Redox Reactions

What this chapter is about

This chapter introduces the concept of redox reactions — chemical processes involving the transfer of electrons between substances. You will learn to identify oxidation and reduction happening simultaneously, assign oxidation numbers to atoms in compounds, and use these skills to balance complex redox equations. Understanding redox is essential because these reactions power batteries, cause rusting, enable photosynthesis, and drive cellular respiration in living organisms.

At the Class 11 level, you build on your earlier understanding of chemical reactions and move towards a more systematic, electron-based view of chemistry. You will see how the classical definitions (involving oxygen and hydrogen) connect with the modern electronic definition, and why oxidation and reduction always occur together. By the end, you should be able to identify the oxidising agent and reducing agent in any reaction, calculate oxidation numbers confidently, and balance redox equations using either the oxidation number method or the half-reaction (ion-electron) method.

This knowledge forms the foundation for electrochemistry in Class 12, where you will study galvanic cells, electrolysis, and the quantitative relationship between electrical energy and chemical change.

Key ideas

• Oxidation is the loss of electrons; reduction is the gain of electrons. Both processes always occur together in a redox reaction.
• The oxidation number (or oxidation state) of an atom is a hypothetical charge it would have if all bonds were completely ionic; it helps track electron transfer.
• An oxidising agent (oxidant) accepts electrons and is itself reduced; a reducing agent (reductant) donates electrons and is itself oxidised.
• Standard rules fix oxidation numbers: free elements have 0; monoatomic ions equal their charge; oxygen is usually −2 (except in peroxides, −1); hydrogen is usually +1 (except in metal hydrides, −1); halogens are typically −1 when combined with metals.
• The algebraic sum of oxidation numbers in a neutral molecule is zero; in a polyatomic ion, it equals the ion's charge.
• Disproportionation occurs when one substance is simultaneously oxidised and reduced, producing two different products.
• Redox equations can be balanced by the oxidation-number method (equalise increase and decrease in oxidation numbers) or the half-reaction method (write separate oxidation and reduction half-reactions, balance atoms and charges, then add).

Formulas and facts to remember

1. Oxidation number of an element in its standard state is 0. Examples: O₂, N₂, Fe, S₈ all have oxidation number zero.
2. Sum rule for molecules: Σ(oxidation numbers of all atoms) = 0.
3. Sum rule for ions: Σ(oxidation numbers) = charge on the ion.
4. Common oxidation states: Alkali metals +1; alkaline earth metals +2; aluminium +3; fluorine always −1.
5. Half-reaction balancing steps (acidic medium): Balance atoms other than O and H → balance O by adding H₂O → balance H by adding H⁺ → balance charge by adding electrons → equalise electrons in both half-reactions → add.
6. For basic medium: After balancing as in acidic medium, add OH⁻ to both sides to neutralise H⁺, forming H₂O.
7. Disproportionation indicator: Same element shows both an increase and a decrease in oxidation number in a single reaction.
8. Electrochemical series link: A substance higher in the series (more negative standard potential) is a stronger reducing agent.

Worked examples

Example 1: Assigning oxidation numbers

Problem: Find the oxidation number of manganese in KMnO₄.

Solution:

Let the oxidation number of Mn be x.

Potassium (K) is an alkali metal, so its oxidation number = +1.

Oxygen (O) has oxidation number = −2 (standard rule).

For the neutral molecule: (+1) + x + 4(−2) = 0

+1 + x − 8 = 0

x = +7

Answer: Mn has an oxidation number of +7 in KMnO₄.

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Example 2: Identifying oxidising and reducing agents

Problem: In the reaction Zn + CuSO₄ → ZnSO₄ + Cu, identify the species oxidised, species reduced, oxidising agent, and reducing agent.

Solution:

Write oxidation numbers:

• Zn (metal) changes from 0 to +2 in ZnSO₄. It loses 2 electrons → oxidation.
• Cu in CuSO₄ is +2; in Cu metal it is 0. It gains 2 electrons → reduction.

Species oxidised: Zn (loses electrons).

Species reduced: Cu²⁺ (gains electrons).

Reducing agent: Zn (donates electrons, causes reduction of Cu²⁺).

Oxidising agent: Cu²⁺ in CuSO₄ (accepts electrons, causes oxidation of Zn).

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Example 3: Balancing a redox equation by the half-reaction method (acidic medium)

Problem: Balance the reaction: Fe²⁺ + Cr₂O₇²⁻ → Fe³⁺ + Cr³⁺ (in acidic solution).

Solution:

Step 1 – Write unbalanced half-reactions:

Oxidation: Fe²⁺ → Fe³⁺

Reduction: Cr₂O₇²⁻ → Cr³⁺

Step 2 – Balance atoms other than O and H:

Cr₂O₇²⁻ → 2 Cr³⁺

Step 3 – Balance oxygen by adding H₂O:

Cr₂O₇²⁻ → 2 Cr³⁺ + 7 H₂O

Step 4 – Balance hydrogen by adding H⁺:

14 H⁺ + Cr₂O₇²⁻ → 2 Cr³⁺ + 7 H₂O

Step 5 – Balance charge by adding electrons:

Left side charge: +14 + (−2) = +12. Right side charge: 2(+3) = +6.

Add 6 e⁻ to left: 6 e⁻ + 14 H⁺ + Cr₂O₇²⁻ → 2 Cr³⁺ + 7 H₂O

For oxidation half-reaction: Fe²⁺ → Fe³⁺ + e⁻

Step 6 – Equalise electrons:

Multiply oxidation half-reaction by 6: 6 Fe²⁺ → 6 Fe³⁺ + 6 e⁻

Step 7 – Add half-reactions:

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

Balanced equation verified: atoms and charges tally on both sides.

Common mistakes

• Forgetting that oxidation and reduction always happen together → remember redox is a coupled process; one cannot occur alone.
• Assigning oxygen an oxidation number of −2 in peroxides like H₂O₂ → in peroxides, oxygen is −1; apply the sum rule to check.
• Confusing the oxidising agent with the species oxidised → the oxidising agent itself gets reduced; it causes oxidation in another substance.
• Failing to balance charge in half-reactions → after balancing atoms, always count total charge on each side and add electrons to equalise.
• Ignoring the medium (acidic or basic) when balancing → use H⁺ and H₂O in acidic medium; add OH⁻ afterwards for basic medium.

Quick revision

• Oxidation = loss of electrons; reduction = gain of electrons (remember: OIL RIG — Oxidation Is Loss, Reduction Is Gain).
• Sum of oxidation numbers in a neutral compound is zero; in an ion, it equals the ion's charge.
• The substance that donates electrons is the reducing agent; the one that accepts electrons is the oxidising agent.
• Balance redox equations by equalising electron loss and gain, then check atoms and charge.
• Disproportionation: one element is both oxidised and reduced in the same reaction.