NCERT Class 11 Chemistry Chapter 8 Redox Reactions

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NCERT Class 11 Chemistry Chapter 8 Redox Reactions

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Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. NCERT Class 11 Chemistry Chapter 8 Redox Reactions Solutions are part of All Subject Solutions. Here we have given NCERT Class 11 Chemistry Part: I, Part: II Notes. NCERT Class 11 Chemistry Chapter 8 Redox Reactions Notes, NCERT Class 11 Chemistry Textbook Solutions for All Chapters, You can practice these here.

Redox Reactions

Chapter: 8

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

(a)

Ans: P in Na₂HPO₄

+ 1 + 1x -2  

NaH₂PO₄

(+1) + (+2) + x + 4 (-2) = 0

Or x + 3 -8 = 0

Or x = +5

(b) 

Ans: S in NaHSO₄

+ 1 + 1 x -2 

NaHSO₄

(+1) = (+1) + x + 4 (-2) = 0

Or  X – 6 = 0

Or  X = +6

(c)

Ans: P in H₄P₂O₇

+ 1 x-2

H₄P₂O₇

∴ 4(+1) + 2(x) + 7 (-2) = 0

X = + 5

Thus the oxidation number of P in H₄P₂O₇  = + 5

(d) 

Ans: Mn in K₂MnO₄ 

+ 1 x-2 

K₂MnO₄ 

∴ 2(+1) + x +(-2) = 0 

Or x – 6 = 0 

Or x = + 6

Thus, oxidation number of Mn in in K₂MnO₄  = + 6

(e) 

Ans: O in CaO₂ 

+ 2 x 

CaO₂ 

∴ (+2) + 2(x) = 0 

Or x = -1

(f) 

Ans: B in NaBH₄ 

+ 1 x-1 

NaBH₄

∴ + 1 x x + 4 (-1) = 0

X – 3 = 0 

X = + 3

Thus, the Oxidation number of B in NaBH₄ = + 3

(g) 

Ans: S in H₂S₂O₇

+ 1 x-2

H₂S₂O₇

2(+1) + 2 (+x) = 7 (-2) = 0

X – 12 =0

Or = + 6 

Thus, the Oxidation number of S in H₂S₂O₇ = + 6 

(h)

Ans: S in KAl(SO₄)₂.12 H₂O 

+ 1 + 3 x – 2

KAl(SO₄) 2.12 H₂O 

= + 1 + 3 + 2x + 8 (-2) + 12(0) = 0

Or 2x – 12 = 0 

Or x = +6

Thus, the oxidation number of S in KAl(SO₄)₂.12 H₂O = + 6 .

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

Ans: 1 + 3x = 0 

3x = -1

X = – 1/3

Ans: 2 + 4x – 12 = 0 

Or 4x – 10 = 0

X = 10/4 

5/ 2 

Ans: 3x -8 = 0 

Or x = 8 / 3

Ans: + 6 – 2 = 0 

2x = -4

X = -2

Ans: = 0 

Or 2x + 4 – 4 = 0 

2x = 0 

X = 0

3. Justify that the following reactions are redox reactions: 

(a) CuO(s) + H₂(g) → Cu(s) + H₂O(g) 

Ans: As a result, CuO is reduced to Cu, and H₂ is oxidized to form H₂O. Therefore, this reaction exemplifies a redox process.

(b) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

Ans: 

The oxidation number of Fe²⁺ decreases from +3 in Fe₂O₃ to 0 in Fe, while the oxidation number of carbon increases from +2 in CO to +4 in CO₂. Thus, iron undergoes reduction, and carbon undergoes oxidation, confirming that this is indeed a redox reaction.

(c) 4BCl₃(g) + 3LiAlH₄(s) → 2B₂H₆(g) + 3LiCl(s) + 3 AlCl₃ (s) 

Ans: 

Boron is reduced, while lithium and aluminum are oxidized. Therefore, this reaction qualifies as a redox process.

(d) 2K(s) + F₂(g) → 2K⁺ F^– (s) 

Ans: Potassium is oxidized to K⁺, and F₂ is reduced to F⁻. Thus, this is a redox reaction.

(e) 4 NH₃(g) + 5 O₂(g) → 4NO(g) + 6H₂O(g) 

Ans: Here, O.N. of N increases from -3 in NH₃ to +2 in NO while that of O decreases from O to oxidised while O2 is reduced.

Additionally, hydrogen is removed from NH₃ and added to O₂. As a result, NH₃ is oxidized, while O₂ is reduced. Hence, this reaction is a redox process.

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: 

In this reaction, the oxidation number of fluorine decreases from 0 in F₂ to -1 in HF and increases from 0 in F₂ to +1 in HOF. Consequently, F₂ is both reduced and oxidized. Therefore, this is a redox reaction and, more specifically, a disproportionation reaction.

5. Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇ ^2– and NO₃ ^–. Suggest structure of these compounds. Count for the fallacy. 

Ans: The O.N of in H₂SO₅:

2(+1) + x = 5(-2) 0

Or x = + 8 

The structure of peroxy monosulfuric acid (H₂SO₅) features sulphur bonded to two hydroxyl groups and one peroxide group. Each hydroxyl group contributes -1 for oxygen, while the peroxide group adds -1. This arrangement maintains sulphur’s oxidation state within its valence constraints, fitting within the +6 limit.The structure of (H₂SO₅):

2 x (+1) + x +2 (-1) +3 x (-2) = 0

Or

X = + 6

(for H) (for S) (for o-o) (for other O atoms) 

(ii) O.N. of Cr in CrO₅ 

O.N of Cr is: 

X + 5 (-2) = 0 

Or

X = + 10

In chromium trioxide (CrO^3​), chromium is surrounded by three double-bonded oxygen atoms. The structure adheres to chromium’s oxidation number limitation of +6, consistent with its 3d⁵4s¹ electron configuration. This method of calculation ensures accuracy in its maximum oxidation state.The structure of  (CrO^3​):^​

From the structure of chromium trioxide (3CrO³​), the oxidation number (O.N.) of chromium can be calculated as:

(iii) O.N. of N in NO₃^–

According to conventional bonding method, 

Thus, there is no fallacy regarding the oxidation number of nitrogen in NO₃⁻​; it remains consistent whether calculated using the conventional method or the chemical bonding approach.

6. Write formulas for the following compounds: 

(a) Mercury(II) chloride.

Ans: Hg(II)Cl₂

(b) Nickel(II) sulphate. 

Ans: Ni(II)SO₄

(c) Tin(IV) oxide. 

Ans: Sn(IV)O₂

(d) Thallium(I) sulphate. 

Ans: Tl_2​(I)SO₄​

(e) Iron(III) sulphate. 

Ans: Fe₂(III)_​(SO_4​)₃​

(f) Chromium(III) oxide. 

Ans: Cr₂​(III)O₃​

7. Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5. 

Ans: 

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: (i) In sulphur dioxide (SO₂​), sulphur has an oxidation number of +4, with a range from +6 to -2. This allows SO₂​ to act both as an oxidizing and a reducing agent, as sulphur can either gain or lose electrons.

(ii) In hydrogen peroxide (H₂O₂​), oxygen has an oxidation number of -1. Oxygen can vary from 0 to -2 and can also take on +1 and +2 states. Thus, H₂O₂ can act as both an oxidizing and reducing agent.

(iii) In O₃, the O.N. of O zero. It can only decrease its O.N. from zero to -1 or -2, but cannot increase to = 2. Therefore, O₃ acts only as an oxidant.

(iv) In nitric acid (HNO₃​), nitrogen has an oxidation number of +5, with a range from +5 to -3. Thus, HNO₃ can only act as an oxidizing agent.

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: Photosynthesis, though complex, can be broadly understood in two steps. First, water (H₂O) decomposes into hydrogen (H₂​) and oxygen (O₂​) in the presence of chlorophyll. In the second step, the hydrogen (H2​) reduces carbon dioxide (CO₂​) to glucose (C₆H₁₂O₆​).During the second step H₂O molecules are also produced as shown below: 

Thus, the photosynthesis equation should be written as 6CO₂ + 12H₂O → C₆H₁₂O₆ + 6O₂ + 6H₂O. This highlights that 12 molecules of water are used and 6 molecules are produced per carbohydrate molecule formed.

(b) Writing O_2​ twice indicates that oxygen is obtained from both reactants: one part comes from the decomposition of water and the other from the reduction of carbon dioxide.

To determine the path of reactions, use H₂¹⁸OH or D₂O in reaction (a) and H₂¹⁸O₂​ or O₃^18​ in reaction 

(b) These isotopic tracers help identify reaction pathways by tracking the movement of the isotopic labels through the reactions.

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

Ans: Silver in the +2 oxidation state is unstable. When AgF₂ forms, silver quickly gains an electron to revert to the more stable +1 state, resulting in the formation of Ag⁺. This electron acceptance makes AgF₂ a very strong oxidising agent.

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: (i) Carbon acts as a reducing agent and O​ as an oxidizing agent. When excess carbon burns in limited O₂​, it forms CO with C in the +2 oxidation state. With excess O₂​, CO is further oxidized to CO₂, where C has an oxidation state of +4.

(ii) Phosphorus (P₄​) is a reducing agent, and chlorine (Cl₂​) is an oxidizing agent. With excess P₄​, PCl₃​ forms with phosphorus in the +3 oxidation state. With excess Cl₂​ is further oxidized to PCl₅​, where phosphorus has an oxidation state of +5.

(iii) Sodium (Na) is a reducing agent, and oxygen (O₂​) is an oxidizing agent. With excess sodium, sodium oxide (Na₂O) forms, where oxygen is in the -2 oxidation state. With excess O2​, Na₂O_2​ forms, where oxygen is in the -1 oxidation state, which is higher than -2.

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.

Ans: Toluene can be oxidized to benzoic acid using potassium permanganate in acidic, basic, and neutral media, with the oxidation process being selective and controlled by the specific medium used.

(i) Acidic Medium:

(ii) Basic and neutral media:

In industry, alcoholic KMnO₄ is preferred for oxidizing toluene to benzoic acid because it provides better control, prevents over-oxidation, and ensures higher selectivity and purity of the product.

(i) The cost of adding acid or base is avoided, as the reaction itself generates OH ions in a neutral medium, eliminating the need for external addition.

(ii) Reactions proceed faster in a homogeneous medium compared to a heterogeneous one, making homogeneous mediums more efficient for quicker reaction rates.

(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: When concentrated H₂SO₄ is added to an inorganic mixture containing chloride, it produces a pungent-smelling gas (HCl). This occurs because the stronger acid (H₂SO₄) displaces the weaker acid (HCl) from its salt, leading to the release of hydrogen chloride gas. 

Since HCI is a very weak reducing agent, it cannot reduce H₂SO₄ to SO₂ and hence HCI is not oxidised to Cl₂. Stronger acid.

However, when the mixture contains bromide ion, the initially produced HBr being a stronger reducing agent than HCI, reduces H₂SO₄ and is itself oxidised to produce red vapour of Br₂.

2NaBr + 2H₂SO₄ → 2NaHSO₄ + 2HBr

2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O

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) + 2 Cu²⁺(aq) + 5OH^– (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: 

14. Consider the reactions:

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

Ans: The different reactions of thiosulfate with Br₂ and I₂ are due to their oxidizing strengths. Br₂, a stronger oxidizer, oxidizes S in S₂O₃²⁻ from +2 to +6, forming SO₄²⁻. In contrast, I₂, a weaker oxidizer, converts S in S₂O₃²⁻ to +2.5, forming S₄O₆²⁻.

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

Ans: Halogens are strong oxidizing agents due to their high tendency to accept electrons. Their oxidizing power correlates with their electrode potentials, which decrease as follows: F₂ (+2.87 V) > Cl₂, (+1.36 V) > Br₂ (+1.09 V) > I₂ (+0.54 V), Thus, their oxidizing power decreases in this order.

Fluorine (F₂) is the strongest oxidizing agent among halogens due to its highest electrode potential. Conversely, halide ions act as reducing agents, with their reducing power decreasing as follows: I⁻ (-0.54 V) > Br⁻ (-1.09 V) > Cl⁻ (-1.36 V) > F⁻ (-2.87 V). Therefore, hydriodic acid (HI) is the strongest reductant. This is evidenced by HI and HBr reducing sulfuric acid (H₂SO₄) to sulfur dioxide (SO₂), while HCl and HF do not.

For example:

HI andHBr reduce H₂SO₄ to SO₂ While HCI And HF do not.

2Hbr + H₂SO₄ → Br₂ + SO + 2H₂O:

2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O

Additionally, I⁻ reduces Cu²⁺ to Cu⁺, while Br⁻ does not.

16. Why does the following reaction occur? 

XeO₆^4– (aq) + 2F– (aq) + 6H+(aq) → XeO₃(g)+ F₂(g) + 3H₂O(l) 

What conclusion about the compound Na₄XeO₆ (of which XeO₆ ^4– is a part) can be drawn from the reaction. 

Ans: 

Thus, HI is a stronger reductant than HBr. Among HCl and HF, HCl is a stronger reducing agent than HF because HCl can reduce MnO₂ to Mn²⁺, whereas HF cannot.

The formation of O₂F₂​ is not observed. Consequently, the reducing character of hydrohalic acids decreases in the order: HI > HBr > HCl > HF.

17. Consider the reactions:

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

Ans: Reactions (a) and (b) show that H₃PO₂ (hypophosphorous acid) acts as a reducing agent, reducing AgNO₃ and CuSO₄ to Ag and Cu, respectively. Conversely, AgNO₃ and CuSO₄ act as oxidizing agents, oxidizing H₃PO₂ to H₃PO₄ (phosphoric acid).

Reaction (c) suggests that [Ag(NH₃)₂]+ oxidises C₆H₃CHO (benzaldehyde) to CH₃COO^– (benzoate ion) but reaction (d) indicates that Cu²⁺ ions cannot oxidise C₆H₃CHO to C₆H₃COO^–. Therefore, from the above reactions, we conclude that Ag⁺ ion is a stronger oxidising agent than Cu²⁺ ion.

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

(a) MnO₄ ^– (aq) + I^– (aq) → MnO₂ (s) + I2(s) (in basic medium)

Ans: The skeleton equation is:

MnO₄−​(aq) + I−(aq) → MnO_2​(s) + I2​(s)

Step (I) Separation of the equation in two half reaction is

(ii)  Identify the atoms which undergon change in O.N. 

(iii) Identify the species involved in the oxidation and reduction half-reactions:

Step (II) Balancing the oxidation half reaction

21-  I2 + 2e-

Step (III) Balancing the reduction half reaction.

The reduction half-reaction is:

MnO₄⁻ → MnO₂.

(i) The oxidation number decreases by 3, so add 3 electrons to the reactant side: 

MnO₄⁻+3e⁻ → MnO₂

(ii) To balance oxygen, add 2 H2O molecules to the product side:

MnO4⁻ ​+ 3e⁻ → MnO₂​ + 2H₂​O

(iii) To balance the charges, add 4 OH to the product side. Then, add 4H₂​O to the reactant side: 

MnO₄⁻​+3e⁻+4H₂​O → MnO₂​+4OH⁻+2H₂​O

Thus, the reduction half-reaction is balanced. 

Step IV. Adding the two half reactions 

In order to equate the electrons, multiply eqn. 

(i) by 3 and eqn. 

(ii) by 2 Add the two equations.

(b) MnO₄ ^– (aq) + SO₂ (g) → Mn₂+ (aq) + HSO₄ ^– (aq) (in acidic solution) 

Ans: The Skeleton equation is: 

MnO₄^– (aq) + SO₂ → (Mn)₂⁺ + (HSO₄)^–

Step (I) Separation of the equation in two half reaction:

(i) Write the O.N. of the atoms involved in the reaction 

(ii) Identify the atoms that experience a change in oxidation number (O.N.).

(iii) Determine the species that are involved in the oxidation and reduction half-reactions.

Oxidation half reaction: SO₂ → HSO₄^–

Reduction half reaction: MnO₄ → Mn²⁺

Step (II) Balancing the Oxidation half reaction is: 

The oxidation half reaction is: 

SO₂ → HSO₄^–

(i) Since the oxidation number increases by 2, add 2 electrons to the product side to balance the change in oxidation number.

SO₂ → HSO₄^–

(ii) To balance the oxygen atoms, add 2H₂O molecules to the reactant side. Then, to balance the hydrogen atoms, add 3H⁺ ions to the product side.

SO₂ + 2H₂O → HSO₄^– + 3H⁺ + 2e^–

Step (III) Balancing the reduction half reaction:

(i) Since the oxidation number decreases by 5, add 5 electrons to the reactant side to balance the change in oxidation number.

(ii) To balance the number of oxygen atoms, add four H₂O molecules to the product side. Then, to balance the hydrogen atoms, add eight H+ ions to the reactant side.

MnO₄^– + 8H+ 5e- → Mn₂+ + 4H₂O

Step IV. Adding the two halfreations:

To equalize the number of electrons, multiply equation (i) by 5 and equation (ii) by 2. Then, add the two equations together.

Therefore, N2H₄ acts the reducing agent 

While ClO_3- Acts as the oxidising agent.

(c) H₂O₂ (aq) + Fe₂+ (aq) → Fe₃+ (aq) + H₂O (l) (in acidic solution) 

Ans: In the oxidation half-reaction, iron(II) ions are oxidized to iron(III) ions:

Fe²⁺(aq) → Fe³⁺(aq)+e^–

Combining these half-reactions, the balanced redox equation is:

H₂​O₂​(aq) + 2Fe²⁺(aq) + 2H⁺(aq) → 2Fe³⁺(aq) + 2H₂​O(l)

This equation represents the complete redox process, with all species balanced for mass and charge.

(d) Cr2O₇ ^2– + SO₂(g) → Cr₃+ (aq) + SO₄ ^2– (aq) (in acidic solution) 

Ans: The oxidation half-reaction, sulphur dioxide is oxidised to sulphate ions: 

SO₂​(g) + 2H₂​O(l) → SO₄²⁻​(aq) + 4H⁺ (aq) + 2e⁻

In the reduction half-reaction, dichromate ions are reduced to chromium(III) ions:

Cr2​O₇²⁻ ​(aq)+14H⁺(aq) + 6e− → 2Cr³⁺(aq) + 7H₂​O(l)

The balanced redox equation is:

Cr2​O₇²⁻ ​(aq) + 3SO²​(g) + 2H₂​O(l) → 2Cr³⁺(aq) + 3SO₄²⁻​(aq) + 10H + (a)

This represents the overall reaction, with all elements balanced.

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) + HPO₂ – (aq) 

Ans: 

P4 acts both as an oxidising as well as a reducing agent.

Oxidation Number Method:

Decrease in O.N. of P in PH₃: 12 (since 3 × 4 = 12)

Increase in O.N. of P in H₂PO₂⁻: 4 (since 1 × 4 = 4)

Therefore, to balance increase/decreases 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)

Balance H by adding 3H₂O to the left and 3OH⁻ to the right:

P4(s) + 6OH⁻(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂⁻(aq) + 3OH⁻(aq)

Or

P4​(s) + 3OH⁻(aq) + 3H₂​O(l) → PH_3​(g) + 3H_2​PO₂⁻​(aq) ……………..(i)

Thus Eq. (i) represents the correct balanced equation.

Ion-Electron Method:

Oxidation half-reaction:

P4​(s) → H₂PO₂^–(aq) …………(ii)

Balance p atoms, we have,

BalanceO.N. by adding electrons, 

P4​(s) → 4H₂​PO₂⁻​(aq) + 4e⁻

Balanced with 8 OH⁻: 

P₄(s)+8OH⁻(aq)→ 4H₂PO₂⁻(aq)………..(iii)

O and H get automatically balanced. Thus, Eq.

(iii) Represent the balanced oxidation half reaction. 

Reduction half reaction: 

………..(iv)

Balancing P atom, we have,

P4(s) + 12e^– → 4PH₃(g)

Balance O.N. by adding electrons,

P4(s) + 12e^– → 4PH₃(g) 

Balance charge by adding 12OH- ions,

P4(s) 12e- → 4PH₃(g) + 12OH-(eq) 

Balance O atoms. By Adding 12H₂O to L.H.S. of above equation.

P4(s) + 12H₂O(I) + 12e^– → 4PH₃(g) + 12OH^–(aq)………(v)

To cancel out electrons, multiply Eq. (iii) by 3 and add it to Eq. (v), we have,

4P₄(s) + 24OH^– (aq) + 12H₂O(I) → 4PH₃ (aq) + 12H₂PO₂^–(aq) + 12OH^–(aq)

Or

4P₄(g) + 3OH^–(aq) + 3H₂O(I) → PH3(aq) + 3H₂PO₂^–(aq)………..iv)

Thus eq. (vi) represents the correct balance equation.

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

Ans:

Reduction half reaction: 

Balance O.N. by adding electrons,

CIO₃^–(aq) + 6E^– → CI^–(aq)

Balance O.N. by adding electron,

CIO₃^–(aq) + 6E^– → CI^–(aq) + 6OH^–(aq)

Balance O atoms by adding 3H₂O.

CIO₃^– (aq) + 3H₂O(I) + 6e^–→ CI^–(aq) + 6OH^–(aq) …………..(iii)

Thus, Eq. (iii) represents the correct balance reduction half equation,

To cancel out electrons gained and lost, multiply Eq. (ii) by 3 and Eq. (iii) ny 4 and add, we have, 

3N₂H₂(I) + 4CIO^–₃(aq) → 6NO(g) + 4CI-(aq) + 6H₂O(I)…………(iv)

(c) Cl₂O₇ (g) + H₂O₂(aq) → ClO₂ – (aq) + O₂(g) + H⁺

Ans: 

Oxidation number method: 

Total decrease in O.N. of CI₂O7 = 4 × 2 = 8

Total increase in O.N. of H₂O₂ = 2 × 1 =2

∴ To balance increase/ decrease in O.N. multiply H₂O₂ and O₂ by 4, we have, 

CI₂O₇ (g) + 4H₂O₂(aq) → CIO₂– (aq) + 40(g) 

To balance CI atoms, multiply CIO2 by 2, we have

CI₂O₇(g) + 4H₂O₂(aq) →2CIO₂^–(aq) + 4O₂(g) 

To Balance O atoms, add 3H₂O to R.H.S., we have,

CI₂O₇(g) + 4H₂O₂(aq) → 2CIO₂– (aq) + 4O₂(g) + 3H₂O(I) 

To balance H atoms, add 2H₂O to R.H.S. and 2OH^– to L.H.S, we have,

CI₂O₇(g) + 4H₂O₂(g) + 2OH-(aq) → 2CIO₂^–(aq) +4O₂(g) + 5H₂O

This represents the balanced redox equation. 

Ion electron method: 

Oxidation half reaction: 

Balance O.N by adding electron, 

H₂O(aq) → O₂(g) +2e^–

Balance charge by adding 2 OH- ions,

H₂O₂(aq) + 2OH^–(aq) → O₂(g) + 2H₂O(I) + 2e^–……………(i) 

Reduction half reaction: 

Balance CI atoms;

CI₂O₇(g) + 8e^– → 2CIO₂^–(aq)

Balance O.N. by adding electrons, 

CI₂O₇(g) + 8e^– → 2CIO₂^–(aq) + 6OH^–

Balance O atoms by adding 3H₂O to L.H.S., we have,

CI₂O₇(g) + 3H₂O(I) + 8e^– → 2CIO₂^–(aq) + 6OH^–(aq) ………..(iii)

To cancel out electrons, multiply Eq. (i) by 4 and add it to Eq. (ii), we have, 

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

(CN)₂(g) + 2OH^– (aq) → CN– (aq) + CNO^– (aq) + H₂O(l) 

Ans: The following four informations can be drawn om the above reaction:

(i) The reaction involves decomposition of cyanogen, (CN)₂ in the alkaline medium to cyanide ion, CN and cyanate ion, CNO.

(ii) The oxidation number of carbon decreases from +3 in (CN)₂ to +2 in the cyanide ion (CN⁻) and increases from +3 in (CN)_2​ to +4 in the cyanate ion (CNO⁻). This indicates that cyanogen is simultaneously reduced to the cyanide ion and oxidized to the cyanate ion.

(iii) It is an example of a redox reaction in general and a disproportionation reaction in particular.

(iv) Cyanogen acts as a pseudohalogen (exhibiting behavior similar to halogens), while the cyanide ion functions as a pseudohalide ion (similar to a halide ion).

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 skeletal equation is:

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

Oxidation half equation:

Balance O.N. by adding electrons, 

Mn³⁺ (aq) → MnO₂(s) +e^– 

Balance charge by adding 4H⁺ ions, 

Balance O atoms by adding 2H₂O:

Reduction half  equation:

Balance O.N. by adding electrons: 

Mn³⁺ (aq) + e^– → Mn²⁺ (aq) ..(ii)

Adding eq. (i) and Eq. (ii) the balance equation for the disproportionation reaction is 

2Mn³⁺ (aq) + 2H₂O(l) →

MnO²(s) + Mn²⁺ (aq) + H⁺ (aq)

22. Consider the elements: Cs, Ne, I and F. 

(a) Identify the element that exhibits -ve oxidation state.

Ans: Fluorine, being the most electronegative element, exhibits only a -1 oxidation state.

(b) Identify the element that exhibits +ve oxidation state. 

Ans: Cesium is Alkali metals, due to having a single electron in their valence shell, consistently show an oxidation state of +1.

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

Ans: I. Because of the presence of seven electrons in the valence shell, I shows an oxidation state of 1 (in compounds of I with more electropositive elements such as H, Na, K, Ca, etc.) or an oxidation state of + 1 (in compounds of I with more electronegative elements, i.e., O, F, etc.) and because of the presence of d-orbitals it also exhibits +ve oxidation states of +3, +5 and + 7.

(d) Identify the element which exhibits -ve nor +ve nor does the positive oxidation state.

Ans: Neon (Ne): As an inert gas with high ionization energy and positive electron gain enthalpy, neon does not exhibit either negative or positive oxidation states.

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: Reduction half-reaction:

Cl₂​(aq) + 2e⁺ → 2Cl−(aq)

Oxidation half-reaction:

SO₂​(aq) + 2H₂​O(l) → SO₄²⁻​(aq) + 4H⁺(aq) + 2e⁻

Combine the half-reactions:

Adding the reduction and oxidation half-reactions gives:

Cl₂​(aq) + SO₂​(aq) + 2H₂​O(l) → 2Cl⁻(aq) + SO₄²⁻​(aq) + 4H⁺(aq)

Therefore, the equation Cl₂​ + SO₂​ + 2H_2​O → 2Cl⁻ + SO₄²⁻​+ 4H⁺ accurately represents the redox reaction between chlorine and sulphur dioxide in water.

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. 

Ans: The non – metal are: P₄, Cl₂ and S₈.

(iii) 

(b) Select three metals that can show disproportionation reaction. 

Ans: The metals that can show disproportionation reactions are: Cu²⁺, Ga⁺, In⁺, Mn³⁺, etc.

2Cu + (aq) → Cu²⁺ (aq) + Cu(s)

3Ga + (aq) → Ga³⁺ (aq) +2 ga(a) 

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 is:

Here, 68g of NH₃ will react with react with O₂ = 160g 

∴ 10g of NH₃ will react with 

However, the amount of O₂​ actually available is 20.0 g, which is less than the amount required for the reaction.

Therefore, O₂ is the limiting reagent and hence calculations must be based upon the amount of O₂ taken and not on the amount of NH₃ taken.

20 g of oxygen will produce No. 120/160 × 20 = 15g

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).

Ans: The possible reaction between Fe³⁺(aq) and I^– (aq) 

2Fe³⁺ (aq) + 21^– (aq) → 2Fe³⁺ (aq) + I₂ (s)

The given redox reaction can be separated into two half-reactions: one for oxidation and one for reduction. Using Table 8.1, we write the electrode potentials for each half-reaction to determine the overall cell potential. The oxidation half-reaction involves the loss of electrons, while the reduction half-reaction involves the gain of electrons.

Oxidation: 

Reduction: 

Overall reaction: 

E° for the overall reaction is positive. Thus, the reaction between Fe³⁺(aq) and I^– (aq) is feasible.

(b) Ag⁺(aq) and Cu(s)

Ans: The reaction Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s) can be split into two half-reactions. Using Table 8.1, we find the electrode potentials for can be split into two half-reactions. 

Since the E° for the overall reaction is positive, the reaction between Ag⁺(aq) and Cu(s) is feasible.

(c) Fe³⁺ (aq) and Cu(s) 

Ans: Suppose the reaction between Fe³⁺ (aq) and Cu(s)  occurs according to the following equation. Cu(s) + 2Fe³⁺ (aq) → 3Cu²⁺ (aq) + 2Fe²⁺ (aq) 

By referring to Table 8.1 for standard electrode potentials, you can write these half-reactions with their respective standard electrode potentials to analyze the overall reaction.

Since the E° for the overall reaction is positive, the reaction between Fe³⁺ (aq) and Cu(s)  is feasible.

(d) Ag(s) and Fe³⁺(aq) 

Ans: The reaction between Ag⁺(aq) and Cu(s) is Cu(s) + 2Ag + (aq) → Cu²⁺(aq) + 2Ag(s) This redox reaction can be divided into two half-reactions. Using Table 8.1, we write:

Since the E° for the overall reaction is negative, the reaction between Ag(s) and Fe³⁺(aq)  is not feasible.

(e) Br₂(aq) and Fe²⁺(aq)

Ans: For the reaction Br₂(aq) + 2Fe²⁺(aq) → 2Br⁻(aq) + 2Fe³⁺ split into half-reactions, the electrode potentials from Table 8.1 are used to determine the overall cell potential.

Here, E° for the overall reaction is positive. Hence, the reaction between Br₂(aq) and Fe²⁺(aq)  is feasible.

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

(i) An aqueous solution of AgNO₃ with silver electrodes.

Ans: For an aqueous solution of silver nitrate (AgNO₃​) with silver electrodes:

At the Cathode: Silver ions (Ag⁺) have a lower discharge potential compared to hydrogen ions (H⁺). Therefore, silver ions are preferentially reduced and deposited onto the electrode.

At the Anode: The silver anode dissolves, releasing silver ions into the solution:

Ag → Ag⁺ + e⁻

(ii) An aqueous solution AgNO₃ with platinum electrodes.

Ans: If electrolysis of an AgNO₃ solution is performed using platinum electrodes instead of silver electrodes, oxidation of water occurs at the anode. Platinum, being a noble metal, does not easily undergo oxidation. Consequently, oxygen gas (O₂​) is liberated at the anode according to the following reaction:

2H₂​O → O₂​ + 4H ⁺ + 4e⁻

Thus, during the electrolysis of an aqueous AgNO₃ solution with platinum electrodes, Ag⁺ ions from the solution are deposited on the cathode, while O₂ gas is evolved at the anode.

(iii) A dilute solution of H₂SO₄ with platinum electrodes. 

Ans: Thus, when electricity is passed through the dilute 4H₂​SO₄​ solution, H⁺(aq) ions move toward the cathode, while SO₄²⁻(aq) ions move toward the anode. In other words, at the cathode, either H⁺(aq) ions or H₂O molecules are reduced.

Their standard electrode potentials are:

2H⁺(aq) + 2e⁻ → H2​(g); E^∘ = 0.0 V

H₂​O(l) + 2e⁻ → H₂​(g) + 2OH⁻(aq); E^∘ = −0.83 V

Since the electron potential (i.e., reduction potential) of H⁺(aq) ions is higher than that of H₂​O therefore, at the cathode, it is H⁺(aq) ions (rather than H_2​O molecules) which are reduced to evolve H₂ gas.

Similarly, at the anode, either SO₄²⁻​(aq) ions or H₂​O molecules are oxidized. Given that the oxidation potential of SO₄²⁻ is expected to be much lower (as it involves the cleavage of many bonds compared to those in H₂​O), H₂​O molecules are preferentially oxidized to produce O₂​ gas at the anode.

From the above discussion, it follows tha during electrolysis often aqueous solution of H₂​SO₄, only the electrolysis of H₂​O occurs liberating the cathode and O. at the anode. H₂

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

Ans: For the electrolysis of an aqueous CuCl₂ solution with platinum electrodes:

At the Cathode: Copper(II) ions (Cu₂⁺) are reduced to copper metal (Cu) in preference to protons:

Cu₂⁺ +2e⁻ → Cu

At anode: Chloride ions will be oxidized in preference to hydroxide ions

2Cl⁻ → Cl₂​ + 2e⁻

Thus, during the electrolysis of an aqueous CuCl₂​ solution, copper metal is deposited at the cathode, and chlorine gas is evolved at the anode.

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: From the Table 8.1

A metal with a more negative electrode potential is a stronger reducing agent compared to a metal with a less negative or positive electrode potential. Therefore, magnesium (Mg) can displace all metals listed above from their aqueous solutions. Aluminum (Al) can displace all metals except magnesium from their salt solutions. Similarly, zinc (Zn) can displace all metals except magnesium and aluminum from their aqueous salt solutions, while iron (Fe) can only displace copper (Cu) from its salt solutions. Thus, the metals can be arranged in the order of their ability to displace each other from salt solutions as follows: Mg, Al, Zn, Fe, Cu.

29. Given the standard electrode potentials, K⁺/K = –2.93V, Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V, Mg²⁺/Mg = –2.37V, Cr³⁺/Cr = –0.74V arrange these metals in their increasing order of reducing power.   

Ans: To arrange the metals in increasing order of reducing power based on their standard electrode potentials, you need to consider the electrode potentials of their reduction reactions. A more negative electrode potential indicates a stronger reducing agent.

Since the electrode potentials increase in the order;

K⁺/K = –2.93 V

Ag⁺/Ag = 0.80 V

Hg₂²⁺/Hg = 0.79 V

Mg²⁺/Mg = –2.37 V

Cr³⁺/Cr = –0.74 V

To determine the reducing power, note that a metal with a more negative standard electrode potential will have a higher reducing power.

Thus, the order in increasing reducing power is: Ag < Hg < Cr < Mg < K.

30. Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn²⁺(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 given redox reaction is:

Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)

Since zinc is oxidized to Zn2⁺ ions and Ag⁺ is reduced to silver metal, oxidation occurs at the zinc electrode and reduction occurs at the silver electrode. Thus, the galvanic cell corresponding to the above redox reaction can be represented as:

Zn∣Zn²⁺(aq)∥Ag⁺(aq)∣Ag(s)

(i) Since oxidation occurs at the zinc electrode, electrons are released and accumulate at the zinc electrode. Consequently, the zinc electrode becomes negatively charged.

(ii) In the galvanic cell, ions in the electrolyte solutions carry the current. Electrons flow from the zinc electrode to the silver electrode through the external circuit, while the current flows from the silver electrode to the zinc electrode.

(iii) The reactions occurring at the two electrodes are:

Ag⁺(aq) + Ag(s).

Zn(s) Zn²⁺(aq) + 2e^–.