
The answer is the option (iii) F
Explanation: Fluorine is the most electronegative element; therefore, it does not show a disproportionation tendency.
Question:12
Which of the following statement(s) is/are not true about the following decomposition reaction.
(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:
The answer is the option (i) and (iv)
Explanation: The statements above are not true because it is evident in the reaction that, potassium remains in the same oxidation state, and oxygen is being oxidised
Question:13
Identify the correct statement (s) in relation to the following reaction:
(i) Zinc is acting as an oxidant
(ii) Chlorine is acting as a reductant
(iii) Hydrogen ion is acting as an oxidant
(iv) Zinc is acting as a reductant
Answer:
The answer is the option (iii) and (iv)
Explanation: The above statement is correct because you according to the reaction zinc is being oxidised and hydrogen is reduced.
Question:14
The answer is the option (iii) and (iv)
Explanation: Atoms having electronic configurations such as in Option (iii) and (iv) will exhibit more than one oxidation state in the compound because in option (iii) electron can be removed from 4s as well as 3d. Similarly, in option (iv) electron can be removed from 3p and 3s both.
Question:15
Identify the correct statements with reference to the given reaction
(i) Phosphorus is undergoing reduction only.
(ii) Phosphorus is undergoing oxidation only.
(iii) Phosphorus is undergoing oxidation as well as reduction.
(iv) Hydrogen is undergoing neither oxidation nor reduction.
Answer:
The answer is the option (iii) and (iv)
Explanation: This is a kind of disproportionation reaction in which phosphorous is being reduced as well as oxidised, as given in opt (iii); whereas hydrogen remains same in +1 oxidation state as in opt (iv). Therefore, option (iii) and (iv) are correct.
Question:16
Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?
Answer:
The answer is the option (i) and (ii)
Explanation: They will act as anodes because both of the electrodes has a negative value of standard reduction potential
Question:17
The hypochlorite ion formed in the reaction oxidises the colour-bearing stains of the substances to colourless compounds; therefore, in the above reaction, the hypochlorite ion acts as a bleach
Question:18
undergoes disproportionation reaction in acidic medium but does not. Give reason.
Answer:
In , Mn has the highest oxidation state (+7); therefore, it cannot undergo oxidation which is required for disproportionation here. Hence doesn’t undergo disproportionation reaction. Whereas in manganese is in +6 (i.e., less than that of in ) oxidation state which can be oxidised as well as reduced, hence it undergoes disproportionation.
Question:19
Lead is present in +4 oxidation state and the stable oxidation state of lead in PbO is +2. thus it can act as an oxidising agent. Therefore, it can oxidise ions of HCl into Chlorine.
Question:20
Nitric acid is an oxidising agent and reacts with PbO but it does not react with . Explain why?
Answer:
We know that HNO3 is and oxidising agent and is a basic oxide. Therefore, it is unlikely that any reaction will occur between them, however, the acid base reaction occurs between these two compounds, as shown below:
Question:22
Calculate the oxidation number of phosphorus in the following species.
and
Answer:
Let us consider the oxidation number of phosphorus to be x.
Thus, O.S. of phosphorous is +3.
Thus, O.S. of phosphorous in this ion is +5.
Question:23
Calculate the oxidation number of each sulphur atom in the following compounds:
Answer:
(a) Consider oxidation number of sulphur atom x.
Question:24
Balance the following equations by the oxidation number method.
Answer:
2+ 6+ 2- +3 +3
(a) Balance the inceases and decreases in O.N.
2+ 6+ 2- +3 +3
(b) Balancing H and O atoms by adding molecules
O.N. increases by +5
Now, we know that the total increase in O.N. = and the reduction in O.N. = 1
In order to equalize O.N., we will multiply with 10, which will give us,
Now, we will be balancing the atoms other than O and H, to give us
Now, on balancing O and H we will get,
0 +2 -1 +2.5
Total increase in O.N.
Total decrease in O.N.
To equalise O.N. multiply and by 2.
+4 +3 +2 +4
Total increase in O.N.
Total decrease in O.N.
Now, in order to equalize O.N., we will multiply with 2 to give us,
Now, we will balance H and O by adding 2H2O on the right side, and 4H+ on the left side of equation.
Question:25
In the above reaction, it can be seen that Cl oxidizes from -1 to 0 and Nitrogen reduces from +5 to +3. Thus, Cl oxidizes to act as a reducing agent and the nitric acid works as the oxidizing agent.
Therefore, in the above reaction, the reducing agent is HCl and the oxidizing agent is
The above mentioned reaction is the perfect example of a displacement reaction. As a result, none of the components undergo the process of reduction or oxidation. This is because the oxidation states of Hg, Cl, K, & I are unchanged in both the products and reactants.
In the above reaction, the oxidising agent is and the reducing agent is CO.
In the above reaction, oxidising agent is and the reducing agent is .
Question:26
Balance the following ionic equations
As step (i), we will be generating the unbalanced skeleton i.e.,
As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.
Oxidation is and Reduction is
As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.
Oxidation is and Reduction is
Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.
The oxidation is and reduction is
As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.
Hence, oxidation is and
reduction is
Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction. We will now multiply the oxidation reaction by 3, which will give us,
The oxidation as and
the reduction as
Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.
Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
Now, we know that the charge on the LHS is +6 which is same as the RHS of the equation, thereby enduring that the reaction is balanced.
As step (i), we will be generating the unbalanced skeleton i.e.,
As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.
Thus, oxidation is and the reduction is
As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.
Thus, oxidation is and the reduction is
Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.
Thus, oxidation is and
reduction is
As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.
Thus, oxidation is and
reduction is
Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.
Thus, oxidation is and
reduction is
Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.
Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
We will now verify whether all the charges are balanced.
So, the LHS = and the RHS =
Thus the sum total is same on both the sides, therefore the solve reaction is right.
As step (i), we will be generating the unbalanced skeleton i.e.,
Here in Mn undergoes reduction and S undergoes oxidation.
As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.
Thus, oxidation is and reduction is
As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples. But as we can see that the atoms are balanced, so the reaction will be the same as the previous step.
Thus, oxidation is and reduction is
Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.
Thus, oxidation is and
reduction is
As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.
Thus, oxidation is and
reduction is
Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.
Thus, oxidation is and
reduction is
Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.
Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
We will now verify whether all the charges are balanced.
So, LHS = and the RHS =
Equal charges on both sides imply towards a balanced equation.
As step (i), we will be generating the unbalanced skeleton i.e.,
As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.
Thus, oxidation is and reduction is
As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.
As we can see that the atoms are balanced, therefore the reaction is going to be the same as the previous step.
Thus, oxidation is and reduction is
Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.
Thus, oxidation is and
reduction is
As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.
Thus, the oxidation is and
reduction is
Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.
Thus, oxidation is and
reduction is
Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.
Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
Thus, the charge on LHS and RHS is equal.
Question:27
Match Column I with Column II for the oxidation states of the central atoms.
COLUMN I | COLUMN II |
(i) | (a)+3 |
(ii) | (b)+4 |
(iii) | (c)+5 |
(iv) | (d)+6 |
(e)+7 |
Answer:
(i) → (d); (ii) → (e); (iii) → (c); (iv) → (a)
Question:28
Match the items in Column I with relevant items in Column II.
COLUMN I | COLUMN II |
(i) Ions having positive charge | (a) +7 |
(ii) The sum of oxidation number of all atoms in a netural molecules | (b) -1 |
(iii) Oxidation number of hydrogen ion | (c)+1 |
(iv) Oxidation number of fluorine in NaF | (d) 0 |
(v) Ions having negative charge | (e)Cation |
(f) Anion |
Answer:
(i) → (e); (ii) → (d); (iii) → (c); (iv) → (b); (v) → (f)
Question:29
In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : Among halogens fluorine is the best oxidant.
Reason (R) : Fluorine is the most electronegative atom.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Answer:
The answer is the option (ii) Both A and R are true but R is not the correct explanation of A.
Explanation: Fluorine is considered as the best oxidant among halogens because it is the most electronegative element. There is no other element that can oxidize fluorine. Its electronegativity is 3.98, which is the highest in the periodic table. This results in good oxidant properties.
Question:30
In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): In the reaction between potassium permanganate and potassium iodide, permanganate ions act as oxidising agent.
Reason (R) : Oxidation state of manganese changes from +2 to +7 during the reaction.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Answer:
The answer is the option (iii) A is true, but R is false.Explanation: Due to the change of permanganate ion changes to
As per this equation, the maximum oxidation state that Mn is capable of reaching as per its electronic configuration is +7. Therefore, being an oxidising agent, it cannot undergo oxidation and hence undergoes reduction. On the other hand, Iodine undergoes oxidation from -1 to 0.
Question:31
In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionation reaction.
Reason (R) : The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in and –2 oxidation state in .
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Answer:
The answer is the option (i) Both A and R are true, and R is the correct explanation of A.
Explanation:
Here the oxygen of peroxide is present in -1 state, and it is converted to zero oxidation state in undergoing oxidation and decreases to -2 oxidation state in undergoing reduction.
Question:32
In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : Redox couple is the combination of oxidised and reduced form of a substance involved in an oxidation or reduction half cell.
Reason (R) : In the representation and and are redox couples.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Answer:
The answer is the option (ii) Both A and R are true, but R is not the correct explanation of A.
Explanation: A redox couple is defined as a pair of compounds or elements having together the oxidised and reduced forms of it and taking part in an oxidation or reduction half-reaction.
Question:33
Explain redox reactions on the basis of electron transfer. Give suitable examples.
Answer:
When a redox reaction takes place, the species which loses the electrons is regarded as undergoing the oxidation reaction. It therefore acts as an oxidizing agent or oxidant. For the species that accepts electrons is regarded as undergoing reduction and therefore it behaves as a reductant.
The above mentioned transfer is largely based on the relative electronegativity of difference between two interacting species. The higher electronegative element attracts more electrons while on the other hand the electropositive element loses electrons.
Splitting of the reaction into two halves here proves that sodium is oxidised, and hydrogen is reduced. Here, the sodium atom is oxidised, and the hydrogen atom is reduced; therefore, it is a redox reaction.
is a redox change.
and the other half-reaction is:
Question:34
If a reaction might take place or not is dependent on the net cell EMF of the cell. The equation for the same is
In option (ii) it is clearly visible that, the reaction can take place as Mg has a more negative value of . Hence, Mg is oxidised by losing electron and iron is reduced by gaining an electron.
We can say that Fe undergoes reduction and Mg undergoes oxidation,
Question:35
Why does fluorine not show disproportionation reaction?
Answer:
We already know that; fluorine has the highest electronegativity in the entire periodic table with EN value of 3.98. Therefore, it is the most electronegative element in its period as well as in the group. It shows only -1 oxidation state and is smallest in size of all the halogens. We know that fluorine has the greatest reduction potential therefore, it is not able to undergo oxidation by itself. Hence, fluorine shows a disproportionation reaction.
Question:37
For all the chlorine atoms, we will assume the oxidation state to be x, and for atoms Na, K = +1, O = -2. Now, we know that, the sum of oxidation numbers of all the atoms in a compound is the same as the charge on that compound for an ion or zero.
Hence,
In the structure there are two atoms of chlorine with varying types of O ? O, Cl ? Cl. The structure is explained as follows: -
As chlorine is in its naturalthe oxidation number is zero.
Therefore the ascending order is,
Question:38
In order to determine the strength of a reductant/oxidant, we can use the titration method; it can be performed with the help of a redox-sensitive indicator.
The usage indicators in redox titration are as shown below:
In one scenario, the reagent itself is intensely coloured, e.g., permanganate ion, . In here, acts as a self-indicator. In this case, the visible endpoint is attained after the last of the reductantis oxidised. Now, the first lasting tinge of pink colour appears as concentration as low as mol . This make sure of a minimal ‘overshoot’ in colour beyond the equivalence point, which is the point from where the reductant and the oxidant are equal in terms of their mole stoichiometry.
The NCERT exemplar Class 11 Chemistry solutions chapter 8 pdf download is a best-suited format offered to students who want an all-time access to the topics’ questions and answers and want to go back to them as per their requirement.
Also, check - NCERT Solutions for Class 11
Class 11 Chemistry NCERT Exemplar Solutions Chapter 8 include the following topics:
1. Classical Idea of Redox Reactions – Oxidation And Reduction Reactions
2. Redox Reactions In Terms Of Electron Transfer Reactions
2.1 Competitive Electron Transfer Reactions
3. Oxidation Number
3.1 Types Of Redox Reactions
3.2 Balancing Of Redox Reactions
3.3 Redox Reactions As The Basis For Titrations
3.4 Limitations Of Concept Of Oxidation Number
3.5 Redox Reactions And Electrode Processes.
After studying Class 11 Chemistry NCERT exemplar solutions chapter 8, students will have been enlightened upon the concept of the occurrence of a specific type of chemical reaction named as redox reactions.
They will have the complete knowledge as to the reasons and factors that influence the occurrence of them and how we can manage these factors.
After getting a hold on the theoretical aspect of redox reactions, students will learn to evaluate them after gaining the expertise to balance their equations so that they can draw a conclusion about the reaction.
NCERT Exemplar Class 11 Chemistry solutions chapter 8 also imparts knowledge about the practical applications of redox reactions, namely, titrations.
NCERT exemplar solutions for Class 11 Chemistry chapter 8 makes the students familiar with the concept of oxidation and oxidation number after which they engage in learning about the concluding topic of the chapter that is redox reactions and electrode processes.
Chapter-1 - Some Basic Concepts of Chemistry
Chapter-2 - Structure of Atom
Chapter-3 - Classification of Elements and Periodicity in Properties
Chapter-4 - Chemical Bonding and Molecular Structure
Chapter-5 - States of Matter
Chapter-6 - Thermodynamics
Chapter-7 - Equilibrium
Chapter-8 - Redox Reaction
Chapter-9 - Hydrogen
Chapter-10 - The S-Block Elements
Chapter-11 - The P-Block Elements
Chapter-12 - Organic chemistry- some basic principles and techniques
Chapter-14 - Hydrocarbons
Chapter-15 - Environmental Chemistry
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