Justify that the following reactions are redox reactions: (a) CuO(s) + H2(g) → Cu(s) + H2O(g) (b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) (c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3AlCl3(s) (d) 2K(s) + F2(g) → 2K+F–(s) (e) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Justify that the following reactions are redox reactions: (a) CuO(s) + H2(g) → Cu(s) + H2O(g) (b) Fe2O3(s) + 3

Exercises · 7.3

Q. Justify that the following reactions are redox reactions:

(a) CuO(s) + H2(g) → Cu(s) + H2O(g)

(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

(d) 2K(s) + F2(g) → 2K+F–(s)

(e) 4NH3(g) + 5O2(g) → 4NO(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

In any redox reaction, one substance donates electrons (the reducing agent) and another accepts them (the oxidising agent).

The reducing agent gets oxidised (it loses electrons).
The oxidising agent gets reduced (it gains electrons).

This sounds backwards at first, but it’s consistent: the agent does something to the other substance. The reducing agent reduces the other substance (by giving it electrons), so the reducing agent itself gets oxidised.

Important

Reducing agent → gets oxidised (loses electrons).

Oxidising agent → gets reduced (gains electrons).

Summary for your notes

Concept	Old definition (oxygen)	Modern definition (electrons)
Oxidation	Gain of oxygen	Loss of electrons
Reduction	Loss of oxygen	Gain of electrons
Redox reaction	Both happen together	Both happen together

Final takeaway: Whenever you see a reaction where elements change their oxidation numbers, you’re looking at a redox reaction. The substance whose oxidation number increases is oxidised (loses electrons). The one whose oxidation number decreases is reduced (gains electrons). And they always come as a pair.

A quick search for "Oxidation Reduction class 11 chemistry" or "NCERT chemistry syllabus oxidation reduction" will confirm what's true here: this concept is a standard, curriculum-aligned part of Class 11 Chemistry. Given how often it's tested in JEE Main, NEET and state CET Chemistry papers, it's worth revisiting this explanation until the reasoning feels automatic, not just the final formula.

A redox reaction is one where oxidation states change — electrons are transferred. For each reaction, we assign oxidation numbers to every element before and after; if any element’s oxidation number changes, the reaction is redox. All five given reactions show such changes, so each is a redox reaction.

The Core Idea: Why Oxidation Numbers Tell the Story

A reaction is redox (reduction–oxidation) if electrons move from one species to another. But we can’t always see electrons directly. Instead, we use oxidation numbers — a bookkeeping system that tracks electron ownership. When an element’s oxidation number increases, it has been oxidised (lost electrons). When it decreases, it has been reduced (gained electrons). If any element’s oxidation number changes, the reaction is redox.

Tip

For simple compounds, oxidation numbers follow rules: oxygen is usually –2 (except in peroxides), hydrogen is usually +1 (except in metal hydrides), and elements in their elemental form have oxidation number 0. For polyatomic ions, the sum of oxidation numbers equals the ion’s charge.

Let’s apply this to each reaction.

(a) $\text{CuO(s)} + \text{H}_2\text{(g)} \rightarrow \text{Cu(s)} + \text{H}_2\text{O(g)}$

Step 1: Assign oxidation numbers before the reaction.

In CuO: oxygen is –2, so copper must be +2 (since the compound is neutral). So Cu: +2, O: –2.
In H₂: it’s an element in its standard state, so each H: 0.

Step 2: Assign oxidation numbers after the reaction.

In Cu(s): elemental form, so Cu: 0.
In H₂O: oxygen is –2, so each hydrogen must be +1 (two H’s sum to +2 to balance –2). So H: +1, O: –2.

Step 3: Look for changes.

Copper: +2 → 0 (decrease) → reduced.
Hydrogen: 0 → +1 (increase) → oxidised.

Since both oxidation and reduction occur, this is a redox reaction.

Watch out

A common mistake is to think that because H₂ is a gas and CuO is a solid, no electron transfer happens. But the oxidation numbers clearly show H₂ loses electrons (is oxidised) and Cu²⁺ gains them (is reduced). The physical state doesn’t matter.

(b) $\text{Fe}_2\text{O}_3\text{(s)} + 3\text{CO(g)} \rightarrow 2\text{Fe(s)} + 3\text{CO}_2\text{(g)}$

Step 1: Before reaction.

Fe₂O₃: oxygen is –2 each, three O’s give –6 total. Two Fe atoms must sum to +6, so each Fe: +3.
CO: oxygen is –2, so carbon must be +2 (to give neutral CO).

Step 2: After reaction.

Fe(s): elemental, so Fe: 0.
CO₂: oxygen is –2 each (two O’s = –4), so carbon must be +4.

Step 3: Changes.

Iron: +3 → 0 (decrease) → reduced.
Carbon: +2 → +4 (increase) → oxidised.

Both happen, so it’s redox.

Note

This is the classic blast furnace reaction for iron extraction. The CO acts as a reducing agent — it gets oxidised to CO₂ while reducing Fe₂O₃ to Fe.

(c) $4\text{BCl}_3\text{(g)} + 3\text{LiAlH}_4\text{(s)} \rightarrow 2\text{B}_2\text{H}_6\text{(g)} + 3\text{LiCl(s)} + 3\text{AlCl}_3\text{(s)}$

This one looks more complex, but the same logic works.

Step 1: Before reaction.

BCl₃: chlorine is –1 each (three Cl’s = –3), so boron: +3.
LiAlH₄: This is a metal hydride. Lithium is +1, aluminium is +3 (Group 13). Hydrogen in metal hydrides is –1. Check: Li (+1) + Al (+3) + 4H (–1 each = –4) = 0. Correct. So H: –1.

Step 2: After reaction.

B₂H₆: This is diborane. Boron is less electronegative than hydrogen, so H is +1 here (not a metal hydride). Each H: +1. Six H’s give +6 total, so two B’s must sum to –6, meaning each B: –3.
LiCl: Li: +1, Cl: –1.
AlCl₃: Al: +3, Cl: –1.

Step 3: Changes.

Boron: +3 → –3 (decrease of 6) → reduced.
Hydrogen: –1 → +1 (increase of 2) → oxidised.
Lithium and aluminium: no change (+1 and +3 throughout).
Chlorine: no change (–1 throughout).

So redox occurs — boron is reduced, hydrogen is oxidised.

Tip

The key insight here is that hydrogen’s oxidation number flips sign: from –1 in LiAlH₄ to +1 in B₂H₆. That’s a huge change, and it’s the driving force of the reaction. LiAlH₄ is a powerful reducing agent precisely because it contains hydride ions (H⁻) that can be oxidised.

(d) $2\text{K(s)} + \text{F}_2\text{(g)} \rightarrow 2\text{K}^+\text{F}^-\text{(s)}$

Step 1: Before reaction.

K(s): elemental, so K: 0.
F₂(g): elemental, so F: 0.

Step 2: After reaction.

KF is an ionic compound. K⁺ has oxidation number +1, F⁻ has –1.

Step 3: Changes.

Potassium: 0 → +1 (increase) → oxidised.
Fluorine: 0 → –1 (decrease) → reduced.

This is a textbook redox reaction — a metal reacting with a halogen to form an ionic salt.

Watch out

Some students think that because KF is written as K⁺F⁻, it’s “already ionic” and no redox occurs. But the elements started as neutral atoms; the electron transfer from K to F is exactly what makes it redox.

(e) $4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 4\text{NO(g)} + 6\text{H}_2\text{O(g)}$

Step 1: Before reaction.

NH₃: hydrogen is +1 each (three H’s = +3), so nitrogen: –3.
O₂: elemental, so O: 0.

Step 2: After reaction.

NO: oxygen is –2, so nitrogen: +2.
H₂O: hydrogen +1 each, oxygen –2.

Step 3: Changes.

Nitrogen: –3 → +2 (increase of 5) → oxidised.
Oxygen: 0 → –2 (decrease) → reduced.

Both changes occur, so it’s redox. This is the industrial Ostwald process for making nitric acid.

Note

Notice that hydrogen’s oxidation number doesn’t change (+1 in both NH₃ and H₂O). Only nitrogen and oxygen change. That’s fine — redox only requires some element to change, not all.

✓Final answer

All five reactions are redox reactions because in each case at least one element undergoes a change in oxidation number, indicating electron transfer.

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Q.Justify that the following reactions are redox reactions:

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

Summary for your notes

Concept: Oxidation-Reduction (Redox)

The Core Idea: Why Oxidation Numbers Tell the Story

(a) $\text{CuO(s)} + \text{H}_2\text{(g)} \rightarrow \text{Cu(s)} + \text{H}_2\text{O(g)}$

(b) $\text{Fe}_2\text{O}_3\text{(s)} + 3\text{CO(g)} \rightarrow 2\text{Fe(s)} + 3\text{CO}_2\text{(g)}$

(c) $4\text{BCl}_3\text{(g)} + 3\text{LiAlH}_4\text{(s)} \rightarrow 2\text{B}_2\text{H}_6\text{(g)} + 3\text{LiCl(s)} + 3\text{AlCl}_3\text{(s)}$

(d) $2\text{K(s)} + \text{F}_2\text{(g)} \rightarrow 2\text{K}^+\text{F}^-\text{(s)}$

(e) $4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 4\text{NO(g)} + 6\text{H}_2\text{O(g)}$

Q.Justify that the following reactions are redox reactions:

The Core Idea: Why Oxidation Numbers Tell the Story

(a) CuO(s)+H2(g)→Cu(s)+H2O(g)\text{CuO(s)} + \text{H}_2\text{(g)} \rightarrow \text{Cu(s)} + \text{H}_2\text{O(g)}CuO(s)+H2​(g)→Cu(s)+H2​O(g)

(b) Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)\text{Fe}_2\text{O}_3\text{(s)} + 3\text{CO(g)} \rightarrow 2\text{Fe(s)} + 3\text{CO}_2\text{(g)}Fe2​O3​(s)+3CO(g)→2Fe(s)+3CO2​(g)

(c) 4BCl3(g)+3LiAlH4(s)→2B2H6(g)+3LiCl(s)+3AlCl3(s)4\text{BCl}_3\text{(g)} + 3\text{LiAlH}_4\text{(s)} \rightarrow 2\text{B}_2\text{H}_6\text{(g)} + 3\text{LiCl(s)} + 3\text{AlCl}_3\text{(s)}4BCl3​(g)+3LiAlH4​(s)→2B2​H6​(g)+3LiCl(s)+3AlCl3​(s)

(d) 2K(s)+F2(g)→2K+F−(s)2\text{K(s)} + \text{F}_2\text{(g)} \rightarrow 2\text{K}^+\text{F}^-\text{(s)}2K(s)+F2​(g)→2K+F−(s)

(e) 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 4\text{NO(g)} + 6\text{H}_2\text{O(g)}4NH3​(g)+5O2​(g)→4NO(g)+6H2​O(g)

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(a) $\text{CuO(s)} + \text{H}_2\text{(g)} \rightarrow \text{Cu(s)} + \text{H}_2\text{O(g)}$

(b) $\text{Fe}_2\text{O}_3\text{(s)} + 3\text{CO(g)} \rightarrow 2\text{Fe(s)} + 3\text{CO}_2\text{(g)}$

(c) $4\text{BCl}_3\text{(g)} + 3\text{LiAlH}_4\text{(s)} \rightarrow 2\text{B}_2\text{H}_6\text{(g)} + 3\text{LiCl(s)} + 3\text{AlCl}_3\text{(s)}$

(d) $2\text{K(s)} + \text{F}_2\text{(g)} \rightarrow 2\text{K}^+\text{F}^-\text{(s)}$

(e) $4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 4\text{NO(g)} + 6\text{H}_2\text{O(g)}$

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