This chapter is the foundation of electrochemistry—it explains how batteries work, why iron rusts, and how your body extracts energy from food through the “hot potato” game of electrons.

The Electron Tug-of-War: Mastering Redox Reactions

In chemistry, most reactions are actually “trading deals.” One atom gives up electrons, and another atom takes them. This is Redox (Reduction-Oxidation). If you understand where the electrons are moving, you can predict everything from the power of a lithium-ion battery to the corrosive nature of bleach.

The Core Pillars of Redox

1. Oxidation and Reduction (OIL RIG)

The easiest way to remember this is the acronym OIL RIG:

• Oxidation Is Loss (of electrons).
• Reduction Is Gain (of electrons).
• Oxidizing Agent: The “thief” that takes electrons (it gets reduced).
• Reducing Agent: The “donor” that gives away electrons (it gets oxidized).

2. Oxidation Number: The Chemical Ledger

Oxidation numbers are “bookkeeping” tools. They aren’t always real charges, but they tell us who is winning the electron tug-of-war.

• Free elements (O₂, Fe, H₂) always have an oxidation number of 0.
• Oxygen is usually -2 (except in peroxides where it is -1).
• Hydrogen is usually +1 (except in metal hydrides where it is -1).

3. Balancing by the Ion-Electron Method

Redox equations are too complex to balance by simple “hit and trial.” You must split them into two Half-Reactions:

1. Balance atoms (except O and H).
2. Balance O by adding H₂O.
3. Balance H by adding H⁺.
4. Balance charge by adding electrons (e⁻).

The Gauntlet: 10 Challenging Aptitude Questions

Question 1: The Disproportionation Trap

In the reaction 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O, identify which species is oxidized and which is reduced. What makes this a “disproportionation” reaction?

Question 2: The Oxidation State Puzzle

Calculate the oxidation number of Sulfur in the following species:

1. H₂SO₄
2. H₂S₂O₇ (Oleum)
3. Na₂S₄O₆ (Sodium tetrathionate)

Question 3: The Balancing Act (Acidic)

Balance the following skeleton equation in an acidic medium:

MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂

Question 4: The Paradox of CrO₅

Using standard rules, the oxidation number of Chromium in CrO₅ (Chromium pentoxide) appears to be +10. Why is this chemically impossible, and what is the actual oxidation state?

Question 5: Agent Identification

In the reaction CuO + H₂ → Cu + H₂O, identify the Oxidizing Agent and the Reducing Agent. Which substance undergoes reduction?

Question 6: The “n-factor” Calculation

What is the n-factor (change in oxidation state per molecule) of KMnO₄ when it acts as an oxidizing agent in:

1. Strongly Acidic Medium
2. Neutral/Weakly Alkaline Medium
3. Strongly Alkaline Medium

Question 7: Stock Notation

Represent the following compounds using Stock Notation (Roman numerals for oxidation states):

1. Tl₂O
2. FeO
3. MnO₂

Question 8: Balancing in Basic Medium

Balance the following reaction in a basic medium:

P₄ + OH⁻ → PH₃ + HPO₂⁻

Question 9: The Iodine Titration

In the reaction I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, which species acts as the reducing agent? This is a classic reaction used in “Iodometry.”

Question 10: Displacement Logic

Based on the activity series, will a reaction occur if you place a Copper strip into a solution of Silver Nitrate (AgNO₃)? Explain why or why not.

Detailed Explanations & Solutions

1. Disproportionation

Chlorine (Cl₂) starts at 0. In Cl⁻ it is -1 (Reduced). In ClO₃⁻ it is +5 (Oxidized).

Result: Chlorine is both oxidized and reduced. This is the definition of disproportionation.

2. Sulfur Oxidation States

1. H₂SO₄: +1(2) + x + -2(4) = 0 → x = +6.
2. H₂S₂O₇: +1(2) + 2x + -2(7) = 0 → 2x = 12 → x = +6.
3. Na₂S₄O₆: Due to the S-S bonds, two sulfurs are 0 and two are +5. Average = +2.5.

3. Balancing MnO₄⁻/C₂O₄²⁻

• Mn goes from +7 to +2 (Gains 5e⁻).
• C goes from +3 to +4 (Loses 1e⁻ per C, so 2e⁻ per C₂O₄²⁻).
• Cross-multiply: 2MnO₄⁻ + 5C₂O₄²⁻.Result: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.

4. The CrO₅ Exception

CrO₅ has a “butterfly” structure with four oxygen atoms in peroxide bonds (Ox. No = -1) and one oxygen in a double bond (Ox. No = -2).

Result: x + 4(-1) + 1(-2) = 0 → x = +6. (Maximum Ox. state for Cr).

5. Agent ID

Hydrogen (H₂) gains oxygen/loses electrons → It is the Reducing Agent.

Copper Oxide (CuO) loses oxygen/gains electrons → It is the Oxidizing Agent.

6. KMnO₄ n-factors

1. Acidic: Mn⁺⁷ → Mn²⁺ (n = 5).
2. Neutral: Mn⁺⁷ → MnO₂ (+4) (n = 3).
3. Strongly Basic: Mn⁺⁷ → MnO₄²⁻ (+6) (n = 1).

7. Stock Notation

1. Tl₂O: Thallium(I) oxide.
2. FeO: Iron(II) oxide.
3. MnO₂: Manganese(IV) oxide.

8. Basic Medium Balancing

P₄ acts as both oxidizing and reducing agent.

Result: P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻.

9. Iodine/Thiosulfate

Sulfur in S₂O₃²⁻ goes from +2 to +2.5 (Oxidized).

Result: Thiosulfate (S₂O₃²⁻) is the Reducing Agent.

10. Metal Displacement

Copper is more reactive than Silver (it is higher in the activity series). It will “push” the Silver out of the solution.

Result: Yes, reaction occurs. The solution turns blue (Cu²⁺) and silver crystals form.

Pro-Tip: The “Hydrogen Rule” for Redox

In organic chemistry, a simple shortcut is:

• Oxidation: Adding Oxygen or Removing Hydrogen.
• Reduction: Removing Oxygen or Adding Hydrogen.