Oxidation and Reduction Reactions

Introduction

Every time you charge your phone, start your car, or see iron rust, redox reactions are happening! Oxidation-reduction (redox) reactions are among the most important chemical processes in nature and technology. From the batteries powering electric vehicles to the metabolic processes keeping you alive, redox chemistry is everywhere.

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Interactive: Redox Reaction Visualization

Watch electrons transfer between atoms in a redox reaction:

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What is Oxidation and Reduction?

Classical vs Modern Definitions

Concept	Classical Definition	Modern Definition (Electronic)
Oxidation	Addition of oxygen / Removal of hydrogen	Loss of electrons
Reduction	Removal of oxygen / Addition of hydrogen	Gain of electrons

The OIL RIG Memory Trick

Example: Formation of NaCl

$$\text{2Na} + \text{Cl}_2 \rightarrow \text{2NaCl}$$

• Oxidation: $\text{Na} \rightarrow \text{Na}^+ + e^-$ (Na loses electron)
• Reduction: $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$ (Cl gains electrons)

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Oxidation Number (Oxidation State)

The oxidation number is the hypothetical charge an atom would have if all bonds were 100% ionic.

Rules for Assigning Oxidation Numbers

1. Free element: Oxidation state = 0

   • Examples: Na, O₂, P₄, S₈ all have O.S. = 0

2. Monoatomic ion: O.S. = charge on ion

   • Na⁺ → +1, Cl⁻ → -1, Fe³⁺ → +3

3. Oxygen: Usually -2

   • Exceptions:
     • Peroxides (H₂O₂, Na₂O₂): -1
     • Superoxides (KO₂): -1/2
     • OF₂: +2 (fluorine more electronegative)

4. Hydrogen: Usually +1

   • Exception: Metal hydrides (NaH, CaH₂): -1

5. Fluorine: Always -1 (most electronegative)

6. Other halogens (Cl, Br, I): Usually -1

   • Exception: When bonded to O or more electronegative halogen

7. Alkali metals (Group 1): Always +1

8. Alkaline earth metals (Group 2): Always +2

9. Aluminum: Always +3 in compounds

10. Sum of oxidation states:

    • Neutral molecule: = 0
    • Polyatomic ion: = charge on ion

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Calculating Oxidation States

Example 1: Sulfur in H₂SO₄

Let O.S. of S = x

$$2(+1) + x + 4(-2) = 0$$ $$2 + x - 8 = 0$$ $$\boxed{x = +6}$$

Example 2: Chromium in Cr₂O₇²⁻

Let O.S. of Cr = x

$$2x + 7(-2) = -2$$ $$2x - 14 = -2$$ $$\boxed{x = +6}$$

Example 3: Carbon in CH₃COOH

Important: Different carbon atoms can have different oxidation states!

• Methyl carbon (CH₃): O.S. = -3
• Carboxyl carbon (COOH): O.S. = +3
• Average O.S. of carbon: 0

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Types of Redox Reactions

1. Combination Reactions

Two or more reactants combine:

$$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$$

(C: 0 → +4, O: 0 → -2)

2. Decomposition Reactions

Single reactant breaks down:

$$2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$$

(O in H₂O₂: -1 → -2 and 0)

3. Displacement Reactions

One element displaces another:

$$\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$$

(Zn: 0 → +2, Cu: +2 → 0)

4. Disproportionation Reactions

Same element simultaneously oxidized and reduced:

$$3\text{Cl}_2 + 6\text{OH}^- \rightarrow 5\text{Cl}^- + \text{ClO}_3^- + 3\text{H}_2\text{O}$$

Chlorine in Cl₂: 0 → -1 (reduction) and +5 (oxidation)

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Balancing Redox Equations

Method 1: Oxidation Number Method

Steps:

1. Assign oxidation numbers to all atoms
2. Identify atoms undergoing oxidation/reduction
3. Calculate total increase and decrease in O.S.
4. Multiply to equalize total change
5. Balance other atoms and charges

Example: Balance in acidic medium:

$$\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$$

Step 1: Oxidation states

• Mn: +7 → +2 (reduction by 5)
• Fe: +2 → +3 (oxidation by 1)

Step 2: Equalize total change

• Need 5 Fe²⁺ for every 1 MnO₄⁻

Step 3: Balance:

$$\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$$

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

Steps:

1. Write oxidation and reduction half-reactions
2. Balance atoms except O and H
3. Balance O by adding H₂O
4. Balance H by adding H⁺ (acidic) or OH⁻ (basic)
5. Balance charge by adding electrons
6. Multiply half-reactions to equalize electrons
7. Add half-reactions and cancel common terms

Example: Balance in acidic medium:

$$\text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} \rightarrow \text{Cr}^{3+} + \text{Fe}^{3+}$$

Oxidation half:

$$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$

Reduction half:

$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$

Multiply oxidation by 6:

$$6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^-$$

Add:

$$\boxed{\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}}$$

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Balancing in Basic Medium

Additional step: After balancing in acidic medium, add OH⁻ to both sides to neutralize H⁺

Example: Balance:

$$\text{MnO}_4^- + \text{I}^- \rightarrow \text{MnO}_2 + \text{I}_2$$

(basic medium)

Step 1: Balance as if acidic:

$$\text{MnO}_4^- + 4\text{H}^+ + 3e^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$$ $$2\text{I}^- \rightarrow \text{I}_2 + 2e^-$$

Step 2: After equalizing and adding, neutralize H⁺:

$$2\text{MnO}_4^- + 8\text{H}^+ + 6\text{I}^- \rightarrow 2\text{MnO}_2 + 3\text{I}_2 + 4\text{H}_2\text{O}$$

Add 8 OH⁻ to both sides:

$$\boxed{2\text{MnO}_4^- + 6\text{I}^- + 4\text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + 3\text{I}_2 + 8\text{OH}^-}$$

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Oxidizing and Reducing Agents

Definitions

• Oxidizing agent (Oxidant): Substance that gets reduced (gains electrons, causes oxidation)
• Reducing agent (Reductant): Substance that gets oxidized (loses electrons, causes reduction)

Common Oxidizing Agents

Agent	Reaction	Strength
KMnO₄ (acidic)	MnO₄⁻ → Mn²⁺	Very strong
K₂Cr₂O₇ (acidic)	Cr₂O₇²⁻ → Cr³⁺	Strong
H₂O₂	H₂O₂ → H₂O	Moderate
Cl₂, Br₂	X₂ → 2X⁻	Strong
O₂	O₂ → O²⁻	Strong

Common Reducing Agents

Agent	Reaction	Strength
H₂	H₂ → 2H⁺	Moderate
Metals (Na, Zn)	M → M^n+	Strong
H₂S	H₂S → S	Moderate
SO₂	SO₂ → SO₄²⁻	Moderate
FeSO₄	Fe²⁺ → Fe³⁺	Moderate

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Practice Problems

Level 1: JEE Main

Q1. Determine the oxidation state of:

• (a) N in NH₄⁺
• (b) S in S₂O₃²⁻
• (c) Cr in CrO₅

Q2. Identify the oxidizing and reducing agents:

$$\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$$

Q3. Which of the following is a disproportionation reaction?

• (a) 2Cu⁺ → Cu²⁺ + Cu
• (b) 2H₂ + O₂ → 2H₂O
• (c) Zn + CuSO₄ → ZnSO₄ + Cu

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Level 2: JEE Main/Advanced

Q4. Balance the following in acidic medium:

$$\text{MnO}_4^- + \text{H}_2\text{O}_2 \rightarrow \text{Mn}^{2+} + \text{O}_2$$

Q5. Balance in basic medium:

$$\text{Cr(OH)}_3 + \text{H}_2\text{O}_2 \rightarrow \text{CrO}_4^{2-}$$

Q6. The average oxidation state of Fe in Fe₃O₄ is +8/3. Explain the actual structure.

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Level 3: JEE Advanced

Q7. When 0.1 mol of MnO₄⁻ oxidizes Fe²⁺ to Fe³⁺ in acidic medium, how many moles of Fe²⁺ are oxidized?

Q8. A mixture of FeO and Fe₂O₃ is treated with acidified KMnO₄. If 1 mole of mixture requires 0.5 moles of KMnO₄, find the mole ratio of FeO:Fe₂O₃.

Q9. In the reaction:

$$x\text{MnO}_4^- + y\text{C}_2\text{O}_4^{2-} + z\text{H}^+ \rightarrow x\text{Mn}^{2+} + 2y\text{CO}_2 + \frac{z}{2}\text{H}_2\text{O}$$

Find the values of x, y, and z.

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Solutions to Practice Problems

A1.

• (a) N = -3, (b) S = +2, (c) Cr = +6

A2. Zn is reducing agent (oxidized), Cu²⁺ is oxidizing agent (reduced)

A3. (a) Disproportionation - Cu⁺ goes to both +2 and 0

A7. 0.5 moles (each MnO₄⁻ accepts 5e⁻, each Fe²⁺ donates 1e⁻)

A9. x=2, y=5, z=16

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Memory Tricks for JEE

Oxidation State Quick Reference

Balancing Priority

1. Easy method: Use oxidation number method when changes are clear
2. Foolproof method: Use ion-electron method for complex reactions
3. Always verify: Check atoms AND charges balance!

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Related Topics

Within Electrochemistry

• Electrochemical Cells — Where redox reactions generate electricity
• Electrode Potentials — Predicting spontaneity of redox reactions
• Electrolysis — Forcing non-spontaneous redox reactions

Cross-Chapter Connections

• Chemical Kinetics — Rate of redox reactions
• Thermodynamics — Energy changes in redox (ΔG, ΔH)
• Equilibrium — Redox equilibria and Le Chatelier’s principle

Physics Connections

• Current Electricity — Electron flow in circuits
• Electromagnetism — Electromagnetic effects in electrochemistry

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