Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 8.10:

The compound AgF₂ is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?

Answer:

The oxidation state of Ag in AgF₂ is +2. But, +2 is an unstable oxidation state of Ag. Therefore, whenever AgF₂ is formed, silver readily accepts an electron to form Ag⁺. This helps to bring the oxidation state of Ag down from +2 to a more stable state of +1. As a result, AgF₂ acts as a very strong oxidizing agent.

Question 8.11:

Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Answer:

Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. This can be illustrated as follows:

(i)P₄ and F₂ are reducing and oxidising agents respectively.

If an excess of P₄ is treated with F₂, then PF₃ will be produced, wherein the oxidation number (O.N.) of P is +3.

However, if P₄ is treated with an excess of F₂, then PF₅ will be produced, wherein the O.N. of P is +5.

(ii)K acts as a reducing agent, whereas O₂ is an oxidising agent.

If an excess of K reacts with O₂, then K₂O will be formed, wherein the O.N. of O is –2.

However, if K reacts with an excess of O₂, then K₂O₂ will be formed, wherein the O.N. of O is –1.

(iii)C is a reducing agent, while O₂ acts as an oxidising agent.

If an excess of C is burnt in the presence of insufficient amount of O_2, then CO will be produced, wherein the O.N. of C is +2.

On the other hand, if C is burnt in an excess of O₂, then CO₂ will be produced, wherein the O.N. of C is +4.

Question 8.12:

How do you count for the following observations?

(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?

Answer:

(a) In the manufacture of benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant because of the following reasons.

(i) In a neutral medium, OH^– ions are produced in the reaction itself. As a result, the cost of adding an acid or a base can be reduced.

(ii) KMnO₄ and alcohol are homogeneous to each other since both are polar. Toluene and alcohol are also homogeneous to each other because both are organic compounds. Reactions can proceed at a faster rate in a homogeneous medium than in a heterogeneous medium. Hence, in alcohol, KMnO₄ and toluene can react at a faster rate.

The balanced redox equation for the reaction in a neutral medium is give as below:

(b) When conc. H₂SO₄ is added to an inorganic mixture containing bromide, initially HBr is produced. HBr, being a strong reducing agent reduces H₂SO₄ to SO₂ with the evolution of red vapour of bromine.

But, when conc. H₂SO₄ is added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, being a weak reducing agent, cannot reduce H₂SO₄ to SO_2.

Question 8.13:

Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:

(a) 2AgBr (s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr (aq) + C₆H₄O₂(aq)

(b) HCHO(l) + 2[Ag (NH₃)₂]⁺(aq) + 3OH^–(aq)

→ 2Ag(s) + HCOO^–(aq) + 4NH₃(aq) + 2H₂O(l)

(c) HCHO (l) + 2Cu²⁺⁽aq) + 5 OH^–(aq) → Cu₂O(s) + HCOO^–(aq) + 3H₂O(l)

(d) N₂H₄(l) + 2H₂O₂(l) → N₂(g) + 4H₂O(l)

(e) Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

Answer:

(a) Oxidised substance → C₆H₆O₂

Reduced substance → AgBr

Oxidising agent → AgBr

Reducing agent → C₆H₆O₂

(b)Oxidised substance → HCHO

Reduced substance →

Oxidising agent →

Reducing agent → HCHO

(c) Oxidised substance → HCHO

Reduced substance → Cu²⁺

Oxidising agent → Cu²⁺

Reducing agent → HCHO

(d) Oxidised substance → N₂H₄

Reduced substance → H₂O₂

Oxidising agent → H₂O₂

Reducing agent → N₂H₄

(e) Oxidised substance → Pb

Reduced substance → PbO₂

Oxidising agent → PbO₂

Reducing agent → Pb

Question 8.14:

Consider the reactions:

(aq) + I₂(s) →  (aq) + 2I^–(aq)

aq) + 2Br₂(l) + 5 H₂O(l) →  (aq) + 4Br^–(aq) + 10H⁺(aq)

Why does the same reductant, thiosulphate react differently with iodine and bromine?

Answer:

The average oxidation number (O.N.) of S in  is +2. Being a stronger oxidising agent than I₂, Br₂ oxidises   to  , in which the O.N. of S is +6. However, I₂ is a weak oxidising agent. Therefore, it oxidises  to

, in which the average O.N. of S is only +2.5. As a result,  reacts differently with iodine and bromine.

Question 8.15:

Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Answer:

F₂ can oxidize Cl^– to Cl₂, Br^– to Br_2, and I^– to I₂ as:

On the other hand, Cl₂, Br₂, and I₂ cannot oxidize F^– to F₂. The oxidizing power of halogens increases in the order of I₂ < Br₂ < Cl₂ < F₂. Hence, fluorine is the best oxidant among halogens.

HI and HBr can reduce H₂SO₄ to SO₂, but HCl and HF cannot. Therefore, HI and HBr are stronger reductants than HCl and HF.

Again, I^– can reduce Cu²⁺ to Cu⁺, but Br^– cannot.

Hence, hydroiodic acid is the best reductant among hydrohalic compounds.

Thus, the reducing power of hydrohalic acids increases in the order of HF < HCl < HBr < HI.

Question 8.16:

Why does the following reaction occur?

(aq) + 2F^– (aq) + 6H⁺(aq) → XeO₃(g) + F₂(g) + 3H₂O(l)

What conclusion about the compound Na₄XeO₆ (of which   is a part) can be drawn from the reaction.

Answer:

The given reaction occurs because  oxidises

 and  reduces .

In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in

to +6 in XeO₃ and the O.N. of F increases from –1 in F^– to O in F_2.

Hence, we can conclude that  is a stronger oxidising agent than F^–_.

Question 8.17:

Consider the reactions:

(a) H₃PO₂(aq) + 4 AgNO₃(aq) + 2 H₂O(l) → H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)

(b) H₃PO₂(aq) + 2CuSO₄(aq) + 2 H₂O(l) → H₃PO₄(aq) + 2Cu(s) + H₂SO₄(aq)

(c) C₆H₅CHO(l) + 2[Ag (NH₃)₂]⁺(aq) + 3OH^–(aq) → C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃ (aq) + 2 H₂O(l)

(d) C₆H₅CHO(l) + 2Cu²⁺⁽aq) + 5OH^–(aq) → No change observed.

What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions?

Answer:

Ag⁺ and Cu²⁺ act as oxidising agents in reactions (a) and (b) respectively.

In reaction (c), Ag⁺ oxidises C₆H₅CHO to C₆H₅COO^–, but in reaction (d), Cu²⁺ cannot oxidise C₆H₅CHO.

Hence, we can say that Ag⁺ is a stronger oxidising agent than Cu²⁺_.