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

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

Introduction

Mastering redox balancing unlocks 10-15 marks in IIT JEE (Main + Advanced)—enough to push you from a 90 to a 100+ percentile. It’s also the key to real-world applications like batteries, corrosion prevention, and industrial electroplating. If you can balance redox reactions fast and error-free, you’ll save 5+ minutes per question in the exam.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Oxidation states – How to assign them to elements in compounds.
2. Half-reactions – The concept of oxidation (loss of electrons) and reduction (gain of electrons).
3. Basic balancing – Balancing atoms other than H and O in chemical equations.

If any of these are unclear, stop now and review them first.

KEY TERMS & FORMULAS

Key Terms

Formulas & Rules

1. Oxidation Number Rules (MEMORISE THIS)
2. Free elements: 0 (e.g., Na, O₂, Cl₂).
3. Monatomic ions: Charge of the ion (e.g., Na⁺ = +1, Cl⁻ = -1).
4. Oxygen: -2 (except in peroxides like H₂O₂ where it’s -1).
5. Hydrogen: +1 (except in metal hydrides like NaH where it’s -1).
6. Fluorine: -1 (always).
7. Sum of oxidation numbers in a neutral compound = 0.

8. Sum of oxidation numbers in a polyatomic ion = charge of the ion.

9. Change in Oxidation Number (ΔON)

10. ΔON = Final ON – Initial ON.
11. If ΔON is positive, the species is oxidized.

12. If ΔON is negative, the species is reduced.

13. Balancing Half-Reactions (Ion-Electron Method)

14. Acidic Medium:
    1. Balance atoms other than H and O.
    2. Balance O by adding H₂O.
    3. Balance H by adding H⁺.
    4. Balance charge by adding e⁻.
15. Basic Medium:
    1. Follow steps 1-4 for acidic medium.
    2. Add OH⁻ to both sides to neutralize H⁺.
    3. Combine H⁺ + OH⁻ → H₂O and simplify.

STEP-BY-STEP METHOD

Method 1: Oxidation Number Method

Use when: The reaction is in molecular form (not ionic).

Steps:

1. Assign oxidation numbers to all elements.
2. Identify the species oxidized and reduced by checking ΔON.
3. Calculate the total change in oxidation number for both oxidation and reduction.
4. Balance the changes by multiplying the species by the LCM of the changes.
5. Balance the remaining atoms (other than H and O).
6. Balance O by adding H₂O (if needed).
7. Balance H by adding H⁺ (if in acidic medium) or OH⁻ (if in basic medium).
8. Verify charge balance (if in ionic form).

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

Use when: The reaction is in ionic form or involves polyatomic ions.

Steps (Acidic Medium):

1. Split the reaction into two half-reactions (oxidation and reduction).
2. Balance atoms other than H and O in each half-reaction.
3. Balance O by adding H₂O to the side deficient in O.
4. Balance H by adding H⁺ to the side deficient in H.
5. Balance charge by adding e⁻ to the more positive side.
6. Multiply half-reactions to equalize e⁻ transfer.
7. Add the half-reactions and cancel common terms.
8. Verify atom and charge balance.

Steps (Basic Medium):

1. Follow steps 1-5 for acidic medium.
2. Add OH⁻ to both sides to neutralize H⁺ (H⁺ + OH⁻ → H₂O).
3. Cancel H₂O molecules if they appear on both sides.
4. Verify atom and charge balance.

WORKED EXAMPLES

Example 1 – Basic (Oxidation Number Method)

Problem: Balance the following reaction in acidic medium: KMnO₄ + HCl → KCl + MnCl₂ + Cl₂ + H₂O

Step-by-Step Solution:

1. Assign oxidation numbers:
2. KMnO₄: K = +1, Mn = +7, O = -2
3. HCl: H = +1, Cl = -1
4. KCl: K = +1, Cl = -1
5. MnCl₂: Mn = +2, Cl = -1
6. Cl₂: 0 (free element)

7. H₂O: H = +1, O = -2

8. Identify oxidized and reduced species:

9. Mn changes from +7 → +2 (reduction, ΔON = -5).

10. Cl changes from -1 → 0 (oxidation, ΔON = +1).

11. Calculate total change:

12. Reduction: 1 Mn gains 5 e⁻.

13. Oxidation: 1 Cl loses 1 e⁻ → 5 Cl lose 5 e⁻ (to balance e⁻ transfer).

14. Balance Mn and Cl:

15. KMnO₄ + 5 HCl → KCl + MnCl₂ + 5/2 Cl₂ + H₂O

16. Balance K and remaining Cl:

17. KMnO₄ + 5 HCl → KCl + MnCl₂ + 5/2 Cl₂ + H₂O

18. Balance Cl on the right: KMnO₄ + 8 HCl → KCl + MnCl₂ + 5/2 Cl₂ + 4 H₂O

19. Balance H and O:

20. H: 8 H on left → 4 H₂O on right (already balanced).

21. O: 4 O on left → 4 O in H₂O (balanced).

22. Final balanced equation: 2 KMnO₄ + 16 HCl → 2 KCl + 2 MnCl₂ + 5 Cl₂ + 8 H₂O

What we did and why: - We balanced Mn (reduction) and Cl (oxidation) first. - Multiplied Cl by 5 to match the 5 e⁻ gained by Mn. - Balanced K, Cl, H, O last to ensure all atoms are equal.

Example 2 – Medium (Ion-Electron Method, Acidic Medium)

Problem: Balance the following reaction in acidic medium: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺

Step-by-Step Solution:

1. Split into half-reactions:
2. Oxidation: Fe²⁺ → Fe³⁺

3. Reduction: Cr₂O₇²⁻ → Cr³⁺

4. Balance atoms (other than H and O):

5. Oxidation: Fe²⁺ → Fe³⁺ (already balanced).

6. Reduction: Cr₂O₇²⁻ → 2 Cr³⁺ (balance Cr).

7. Balance O by adding H₂O:

8. Reduction: Cr₂O₇²⁻ → 2 Cr³⁺ + 7 H₂O

9. Balance H by adding H⁺:

10. Reduction: Cr₂O₇²⁻ + 14 H⁺ → 2 Cr³⁺ + 7 H₂O

11. Balance charge by adding e⁻:

12. Oxidation: Fe²⁺ → Fe³⁺ + 1 e⁻

13. Reduction: Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O

14. Equalize e⁻ transfer:

15. Multiply oxidation by 6: 6 Fe²⁺ → 6 Fe³⁺ + 6 e⁻

16. Add half-reactions:

17. Cr₂O₇²⁻ + 14 H⁺ + 6 Fe²⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O

18. Verify balance:

19. Atoms: Cr (2), O (7), H (14), Fe (6) → Balanced.
20. Charge: Left = -2 + 14 + 12 = +24 | Right = 6 + 18 = +24 → Balanced.

What we did and why: - Split into oxidation (Fe²⁺ → Fe³⁺) and reduction (Cr₂O₇²⁻ → Cr³⁺). - Balanced Cr, O, H, charge in the reduction half. - Multiplied Fe²⁺ by 6 to match 6 e⁻ in reduction. - Combined and verified atom + charge balance.

Example 3 – Exam-Style (Disproportionation, Basic Medium)

Problem: Balance the following reaction in basic medium: Cl₂ + OH⁻ → Cl⁻ + ClO₃⁻ + H₂O

Step-by-Step Solution:

1. Assign oxidation numbers:
2. Cl₂: 0 (free element).
3. Cl⁻: -1.

4. ClO₃⁻: Cl = +5, O = -2.

5. Identify disproportionation:

6. Cl₂ → Cl⁻ (reduction, ΔON = -1).

7. Cl₂ → ClO₃⁻ (oxidation, ΔON = +5).

8. Split into half-reactions:

9. Reduction: Cl₂ → Cl⁻

10. Oxidation: Cl₂ → ClO₃⁻

11. Balance atoms (other than H and O):

12. Reduction: Cl₂ → 2 Cl⁻

13. Oxidation: Cl₂ → 2 ClO₃⁻

14. Balance O by adding H₂O:

15. Oxidation: Cl₂ + 6 H₂O → 2 ClO₃⁻

16. Balance H by adding H⁺:

17. Oxidation: Cl₂ + 6 H₂O → 2 ClO₃⁻ + 12 H⁺

18. Balance charge by adding e⁻:

19. Reduction: Cl₂ + 2 e⁻ → 2 Cl⁻

20. Oxidation: Cl₂ + 6 H₂O → 2 ClO₃⁻ + 12 H⁺ + 10 e⁻

21. Equalize e⁻ transfer:

22. Multiply reduction by 5: 5 Cl₂ + 10 e⁻ → 10 Cl⁻

23. Add half-reactions:

24. 6 Cl₂ + 6 H₂O → 10 Cl⁻ + 2 ClO₃⁻ + 12 H⁺

25. Convert to basic medium (add OH⁻):

    • Add 12 OH⁻ to both sides: 6 Cl₂ + 6 H₂O + 12 OH⁻ → 10 Cl⁻ + 2 ClO₃⁻ + 12 H₂O

26. Simplify H₂O:

    • 6 Cl₂ + 12 OH⁻ → 10 Cl⁻ + 2 ClO₃⁻ + 6 H₂O

27. Divide by 2 for simplest form:

    • 3 Cl₂ + 6 OH⁻ → 5 Cl⁻ + ClO₃⁻ + 3 H₂O

What we did and why: - Recognized disproportionation (Cl₂ is both oxidized and reduced). - Balanced Cl, O, H, charge in both half-reactions. - Converted to basic medium by adding OH⁻. - Simplified to the smallest whole-number coefficients.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your redox balancing cheat sheet in 60 seconds.

1. Oxidation Number Method:
2. Assign ON → Find ΔON → Balance e⁻ → Balance atoms → Add H₂O/H⁺/OH⁻.

3. Pro tip: If Cl₂ → Cl⁻ + ClO₃⁻, it’s disproportionation—split into two half-reactions!

4. Ion-Electron Method:

5. Split into oxidation (loses e⁻) and reduction (gains e⁻).
6. Balance atoms → O (H₂O) → H (H⁺/OH⁻) → charge (e⁻).
7. Acidic medium? Add H⁺. Basic medium? Add OH⁻ to neutralize H⁺.

8. Multiply half-reactions to cancel e⁻, then add.

9. Exam Traps:

10. Disproportionation? Same element in two products.
11. Hidden medium? OH⁻ = basic, H⁺ = acidic.
12. Fractions? Multiply the whole equation by 2.

Final check: Count atoms and charge on both sides. If they match, you’re golden. Now go crush that redox question!