Explain redox reactions on the basis of electron transfer. Give suitable examples.

NCERT Exemplar · Q33

Q. Explain redox reactions on the basis of electron transfer. Give suitable examples.

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Concept understanding — Oxidation Reduction

Let’s start with something you already know from everyday life.

The intuition: what does “oxidation” really mean?

Think of a piece of iron left out in the rain. Over time, it turns into reddish-brown rust. Or think of a slice of apple turning brown when you leave it on the table. Or a fire burning wood to ash. In all these cases, something is combining with oxygen — that’s the original meaning of “oxidation.” The iron combines with oxygen from the air to form iron oxide (rust). The apple’s chemicals react with oxygen in the air. The wood burns because carbon in the wood combines with oxygen.

So the first, simplest idea: oxidation = adding oxygen. And the reverse — taking oxygen away — was called reduction. For example, if you heat iron oxide with carbon, the carbon steals the oxygen away, leaving pure iron. That’s reduction: removing oxygen.

But chemists soon realised this was too narrow. Many reactions that look like oxidation-reduction don’t involve oxygen at all. For instance, when sodium metal reacts with chlorine gas to make table salt, no oxygen is involved — yet the sodium clearly “rusts” in a sense, and the chlorine “steals” something from it.

So the definition had to be broadened.

The precise modern definition: electron transfer

Here’s the clean, exam-ready statement:

Oxidation is the loss of electrons by a substance.

Reduction is the gain of electrons by a substance.

They always happen together — you cannot have one without the other. That’s why we call them redox reactions (short for reduction-oxidation).

Let’s see this with the sodium-chlorine example:

Sodium atom ( $\text{Na}$ ) loses one electron to become $\text{Na}^+$ . That’s oxidation.
Chlorine atom ( $\text{Cl}$ ) gains that electron to become $\text{Cl}^-$ . That’s reduction.

You can write the two halves separately:

\text{Na} \rightarrow \text{Na}^+ + e^- \quad (\text{oxidation})

\text{Cl} + e^- \rightarrow \text{Cl}^- \quad (\text{reduction})

Add them together:

\text{Na} + \text{Cl} \rightarrow \text{Na}^+ + \text{Cl}^-

That’s table salt.

Tip

A handy mnemonic: OIL RIG — Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

How to spot a redox reaction without seeing electrons

You can’t watch electrons move directly. So chemists use oxidation numbers (also called oxidation states) — a bookkeeping system that tracks electrons.

Rules (simplified for first-time learners):

An atom in its elemental form has oxidation number 0.
A monatomic ion has oxidation number equal to its charge (e.g., $\text{Na}^+$ is +1, $\text{Cl}^-$ is -1).
Oxygen is usually -2 (except in peroxides).
Hydrogen is usually +1 (except in metal hydrides).
The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the ion’s charge.

Then:

Oxidation = increase in oxidation number.
Reduction = decrease in oxidation number.

Example: Rusting of iron.

4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3

Fe starts at 0 (elemental). In $\text{Fe}_2\text{O}_3$ , each Fe is +3. So Fe’s oxidation number goes up from 0 to +3 → oxidation.
O starts at 0 (in $\text{O}_2$ ). In $\text{Fe}_2\text{O}_3$ , each O is -2. So O’s oxidation number goes down from 0 to -2 → reduction.

Watch out

A common mistake: thinking that “reduction” means something becomes smaller or less. It doesn’t — it’s about gaining electrons (or losing oxygen, in the old sense). The name comes from metallurgy: when you “reduce” iron ore to iron, you’re taking away oxygen, so the mass reduces.

One more way to think about it …

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Redox reactions involve the simultaneous transfer of electrons from one species (which gets oxidised) to another (which gets reduced). Oxidation is electron loss; reduction is electron gain.

The electron-transfer picture of redox

At their heart, redox reactions are about electrons changing ownership. When a substance loses electrons, we say it is oxidised; when it gains electrons, it is reduced. These two processes always occur together—electrons don't vanish into thin air, so whatever one species loses, another must gain.

The mnemonic OIL RIG captures this: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

Every redox reaction can be split conceptually into two half-reactions: one showing oxidation, the other showing reduction. By tracking electron movement in each half, we see the complete electron-transfer story.

Step-by-step: identifying redox through electron transfer

Assign oxidation states to each atom in the reactants and products. The change in oxidation number signals electron transfer: an increase means oxidation (electron loss), a decrease means reduction (electron gain).
Write the oxidation half-reaction. Identify the species that loses electrons. Show the electrons as products on the right-hand side.
Write the reduction half-reaction. Identify the species that gains electrons. Show the electrons as reactants on the left-hand side.
Balance electrons between the two half-reactions so that the number lost equals the number gained, then add them to recover the overall balanced equation.
Identify the oxidising and reducing agents. The substance that gets reduced (gains electrons) is the oxidising agent; the substance that gets oxidised (loses electrons) is the reducing agent.

Example 1: Reaction of zinc with copper(II) sulphate

Consider the reaction

\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}.

Oxidation states:

Zinc starts at $0$ and ends at $+2$ in $\ce{ZnSO4}$ ; copper starts at $+2$ in $\ce{CuSO4}$ and ends at $0$ .

Oxidation half-reaction (zinc loses two electrons):

\ce{Zn -> Zn^{2+} + 2e^-}

Reduction half-reaction (copper(II) gains two electrons):

\ce{Cu^{2+} + 2e^- -> Cu}

Adding these half-reactions cancels the electrons and gives the net ionic equation

\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}.

Here zinc is oxidised (it is the reducing agent) and copper(II) is reduced (it is the oxidising agent). The two electrons transferred from zinc to copper(II) drive the entire reaction.

Example 2: Combustion of magnesium in oxygen

\ce{2Mg(s) + O2(g) -> 2MgO(s)}

Oxidation states:

Magnesium goes from $0$ to $+2$ ; oxygen goes from $0$ to $-2$ .

Oxidation half-reaction (each magnesium atom loses two electrons):

\ce{Mg -> Mg^{2+} + 2e^-}

Multiply by 2 to match the stoichiometry:

\ce{2Mg -> 2Mg^{2+} + 4e^-}

Reduction half-reaction (each oxygen atom gains two electrons, and $\ce{O2}$ contains two atoms):

\ce{O2 + 4e^- -> 2O^{2-}}

Adding these gives

$\ce{2Mg + O2 -> 2Mg^{2+} + 2O^{2-}},$ …

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Q.Explain redox reactions on the basis of electron transfer. Give suitable examples.

Concept understanding — Oxidation Reduction

The intuition: what does “oxidation” really mean?

The precise modern definition: electron transfer

How to spot a redox reaction without seeing electrons

One more way to think about it …

Redox Reactions Based on Electron Transfer

Essential Reasoning

The electron-transfer picture of redox

Step-by-step: identifying redox through electron transfer

Example 1: Reaction of zinc with copper(II) sulphate

Example 2: Combustion of magnesium in oxygen

Q.Explain redox reactions on the basis of electron transfer. Give suitable examples.

The electron-transfer picture of redox

Step-by-step: identifying redox through electron transfer

Example 1: Reaction of zinc with copper(II) sulphate

Example 2: Combustion of magnesium in oxygen

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