PSEB 11th Class Chemistry Solutions Chapter 8 Redox Reactions

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 8 Redox Reactions Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

PSEB 11th Class Chemistry Guide Redox Reactions InText Questions and Answers

Question 1.
Assign oxidation numbers to the underlined elements in each of the following species:

Answer:
(a) Let the oxidation number of P be x.
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = – 2
In neutral compounds, the sum of the oxidation numbers of all the atoms is zero.
1 (+1) + 2 (+1) +1 00 + 4 (-2) = 0
1 + 2 + x – 8 = 0
3 + x + (-8) – 0
x = 8 – 3
⇒ x = + 5
Hence, the oxidation number of P is +5.

(b) Let the oxidation number of S be x.

1 (+1) +1 (+1) +1 (x) + 4 (-2) = 0
⇒ 1 + 1 + x – 8 = 0
⇒ x = + 6
Hence, the oxidation number of S is +6.

(c) Let the oxidation number of P be x.

4 (+1) + 2 (x) + 7 (-2) = 0
⇒ 4 + 2x – 14 = 0
⇒ 2x = +10
⇒ x = + 5
Hence, the oxidation number of P is + 5.

(d) Let the oxidation number of Mn is x.

2 (+1) + x + 4 (-2) = 0
⇒ 2 + x — 8 = 0
⇒ x = + 6
Hence, the oxidation number of Mn is + 6.

(e) Let the oxidation number of O be x.

1 (+ 2) + 2 (x) = 0
⇒ 2 + 2x = 0
⇒ x = – 1
Hence, the oxidation number of O is -1.

(f) Let the oxidation number of B be x.

1 (+1)+1 (x) + 4 (-1) = 0
⇒ 1 + x – 4 = 0
⇒ x = + 3
Hence, the oxidation number of B is + 3.

(g) Let the oxidation number of S is x.

2 (+1) + 2 (x) + 7 (-2) = 0
⇒ 2 + 2x -14 = 0
⇒ 2x = +12
x = +6
Hence, the oxidation number of S is + 6.

(h) Let the oxidation number of S be x.

1 (+1) +1 (+ 3) + 2 (x) + 8 (-2) +12 (2 x 1 + (-2)) = 0
⇒ 1 + 3 + 2x -16 + 24 – 24 = 0
⇒ 2x = 12
x = + 6
Hence, the oxidation number of S is + 6.

Question 2.
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?


Answer:
(a) In

the oxidation number (O. N.) of K is +1. Hence, the average oxidation number of I is \(\frac{-1}{3}\). However, O.N. cannot be fractional.
Therefore, we will have to consider the structure of KI₃ to find the oxidation states.
In a KI₃ molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a KI₃ molecule, the O.N. of the two I atoms forming the I₂ molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is -1.

(b) Let the oxidation number of S be x.

2 (+1) + 4 (x) + 6 (-2) = 0
=» 2 + 4x – 12 = 0
⇒ 4x = +10
⇒ x = + 2 \(\frac{1}{2}\)
However, O.N. cannot be fractional. Hence S must be present in different oxidation states in the molecule.

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

(c) Let the oxidation number of Fe be x.

3(x) + 4(-2) = 0
3x – 8 = 0
x = \(+\frac{8}{3}\)
However O.N. cannot’be fractional. Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of+3

(d) Let oxidation number of C be x.

2 (x) + 6 (+1) + 1 (-2) = 0
2x + 4 = 0
x = -2
Hence, the O.N. of C is – 2

(e) Let the oxidation number of C be x.

2 (x) + 4 (+1) + 2 (-2) = 0
2x = 0
x = 0
Therefore, the average oxidation number of C is zero.
Let us consider the structure of CH₃COOH

Oxidation number of atom = 1(+1) + x+1(-2) +1(-1) = 0
x = +2
Similarly, oxidation number of C₂ atom
3(+1) + x+ 1(-1) = 0
x = -2.

Question 3.
Justify that the following reactions are redox reactions:
(a) CuO (s) + H₂(g) → Cu(s) + H₂O(g)
(b) Fe₂O₃ (s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
(c) 4BCl₃ (g) + 3LiAlH₄ (s) → 2B₂H₆(g) + 3LiCl (s) + 3 AlCl₃(S)
(d) 2K (s) + F₂ (g) → 2K⁺F^–(s)
(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O (g)
Answer:
(a) CuO (s) + H₂(g) → Cu(s) + H₂O(g)
Let us write the oxidation number of each element involved in the given reaction as :

Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in H₂ to +1 in H₂O i.e., H₂ is oxidized to H₂O. Hence, this reaction is a redox reaction.

(b) Fe₂O₃ (s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
Let us write the oxidation number of each element in the given reaction as:

Here, the oxidation number of Fe decreases from +3 in Fe₂O₃ to 0 in Fe i.e., Fe₂O₃ is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO₂ i.e., CO is oxidized to CO₂.
Hence, the given reaction is a redox reaction.

(c) 4BCl₃ (g) + 3LiAlH₄ (s) → 2B₂H₆(g) + 3LiCl (s) + 3 AlCl₃(S)
The oxidation number of each elements in the given reaction can be represented as:

In this reaction, the oxidation number of B decreases from +3 in BCl₃ to -3 in B₂H_6- i.e., BCl₃ is reduced to B₂H₆. Also, the oxidation number of H increases from -1 in LiAlH₄ to +1 in B₂H₆ i.e., LiAlH₄ is oxidized to B₂H₆. Hence, the given reaction is a redox reaction.

(d) 2K (s) + F₂ (g) → 2K⁺F^–(s)
The oxidation number of each element in the given reaction can be represented as:

In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F₂ to -1 in KF i.e., F₂ is reduced to KF.
Hence, the above reaction is a redox reaction.

(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O (g)
The oxidation number of each elements in the given reaction can be represented as:

Here, the oxidation number of N increases from -3 in NH₃ to +2 in NO. On the other hand, the oxidation number of O₂ decreases from 0 in O₂ to -2 in NO and H₂O i.e.,O₂ is reduced. Hence, the given reaction is a redox reaction.

Question 4.
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.
Ans. Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

Here, we have observed that the oxidation number of F increases from 0 in F₂ to +1 in HOF. Also, the oxidation number decreases from 0 in F₂ to -1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.

Question 5.
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) and \(\mathrm{NO}_{3}^{-}\). Suggest structure of these compounds. Count for the fallacy.
Answer:

2 (+1) +1 (x) + 5 (-2) = 0
⇒ 2 + x -10 = 0
⇒ x = + 8
However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6. The structure of H2S05 is shown as follows :

Now, 2 (+1) +1 (x) + 3 (-2) + 2 (-1) – 0
⇒ 2 + x – 6 – 2 = 0
⇒ x = + 6
Therefore, the O.N. of S is +6.

2 (x) + 7 (-2) = – 2 ⇒ 2x -14 = – 2
⇒ x = + 6
The structure of Cr2Oy is shown as follows:

Let the oxidation number of each Cr atom be
4(-2) + (-2) +1(-2) + 2x = 0
– 8 – 2 – 2 + 2x = 0
2x = +12
x = +6

Oxidation number of Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is same.
Hence, there is no fallacy.

1 (x) + 3(-2) = -1
= x – 6 = -1
x = +5
The structure of \(\mathrm{NO}_{3}^{-}\) is shown as follows:

Let the oxidation number of N be x.
1(-1) + x + 1(-2) + 1(-2) = 0
x = + 5
Oxidation number of N in \(\mathrm{NO}_{3}^{-}\) ion is same.
Hence, there is no fallacy.

Question 6.
Write formulas for the following compounds:
(a) Mercury (II) chloride
(b) Nickel (II) sulphate
(c) Tin (IV) oxide
(d) Thallium (I) sulphate
(e) Iron (III) sulphate
(f) Chromium (III) oxide
Answer:
(a) Mercury (II) chloride: Hg (II) Cl₂
(b) Nickel (II) sulphate: Ni (II) SO₄
(c) Tin (IV) oxide : Sn (IV) O₂
(d) Thallium (I) sulphate: Tl₂ (I) SO₄
(e) Iron (III) sulphate: Fe₂(III) (SO₄)₃
(f) Chromium (III) oxide: Cr₂(III)O₃

Question 7.
Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.
Answer:
The substances where carbon can exhibit oxidation states from -4 to +4 are listed in the following table :

Substance	Formula	Oxidation State of Carbon
Methane	CH4	-4
Ethane	C2H6	-3
Ethene	C2H4	-2
Ethyne	C2H2	-1
Dichloromethane	CH2Cl2	0
Hexachlorobenzone	C6Cl6	+1
Carbon monoxide	CO	+2
Oxalic acid	(COOH)2	+3
Carbon dioxide	CO2	+4

The substances where nitrogen can exhibit oxidation sates from -3 to +5 are listed in in the following table.

Substance	Formula	Oxidation State of Nitrogen
Ammonia	NH3	-3
Hydrazine	N2H4	-2
Hydride	N2H2	-1
Dinitrogen gas	N2	0
Nitrous oxide	N2O	+1
Nitric oxide	NO	+2
Dinitrogen trioxide	N2O3	+3
Nitrogen dioxide	NO2	+4
Nitrogen pentoxide	N2O5	+5

Question 8.
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
Answer:
(i) In sulphur dioxide (SO₂), the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to -2.
Therefore, SO₂ can act as an oxidising as well as reducing agent.
(ii) In hydrogen peroxide (H₂O₂), the O.N. of O is -1 and the range of the O.N. that O can have is from 0 to -2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, H₂O₂ can act as an oxidising as well as reducing agent.
(iii) In ozone (O₃) the O.N. of O is zero and the range of the O.N. that O can have is from 0 to -2. Therefore, the O.N. of O can only decrease in this case. Hence,O₃ acts only as an oxidant.
(iv) In nitric acid (HNO₃) the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to -3. Therefore, the O.N. of N can only decrease in this case. Hence, HNO₃ acts only as an oxidizing agent.

Question 9.
Consider the reactions:
(a) 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆ (aq) + 6O₂(g)
(b) O₃(g) + H₂O₂(Z) → H₂O(1) + 2O₂(g)
Why it is more appropriate to write these reactions as:
(a) 6CO₂ (g) + 12H₂O(1) → C₆H₁₂O₆ (aq) + 6H₂O(l) + 6O₂ (g)
(b) O₃(g) + H₂O₂(Z) → + H₂O(Z) + O₂(g) + O₂(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Answer:
(a) The process of photosynthesis involves two steps :
Step 1: H₂O decomposes to give H₂ and O₂.
2H₂O(l) → 2H₂(g) + O₂(g)
Step 2: The H₂ produced in step 1 reduces CO₂ thereby producing glucose (C₆H₁₂O₆) and H₂O.
6CO₂(g) + 12H₂(g) → C₆H₁₂O₆(S) + 6H₂O(l)
Now, the net reaction of the process is given as:

It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis.
The path of this reaction can be investigated by using radioactive H₂O¹⁸ in place of H₂O

(b) O₂ is produced from each of the two reactants O₃ and H₂O₂. For this reason O₂ is written twice.
The given reaction involves two steps. First O₃ decomposes to form O₂ and O. In the second step H₂O₂ reacts with the O produced in the first step thereby producing H₂O and O₂.

The path of this reaction can be investigating by using \(\mathrm{H}_{2} \mathrm{O}_{2}^{18}\) or \(\mathrm{O}_{3}^{18}\).

Question 10.
The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?
Answer:
AgF₂ → Ag + F₂
The oxidation state of Ag in Ag F₂ is +2. But +2 is an unstable oxidation state of Ag. Therefore, whenever Ag F₂ 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, Ag F₂ acts as a very strong oxidizing agent.

Question 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 two 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, where in 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.

Question 12.
How do you account 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) Oxidation of toluene to benzoic acid in acidic medium.

Oxidation of toluene to benzoic acid in basic and neutral medium.

On industrial scale, alcoholic potassium permanganate is preferred to acidic or alkaline potassium permanganate because in the presence of alcohol, both the reactants KMnO₄ and C₆H₅CH₃ are mixed very well and form homogeneous solution and in homogeneous medium reaction takes place faster than in heterogeneous medium. Further more in neutral medium, OH^– ions are produced in the reaction itself.

(b)

HCl is a weak reducing agent. k cannot reduce H₂SO₄ to SO₂ that’s why pungent smelling gas HCl is obtained.

HBr is a strong reducing agent, it reduces H₂S0₄ to SO₂ and is itselfoxidised to Br₂. That’s why we get red vapours of bromine when conc.
H₂SO₄ reacts with inorganic mixture containing bromide salt.

Question 13.
Identify the substance oxidised, reduced, oxidising agent andreducing 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^– → Cu₂O (s) +HCOO^–(aq) + 3H₂O(l)
(d) N₂H₄ (Z) + 2H₂O₂ (1) → N₂ (g) + 4H₂O (l)
(e) Pb (s) +PbO₂ (s) + 2H₂SO₄ (ag) → 2PbSO₄(s) + 2H₂O(l)
Answer:
(a) Oxidised substance -C₆H₆O₂
Reduced substance – AgBr
Oxidising agent – AgBr
Reducing agent -C₆H₆O₂

(b) Oxidised substance – HCHO
Reduced substance – [Ag (NH₃)₂]⁺
Oxidising agent – [Ag (NH₃)₂]⁺
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 – Pb O₂
Reducing agent – Pb

Question 14.
Consider the reactions:

Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer:
The average oxidation number (O.N.) of S in \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is +2. Being a stronger oxidising agent than I₂, Br₂ oxidises \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\) in which the O.N. of S is +6. However I₂ is a weak oxidising agent. Therefore, it oxidises \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) in which the average O.N. of S is only +2.5. As a result, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) reacts differently with iodine and bromine. ,

Question 15.
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer:
(i) F₂ can oxidize Cl^– to Cl₂, Br^– to Br₂ and I^– to I₂ as:
F₂(aq) + 2Cl^–(s) → 2F^–(aq) + Cl₂(g)
F₂(aq) + 2Br^–(aq) → 2F^–(aq) + Br₂(7)
F₂(aq) + 2I^–(aq) → 2F^–(aq) + I₂(s)

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.

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

2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O
2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O

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

4I^–(aq) + 2Cu²⁺(aq) → Cu₂I₂(s) + I₂(aq)

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 16.
Why does the following reaction occur?
\(\mathrm{XeO}_{6}^{4-}\)(aq) + 2F^– (aq) + 6H⁺(aq) → XeO₃(g) +F₂(g) + 3H₂O(i)
What conclusion about the compound Na₄XeO₆ (of which \(\mathrm{XeO}_{6}^{4-}\) is a part) can be drawn from the reaction.
Answer:
The given reaction occurs because \(\mathrm{XeO}_{6}^{4-}\) oxidizes being an oxidizing agent and F^– reduces \(\mathrm{XeO}_{6}^{4-}\)

In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in \(\mathrm{XeO}_{6}^{4-}\) to +6 in XeO₃ and the O.N. of F increases from -1 in F^– to 0 in F₂.
Hence, we can conclude that Na₄XeO₆ is a strong oxidizing agent thanF^–.

Question17.
Consider the reactions:
(a) H₃PO₂ (aq) + 4 AgNO₃ (aq) + 2H₂O(l) → H₃PO₄ (aq) + 4Ag(s) + 4HNO₃ (aq)
(b) H₃PO₂ (oqr) + 2CuSO₄ (aq) + 2 H₂O(0 → H₃PO₄ (aq) + 2Cu(s) +H₂SO₄ (aq)
(c) C₆H₅CHO(l) + 2[Ag (NH₃)₂]+ (aq) + 3OH^– (aq) → C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃ (aq) + 2H₂O(l)
(d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH^–(aqf) → No change observed.
What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions?
Answer:
Ag⁺ and Cu²⁺ act as an oxidizing agents in reactions (a) and (b) respectively.
In reaction (c), Ag⁺ oxidizes C₆H₅CHO to C₆H₅COO^–, but in reaction (d) Cu²⁺ cannot oxidize C₆H₅CHO. Hence, we can say that Ag⁺ is a stronger oxidising agent than Cu²⁺.

Question 18.
Balance the following redox reactions by ion-electron method:


Answer:
(a) Step 1 : The two half reactions involved in the given reaction are:
Oxidation half reaction :

Reduction half reaction:

Step 2 : Balancing I in the oxidation half reaction, we have
2I^– (aq) → I₂(s)
Now, to balance the charge, we add 2 e“ to the RHS of the reaction.
2I^–(aq) → I₂(s) + 2e^–

Step 3 : In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction.
\(\mathrm{MnO}_{4}^{-}\) + 3e^– → MnO₂(aq)
Now, to balance the charge, we add 4 OH“ ions to the RHS of the reaction as the reaction is taking place in a basic medium.
\(\mathrm{MnO}_{4}^{-}(a q)^{-}\) + 3e^– → MnO₂(aq) + 4OH^–

Step 4 : In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.
\(\mathrm{MnO}_{4}^{-}\) (aq) + 2H₂O + 3e^– → MnO₂(aq) + 4OH^–

Step 5 : Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:
6I^–(aq) → 3I₂(s) + 6e^–
2 \(\mathrm{MnO}_{4}^{-}\)(aq) + 4H₂O + 6e^– → 2MnO₂(s) + 8OH^–(aq)

Step 6 : Adding the two half reactions, we have the net balanced redox reaction as:
6I^–(aq) + 2\(\mathrm{MnO}_{4}^{-}\)(aq) + 4H₂O(l) → 3I₂(s) + 2MnO₂(s) + 8OH^–(aq)

(b) Following the steps as in part (a) we have the oxidation half reaction as:
SO₂(g) + 2H₂O(Z) → \(\mathrm{HSO}_{4}^{-}\)(aq) + 3H⁺ (aq) + 2e^– (aq)
And the reduction half reaction as:
\(\mathrm{MnO}_{4}^{-}\)(aq) + 8H⁺(aq) + 5e^– → Mn²⁺(aq) + 4H₂O(l)

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as :

2Mno₄^–(aq) + 5SO₂(g) + 2H₂O(Z) + H⁺(aq) → 2Mn(aq) + 5HSO₄^–((aq)

(c) Following the steps as in part (a), we have the oxidation half reaction as:
Fe²⁺(aq) → Fe³⁺(aq) + e^–

And the reduction half reaction as:
H₂O₂(aq) + 2H⁺ (aq) + 2e^– → 2H₂O(l)

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
H₂O₂(aq) + 2Fe²⁺(aq) + 2H⁺(aq) → 2Fe³⁺ (aq) + 2H₂O(Z)

(d) Following the steps as in part (a), we have the oxidation half reaction as:

Question 19.
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P₄(s) + OH^–(aq) → PH₃(g) + H₂PO₂^–(aq)
(b) N₂H₄(l) + CIO₃^–(aqr) → NO(g) + Cl^–(g)
(c) Cl₂O₇(g) + H₂O₂(aqr) → ClO^–₂(aq) + O₂(g) + H⁺
Answer:

O.N. increases by 1 per P atom.
P₄ acts both as an oxidising as well as a reducing agent.

Oxidation number method:
Total decrease in O.N. of P₄ in PH₃ = 3 x 4 = 12
Total increase in O.N. of P₄ in H₂PO₂ = 1 x 4 = 4
Therefore, to balance increase/decrease in O.N. multiply PH₃ by 1 and H₂PO₂^– by 3, we have,

P₄(s) + OH^–(aq) → PH₃(g) + 3H₂PO₂^–(aq)

To balance O atoms, multiply OH- by 6, we have,

P₄(s) + 6OH^–(aq) → PH₃(g) + 3H₂PO₂^–(aq)

To balance H atoms, add 3H₂O to L.H.S. and 3OH^– to the R.H.S. we have,

P₄ (s) + 6OH^–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^–(aq) + 3OH^–(aq)
or P₄(s) + 3OH^–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^–(aq) …(1)

Thus, Eq. (1) represents the correct balanced equation.

Ion-electron method : The two half reactions are:

Oxidation number method :
Total increase in O.N. of N = 2 x 4 = 8
Total decrease in O.N. of Cl = 1 x 6 = 6
Therefore, to balance increase/decrease in O.N. multiply N₂H₂ by 3 and \(\mathrm{ClO}_{3}^{-}\) by 4, we have,

3N₂H₄(l) + 4ClO₃^– (aq) → NO(g) + Cl^–(aq)

To balance N and Cl atoms, multiply NO by 6 and Cl^– by 4, we have,

3N₂H₄(l) + 4ClO₃^–(aq) → 6NO(g) + 4Cl^–(aq)

Balance O atoms by adding 6H₂O, in R.H.S.

3N₂H₄(l) + 4ClO₃^–(aq) → 6NO(g) + 4Cl^–(aq) + 6H₂O (l) …(1)

H atoms get automatically balanced and thus Eq. (1) represents the correct balanced equation.

Ion electron method :

Oxidation number method :
Total decrease in O.N. of Cl₂O₇ = 4 x 2 = 8
Total increase in O.N. of H₂O₂ = 2 x 1 = 2
∴ To balance increase/decrease in O.N. multiply H₂O₂ and O₂ by 4, we have,

Question 20.
What type of information can you draw from the following reaction?
(CN)₂(g) + 2OH^–(aq) → CN^–(aq) + CNO^–(aq) + H₂O(l)
Answer:
The oxidation number of carbon in (CN)₂ CN^– and CNO^– is +3, +2 and +4 respectively. These are obtained as shown below:
Let the oxidation number of C be x.
(i) (CN)₂
2 (x – 3) = 0
∴ x = +3

(ii) CN^–
x -3 = -1
∴ x = +2

(iii) CNO^–
x – 3 – 2 = -1
∴ x = + 4

The oxidation number of carbon in the various species is:

The following information we can drawn from the above reaction :
(i) Decomposition of cyanogen in the cyanide ion (CN^–) and cyanate ion (CNO^–) occurs in basic medium.
(ii) Cyanogen (CN)₂ acts as both reducing agent as well as oxidising agent.
(iii) The reaction is an example of disproportionation reaction.
(iv) Cyanogen (CN)₂ is called pseudohalogen while CN^–, CNO^– ions are called pseudohalide ions.

Question 21.
The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂ and H⁺ ion. Write a balanced ionic equation for the reaction.
Answer:
The given reaction can be represented as:

Mn³⁺ (aq) → Mn²⁺(aq) + MnO₂(s) + H⁺(aq)

The oxidation half equation is:

The oxidation number is balanced by adding one electron as:
Mn³⁺(aq) → MnO₂(s) + e^–

The charge is balanced by adding 4H+ ion as :
Mn³⁺(aq) → MnO₂(s) + 4H⁺ (aq) + e^–

The O atoms and H+ ions are balanced by adding 2H20 molecules as:
Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + 4H⁺ (aq) + e^– …(1)

The reduction half equation is:
Mn³⁺(aq) → Mn²⁺(aq)

The oxidation number is balanced by adding one electron as :
Mn³⁺ (aq) + e^– → Mn²⁺ (aq) … (2)

The balanced chemical equation can be obtained by adding equation (1) and (2) as :
2Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + Mn²⁺(aq) + 4H⁺(aq)

Question 22.
Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor the positive oxidation state.
Answer:
(a) F exhibits only negative oxidation state of-1.
(b) Cs exhibits only positive oxidation state of +1
(c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of-1, 0, +1, + 3, + 5, and + 7.
(d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states.

Question 23.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
Answer:
The given redox reaction can be represented as:

Question 24.
Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non-metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.
Answer:
In disproportionation reaction, one of the reacting substances always contains an element that can exist in at least three oxidation states.

Question 25.
In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with lO.OOg of ammonia and 20.00 g of oxygen?
Answer:
The balanced chemical equation for the given reaction is given as:

68 g of NH₃ reacts with 160 g of O₂
Therefore 10 g of NH₃ reacts with \(\frac{160 \times 10}{68}\). g of O₂ or 23.53 g of O₂
But the available amount of O₂ is 20 g which is less than the amount required to react with 10 g NH₃. So, O₂ is the limiting reagent and it limits the amount of NO produced. From the above balanced equation.
160 g of O₂; produces 120 g NO.
Therefore, 20 g of O₂; produces = \(\frac{120 \times 20}{160}\) = 15 g NO

Question 26.
Using the standard electrode potentials given in the table 8.1,
predict if the reaction between the following is feasible: *
(a) Fe³⁺ (aq) and I^– (aq)
(b) Ag⁺ (aq) and Cu (s)
(c) Fe³⁺ (aq) and Cu(s)
(d) Ag (s) and Fe³⁺ (aq)
(e) Br₂ (aq) and Fe²⁺(aqr)
Answer:

Question 27.
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes
(iii) A dilute solution of H₂SO₄ with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes,
Answer:
(i) AgNO₃ ionizes in aqueous solutions to form Ag⁺ and \(\mathrm{NO}_{3}^{-}\) ions.
On electrolysis, either Ag⁺ ions or H₂O molecules can be reduced at the cathode. But the reduction potential of Ag⁺ ions is higher than that of H₂O.

Hence, Ag+ ions are reduced at the cathode. Similarly, Ag metal or H20 molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H20 molecules.

Ag(s) → Ag⁺(aq) + e^– ;\(E^{\ominus}\) = -0.80V

2H₂O₂(g) → O₂(g) + 4H⁺(aq) + 4e^– ;\(E^{\ominus}\) = -1.23V
Therefore, Ag metal gets oxidized at the anode.

(ii) Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O₂. At the cathode, Ag⁺ ions are reduced and get deposited.

(iii) H₂SO₄ ionizes in aqueous solutions to give H⁺ and \(\mathrm{SO}_{4}^{2-}\) ions.

H₂SO₄ (aq) → 2H⁺(aq) + \(\mathrm{SO}_{4}^{2-}\) (aq)

On electrolysis, either of H⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of H⁺ ions is higher than that of H₂O molecules.

2H⁺(aq) + 2e^– → H₂(g); \(E^{\ominus}\) = 0.0 V
2H₂O(aq) + 2e^– → H₂(g) + 2OH^–(aq): \(E^{\ominus}\) = – 0.83V

Hence, at the cathode, H⁺ ions are reduced to liberate H₂ gas.
On the other hand, at the anode, either of \(\mathrm{SO}_{4}^{2-}\) ions or H₂O molecules can get oxidized. But the oxidation of \(\mathrm{SO}_{4}^{2-}\) involves breaking of more bonds than that of H₂O molecules. Hence, \(\mathrm{SO}_{4}^{2-}\) ions have a lower oxidation potential than H₂O. Thus, H₂O is oxidized at the anode to liberate O₂ molecules.

(iv) In aqueous solutions, CuCl₂ ionizes to give Cu²⁺ and Cl^– ions as
CuCl₂ (aq) → Cu²⁺ (aq) + 2Cl^– (aq)

On electrolysis either of Cu²⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of Cu²⁺ is more than that of H₂O molecules.

Cu²⁺(aq) + 2e^– → Cu(aq) ;\(E^{\ominus}\) = + 0.34V ;
H₂O(l) + 2e^– → H₂(g) + 2OH^– ;\(E^{\ominus}\) = – 0.83V

Hence, Cu²⁺ ions are reduced at the cathode and get deposited.
Similarly, at the anode, either of Cl^– or H₂O is oxidized. The oxidation , potential of H₂O is higher than that of Cl^–

2Cl^–(aq) → Cl₂(g) + 2e^– ;\(E^{\ominus}\) = -1.36V :K
2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e^– ;\(E^{\ominus}\) = -1.23V

But oxidation of H₂O molecules occurs at a lower electrode potential . than that of Cl^– ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl^– ions are oxidized at the anode to liberate Cl₂ gas. :

Question 28.
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn. ‘
Answer:
A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt.
The order of the increasing reducing power of the given metals is
Cu < Fe < Zn < Al < Mg. Hence, we can say that Mg can displace Al from its salt solution, but Al cannot displace Mg. Thus, the order in which the given metals displace each other from the solution of their salts is given below:
Mg > Al > Zn > Fe > Cu

Question 29.
Given the standard electrode potentials:
K⁺ / K = – 2.93V, Ag⁺ / Ag = 0.80 V, Hg²⁺ /Hg = 0.79 V
Mg²⁺/ Mg = -2.37 V, Cr³⁺/Cr = -0.74 V Arrange these metals in their increasing order of reducing power.
Answer:
The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metal is
Ag < Hg < Cr < Mg < K.

Question 30.
Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aqr) → Zn²⁺(aqr) + 2Ag(s) takes place, further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode.
Answer:
The galvanic cell corresponding to the given redox reaction can be represented as:
Zn / Zn²⁺ (aq) | | Ag⁺(aq) / Ag
(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn²⁺ and the leaving electrons accumulate on this electrode.
(ii) Ions are the carriers of current in the cell.
(iii) The reaction taking place at Zn electrode can be represented as:
Zn(s) → Zn²⁺ (aq) + 2e^–
and the reaction taking place at Ag electrode can be represented as:
Ag⁺(aq) + e^– → Ag(s)