Redox Reactions

Let x be the O.N. of C.
O.N. of C in cyanogen, (CN)₂ = 2 (x – 3) = 0 or x = +3

O.N. of C in cyanide ion, CN^- = x – 3 = -1 or x = +2

O.N. of C in cyanate ion, CNO =x-3-2 = -l or x: = +4

The four information about the reaction are:
(i) The reaction involves decomposition of cyanogen, (CN)₂ in the alkaline medium to cyanide ion, CN and cyanate ion, CNO^–.

(ii) The O.N. of C decreases from +3 in (CN)₂ to +2 in CN^–ion and increases from +3 in (CN)₂ to +4 in CNO^– ion. Thus, cyanogen is simultaneously reduced to cyanide ion and oxidised to cyanate ion.

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

(iv) Cyanogen is a pseudohalogen (behaves like halogens) while cyanide ion is a pseudohalide ion (behaves like halide ion).

(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

(a) Toluene can be oxidised to benzoic acid in acidic, basic and neutral media according to the following redox equations:

In the laboratory, benzoic acid is usually prepared by alcoholic KMnO₄ oxidation of toluene. However, in industry alcoholic KMnO₄ is preferred over acidic or alkaline KMnO₄ because of the following reasons:
(i) The cost of adding an acid or the base is avoided because in the neutral medium, the base (OH^- ions) are produced in the reaction itself.
(ii) Since reactions occur faster in homogeneous medium than in heterogeneous medium, therefore, alcohol helps in mixing the two reactants, i.e., KMnO₄  (due to its polar nature) and toluene (because of its being an organic compound).

(b) KBr + H₂? SO₄? → KHSO₄? + HBr
HBr is a strong reducing agent which further reduces H₂? SO₄?  as follows
2HBr + H₂? SO₄? → Br₂? + SO₂? + H₂? O

But in case of mixture containing Chloride the reaction ends up on forming HCl as HCl is a weak reducing agent which can't further reducethe sulphuric acid further.
KCl+H₂? SO₄? →KHSO₄? +HCl? (colourless pungent smell gas)

The three examples are:

(i) When excess P_4?   (reducing agent) reacts with F_2? (oxidizing agent),  PF_3?  is produced in which P has +3 oxidation number.
                        P_4? (excess) + F₂? → PF₃?
But if fluorine is in excess,  PF₅?  is formed in which P has oxidation number of +5.
                        P_4? ? + F₂? (excess) → PF₅?

(ii) Oxidizing agent is oxygen and reducing agent is K. When excess K reacts with oxygen,  K_2? O is formed in which oxygen has oxidation number of -2.
                        4K (excess) + O₂? → 2K₂? O

But if oxygen is in excess, then K_2? O₂?  is formed in which O has oxidation number of -1.
                        2K + O₂? (excess) → K₂? O₂?

(iii) The oxidizing agent is oxygen and the reducing agent is C. When an excess of C reacts with oxygen, CO is formed in which C has +2 oxidation number.
                        C (excess) + O_2? → CO

When excess of oxygen is used,  CO₂?  is formed in which C has +4 oxidation number.
                        C + O₂? (excess) → CO₂