Important Questions for CBSE Class 11 Chemistry Chapter 7 - Redox Reactions

1 Mark Questions

1. Define oxidation reaction?

Ans: A reaction in which oxygen gets added, or removal of a hydrogen atom takes place is called an oxidation reaction.

2. Define reduction reaction?

Ans: A reaction in which oxygen gets removed, or the addition of a hydrogen atom takes place is called an oxidation reaction.

3. In the reactions given below, identify the species undergoing oxidation and reduction.${{H}_{2}}S(g)+C{{l}_{2}}\to 2HCl(g)+S(s)$

Ans: Chlorine, being an electronegative element, is added to hydrogen, so ${{H}_{2}}S$ is oxidised. Hydrogen is added to chlorine, hence chlorine reduces while sulphur gets oxidized.

4. What are the most essential conditions that must be satisfied in a redox reaction?

Ans: A total number of electrons lost should be equal to the total number of electrons gained by the oxidising agent.

5. In the reaction $Mn{{O}_{2}}+4HCl\to MnC{{l}_{2}}+C{{l}_{2}}+2{{H}_{2}}O$, which species is oxidized?

Ans: As the addition of chlorine occurs in $HCl$, hence, it is the oxidising agent.

6. Why is the following reaction an example of an oxidation reaction? $C{{H}_{4}}(g)+2{{O}_{2}}(g)\to C{{O}_{2}}(g)+2{{H}_{2}}O$

Ans: It is so because the addition of oxygen is observed in $C{{H}_{4}}$. The addition of oxygen signifies oxidation.

7. Define oxidation in terms of electron transfer.

Ans: Loss of electrons performed by the reducing agent is called oxidation. If the oxidation number of an element changes from 0 to +1, then it is said to be oxidised.

8. What is meant by reduction?

Ans: Gain of electrons performed by the oxidising agent is called oxidation. If the oxidation number of an element changes from 0 to -1, then it is said to be reduced.

9. Define an oxidizing agent. Name the best reducing agent.

Ans: A substance that can easily gain electrons is called an oxidising agent. Fluorine molecules are the best oxidising agent.

10. What is meant by reducing? Name the best reducing agent.

Ans: A substance that can easily lose electrons is called a reducing agent. Lithium is the best reducing agent.

11. What is the oxidation number of manganese in $KMn{{O}_{4}}$?

Ans: If ‘x’ is the oxidation number of manganese, then:

$  1+x+4(-2)=0 $ 

$ \Rightarrow x=+7 $

12. What happens to the oxidation number of an element in oxidation?

Ans: Oxidation number increases during oxidation. If the oxidation number of an element changes from 0 to +1, then it is said to be oxidised.

13. Name one compound in which the oxidation number of Cl is + 4.

Ans: $Cl{{O}_{2}}$, here the oxidation number of chlorine is +4. It can be found out by taking the oxidation number as “x”.

14. Indicate the oxidizing and reducing agents in the following reaction : $2C{{u}^{2+}}+4{{I}^{-}}\to 2CuI+{{I}_{2}}$.

Ans: $C{{u}^{2+}}$ is an oxidising agent and ${{I}^{-}}$ is a reducing agent. The copper ion accepts the electron and itself gets reduced while the iodide ion loses an electron.

15. A metal ion ${{M}^{3+}}$ loses 3 electrons. What will be its oxidation number?

Ans: The oxidation number will be ($3+3=+6$). Losing an electron means a more positive charge on the atom, signifying that the element is oxidised.

16. Name the different types of redox reaction

Ans:

• Combination reactions

• Decomposition reactions

• Displacement reactions

• Disproportionation reactions

17. Identify the type of redox reaction this reaction follows. $3Mg(s)+{{N}_{2}}(g)\xrightarrow{\Delta }M{{g}_{3}}{{N}_{2}}(s)$

Ans: As in the reaction, 2 reactants form a single product on heating; it is a combination reaction.

18. The displacement reactions of Cl, Br, I using fluorine are not generally carried out in an aqueous solution. Give a reason.

Ans: Fluorine being a reactive element replaces chloride bromide and iodide ions in solution and it reacts with water and displaces the oxygen present there.

19. Which is the strongest oxidizing agent?

Ans: ${{F}_{2}}$ is the strongest oxidising agent. It is the most electronegative element and undergoes reduction by accepting an electron.

20. Why ${{F}^{-}}$ ions Cannot be converted to ${{F}_{2}}$ by chemical means?

Ans: It is chemically impossible as fluorine is an oxidising agent, it does not lose electrons.

21. Define disproportionation reaction.

Ans: In a disproportionation reaction an element in one oxidation state is oxidized and reduced simultaneously.

22. Identify the reaction $2{{H}_{2}}{{O}_{2}}(aq)\to 2{{H}_{2}}O(g)+{{O}_{2}}(g)$

Ans: It is a disproportionate reaction. It is so because hydrogen peroxide is getting both oxidised and reduced simultaneously.

23. Which gas is produced when less reactive metals like Mg and Fe react with steam?

Ans: $Mg+2{{H}_{2}}O\xrightarrow{\Delta }Mg{{(OH)}_{2}}$

${{H}_{2}}Fe+3{{H}_{2}}O\xrightarrow{\Delta }F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}$

Dihydrogen gas is released.

24. All decomposition reactions are not redox reactions. Give a reason.

Ans: It is because in a decomposition reaction both the products or one of the two products should be in elemental form, so all decomposition reactions are not redox reactions. Example- Decomposition of calcium carbonate.

25. Complete the following redox reactions and balance the following equations- 

(i) $C{{r}_{2}}{{O}_{7}}^{2-}+{{C}_{2}}{{O}_{4}}^{2-}\to C{{r}^{3+}}+C{{O}_{2}}$ (in presence of acid) 

Ans: In presence of acid,${{H}^{+}}$ ions are available. The reactions are:

$  C{{r}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O $

$ [{{C}_{2}}{{O}_{4}}^{2-}\to 2C{{O}_{2}}+2{{e}^{-}}]\times 3$ 

We multiply the second equation by 3 so as to balance the number of electrons, and we get the final equation as: 

$C{{r}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+3{{C}_{2}}{{O}_{4}}^{2-}\to2C{{r}^{3+}}+6C{{O}_{2}}+7{{H}_{2}}O$

(ii) $S{{n}^{2+}}+C{{r}_{2}}{{O}_{7}}^{2-}\to S{{n}^{4+}}+C{{r}^{3+}}$ (in presence of acid)

Ans:  In presence of acid,${{H}^{+}}$ ions are available. The reactions are:

$  C{{r}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O $

 $ [S{{n}^{2+}}\to S{{n}^{4+}}+2{{e}^{-}}]\times 3$ 

We multiply the second equation by 3 so as to balance the number of electrons, and we get the final equation as:

$C{{r}_{2}}{{O}_{7}}^{2-}+3S{{n}^{2+}}+14{{H}^{+}}\to2C{{r}^{3+}}+3S{{n}^{4+}}+7{{H}_{2}}O$ 

26. Write correctly the balanced half-reaction and the overall equations for the following skeletal equations.

(i) $N{{O}_{3}}^{-}+Bi(s)\to B{{i}^{3+}}+N{{O}_{2}}$(in acid solution) 

Ans:

(i) In acidic medium, ${{H}^{+}}$ is available. 

The oxidation half is:

\[Bi(s)\to B{{i}^{3+}}+3{{e}^{-}}\]

The reduction half reaction:

\[[N{{O}_{3}}^{-}+2{{H}^{+}}+{{e}^{-}}\to N{{O}_{2}}+{{H}_{2}}O]\times 3\]

The balanced equation is:

\[Bi(s)+3N{{O}_{3}}^{-}+6{{H}^{+}}\to B{{i}^{3+}}+3N{{O}_{2}}+3{{H}_{2}}O\]

(ii) $Fe{{(OH)}_{2}}(s)+{{H}_{2}}{{O}_{2}}\to Fe{{(OH)}_{3}}(s)+{{H}_{2}}O$(in basic medium)

Ans: In a basic medium, $O{{H}^{-}}$is available.

The oxidation half is:

\[[Fe{{(OH)}_{2}}+O{{H}^{-}}\to Fe{{(OH)}_{3}}+{{e}^{-}}]\times 2\]

The reduction half is:

\[{{H}_{2}}{{O}_{2}}+2{{e}^{-}}\to 2O{{H}^{-}}\]

The balanced equation is:

\[2Fe{{(OH)}_{2}}+{{H}_{2}}{{O}_{2}}\to 2Fe{{(OH)}_{3}}\]

27. Define half–cell.

Ans: A half cell consists of conducting electrolyte and electrode structure, separated by a Helmholtz double layer.

28. Set up an electrochemical cell for the redox reaction $N{{i}^{2+}}(aq)+Fe(s)\to Ni(s)+F{{e}^{2+}}(aq)$

Ans: $Fe(s)|F{{e}^{2+}}\,||\,N{{i}^{2+}}(aq)|Ni(s)$

29. Can we store copper sulphate in an iron vessel?

Ans: Iron displaces copper from the copper sulphate solution. It will form iron sulphate.

30. What is the role of a salt bridge in an electrochemical cell?

Ans:

• Provide electrical neutrality

• Prevents the mixing of the electrolytes.

31. Which reaction occurs at the cathode in a galvanic cell?

Ans: At the cathode, reduction happens, whereas oxidation occurs in the anode. A galvanic cell consists of a cathode, anode and electrolyte.

2 Marks Questions

1. Why $Cl{{O}_{4}}^{-}$does not show disproportionation reaction whereas $Cl{{O}^{-}},Cl{{O}_{2}}^{-},Cl{{O}_{3}}^{-}$ shows?

Ans: The chlorine atoms in $Cl{{O}^{-}},Cl{{O}_{2}}^{-},Cl{{O}_{3}}^{-}$ have an oxidation state of +1,+3+5 respectively. However, in $Cl{{O}_{4}}^{-}$, the oxidation state of chlorine is +7, which is maximum. That is why it doesn’t show a disproportionate reaction.

2. How would you know whether a redox reaction is taking place in an acidic/alkaline or neutral medium?

Ans: The presence of an acidic solution can be indicated by the presence of ${{H}^{+}}$ ions.

The presence of a basic or alkaline solution can be indicated by the presence of $O{{H}^{-}}$ ions.

If both of these ions are absent in the chemical reaction, then it is a neutral solution.

3. Write the following redox reactions in the oxidation and reduction half-reaction reactions in the oxidation and reduction half-reactions.

(i) $2K(s)+C{{l}_{2}}(g)\to 2KCl(s)$

Ans:

$K(s)\to {{K}^{+}}(aq)+{{e}^{-}}\,\,(oxidation) $

$C{{l}_{2}}(g)+2{{e}^{-}}\to 2C{{l}^{-}}\,\,(reduction)$

(ii) $2Al(s)+3C{{u}^{2+}}(aq)\to 2A{{l}^{3+}}(aq)+3Cu(s)$

Ans: $  Al(s)\to A{{l}^{3+}}(aq)+3{{e}^{-}}\,\,(oxidation) $ 

$C{{u}^{2+}}+2{{e}^{-}}\to Cu(s)\,\,(reduction)$ 

4. An electrochemical cell is constituted by combining Al electrode (${{E}_{0}}=-1.66V$) and Cu electrode (${{E}_{0}}=+0.34V$). Which of these electrodes will work as a cathode and why?

Ans: Since the electrode potential of Al is lower than that of Cu, therefore, Cu has a higher tendency to get reduced and hence Cu electrode acts as a cathode, as reduction occurs in the cathode.

5. The ${{E}_{0}}$ of $C{{u}^{2+}}/Cu$ is + 0.34V. What does it signify?

Ans: It signifies that $C{{u}^{2+}}$ have more reduction power than that of hydrogen ions.

6. If the reduction potential of an electrode is 1.28V. What will be its oxidation potential?

Ans: The oxidation potential will be ($0-1.28=(-)1.28V$).

7. What is the electrode potential of a standard hydrogen electrode?

Ans: The electrode potential of a standard hydrogen electrode is 0V.

8. Define a redox couple.

Ans: A redox couple is defined as having together reduced and oxidised forms of a substance that takes part in oxidation and reduction half-reaction.

9. Explain why $3F{{e}_{3}}{{O}_{4}}(s)+8Al(s)\to 9Fe(s)+4A{{l}_{2}}{{O}_{3}}(g)$ is an oxidation reaction?

Ans: It is an oxidation reaction because aluminium is getting oxidised, it forms $A{{l}_{2}}{{O}_{3}}$ in the product, indicating that the addition of oxygen has taken place.

5 Marks Questions

1. Balance the following equations by oxidation number method: 

(i) $CuO+N{{H}_{3}}\to Cu+{{N}_{2}}+{{H}_{2}}O$

Ans: Let us observe the chemical equation:

\[\overset{+2}{\mathop CuO}\,+\overset{-3}{\mathop N{{H}_{3}}}\,\to \overset{0}{\mathop Cu}\,+\overset{0}{\mathop {{N}_{2}}}\,+\overset{-2}{\mathop {{H}_{2}}O}\,\]

Oxidation number of copper decreases from +2 to O and that of nitrogen increases from – 3 to 0.

To balance, there should be three atoms of copper and two atoms of nitrogen.

\[3CuO+2N{{H}_{3}}\to 3Cu+{{N}_{2}}+{{H}_{2}}O\]

(ii) ${{K}_{2}}Mn{{O}_{4}}+{{H}_{2}}O\to Mn{{O}_{2}}+KMn{{O}_{4}}+KOH$

Ans: Let us observe the chemical equation:

\[\overset{+5}{\mathop 2{{K}_{2}}Mn{{O}_{4}}}\,+{{H}_{2}}O\to \overset{+4}{\mathop Mn{{O}_{2}}}\,+\overset{+7}{\mathop KMn{{O}_{4}}}\,+KOH\]

The oxidation number of manganese changes from +6 to +4, in one mole, and in the other mole, the oxidation number changes from +6 to +7. 1 mol acquires two electrons while the other loses 1 electron. To balance the oxidation number of manganese, it is multiplied by 2:

\[{{K}_{2}}Mn{{O}_{4}}+2{{K}_{2}}Mn{{O}_{4}}+{{H}_{2}}O\to Mn{{O}_{2}}+2KMn{{O}_{4}}+KOH\]

Further balancing the equation we have:

\[3{{K}_{2}}Mn{{O}_{4}}+2{{H}_{2}}O\to Mn{{O}_{2}}+2KMn{{O}_{4}}+4KOH\]

Important Questions from Redox Reactions (Short, Long & Practice)

Short Answer Type Questions

1. Define Oxidation number and Electrode potential.

2. Define Oxidation and Reduction in terms of electrons.

3. Why are articles made of Iron coated with Zinc to check their rusting?

4. What is the Oxidation number of alkali metals in its compounds?

5. Which reaction occurs at the cathode in a Galvanic cell?

6. What are the maximum and minimum Oxidation numbers of N?

Long Answer Type Questions

1. Starting with the correctly balanced half-reaction, write the overall net ionic equation for the following change:

Chloride ion is oxidised to Cl2 by MnO4- (in acid solution)

2. Give the rules on the basis of which oxidation numbers are assigned to various elements.

3. Write the method used for balancing redox reaction by oxidation number method.

Practice Questions

1. The half cell reactions with their oxidation potentials are:

Pb(s) → Pb2+ + 2e-, E°oxi = + 0.13 v.

Ag(s) → Ag+(aq) + e-, E°oxi = – 0.80 v.

Write the cell reaction and calculate its EMF.

2. How many millimoles of potassium dichromate is required to oxidise 24 cm3 of 0.5 M mohr’s salt solution in an acidic medium.

3. Calculate the cone of hypo (Na2S2O3 5H2O) solution in g dm-3 if 10.0 of this solution decolorized 15 ml of M/40 iodine solution.

4. Determine the volume M/8 KMnO4 solution required to react completely with 25.0 cm3 of M/4 FeSO4 solution in an acidic medium.

5. 16.6 gm of pure KI was dissolved in water and the solution was made up to one litre V cm3 of this solution was acidified with 20 cm3 of 2 MHCl the resulting solution required 10 cm3 of decinormal KIO3 for complete oxidation of I- ions to ICl. Find out the value of V.

Key Features of Important Questions for Class 11 Chemistry Chapter 7 - Redox Reactions

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• Reactions are explained in detail to make students understand concepts in a better way.

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Conclusion

Crucial Chemistry Class 11 Questions Redox Reactions are a valuable resource for students studying for the Class XI Test. We have provided questions and answers for every essential topic covered in NCERT Class 11 Redox Reactions. Students will also get an understanding of the kind of questions that will be asked in the final test and how to answer them.

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