NCERT Solutions Class 11 Chemistry Chapter 7 - Redox Reactions

Q7.1: Assign oxidation numbers to the underlined elements in each of the following species:
(a) NaH₂PO₄ 
 (b) NaHSO₄     
 (c) H₄P₂O₇   
 (d) K₂MnO₄
 (e) CaO₂       
 (f) NaBH₄ 
 (g) H₂S₂O₇ 
 (h) KAl(SO₄)₂.12 H₂O
Ans: (a) 

Let the oxidation number of P be x.

We know that,

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = –2

Then, we have

Hence, the oxidation number of P is + 5.

(b) 

  Then, we have

Hence, the oxidation number of S is +6.

(c) 

  Then, we have

Hence, the oxidation number of P is + 5.

(d) 

   Then, we have

Hence, the oxidation number of Mn is + 6.

(e) 

  Then, we have

Hence, the oxidation number of O is – 1.

(f) 

   Then, we have

Hence, the oxidation number of B is +3.

(g) 

   Then, we have

Hence, the oxidation number of S is +6.

(h) 

  Then, we have

Or,

We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have

Hence, the oxidation number of S is +6.

Q7.2: What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
(a) KI₃
(b) H₂S₄O₆ 
(c) Fe₃O₄ 
(d) CH₃CH₂OH
(e) CH₃COOH
Ans: (a)In KI₃, the oxidation number (O.N.) of K is 1. Hence, the average oxidation number of I is       . 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) H₂S₄O₆

Now, 2(+1) + 4(x) + 6(-2) = 0

 2 + 4x – 12 = 0

 4x = 10

 x = +2½

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) 

On taking the O.N. of O as –2, the O.N. of Fe is found to be +2 ²/₃. However, O.N. cannot be fractional.

Here,one of the three Fe atoms exhibits the O.N. of +2 & the other two Fe atoms exhibit the O.N of +3.

(d)

 2(x)+ 4( +1)+ 1(-2) = 0

 2x +6 – 2 = 0

 x = – 2

Hence, the O.N. of C is –2.

(e)

However, 0 is average O.N. of C. The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and –2 in CH₃COOH.

Q7.3: 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) 4 NH₃(g)+ 5 O₂(g) → 4NO(g)+ 6H₂O(g)

Ans: (a) 

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) 

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) 

The oxidation number of each element in the given reaction can be represented as:

In this reaction, the oxidation number of B decreases from +3 in BCl₃ to –3 in B₂H₆. 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) 

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) 

The oxidation number of each element 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.

Q7.4: 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.

Q7.5: Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, . Suggest structure of these compounds. Count for the fallacy.
Ans: (i) 

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 H₂SO₅ is shown as follows:

Therefore, the O.N. of S is +6.

(ii)

Here, there is no fallacy about the O.N. of Cr in .

The structure of  is shown as follows:

Here, each of the two Cr atoms exhibits the O.N. of +6.

(iii) 

Here, there is no fallacy about the O.N. of N in .

The structure of  is shown as follows:

The N atom exhibits the O.N. of +5.

Q7.6: Write the formulae 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
Ans: (a) Mercury (II) chloride: HgCl₂

(b) Nickel (II) sulphate: NiSO₄

(c) Tin (IV) oxide: SnO₂

(d) Thallium (I) sulphate: Tl₂SO₄

(e) Iron (III) sulphate: Fe₂(SO₄)₃

(f) Chromium (III) oxide: Cr₂O₃

Q7.7: Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.
Ans: The substances where carbon can exhibit oxidation states from –4 to +4 are listed in the following table.
The substances where nitrogen can exhibit oxidation states from –3 to 5 are listed in the following table.

Q7.8: 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?
Ans: 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 a reducing agent.

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, H₂O₂can act as an oxidising as well as a reducing agent.

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.

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 oxidant.

Q7.9: Consider the reactions:

(a) 6 CO₂(g) +6H₂O(l) → C₆ H₁₂ O₆(aq)+ 6O₂(g)

(b) O₃(g) +H₂O₂(l) → H₂O(l) +2O₂(g)

Why it is more appropriate to write these reactions as:

(a) 6CO₂(g) +12H₂O(l) → C₆ H₁₂ O₆(aq)+ 6H₂O(l) +6O₂(g)

(b) O₃(g) +H₂O₂ (l) → H₂O(l) +O₂(g)+ O₂(g)

Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Ans: (a)The process of photosynthesis involves two steps.

Step 1:

H₂O decomposes to give H₂ and O₂.

Step 2:

The H₂ produced in step 1 reduces CO₂, thereby producing glucose (C₆H₁₂O₆) and H₂O.

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 investigated by using  or   .

Q7.10: The compound AgF₂ is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?
Ans: 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.

Q7.11: 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.
Ans: 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.

Q7.12: 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?
Ans: (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.

Q7.13: 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) +2Cu^{2 (}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)

Ans: (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

Q7.14: 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?
Ans: 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.

Q7.15: Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Ans: 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.

Q7.16: Why does the following reaction occur?

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

What conclusion about the compound Na₄XeO₆ (of which   is a part) can be drawn from the reaction.
Ans: 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^–_.

Q7.17: 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) +2Cu^{2 (}aq) +5OH^–(aq) → No change observed.

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

Ans:  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⁺   _.

Q7.18: Balance the following redox reactions by ion-electron method:
MnO^—₄ (aq) + I^—(aq) → MnO₂ (s) + I₂(s) (in basic medium)
MnO^—₄ (aq) + SO₂(g) → Mn²⁺ (aq) + (aq)HSO^—_4(aq) (in acidic solution)
H₂O₂(aq) + Fe²⁺(aq) → Fe³⁺(aq) + H₂O(l) (in acidic solution)
Cr₂O^2-₇+SO₂(g) → Cr³⁺(aq) + SO^2-₄ (aq) (in acidic solution)
Ans: Step 1: The two half reactions involved in the given reaction are:
Oxidation half reaction:

Reduction halfreaction:
Step 2: Balancing I in the oxidation half-reaction, we have:

Now, to balance the charge, we add 2e⁻ to the RHS of the reaction.

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.

Now, to balance the charge, we add 4OH^– ions to the RHS of the reaction as the reaction is taking place in a basic medium.

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.

Step 5: Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:

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

And the reduction half reaction as:

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:

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

And the reduction half reaction as:

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

And the reduction half reaction as:

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

Q7.19: Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

Ans: (a)

The O.N. (oxidation number) of P decreases from 0 in P₄ to – 3 in PH₃ and increases from 0 in P₄ to +2 in  . Hence, P₄ acts both as an oxidizing agent and a reducing agent in this reaction.

Ion–electron method:

The oxidation half equation is:

The P atom is balanced as:

The reduction half equation is:

The P atom is balanced as

By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:

5P₄ (s) + 24H₂O → 12HPO₂^- + 8PH₃(g) + 12H+

As, the medium is basic, add 12OH^– both sides as:

5P₄ (s) + 12H₂O + 12OH− →12HPO₂^-  + 8PH₃(g)

This is the required balanced equation.

Oxidation number method:

Let, total no of P reduced = x
 ∴ Total no of P oxidised = 4 – x

Total decrease in oxidation number of P = x × 3 = 3x

Total increase in oxidation number of P = (4 – x) × 2 = 8 – 2x
 ∵ 3x = 8 – 2x
     x = 8/5
From (i), 

This is the required balanced equation.

(b)

The oxidation number of N increases from – 2 in N₂H₄ to +2 in NO and the oxidation number of Cl decreases from +5 in   to – 1 in Cl^–. Hence, in this reaction, N₂H₄ is the reducing agent and   is the oxidizing agent.

Ion–electron method:

The oxidation half equation is:

The N atoms are balanced as:

The oxidation number is balanced by adding 8 electrons as:

The charge is balanced by adding 8 OH^–ions as:

The O atoms are balanced by adding 6H₂O as:

The reduction half equation is:

The oxidation number is balanced by adding 6 electrons as:

The charge is balanced by adding 6OH^– ions as:

The O atoms are balanced by adding 3H₂O as:

The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:

Oxidation number method:

Total decrease in oxidation number of N = 2 × 4 = 8

Total increase in oxidation number of Cl = 1 × 6 = 6

On multiplying N₂H₄ with 3 and   with 4 to balance the increase and decrease in O.N., we get:

The N and Cl atoms are balanced as:

The O atoms are balanced by adding 6H₂O as:

This is the required balanced equation.

(c)

The oxidation number of Cl decreases from +7 in Cl₂O₇ to +3 in  and the oxidation number of O increases from – 1 in H₂O₂ to zero in O₂. Hence, in this reaction, Cl₂O₇ is the oxidizing agent and H₂O₂ is the reducing agent.

Ion–electron method:

The oxidation half equation is:

The oxidation number is balanced by adding 2 electrons as:

The charge is balanced by adding 2OH^–ions as:

The oxygen atoms are balanced by adding 2H₂O as:

The reduction half equation is:

The Cl atoms are balanced as:

The oxidation number is balanced by adding 8 electrons as:

The charge is balanced by adding 6OH^– as:

The oxygen atoms are balanced by adding 3H₂O as:

The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as: 

Oxidation number method:

Total decrease in oxidation number of Cl₂O₇ = 4 × 2 = 8

Total increase in oxidation number of H₂O₂ = 2 × 1 = 2

By multiplying H₂O₂ & O₂ with 4 to balance the increase & decrease in the oxidation number, we get:

The Cl atoms are balanced as:

The O atoms are balanced by adding 3H₂O as:

The H atoms are balanced by adding 2OH^– and 2H₂O as:

This is the required balanced equation.

Q7.20: What sorts of information can you draw from the following reaction ?

Ans: The oxidation numbers of carbon in (CN)², CN^– and CNO^– are +3, +2 and +4 respectively. These are obtained as shown below:

Let the oxidation number of C be x.

(CN)₂

2(x – 3) = 0

∴x = 3
CN^–

x – 3 = –1

∴x = 2
CNO^–
x – 3 – 2 = –1

∴x = 4
The oxidation number of carbon in the various species is:

It can be easily observed that the same compound is being reduced and oxidised simultaneously in the given equation. Reactions in which the same compound is reduced and oxidised is known as disproportionation reactions. Thus, it can be said that the alkaline decomposition of cyanogen is an example of disproportionation reaction.

Q7.21: 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.
Ans: The given reaction can be represented as:

The oxidation half equation is:

The oxidation number is balanced by adding one electron as:

The charge is balanced by adding 4H ⁺   ions as:

The O atoms and H⁺  ions are balanced by adding 2H₂O molecules as:

The reduction half equation is:

The oxidation number is balanced by adding one electron as:

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

Q7.22: 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 postive oxidation state.

(c) Identify the element that exhibits both positive and negative oxidation states.

(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Ans:  (a) F exhibits only negative oxidation state of –1.

(b) Cs exhibits positive oxidation state of +1.

(c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of – 1, +1, +3, +5, and +7.

(d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states.

Q7.23: 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.
Ans: The given redox reaction can be represented as:

The oxidation half reaction is:

The oxidation number is balanced by adding two electrons as:

The charge is balanced by adding 4H  ions as:

The O atoms and H ⁺   ions are balanced by adding 2H₂O molecules as:

The reduction half reaction is:

The chlorine atoms are balanced as:

The oxidation number is balanced by adding electrons

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

Q7.24: 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.

Ans: In disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states.

(a) P, Cl, and S can show disproportionation reactions as these elements can exist in three or more oxidation states.

(b) Mn, Cu, and Ga can show disproportionation reactions as these elements can exist in three or more oxidation states.

Q7.25: 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 10.00 g. of ammonia and 20.00 g of oxygen?
Ans: The balanced chemical equation for the given reaction is given as:

Thus, 68 g of NH₃ reacts with 160 g of O_2.

Therefore, 10g of NH₃ reacts with  g of O_2, or 23.53 g of O_2.

But the available amount of O₂ is 20 g.

Therefore, O₂ is the limiting reagent (we have considered the amount of O₂ to calculate the weight of nitric oxide obtained in the reaction).

Now, 160 g of O₂ gives 120g of NO.

Therefore, 20 g of O₂ gives g of N, or 15 g of NO.

Hence, a maximum of 15 g of nitric oxide can be obtained.

Q7.26: 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⁺(aq)
Ans:  (a) The possible reaction between  is given by,

E° for the overall reaction is positive. Thus, the reaction between   and   is feasible.

(b) The possible reaction between  is given by,

E° positive for the overall reaction is positive. Hence, the reaction between   &   is feasible.

(c) The possible reaction between  and  is given by,

E° positive for the overall reaction is positive. Hence, the reaction between   and   is feasible.

(d) The possible reaction between  and  is given by,

Here, E° for the overall reaction is negative. Hence, the reaction between  &   is not feasible.

(e) The possible reaction between  and  is given by,

Here, E° for the overall reaction is positive. Hence, the reaction between  and   is feasible.

Q7.27: Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO₃ with silver electrodes

(ii) An aqueous solution AgNO₃ with platinum electrodes

(iii) A dilute solution of H₂SO₄ with platinum electrodes

(iv) An aqueous solution of CuCl₂ with platinum electrodes.

Ans: (i) AgNO₃ ionizes in aqueous solutions to form Ag ⁺   and   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 H₂O molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H₂O molecules.

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  ions.

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.

Hence, at the cathode, H ⁺   ions are reduced to liberate H₂ gas.

On the other hand, at the anode, either of  ions or H₂O molecules can get oxidized. But the oxidation of  involves breaking of more bonds than that of H₂O molecules. Hence,  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 Cu^{2 +}   and Cl^– ions as:

On electrolysis, either of Cu^{2 +}   ions or H₂O molecules can get reduced at the cathode. But the reduction potential of Cu^{2 +}   is more than that of H₂O molecules.

Hence, Cu^{2 +}   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^–.

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.

Q7.28: 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.
Ans: 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

Q7.29: Given the standard electrode potentials,

K ⁺  /K = –2.93V, Ag /Ag = 0.80V,

Hg^{2 +}  /Hg = 0.79V

Mg^{2 +}  /Mg = –2.37V. 

Cr^{3 +}  /Cr = –0.74V

Arrange these metals in their increasing order of reducing power.

Ans: The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metals is Ag < Hg < Cr < Mg < K.

Q7.30: Depict the galvanic cell in which the reaction 

Zn(s)+ 2Ag ⁺  (aq) → Zn^{2 +}  (aq)+ 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.

Ans: The galvanic cell corresponding to the given redox reaction can be represented as:

(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:

And the reaction taking place at Ag electrode can be represented as:

 .