Important Questions for Class 11 Chemistry Chapter 8 - Redox Reaction.

Class 11 chemistry important questions with answers are provided here for Chapter 8 Redox Reactions. These important questions are based on the CBSE board curriculum and correspond to the most recent Class 11 chemistry syllabus. By practising these Class 11 important questions, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 11 Annual examinations as well as other entrance exams such as NEET and JEE.

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Class 11 Redox Reactions Important Questions with Answers

Short Answer Type Questions

Q1. 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.

Answer:

Here, Cl₂ is both oxidised and reduced in ClO^– and Cl^–, respectively. Since Cl^– cannot act as an oxidising agent (O.A.). Therefore, Cl₂ bleaches substances due to oxidising action of hypochlorite ClO^– ion.

Q2. MnO₄^2- undergoes a disproportionation reaction in an acidic medium but MnO₄^– does not. Give a reason.

Answer:

Disproportionation is a redox reaction in which one intermediate oxidation state component transforms into two higher and lower oxidation state compounds.

Oxidation states of manganese ranges from +2 to +7 in its various compounds. MnO₄^– has the maximum oxidation state of +7 hence disproportionation is impossible, but MnO₄^2- has a +6 oxidation state, which can be oxidised as well as reduced.

Q3. 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:

None of the atoms’ O.N. changes in the first reaction. As a result, it cannot be classified as a redox reaction. Because PbO is a basic oxide that combines with HCl acid, it is an acid-base reaction.

PbO₂ is reduced and functions as an oxidising agent in the second process, a redox reaction.

Q4. Nitric acid is an oxidising agent and reacts with PbO, but it does not react with PbO₂. Explain why?

Answer:

PbO is a basic oxide, and it undergoes a straightforward acid-base reaction with HNO₃. In PbO₂, on the other hand, lead is in the +4 oxidation state and cannot be further oxidised. As a result, no reaction occurs. PbO₂ is thus inactive; only PbO interacts with HNO₃.

Q5. Write a balanced chemical equation for the following reactions:

(i) Permanganate ion (MnO₄^–) reacts with sulphur dioxide gas in an acidic medium to produce Mn²⁺ and hydrogensulphate ion.

(Balance by ion electron method)

(ii) Reaction of liquid hydrazine (N₂H₄) with chlorate ion (CIO₃^–) in basic medium produces nitric oxide gas and chloride ion in a gaseous state.

(Balance by oxidation number method)

(iii) Dichlorine heptaoxide (Cl₂O₇) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (ClO₂^–) and oxygen gas.

(Balance by ion electron method)

Answer:

(i) MnO₄^– +SO₂ → Mn²⁺ + HSO₄^– (acidic medium)

Balancing by ion-electron method we get:

2 × {MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O

5 × {SO₂ + 2H₂O → HSO₄^– + 3H⁺ + 2e^–

(ii) N₂H⁴ + ClO₃^– → NO + Cl^– (basic medium)

To make the gain and loss of electrons equal, balancing by oxidation number method we get:

3N₂H₄ + 4ClO₃^– → 6NO + 4Cl^– + 6H₂O

(iii) Cl₂O₇(g) + H₂O₂(aq) → 2ClO₂^– + 3H₂O (acidic medium)

2 × {Cl₂O₇ + 6H⁺ + 8e^– → 2ClO₂^– + 3H₂O}

8× {H₂O₂ → O₂ + 2H⁺ + 2e^–}

Q6. Calculate the oxidation number of phosphorus in the following species.

(a) HPO₃^2- and

(b) PO₄^3-

Answer:

1. Let the oxidation number of phosphorus in HPO₃^2- be x.

H + P + 3O^2-

⇒ +1 + x + (-2)×3 = -2

⇒ +1 + x – 6 = -2

⇒ x – 5 = -2

⇒ x = – 2 + 5

⇒ x = +3

Thus, the oxidation number of phosphorus is +3.

1. Let the oxidation number of phosphorus in PO₄^3- be x.

PO₄^3-

⇒ x + 4 ×(-2) = -3

⇒ x = -3 + 8 = +5

⇒ x = +5

Thus, the oxidation number of phosphorus is +5.

Q7. Calculate the oxidation number of each sulphur atom in the following compounds:

(a) Na₂S₂O₃

(b) Na₂S₄O₆

(c) Na₂SO₃

(d) Na₂SO₄

Answer:

1. The inorganic chemical sodium thiosulphate (also known as sodium thiosulphate) has the formula Na₂S₂O₃ .xH₂O.

We can’t find the oxidation number of each sulphur atom using traditional methods.

Structure will provide a clear explanation. One sulphur molecule possesses +6 O.S., whereas the other has -2.

1. Let the oxidation number of sulphur in Na₂S₄O₆ be x.

⇒ 2 + 4x – 12 = 0

⇒ 4x = +10

⇒ x = 10/4

⇒ x = +2.5

Thus, the oxidation number of sulphur in Na₂S₄O₆ is +2.5.

(c) Let the oxidation number in Na₂SO₃ be x.

Na₂SO₃

⇒ 2×(+1)+x+(−2)×3 = 0

⇒ 2+x−6 = 0

⇒ x = +4

Thus, the oxidation number of sulphur in Na₂SO₃ is +4.

(d) Let the oxidation number of sulphur in Na₂SO₄ be x.

⇒ 2×(+1)+x+(−2)×4 = 0

⇒ 2+x−8 = 0

⇒ x = +6

Thus, the oxidation number of sulphur in Na₂SO₄ is +6.

Q8. Balance the following equations by the oxidation number method.

Answer:

(i) Fe²⁺ + Cr₂O₇^2- + H⁺ → Fe³⁺ + Cr³⁺ + H₂O

The chromium oxidation number drops from +6 in Cr₂O₇² to +3 in Cr³⁺. In Cr₂O₇^2- to Cr³⁺, the total drop for two chromium atoms is 6. Iron’s oxidation number, on the other hand, rises from +2 (in Fe²⁺) to +3 (in Fe³⁺).

Multiply Fe2+ by 6 and Cr₂O₇^2- by 1 to balance the increase and reduction in oxidation numbers. Then there’s,

6Fe²⁺ + Cr₂O₇^2- + H⁺ → Fe³⁺ + Cr³⁺ + H₂O

On both sides of the reaction, Fe and Cr atoms must be balanced.

6Fe²⁺ + Cr₂O₇^2- + H⁺ → 6Fe³⁺ + 2Cr³⁺ + H₂O

Multiply H₂O by 7 to balance O atoms.

6Fe²⁺ + Cr₂O₇^2- + H⁺ → 6Fe³+ + 2Cr³⁺ + 7H₂O

Multiply H+ by 14 to balance H-atoms.

6Fe²⁺ + Cr₂O₇^2- + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

(ii) The given skeleton equation is: I₂ + NO₃^– → NO₂ + IO₃^–

Let us find the oxidation number of both I and N.

LHS,

I₂

⇒ 2x = 0

⇒ x = 0

IO₃⁻

⇒ x + 3 (−2) = −1

⇒ x = +5

For one atom of I, the oxidation number rises by 5. It’s 10 for two atoms.

NO₃⁻

⇒ x + 3 (−2) = −1

⇒ x = +5

NO₂

⇒ x + 2 (−2) = 0

⇒ x = +4

As a result, the oxidation number drops by 1.

Cross multiplying the oxidation number and writing the equation,

I₂ + 10NO₃⁻ →NO₂ + IO³⁻

Balancing the RHS side I and N,

I₂ + 10NO₃⁻ →10NO₂ + 2 IO₃⁻

Now, balancing the oxygen on both the sides by adding H₂O,

I₂ + 10NO₃⁻ →10 NO₂ + 2IO₃⁻ + 4H₂O

Balance the hydrogen by adding H⁺,

I₂ + 10NO₃⁻ + 8H⁺ →10 NO₂ + 2IO₃⁻ + 4H₂O

Hence, the balanced equation is,

I₂ + 10NO₃⁻ + 8H⁺ →10NO₂ + 2IO₃⁻ + 4H₂O

(iii) The unbalanced redox reaction is I₂ + S₂O₃^2- → I^– + S₄O₆^2-

Balancing all atoms other than H and O, we get

I₂ + 2S₂O₃^2- → 2I^– + S₄O₆^2-

I changes its oxidation number from 0 to -1. The oxidation number of one I atom changes by one. For 2 I atoms, the total change in oxidation number is 2.

The oxidation number of S increases from 2 to 2.5, representing a 0.5 increase in the oxidation number of one S atom.

For four S atoms, the total change in oxidation number is 2.

Increases in oxidation number are counterbalanced by decreases in oxidation number.

The atoms of O are in a state of equilibrium. This is the chemical equation that is balanced.

(iv) MnO₂ + C₂O₄^2- → Mn²⁺ + CO₂

Balancing all atoms, we get:

The following is an example of an oxidation-reduction (redox) reaction:

Mn^IV + 2e^– → Mn^II (reduction)

2C^III – 2e^– → 2C^IV (oxidation)

C₂O₄^2- is a reducing agent, while MnO₂ is an oxidising agent.

Q9. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.

Answer:

(i) Writing the O.N. of on each atom, we get:

3HCl + HNO₃ → Cl₂ + NOCl + 2H₂O

Cl’s O.N. goes from -1 (in HCl) to 0 in this case (in Cl₂). As a result of the oxidation of Cl^–, HCl is used as a reducing agent.

HNO₃ functions as an oxidising agent because the O.N. of N reduces from +5 (in HNO₃) to +3 (in NOCl).

Thus, reaction (i) is, therefore, a redox reaction.

(ii) HgCl₂ + 2Kl → Hgl + 2KCl

There is no change in O.N. of any of the atoms, hence this is not a redox reaction.

(iii) Fe₂O₃ + 3CO → 2Fe +3CO₂

Fe₂O₃ works as an oxidising agent because the O.N. of Fe falls from +3 (in Fe₂O₃) to 0 (in Fe).

CO works as a reducing agent because the O.N. of C increases from +2 (in CO) to +4 (in CO₂). As a result, we have a redox reaction.

(iv) PCl₃ + 3H₂O → 3HCl + H₂PO₃

There is no change in O.N. of any of the atoms, hence it is not a redox reaction.

(v) 4NH₃ + 3O₂ → 2N₂ + 6H₂O

Because the O.N. of N in N₂ grows from -3 to 0, NH₃ functions as a reducing agent.

Furthermore, because the O.N. of O drops from 0 to O₂ to -2 in H₂O, O₂ functions as an oxidising agent. As a result, this is a redox reaction.

Q10. Balance the following ionic equations

Answer:

Balanced equations of given unbalanced equations:

(i)

(ii)

(iii)

(iv)

Long Answer Type Questions

Q1. Explain redox reactions on the basis of electron transfer. Give suitable examples.

Answer:

An oxidation-reduction (redox) reaction is a chemical reaction in which electrons are transferred between two substances. Any chemical reaction in which a molecule, atom, or ion’s oxidation number changes by acquiring or losing an electron is known as an oxidation-reduction reaction.

As we know, the reactions,

2Na(s) + Cl₂(g) → 2NaCl(s)

4Na(s) + O₂(g) → 2Na₂(s)

because each of these reactions involves the addition of oxygen or a more electronegative element to sodium, they are redox reactions.

At the same time, chlorine and oxygen are being depleted, while sodium, an electropositive element, has been added. We also know that sodium chloride and sodium oxide are ionic compounds and are perhaps better expressed as Na⁺Cl^–(s) and Na₂⁺O_2, respectively, based on our understanding of chemical bonding(s). The development of charges on the species created leads us to recast the preceding reaction:

Each of the above processes can be thought of as two different phases, one involving electron loss and the other involving electron gain. We can expound on one of these, say sodium chloride production, as an example.

2Na(s) → 2Na⁺(g) + 2e⁻

Cl₂ + 2e⁻ → 2Cl⁻(g)

Each of the preceding steps is referred to as a half-reaction since it clearly demonstrates the involvement of electrons. The total reaction is equal to the sum of the half-reactions:

2Na(s) + Cl²(g) → 2Na + Cl⁻(s) or 2NaCl(s)

The above reactions indicate that 50% of the reactions involving electron loss are oxidation reactions. Similarly, half-reactions involving electron gain are referred to as reduction reactions.

The basic operations of life, such as photosynthesis, respiration, combustion, and corrosion or rusting, are redox reactions.

Q2. On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for E^o value).

Answer:

The cell’s net cell EMF determines whether or not a reaction will occur. The formula is as follows:

Option (ii) clearly shows that the reaction can occur because Mg has a lower Ecirc cell value. As a result, Mg is oxidised by losing an electron, whereas iron is reduced by gaining one.

We can state that Fe goes through reduction and Mg goes through oxidation.

Taking the E^o values,

E^o_cathode = -0.44V

E^o_anode = -2.36V

E^o_cell = -0.44 – (-2.36)V

E^o_cell = +1.92V

Q3. Why does fluorine not show a disproportionation reaction?

Answer:

Disproportionation is a redox reaction in which one intermediate oxidation state component transforms into two higher and lower oxidation state compounds.

The element must be in at least three oxidation states for such a redox reaction to occur. As a result, that element is in the intermediate state during the disproportionation reaction and can transition to both higher and lower oxidation states.

Fluorine is the most electronegative and oxidising element of all the halogens, and it is also the smallest.

It doesn’t have a positive oxidation state (only one) and doesn’t go through the disproportionation reaction.

Q4. Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.

NaClO₄, NaClO₃, NaCIO, KCIO₂, Cl₂O₇, ClO₃, Cl₂O, NaCl, Cl₂, CIO₂.

Which oxidation state is not present in any of the above compounds?

Answer:

Considering oxidation number of chlorine is ‘x’, then 1 + x + 4 × (-2) = 0

∴ x – 7 = 0

∴ x = +7

Na⁽⁺¹⁾Cl⁽⁺⁷⁾O₄^(-2)

Here, oxidation number of chlorine is +7.

Following the above method,

Na⁽⁺¹⁾Cl⁽⁺¹⁾O₃^(-2) Oxidation number of chlorine = +5

Na⁽⁺¹⁾Cl⁽⁺¹⁾O^(-2) Oxidation number of chlorine = +1

K⁽⁺¹⁾Cl⁽⁺³⁾O₂^(-2) Oxidation number of chlorine = +3

Cl₂⁽⁺⁷⁾O₇^(-2) Oxidation number of chlorine = +7

Cl⁽⁺⁶⁾O₃^(-2) Oxidation number of chlorine = +6

Cl₂⁽⁺¹⁾O^(-2) Oxidation number of chlorine = +1

Na⁽⁺¹⁾Cl^(-1) Oxidation number of chlorine = -1

Cl₂ Oxidation number of chlorine = 0

Cl⁽⁺⁴⁾O₂^(-2) Oxidation number of chlorine = +4

The following compounds are listed in increasing order of chlorine oxidation number:

NaCl, Cl₂, Cl₂O, NaCIO, KCIO₂, CIO₂, NaClO₃, ClO₃, Cl₂O₇, NaClO₄.

Q5. Which method can be used to find out the strength of the reductant/oxidant in a solution? Explain with an example.

Answer:

When a reductant (reducing agent) or oxidant (oxidising agent) is linked in a solution using a cell, the relative electrode potential can be measured. The element under discussion can be used as an electrode in a conventional cell with a known electrode potential. The electrode of the given species works as a reductant if it is positive and an oxidant if it is negative.

For example, we want to test Fe³⁺/Fe with a Standard Hydrogen electrode (SHE). For Fe and H, the half-cell reaction is as follows:

H⁺ + e^– → H₂E^o = 0.0V

Fe³⁺ + e^– → Fe²⁺E^o = 0.77

With SHE, any element that needs to be evaluated is used as an electrode. The element’s potential is defined as the quantity of emf it generates in the cell.

E^o_cell = E^o_cathode – E^o_anode

E^o_cell= 0 – E^o_anode

E^o_cell = 0 – 0.77

E^o_cell= -0.77

In comparison to hydrogen, the Fe³⁺ ion has a higher tendency to undergo reduction. As a result, the previously assumed Fe anode configuration can be reversed, and the strength of Fe as a reductant can be established. As a result, the oxidant’s strength can be determined.

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