Chemistry Physical - How to Solve: Redox Reactions (Balancing by Oxidation Number & Ion-Electron Method) – NEET UG Guide

How to Solve: Redox Reactions (Balancing by Oxidation Number & Ion-Electron Method) – NEET UG Guide

Introduction

"Master redox balancing, and you unlock 3-5 marks in NEET Chemistry—every year. Miss it, and you lose easy marks on reactions that power batteries, rusting, and even your own metabolism."

WHAT YOU NEED TO KNOW FIRST

1. Oxidation & Reduction – Loss of electrons (oxidation), gain of electrons (reduction).
2. Oxidation Numbers – Rules for assigning them (e.g., O = -2, H = +1 in most compounds).
3. Half-Reactions – Splitting a redox reaction into oxidation and reduction halves.

KEY TERMS & FORMULAS

1. Oxidation Number (O.N.) Rules

• Elemental form (e.g., O₂, Na): O.N. = 0
• Monatomic ion (e.g., Na⁺, Cl⁻): O.N. = charge
• Oxygen (O): Usually -2 (except in peroxides = -1, superoxides = -½)
• Hydrogen (H): +1 (except in metal hydrides = -1)
• Fluorine (F): Always -1
• Alkali metals (Group 1): +1
• Alkaline earth metals (Group 2): +2
• Sum of O.N. in a neutral compound = 0
• Sum of O.N. in a polyatomic ion = charge of the ion

MEMORISE THIS – These rules are not given in NEET.

2. Ion-Electron Method (Half-Reaction Method)

Steps (given in exam sheet, but memorise the logic):
1. Split into oxidation and reduction half-reactions.
2. Balance atoms other than O and H.
3. Balance O by adding H₂O.
4. Balance H by adding H⁺.
5. Balance charge by adding e⁻.
6. Multiply half-reactions to equalise e⁻.
7. Add half-reactions, cancel common terms.

STEP-BY-STEP METHOD

Method 1: Oxidation Number Method

Step 1: Assign oxidation numbers to all atoms. Step 2: Identify oxidised (O.N. ↑) and reduced (O.N. ↓) elements. Step 3: Calculate total increase in O.N. (oxidation) and total decrease in O.N. (reduction). Step 4: Balance the atoms of oxidised and reduced elements. Step 5: Multiply by LCM of O.N. changes to balance electrons. Step 6: Balance O and H by adding H₂O and H⁺ (acidic medium) or OH⁻ (basic medium). Step 7: Verify charge balance and atom balance.

Method 2: Ion-Electron Method (Half-Reaction Method)

Step 1: Write skeleton ionic equation. Step 2: Split into oxidation and reduction half-reactions. Step 3: Balance atoms other than O and H. Step 4: Balance O by adding H₂O. Step 5: Balance H by adding H⁺ (acidic) or OH⁻ (basic). Step 6: Balance charge by adding e⁻. Step 7: Multiply half-reactions to equalise e⁻. Step 8: Add half-reactions, cancel common terms. Step 9: Verify atom and charge balance.

WORKED EXAMPLES

Example 1 – Basic (Oxidation Number Method)

Problem: Balance: Fe²⁺ + MnO₄⁻ → Fe³⁺ + Mn²⁺ (acidic medium)

Step 1: Assign O.N. - Fe: +2 → +3 (oxidised, Δ = +1) - Mn: +7 → +2 (reduced, Δ = -5)

Step 2: Balance electrons (LCM of 1 and 5 = 5) - Multiply Fe by 5: 5Fe²⁺ → 5Fe³⁺ (+5e⁻) - MnO₄⁻ → Mn²⁺ (-5e⁻)

Step 3: Balance O by adding H₂O - MnO₄⁻ → Mn²⁺ + 4H₂O

Step 4: Balance H by adding H⁺ - MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Step 5: Combine half-reactions - 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

What we did and why: - Balanced electrons first (key step in O.N. method). - Added H₂O and H⁺ to balance O and H. - Verified charge balance (LHS: +2 + (-1) + 8 = +9; RHS: +15 + 2 = +17? No! Wait—correction: LHS = 5(+2) + (-1) + 8(+1) = +17, RHS = 5(+3) + 2 = +17. Correct.)

Example 2 – Medium (Ion-Electron Method)

Problem: Balance: Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻ (acidic medium)

Step 1: Split into half-reactions - Reduction: Cr₂O₇²⁻ → Cr³⁺ - Oxidation: SO₃²⁻ → SO₄²⁻

Step 2: Balance Cr - Cr₂O₇²⁻ → 2Cr³⁺

Step 3: Balance O by adding H₂O - Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O - SO₃²⁻ + H₂O → SO₄²⁻

Step 4: Balance H by adding H⁺ - Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O - SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺

Step 5: Balance charge by adding e⁻ - Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O - SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻

Step 6: Equalise e⁻ (LCM of 6 and 2 = 6) - Multiply oxidation by 3: 3SO₃²⁻ + 3H₂O → 3SO₄²⁻ + 6H⁺ + 6e⁻

Step 7: Add half-reactions - Cr₂O₇²⁻ + 14H⁺ + 3SO₃²⁻ + 3H₂O → 2Cr³⁺ + 7H₂O + 3SO₄²⁻ + 6H⁺

Step 8: Cancel common terms - Cr₂O₇²⁻ + 8H⁺ + 3SO₃²⁻ → 2Cr³⁺ + 4H₂O + 3SO₄²⁻

What we did and why: - Balanced Cr first (2 atoms → 2Cr³⁺). - Added H₂O and H⁺ to balance O and H. - Equalised electrons before combining. - Cancelled H⁺ and H₂O to simplify.

Example 3 – Exam-Style (Disguised Problem)

Problem: A student is given: "Permanganate ion oxidises oxalate ion to CO₂ in acidic medium. Write the balanced equation."

Step 1: Write skeleton equation - MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂

Step 2: Assign O.N. - Mn: +7 → +2 (reduced, Δ = -5) - C: +3 → +4 (oxidised, Δ = +1 per C, +2 total)

Step 3: Balance electrons (LCM of 5 and 2 = 10) - Multiply MnO₄⁻ by 2: 2MnO₄⁻ → 2Mn²⁺ (-10e⁻) - Multiply C₂O₄²⁻ by 5: 5C₂O₄²⁻ → 10CO₂ (+10e⁻)

Step 4: Balance O by adding H₂O - 2MnO₄⁻ → 2Mn²⁺ + 8H₂O - 5C₂O₄²⁻ → 10CO₂

Step 5: Balance H by adding H⁺ - 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O

Step 6: Combine - 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂

What we did and why: - Recognised oxalate (C₂O₄²⁻) → CO₂ is oxidation. - Balanced electrons first (key for O.N. method). - Added H₂O and H⁺ to balance O and H. - Verified charge balance (LHS: -2 + 16 + (-10) = +4; RHS: +4 + 0 = +4. Correct.)

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second redox rescue plan for NEET:

1. Oxidation Number Method:
2. Assign O.N. → Find oxidised/reduced → Balance electrons (LCM) → Balance O/H → Verify charge.

3. Pro tip: If O.N. changes by 1 and 5, multiply by 5 and 1.

4. Ion-Electron Method:

5. Split into half-reactions → Balance atoms → Add H₂O → Add H⁺/OH⁻ → Balance charge → Equalise e⁻ → Combine.

6. Pro tip: In basic medium, add OH⁻ to both sides to neutralise H⁺.

7. Watch out for:

8. Peroxides (O = -1), metal hydrides (H = -1).
9. Basic medium? Add OH⁻ at the end.
10. Always check charge balance—examiners love this trap!

You’ve got this. Now go balance those reactions like a pro!