Ncert Solutions Chemistry Class 11th

(a) Oxidation number method:
The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence,  P₄?  is oxidizing as well as reducing agent.
During reduction, the total decrease in the oxidation number for 4 P atoms is 12.
During oxidation, total increase in the oxidation number for 4 P atoms is 4.

The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying H_2? PO₂⁻? with 3. 

P₄? (s) + OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance O atoms, multiply OH⁻ ions by 6.

P₄? (s) + 6OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.

P₄? (s) + 6OH⁻ (aq) + 3H₂? O (l) → PH₃? (g) + 3H₂? PO₂⁻? (aq) + 3OH⁻ (aq)
Subtract 3 hydroxide ions from both sides.

P_4? (s) + 3OH⁻ (aq) + 3H₂? O (l) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

Ion electron method:
The oxidation half reaction is 

P₄? (s) → H₂? PO₂⁻? (aq).
The P atom is balanced.

P₄? (s) → 4H₂? PO₂⁻? (aq)

The oxidation number is balanced by adding 4 electrons on RHS.

P₄? (s) → 4H₂? PO₂⁻? (aq) + 4e⁻

The charge is balanced by adding 8 hydroxide ions on LHS.

P₄? (s)+8OH⁻ (aq)→4H₂? PO₂⁻? (aq)
The O and H atoms are balanced.

The reduction half reaction is P₄? (s)→PH₃? (g).
The oxidation number is balanced by adding 12 electrons on LHS.

P_4? (s)+12e⁻→PH₃? (g)
The charge is balanced by adding 12 hydroxide ions on RHS.

P₄? (s)+12e⁻→PH_3? (g)+12OH⁻
The oxidation half reaction is multiplied by 3 and the reduction half reaction is multiplied by 2.
The half reactions are then added to obtain balanced chemical equation.

P_4? (s)+3OH⁻ (aq)+3H₂? O (l)→PH₃? (g)+3H₂? PO₂⁻? (aq)

(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 CIO⁻₃ to – 1 in Cl^–. Hence, in this reaction,  N₂H₄ is the reducing agent and CIO⁻₃ is the oxidizing agent.
Ion - electron method:
The oxidation half equation is:

−2               +2

N₂H₄ (l)? NO (g)
The N atoms are balanced as:

N₂H₄ (l)?2NO (g)+8e⁻
The charge is balanced by adding 8 OH^– ions as:

N₂H₄ (I)+8OH⁻ (aq)?2NO (g)+8e⁻
The O atoms are balanced by adding 6H2O as:

N₂H₄ (I)+8OH⁻ (aq)?2NO (g)+6H₂O (I)+8e⁻. (i)
The reduction half equation is:

+5            

(a) The balanced half reaction equations are:
Oxidation half equation:

I⁻ (aq) → I₂ (s)       - (i)

Reduction half reaction equation:

MnO₄⁻ (aq) → MnO₂ (aq) - (ii)

Balance I atoms and charges in the oxidation half reaction.

2I⁻ (aq) → I₂ (s) + 2e⁻

In the reduction half reaction, the oxidation number of Mn changes from +7 to +4. Hence, add 3 electrons to reactant side of the reaction.

MnO₄⁻ (aq) + 3e⁻→ MnO₂ (aq)

Balance charge in the reduction half reaction by adding 4 hydroxide ions to product side.

MnO4⁻ (aq) + 3e⁻ → MnO₂ (aq)+4OH⁻

To balance O atoms, add 2 water molecules to reactant side.

MnO4⁻ (aq)+ 3e⁻ +2H₂O→MnO₂ (aq) + 4OH⁻

To equalize the number of electrons, multiply the oxidation half reaction by 3 and multiply the reduction half reaction by 2.

6I⁻ (aq) → 3I₂ (s)+6e⁻

2MnO₄⁻ (aq) + 6e⁻+4H₂O → 2MnO₂ (aq)+8OH⁻

Add two half-cell reactions to obtain the balanced equation.

2MnO₄⁻? (aq) + 6I⁻ (aq) + 4H₂? O₂? (l) → 2MnO₂? (s) + 3I₂? (s) + 8OH⁻

(b) The balanced half reaction equations are:
Oxidation half equation:

SO₂ (g) + 2H₂O (l) → HS0₄^–  (aq) + 3H⁺ (aq) +2e^–                        … (i)
Reduction half equation:

MnO₄^– (aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (l) ………. (ii)
Multiply Eq. (i) by 3 and Eq. (ii) by 2 and add, we have,

2MnO₄^– (aq) + 5S0₂ (g) + 2H₂0 (l) + H⁺ (aq) → 2Mn²⁺ (aq) + 5HSO₄^– (aq)

(c) Oxidation half equation: Fe²⁺ (aq) → Fe³⁺ (aq) + e^– … (i)
Reduction half equation: H₂O₂ (aq) + 2H⁺ (aq) + 2e^– → 2H₂O (l) … (ii)
Multiply Eq. (i) by 2 and add it to Eq. (ii), we have,

H₂O₂ (aq) +2Fe²⁺ (aq) +2H⁺ (aq) → 2Fe³⁺ (aq) + 2H₂O (l)

(d)Oxidation half equation:

SO₂ (g) + 2H₂O (l) → SO₄^2- (aq) + 4H⁺ (aq) + 2 e^–  … (i)
Reduction half equation:

Cr₂O₇^2– (aq) + 14H⁺ (aq) + 6e^– → 2Cr³⁺ (aq) + 7H₂O (l) … (ii)
Multiply Eq. (i) by 3 and add it to Eq. (ii), we have,

Cr₂O₇^2– (aq) + 3SO₂ (q) + 2H⁺ (aq) → 2Cr³⁺ (aq) + 3SO₄^2- (aq) + H₂O (l)

Fluorine oxidizes chloride ion to chlorine, bromide ion to bromine and iodide ion to iodine respectively.
F₂? + 2Cl⁻ → 2F⁻ + Cl₂?
F₂? + 2Br⁻ → 2F⁻ + Br₂?
F_2? + 2I⁻ → 2F⁻ + I₂?

Chlorine oxidizes bromide ion to bromine and iodide ion to iodine.
Cl_2? + Br⁻ → 2Cl⁻ + Br₂?
Cl₂? + I⁻ → 2Cl⁻ + I₂?

Bromine oxidizes iodide ion to iodine.
Br₂? + I⁻ → 2Br⁻ + I₂?

But bromine and chlorine cannot oxidize fluoride to fluorine. Hence, fluorine is the best oxidizing agent amongst the halogens. The decreasing order of the oxidizing power of halogens is F₂? >Cl₂? >Br₂? >I₂?  

HI and HBr can reduce sulphuric acid to sulphur dioxide but HCl and HF cannot. Thus,  HI and HBr are stronger reducing agents than HCl and HF.
2HI+H₂? SO₄? →I₂? +SO₂? +2H₂? O
2HBr+H₂? SO₄? →Br₂? +SO₂? +2H₂? O

Iodide ion can reduce Cu (II) to Cu (I) but bromide cannot.
4I⁻+2Cu²⁺→Cu₂? I₂? +I₂?

Hence, among the hydrohalic compounds, hydroiodic acid is the best reductant. The reducing power of hydrohalic acids is HF