1. Oxidation and Reduction

Redox reactions involve the simultaneous occurrence of Reduction and Oxidation.

1.1 Classical Concept

Oxidation	Reduction
Addition of Oxygen	Removal of Oxygen
Removal of Hydrogen	Addition of Hydrogen
Addition of electronegative element	Removal of electronegative element

1.2 Electronic Concept (Modern)

Oxidation: Loss of electrons (De-electronation).
Example: Na → Na⁺ + e^-
Reduction: Gain of electrons (Electronation).
Example: Cl₂ + 2e^- → 2Cl^-

Oxidizing Agent (Oxidant): Accepts electrons (gets reduced).
Reducing Agent (Reductant): Donates electrons (gets oxidized).

[Image of oxidation reduction electron transfer chart]

2. Oxidation Number (O.N.)

The residual charge which an atom appears to have when all other atoms are removed from it as ions.

Rules for Calculation

Free State: O.N. of elements in elementary state (H₂, O₂, P₄, S₈) is Zero.
Alkali Metals (Grp 1): Always +1.
Alkaline Earth Metals (Grp 2): Always +2.
Fluorine: Always -1 (Most electronegative).
Hydrogen: Generally +1. (Except in metal hydrides like NaH where it is -1).
Oxygen: Generally -2.
- Peroxides (H₂O₂): -1
- Superoxides (KO₂): -1/2
- With Fluorine (OF₂): +2

Example: KMnO₄
Let Mn be x.
(+1) + x + 4(-2) = 0
1 + x - 8 = 0 → x = +7

3. Types of Redox Reactions

1. Combination Reaction:

Two species combine. (A + B → C)
Example: C(s) + O₂(g) → CO₂(g)

2. Decomposition Reaction:

One compound breaks down.
Example: 2H₂O(l) → 2H₂(g) + O₂(g)

3. Displacement Reaction:

An ion in a compound is replaced by an ion of another element.

Metal Displacement: Zn + CuSO₄ → ZnSO₄ + Cu
Non-metal Displacement: Zn + 2HCl → ZnCl₂ + H₂

4. Disproportionation Reaction:

A single element undergoes both Oxidation and Reduction simultaneously.

2H₂O₂ → 2H₂O + O₂
(O changes from -1 to -2 and 0)

4. Balancing Methods

4.1 Ion-Electron Method (Half-Reaction)

Steps for Acidic Medium:

Separate into Oxidation and Reduction half-reactions.
Balance atoms other than O and H.
Balance O by adding H₂O.
Balance H by adding H⁺.
Balance Charge by adding electrons (e^-).
Equalize electrons in both halves and add them.

Steps for Basic Medium:

Follow acidic steps up to adding H⁺. Then, add OH^- ions to BOTH sides equal to the number of H⁺ ions. (H⁺ + OH^- → H₂O).

5. Applications

1. Corrosion (Rusting):

Oxidation of metal surfaces by atmospheric oxygen and moisture.
Fe → Fe²⁺ + 2e^- (Anode)
O₂ + 4H⁺ + 4e^- → 2H₂O (Cathode)
Rust: Fe₂O₃.xH₂O

2. Rancidity:

Oxidation of fats and oils in food causing unpleasant smell and taste. Prevented by antioxidants (BHA, BHT) or nitrogen packing.

[Image of electrochemical cell schematic]

Redox Titrations

Self Indicator: KMnO₄ (Pink to colorless).
External Indicator: Starch in Iodometric titrations (Blue complex).

Numericals & HOTS

Q1. Structure Based O.N.

Calculate the oxidation number of Chromium in CrO₅ (Chromium Pentoxide).

Solution:
Using the formula method: x + 5(-2) = 0 → x = +10. (Incorrect, max is +6).
Structure Method (Butterfly Structure):
CrO₅ has one double-bonded oxygen (Oxide, -2) and four oxygens in peroxy linkage (-1).

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

Q2. Balancing Redox

Balance the following reaction in acidic medium:
MnO₄^- + Fe²⁺ → Mn²⁺ + Fe³⁺

Solution:
Reduction Half: MnO₄^- → Mn²⁺
Balance O: MnO₄^- → Mn²⁺ + 4H₂O
Balance H: MnO₄^- + 8H⁺ → Mn²⁺ + 4H₂O
Balance Charge: MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O ...(i)

Oxidation Half: Fe²⁺ → Fe³⁺ + e^-
Multiply by 5: 5Fe²⁺ → 5Fe³⁺ + 5e^- ...(ii)

Add (i) and (ii):
MnO₄^- + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Q3. Combined n-factor (HOTS)

Calculate the n-factor of FeC₂O₄ (Ferrous Oxalate) when it is oxidized by KMnO₄ to Fe³⁺ and CO₂.

Solution:
Here, both Cation (Fe²⁺) and Anion (C₂O₄^2-) are oxidized.

1. Fe²⁺ → Fe³⁺
Change in O.N. = +3 - (+2) = 1.

2. C₂O₄^2- → 2CO₂
Carbon O.N. in Oxalate = +3. In CO₂ = +4.
Change per Carbon = 1.
Change for 2 Carbons = 2 × 1 = 2.

Total n-factor = 1 (from Fe) + 2 (from C) = 3

Q4. Identify Reaction Type

Consider the reaction: P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂. Identify the type of reaction and the O.N. changes.

Solution:
P₄ (0) is the reactant.

Product 1: PH₃
x + 3(+1) = 0 → x = -3 (Reduction).

Product 2: NaH₂PO₂ (Sodium Hypophosphite)
(+1) + 2(+1) + x + 2(-2) = 0
1 + 2 + x - 4 = 0
x - 1 = 0 → x = +1 (Oxidation).

Since Phosphorus is oxidized (0 to +1) and reduced (0 to -3), it is a Disproportionation Reaction.

Q5. Quantitative Redox

How many moles of electrons are required to reduce 1 mole of Cr₂O₇^2- to Cr³⁺ in acidic medium?

Solution:
Reaction: Cr₂O₇^2- → 2Cr³⁺

O.N. of Cr in dichromate: 2x - 14 = -2 → 2x = 12 → x = +6.
O.N. of Product Cr = +3.

Change per Cr atom = 6 - 3 = 3.
Since there are 2 Cr atoms in the reactant:
Total change = 2 × 3 = 6.

Answer: 6 moles of electrons.

Q6. Molarity Calculation

20 mL of 0.1 M KMnO₄ is required to oxidize 20 mL of Ferrous Sulphate (FeSO₄) solution in acidic medium. Calculate the molarity of the FeSO₄ solution.

Solution:
Principle: Equivalents of Oxidant = Equivalents of Reductant.
N₁V₁ = N₂V₂
(M₁ × n₁) V₁ = (M₂ × n₂) V₂

For KMnO₄ (Acidic): n₁ = 5.
For FeSO₄ (Fe²⁺ → Fe³⁺): n₂ = 1.

(0.1 × 5) × 20 = (M₂ × 1) × 20
0.5 × 20 = 20 M₂
10 = 20 M₂
M₂ = 0.5 M

Q7. Br₃O₈ Structure

The average oxidation number of Bromine in Br₃O₈ is +16/3. What are the actual oxidation states of the three Br atoms based on its structure?

Solution:
Structure: O=Br(=O)-Br(=O)(=O)-Br(=O)=O
(Terminal Br - Central Br - Terminal Br)

1. Terminal Br (Left): Bonded to 3 Oxygens. O.N. = +6.
2. Central Br: Bonded to 2 Oxygens. O.N. = +4.
3. Terminal Br (Right): Bonded to 3 Oxygens. O.N. = +6.

States: +6, +4, +6

Q8. Representing Compounds

Write the Stock Notation for:
(a) AuCl₃
(b) MnO₂

Solution:
Stock notation uses Roman numerals in brackets to indicate the oxidation state of the metal.

(a) AuCl₃: x + 3(-1) = 0 → x = +3.
Notation: Aurum(III) chloride

(b) MnO₂: x + 2(-2) = 0 → x = +4.
Notation: Manganese(IV) oxide

Q9. Balancing Coefficients

Determine the values of x, y, z in the reaction:
x K₂Cr₂O₇ + y SO₂ + z H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Solution:
Reduction: Cr(+6) → Cr(+3). Change = 3 per atom. Total = 6.
Oxidation: S(+4 in SO₂) → S(+6 in sulphate). Change = 2.

Cross multiply factors: Cr (change 6) and S (change 2). Ratio 1:3.
So, 1 K₂Cr₂O₇ reacts with 3 SO₂.
Therefore, x=1, y=3.
Balance K: 1 K₂SO₄.
Balance SO₄: Total 4 needed on RHS (1 in K salt + 3 in Cr salt).
3 come from SO₂ (after oxidation), so 1 must come from H₂SO₄.
So, z=1.

x=1, y=3, z=1

Q10. Spontaneity Check

Can a solution of 1 M CuSO₄ be stored in a Nickel vessel? Given E°(Ni²⁺/Ni) = -0.25V and E°(Cu²⁺/Cu) = +0.34V.

Solution:
For reaction to occur (dissolution of vessel):
Ni + Cu²⁺ → Ni²⁺ + Cu

E°_cell = E°_cathode (Cu) - E°_anode (Ni)
E°_cell = 0.34 - (-0.25) = +0.59 V.

Since E°_cell is Positive, the reaction is spontaneous.
No, it cannot be stored because Nickel will dissolve and displace Copper.

Important Formulae & Concepts

1. Calculation of Oxidation Number (O.N.)

General Sum Rule:

Σ (O.N. of all atoms) = Charge on species

Range of O.N.:

Max O.N. = Group Number
Min O.N. = Group Number - 8

(Valid for p-block elements)

2. Equivalent Weight (E)

E = Molar Mass / n-factor

n-factor for Redox:

n = Total change in O.N. per molecule

For Disproportionation:

E = E_oxidation + E_reduction

3. Balancing Principle

Total Decrease in O.N. = Total Increase in O.N.

n₁ × M₁ × V₁ = n₂ × M₂ × V₂

(Equivalents of Oxidant = Equivalents of Reductant)

4. Electrode Potential (E°)

Standard Reduction Potential (SRP):

High SRP = Strong Oxidizing Agent (e.g., F₂)
Low SRP = Strong Reducing Agent (e.g., Li)

E°_cell = E°_cathode - E°_anode

20 Golden Facts (NEET)

1. Fluorine's Rule: Fluorine is the most electronegative element. It ALWAYS exhibits an oxidation state of -1 in all its compounds. It cannot have a positive oxidation state.
2. Disproportionation Limit: Elements in their highest or lowest oxidation states CANNOT undergo disproportionation. Example: Cl in ClO₄^- (+7) cannot oxidize further.
3. Fractional O.N.: Oxidation number can be fractional (e.g., in Fe₃O₄, Fe is +8/3). This is an average value; individual atoms have integer states (+2 and +3).
4. Paradox of Structure: In CrO₅ (Butterfly structure), Cr has O.N. of +6, not +10. This is because it has two peroxy linkages (-O-O-) where O is -1.
5. Bleaching Action: SO₂ bleaches by Reduction (Temporary). Cl₂ bleaches by Oxidation (Permanent).
6. KMnO₄ n-factors:
• Acidic Medium: n=5 (+7 → +2)
• Neutral/Faintly Alkaline: n=3 (+7 → +4)
• Strongly Alkaline: n=1 (+7 → +6)
7. K₂Cr₂O₇ n-factor: In acidic medium, Dichromate acts as an oxidant with n=6. (Cr change from +6 to +3 per atom, 2 atoms = 6).
8. Spectator Ions: Ions present in the solution that do not undergo any change in oxidation state during the reaction are called spectator ions.
9. Nitric Acid (HNO₃): It acts only as an Oxidizing Agent because N is in its max oxidation state (+5).
10. Nitrous Acid (HNO₂): N is in +3 state (intermediate). It can act as both an Oxidizing and a Reducing agent.
11. Metal Hydrides: In compounds like LiH or CaH₂, Hydrogen shows an oxidation state of -1 because metals are more electropositive than H.
12. Caro's Acid (H₂SO₅): Sulfur is in +6 state (not +8). It contains one peroxy linkage.
13. Marshell's Acid (H₂S₂O₈): Sulfur is in +6 state. It also contains a peroxy linkage.
14. Electrochemical Series: Li is the strongest reducing agent (Lowest SRP), and F₂ is the strongest oxidizing agent (Highest SRP).
15. Displacement Rule: A metal placed higher in the activity series (lower SRP) can displace a metal lower in the series (higher SRP) from its salt solution. (e.g., Zn displaces Cu).
16. Bleaching Powder: CaOCl₂. The two Chlorine atoms have different oxidation states: -1 (as Chloride Cl^-) and +1 (as Hypochlorite OCl^-).
17. Brown Ring Test: The complex formed [Fe(H₂O)₅NO]²⁺ has Iron in +1 oxidation state, which is rare.
18. H₂O₂ Role: Hydrogen Peroxide acts as both an Oxidizing and Reducing agent in both acidic and basic media.
19. Equivalent Wt of Oxidant: Mol. Mass / No. of electrons gained by one molecule.
20. Neutrality: Redox reactions in solution must obey the principle of electroneutrality. The total charge of reactants must equal the total charge of products.

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