Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in

NCERT Exemplar · Q25

Q. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.

(i) 3HCl(aq) + HNO3(aq) → Cl2(g) + NOCl(g) + 2H2O(l)

(ii) HgCl2(aq) + 2KI(aq) → HgI2(s) + 2KCl(aq)

(iii) Fe2O3(s) + 3CO(g) —Δ→ 2Fe(s) + 3CO2(g)

(iv) PCl3(l) + 3H2O(l) → 3HCl(aq) + H3PO3(aq)

(v) 4NH3 + 3O2(g) → 2N2(g) + 6H2O(g)

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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 simultaneous change in oxidation numbers. Among the given reactions, (i), (iii), and (v) are redox; (ii) and (iv) are not. The oxidising and reducing agents are identified by which species gets reduced (oxidising agent) and which gets oxidised (reducing agent).

The core idea is simple: a redox reaction is one where oxidation numbers of some elements change. Oxidation is an increase in oxidation number (loss of electrons); reduction is a decrease (gain of electrons). The species that gets reduced is the oxidising agent (it causes oxidation by accepting electrons), and the species that gets oxidised is the reducing agent (it causes reduction by donating electrons).

Let’s assign oxidation numbers to each element in every reaction and check for changes.

Reaction (i): $3\text{HCl}(aq) + \text{HNO}_3(aq) \rightarrow \text{Cl}_2(g) + \text{NOCl}(g) + 2\text{H}_2\text{O}(l)$

In HCl: H = +1, Cl = –1. In HNO $_3$ : H = +1, N = +5, O = –2. In Cl $_2$ : Cl = 0. In NOCl: N = +2? Let’s check: O is –2, Cl is –1, so N must be +3 to balance (since NOCl is neutral: N + (–2) + (–1) = 0 → N = +3). In H $_2$ O: H = +1, O = –2.

Changes: Cl goes from –1 (in HCl) to 0 (in Cl $_2$ ) — oxidation (increase). N goes from +5 (in HNO $_3$ ) to +3 (in NOCl) — reduction (decrease). So it’s redox. The oxidising agent is HNO $_3$ (N gets reduced), and the reducing agent is HCl (Cl gets oxidised).
Reaction (ii): $\text{HgCl}_2(aq) + 2\text{KI}(aq) \rightarrow \text{HgI}_2(s) + 2\text{KCl}(aq)$

In HgCl $_2$ : Hg = +2, Cl = –1. In KI: K = +1, I = –1. In HgI $_2$ : Hg = +2, I = –1. In KCl: K = +1, Cl = –1. No oxidation number changes — it’s a double displacement (precipitation) reaction, not redox.
Reaction (iii): $\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \xrightarrow{\Delta} 2\text{Fe}(s) + 3\text{CO}_2(g)$

In Fe $_2$ O $_3$ : Fe = +3, O = –2. In CO: C = +2, O = –2. In Fe: 0. In CO $_2$ : C = +4, O = –2. Fe goes from +3 to 0 — reduction. C goes from +2 to +4 — oxidation. Redox. Oxidising agent: Fe $_2$ O $_3$ (Fe reduced). Reducing agent: CO (C oxidised).
Reaction (iv): $\text{PCl}_3(l) + 3\text{H}_2\text{O}(l) \rightarrow 3\text{HCl}(aq) + \text{H}_3\text{PO}_3(aq)$

In PCl $_3$ : P = +3, Cl = –1. In H $_2$ O: H = +1, O = –2. In HCl: H = +1, Cl = –1. In H $_3$ PO $_3$ : H = +1, O = –2, so P = +3 (since 3(+1) + P + 3(–2) = 0 → P = +3). No change in oxidation numbers — it’s a hydrolysis reaction, not redox. …

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Q.Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.

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