Chapter 2 of 9

Updated 6 August 2026

NCERT Solutions

Redox Reactions — NCERT Solutions

Q: What are the important topics in Redox Reactions for CBSE Class 11 Chemistry?

Key topics in Redox Reactions include Core ideas and definitions, Oxidation number rules and special cases, Reaction types in redox chemistry, Key examples and comparative reactivity. Study these first, then practise questions on each for Class 11 exams.

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Q: How should I revise Redox Reactions for Class 11 exams?

Learn the core ideas first, then work through the 77 practice questions on Redox Reactions. Revise definitions regularly and use flashcards for quick recall before the exam.

CBSE · Class 11 · Chemistry

NCERT Solutions for Redox Reactions, CBSE Class 11 Chemistry: 30 textbook questions solved step by step. Covers Exercises.

77 questions74 flashcards4 formulas & key relations5 concepts

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Exercises

7.1Assign oxidation number to the underlined elements in each of the following species:
(a) NaH₂PO₄ (P)
(b) NaHSO₄ (S)
(c) H₂P₂O₇ (P)
(d) K₂MnO₄ (Mn)
(e) CaO₂ (O)
(f) NaBH₄ (B)
(g) H₂S₂O₇ (S)
(h) KAl(SO₄)₂·12H₂O (S)Show solution

Given: Various compounds; find oxidation number (O.N.) of the underlined element.

Rules used: Sum of O.N. of all atoms = 0 (neutral molecule) or = charge (ion). O.N. of O = −2 (usually), H = +1 (with non-metals), Na = +1, K = +1, Ca = +2, Al = +3.

(a) NaH₂PO₄ — find O.N. of P

Let O.N. of P = $x$ .
$+1 + 2(+1) + x + 4(-2) = 0$
$1 + 2 + x - 8 = 0 \implies x = +5$

O.N. of P = +5

(b) NaHSO₄ — find O.N. of S

Let O.N. of S = $x$ .
$+1 + 1 + x + 4(-2) = 0$
$2 + x - 8 = 0 \implies x = +6$

O.N. of S = +6

(c) H₂P₂O₇ — find O.N. of P

Let O.N. of P = $x$ .
$2(+1) + 2x + 7(-2) = 0$
$2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +5$

O.N. of P = +5

(d) K₂MnO₄ — find O.N. of Mn

Let O.N. of Mn = $x$ .
$2(+1) + x + 4(-2) = 0$
$2 + x - 8 = 0 \implies x = +6$

O.N. of Mn = +6

(e) CaO₂ — find O.N. of O

This is calcium peroxide. Let O.N. of O = $x$ .
$+2 + 2x = 0 \implies x = -1$

O.N. of O = −1 (peroxide linkage O–O)

(f) NaBH₄ — find O.N. of B

In NaBH₄, H is bonded to B (more electronegative than H here? Actually B–H: H is −1 when bonded to metals/metalloids in hydrides). Here H = −1.

Let O.N. of B = $x$ .
$+1 + x + 4(-1) = 0$
$1 + x - 4 = 0 \implies x = +3$

O.N. of B = +3

(g) H₂S₂O₇ — find O.N. of S

Let O.N. of S = $x$ .
$2(+1) + 2x + 7(-2) = 0$
$2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$

O.N. of S = +6

(h) KAl(SO₄)₂·12H₂O — find O.N. of S

Let O.N. of S = $x$ . Consider the formula unit (ignore water of crystallisation for this calculation, or include it — O in water = −2, H = +1).

For the ionic compound: K = +1, Al = +3, each SO₄²⁻ has S with O.N. $x$ :
$x + 4(-2) = -2 \implies x = +6$

O.N. of S = +6

7.2What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
(a) KI₃ (I)
(b) H₂S₄O₆ (S)
(c) Fe₃O₄ (Fe)
(d) CH₃CH₂OH (C)
(e) CH₃COOH (C)Show solution

Concept: Some compounds have elements in non-integral (fractional) or mixed oxidation states, which is rationalised by the actual structure.

(a) KI₃ — O.N. of I

Let O.N. of I = $x$ .
$+1 + 3x = 0 \implies x = -\frac{1}{3}$

This is a fractional value. Rationalisation: KI₃ is actually $\text{K}^+[\text{I}_3]^-$ . In the $\text{I}_3^-$ ion, one I carries −1 and the other two carry 0 (I₂ molecule coordinates to I⁻). So the average is $-1/3$ , but structurally one I is −1 and two are 0.

Average O.N. of I = −1/3

(b) H₂S₄O₆ — O.N. of S

Let O.N. of S = $x$ .
$2(+1) + 4x + 6(-2) = 0$
$2 + 4x - 12 = 0 \implies 4x = 10 \implies x = +2.5$

Rationalisation: Tetrathionate ion has the structure $^{-}O_3S-S-S-SO_3^{-}$ . The two terminal S atoms have O.N. = +5 and the two middle S atoms (S–S bond) have O.N. = 0. Average = $(5+5+0+0)/4 = +2.5$ .

Average O.N. of S = +2.5

(c) Fe₃O₄ — O.N. of Fe

Let O.N. of Fe = $x$ .
$3x + 4(-2) = 0 \implies 3x = 8 \implies x = +\frac{8}{3}$

Rationalisation: Fe₃O₄ is a mixed oxide = FeO·Fe₂O₃. It contains one Fe²⁺ and two Fe³⁺ ions. Average O.N. = $(2+3+3)/3 = 8/3$ .

Average O.N. of Fe = +8/3

(d) CH₃CH₂OH — O.N. of C

For C₁ (–CH₃): Let O.N. = $x_1$ . Each H = +1, bonded to C.
$x_1 + 3(+1) = 0 \text{ (for CH}_3\text{ group, considering C–C bond as zero contribution)}$
Using the formula approach for each carbon:

C of CH₃ group: $x_1 + 3(+1) + (-\text{C}) = 0$ . Using electronegativity: C–H bonds give H = +1; C–C bond: both same, so 0 contribution.

$x_1 + 3(+1) + 0 = 0 \implies x_1 = -3$

C of CH₂OH group: $x_2 + 2(+1) + 0 + (-2+1) = 0$

$x_2 + 2 - 1 = 0 \implies x_2 = -1$

O.N. of C in CH₃ = −3; O.N. of C in CH₂OH = −1

(e) CH₃COOH — O.N. of C

C of CH₃ group: $x_1 + 3(+1) + 0 = 0 \implies x_1 = -3$
C of COOH group: $x_2 + 0 + 2(-2) + 1 = 0 \implies x_2 = +3$

O.N. of C in CH₃ = −3; O.N. of C in COOH = +3

7.3Justify 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)Show solution

Concept: A reaction is a redox reaction if there is a change in oxidation number of at least one element.

(a) CuO(s) + H₂(g) → Cu(s) + H₂O(g)

Species	O.N. of Cu	O.N. of H
CuO	+2	—
Cu	0	—
H₂	—	0
H₂O	—	+1

Cu: +2 → 0 (reduction; CuO is oxidising agent)
H: 0 → +1 (oxidation; H₂ is reducing agent)

This is a redox reaction.

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

Fe in Fe₂O₃: +3 → Fe: 0 (reduction)
C in CO: +2 → C in CO₂: +4 (oxidation)

This is a redox reaction. (Fe₂O₃ is oxidant; CO is reductant)

(c) 4BCl₃(g) + 3LiAlH₄(s) → 2B₂H₆(g) + 3LiCl(s) + 3AlCl₃(s)

B in BCl₃: +3 → B in B₂H₆: −3 (reduction)
H in LiAlH₄: −1 → H in B₂H₆: −1...

Actually: H in LiAlH₄ = −1; H in B₂H₆ = −1 (no change for H). But:

B: +3 → −3 (reduction, gain of 6e⁻)
H: −1 → −1 (no change)

Wait — let us recheck. In B₂H₆, B–H bonds: H = +1 (since B is less electronegative? No — in boranes H bonded to B is −1 by convention since B is more electropositive). Actually in B₂H₆, O.N. of H = −1 and B = +3? Let us use: sum = 0 for B₂H₆: $2x + 6(-1)=0 \Rightarrow x=+3$ . So B stays +3.

Then what changes? Cl in BCl₃: −1; Cl in LiCl: −1; Cl in AlCl₃: −1 — no change. H in LiAlH₄: −1; H in B₂H₆: −1 — no change.

Actually the key change: Al in LiAlH₄: Let O.N. = $x$ : $+1+x+4(-1)=0 \Rightarrow x=+3$ . Al in AlCl₃ = +3. No change.

Hmm — this reaction involves transfer of H⁻ from AlH₄⁻ to BCl₃. The B–Cl bonds are replaced by B–H bonds. Let us reconsider using electronegativity: in BCl₃, Cl is more electronegative, so B = +3, Cl = −1. In B₂H₆, H is less electronegative than B? No — B (2.0) vs H (2.1): H is slightly more electronegative, so H = −1, B = +3. So B doesn't change.

The reaction is actually a metathesis/displacement but NCERT classifies it as redox because the bonding environment changes. In BCl₃, B is bonded to Cl (more electronegative), so B = +3. In B₂H₆, B is bonded to H (less electronegative than Cl but slightly more than B), so B = +3 still. However, H changes: in LiAlH₄ (ionic, H = −1) to B₂H₆ (covalent B–H, H = −1).

NCERT's justification: The O.N. of B changes from +3 (in BCl₃) to −3 (in B₂H₆) if we assign H = +1 in B₂H₆ (treating B–H like a metal hydride in reverse). This is the NCERT approach:

B in BCl₃ = +3; B in B₂H₆ (with H = +1): $2x+6(+1)=0 \Rightarrow x=-3$ → B is reduced (+3 to −3)
H in LiAlH₄ = −1; H in B₂H₆ = +1 → H is oxidised (−1 to +1)

This is a redox reaction. B is reduced (+3→−3); H is oxidised (−1→+1).

(d) 2K(s) + F₂(g) → 2K⁺F⁻(s)

K: 0 → +1 (oxidation)
F: 0 → −1 (reduction)

This is a redox reaction. (K is reducing agent; F₂ is oxidising agent)

(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

N in NH₃: −3 → N in NO: +2 (oxidation)
O in O₂: 0 → O in H₂O: −2 (reduction)

This is a redox reaction. (NH₃ is reducing agent; O₂ is oxidising agent)

7.4Fluorine reacts with ice and results in the change:
H₂O(s) + F₂(g) → HF(g) + HOF(g)
Justify that this reaction is a redox reaction.Show solution

Given: $\text{H}_2\text{O}(s) + \text{F}_2(g) \rightarrow \text{HF}(g) + \text{HOF}(g)$

Assign oxidation numbers:

In H₂O: H = +1, O = −2
In F₂: F = 0
In HF: H = +1, F = −1
In HOF: H = +1, O = −2, F = +1?

Let us find O.N. of F in HOF: H = +1, O = −2, F = $x$ :
$+1 + (-2) + x = 0 \implies x = +1$

Wait — F is the most electronegative element, so it cannot have +1. Let us reconsider: in HOF (hypofluorous acid), O is between H and F. Since F is more electronegative than O, F = −1 and O must be assigned differently.

In HOF: H = +1, F = −1, O = $x$ :
$+1 + x + (-1) = 0 \implies x = 0$

So O.N. of O in HOF = 0.

Changes in oxidation number:

F in F₂: 0 → F in HF: −1 (reduction, gain of electrons)
O in H₂O: −2 → O in HOF: 0 (oxidation, loss of electrons)
F in F₂: 0 → F in HOF: −1 (reduction)

Since F₂ is reduced (0 → −1) and O is oxidised (−2 → 0), this is a redox reaction.

F₂ acts as the oxidising agent and H₂O acts as the reducing agent (O is oxidised from −2 to 0).

7.5Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇²⁻ and NO₃⁻. Suggest structure of these compounds. Count for the fallacy.Show solution

Given: H₂SO₅, Cr₂O₇²⁻, NO₃⁻

(i) H₂SO₅ — O.N. of S

Using the formula (assuming all O = −2, H = +1):
$2(+1) + x + 5(-2) = 0$
$2 + x - 10 = 0 \implies x = +8$

Fallacy: S cannot have O.N. = +8 since its maximum valence is 6 (only 6 electrons in outermost shell available for bonding).

Actual structure: H₂SO₅ is peroxomonosulphuric acid (Caro's acid). It contains a peroxy linkage (–O–O–). Structure: HO–O–S(=O)₂–OH. Two oxygen atoms form the peroxide linkage (O.N. = −1 each) and three oxygen atoms are normal (O.N. = −2).

Recalculating: $2(+1) + x + 2(-1) + 3(-2) = 0$
$2 + x - 2 - 6 = 0 \implies x = +6$

O.N. of S = +6 (correct, within range)

(ii) Cr₂O₇²⁻ — O.N. of Cr

Let O.N. of Cr = $x$ :
$2x + 7(-2) = -2$
$2x - 14 = -2 \implies 2x = 12 \implies x = +6$

O.N. of Cr = +6 (no fallacy; Cr can exhibit +6)

Structure: Dichromate ion has two CrO₄ tetrahedra sharing one oxygen atom. All oxygens are normal (O.N. = −2).

(iii) NO₃⁻ — O.N. of N

Let O.N. of N = $x$ :
$x + 3(-2) = -1$
$x - 6 = -1 \implies x = +5$

O.N. of N = +5 (no fallacy; N can exhibit +5 as it has 5 valence electrons)

Structure: NO₃⁻ is planar with three equivalent N–O bonds (resonance). N is at the centre bonded to three O atoms.

Summary of fallacy: Only H₂SO₅ gives a fallacious result (+8 for S) when all oxygens are assumed to be −2. The correct structure reveals a peroxide linkage, giving S = +6.

7.6Write formulas for the following compounds:
(a) Mercury(II) chloride
(b) Nickel(II) sulphate
(c) Tin(IV) oxide
(d) Thallium(I) sulphate
(e) Iron(III) sulphate
(f) Chromium(III) oxideShow solution

Concept: The Roman numeral in the name gives the oxidation state of the metal. Use it to determine the formula.

(a) Mercury(II) chloride:
Hg²⁺ and Cl⁻ → Formula: $\mathbf{HgCl_2}$

(b) Nickel(II) sulphate:
Ni²⁺ and SO₄²⁻ → Formula: $\mathbf{NiSO_4}$

(c) Tin(IV) oxide:
Sn⁴⁺ and O²⁻ → Formula: $\mathbf{SnO_2}$

(d) Thallium(I) sulphate:
Tl⁺ and SO₄²⁻ → 2 Tl⁺ per SO₄²⁻ → Formula: $\mathbf{Tl_2SO_4}$

(e) Iron(III) sulphate:
Fe³⁺ and SO₄²⁻ → 2 Fe³⁺ per 3 SO₄²⁻ → Formula: $\mathbf{Fe_2(SO_4)_3}$

(f) Chromium(III) oxide:
Cr³⁺ and O²⁻ → 2 Cr³⁺ per 3 O²⁻ → Formula: $\mathbf{Cr_2O_3}$

7.7Suggest a list of the substances where carbon can exhibit oxidation states from −4 to +4 and nitrogen from −3 to +5.Show solution

Carbon (O.N. from −4 to +4):

O.N. of C	Example compound
−4	CH₄ (methane)
−3	C₂H₆ (ethane) — each C: −3
−2	CH₃OH (methanol) — C: −2
−1	C₂H₅OH (ethanol) — CH₂OH carbon: −1
0	HCHO (formaldehyde) — C: 0; also CH₂Cl₂
+1	CHCl₃ (chloroform) — C: +1
+2	CO (carbon monoxide) — C: +2
+3	CHO group in HCOOH — C: +2; CCl₃ group: +3
+4	CCl₄ (carbon tetrachloride) — C: +4; CO₂: +4

Nitrogen (O.N. from −3 to +5):

O.N. of N	Example compound
−3	NH₃ (ammonia)
−2	N₂H₄ (hydrazine)
−1	NH₂OH (hydroxylamine)
0	N₂ (dinitrogen)
+1	N₂O (nitrous oxide)
+2	NO (nitric oxide)
+3	HNO₂ (nitrous acid)
+4	NO₂ (nitrogen dioxide)
+5	HNO₃ (nitric acid), N₂O₅

7.8While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?Show solution

Concept: A substance can act as a reducing agent only if the element in it can be oxidised (i.e., its oxidation state can increase). It can act as an oxidising agent only if the element can be reduced (oxidation state decreases).

SO₂ (S = +4):

S can be oxidised to +6 (e.g., SO₃, H₂SO₄) → SO₂ acts as reducing agent
S can be reduced to lower states (e.g., S = 0 or −2) → SO₂ acts as oxidising agent
Therefore, SO₂ acts as both oxidant and reductant.

H₂O₂ (O = −1):

O can be oxidised to 0 (O₂) → H₂O₂ acts as reducing agent
O can be reduced to −2 (H₂O) → H₂O₂ acts as oxidising agent
Therefore, H₂O₂ acts as both oxidant and reductant.

Ozone O₃ (O = 0 in O₃, but effectively acts as O = −2 after reaction):

O in O₃ is in 0 oxidation state. It can only be reduced to −2 (in H₂O or O²⁻).
O cannot be further oxidised beyond 0 (O is the second most electronegative element; it cannot exhibit positive O.N. under normal conditions).
Therefore, O₃ acts only as an oxidising agent.

Nitric acid HNO₃ (N = +5):

N is in its highest oxidation state (+5). It cannot be oxidised further.
N can only be reduced (to +4, +2, 0, −3 etc.).
Therefore, HNO₃ acts only as an oxidising agent.

Conclusion: SO₂ and H₂O₂ have elements in intermediate oxidation states, so they can both increase and decrease their oxidation states. Ozone and nitric acid have elements at or near their maximum oxidation states (or cannot be oxidised due to electronegativity), so they act only as oxidants.

7.9Consider the reactions:
(a) 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(aq) + 6O₂(g)
(b) O₃(g) + H₂O₂(l) → H₂O(l) + 2O₂(g)
Why is it more appropriate to write these reactions as:
(a) 6CO₂(g) + 12H₂O(l) → C₆H₁₂O₆(aq) + 6H₂O(l) + 6O₂(g)
(b) O₃(g) + H₂O₂(l) → H₂O(l) + O₂(g) + O₂(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.Show solution

Reaction (a):

In the original equation $6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$ , it is not clear whether the oxygen released (O₂) comes from CO₂ or H₂O.

The more appropriate form $6\text{CO}_2 + 12\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{H}_2\text{O} + 6\text{O}_2$ makes it explicit that:

The 6O₂ released comes entirely from water (H₂O is oxidised: O goes from −2 to 0).
The 6H₂O on the product side contains oxygen from CO₂.
C in CO₂ (+4) is reduced to C in glucose (average O.N. = 0).

This correctly represents the path of the reaction (photosynthesis).

Reaction (b):

In $\text{O}_3 + \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + 2\text{O}_2$ , both O₂ molecules appear identical, hiding the fact that they come from different sources.

The more appropriate form $\text{O}_3 + \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O}_2 + \text{O}_2$ shows:

One O₂ comes from O₃ (O: 0 → 0, but ozone is reduced: one O goes to H₂O as O²⁻, and two O atoms form O₂)
One O₂ comes from H₂O₂ (O: −1 → 0, oxidation)

This distinguishes the two O₂ molecules by their origin.

Technique to investigate the path:

Use isotopic labelling (radioactive or stable isotope tracers):

For reaction (a): Use $\text{H}_2^{18}\text{O}$ (water labelled with ¹⁸O). The ¹⁸O₂ released will confirm that O₂ comes from water.
For reaction (b): Use $\text{H}_2^{18}\text{O}_2$ or $^{18}\text{O}_3$ and trace which O₂ molecule contains ¹⁸O.

By mass spectrometry, the labelled products can be identified, confirming the pathway.

7.10The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?Show solution

Given: AgF₂ is unstable but a strong oxidising agent.

Explanation:

In AgF₂, silver is in the +2 oxidation state (Ag²⁺). The normal and stable oxidation state of silver is +1 (Ag⁺).

Since Ag²⁺ is unstable, it has a strong tendency to revert to the more stable Ag⁺ state:
$\text{Ag}^{2+} + e^- \rightarrow \text{Ag}^+$

This means Ag²⁺ readily accepts electrons from other substances, thereby oxidising them. Hence, AgF₂ acts as a very strong oxidising agent.

In other words, the instability of the +2 state of Ag makes AgF₂ a powerful oxidant because it easily gets reduced to the stable +1 state.

7.11Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.Show solution

Concept: The extent of oxidation/reduction depends on the relative amounts of oxidant and reductant.

Illustration 1: Reaction of Na with O₂

Excess O₂ (oxidising agent in excess):

$2\text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O}_2 \quad (\text{Na} = +1, \text{O} = -1, \text{higher O.N. of O product})$

Excess Na (reducing agent in excess):

$4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O} \quad (\text{O} = -2, \text{lower O.N. of O product})$

Illustration 2: Reaction of C with O₂

Excess O₂:

$\text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad (\text{C} = +4, \text{higher oxidation state})$

Limited O₂ (C in excess):

$2\text{C} + \text{O}_2 \rightarrow 2\text{CO} \quad (\text{C} = +2, \text{lower oxidation state})$

Illustration 3: Reaction of P with Cl₂

Excess Cl₂ (oxidising agent in excess):

$2\text{P} + 5\text{Cl}_2 \rightarrow 2\text{PCl}_5 \quad (\text{P} = +5, \text{higher oxidation state})$

Excess P (reducing agent in excess):

$2\text{P} + 3\text{Cl}_2 \rightarrow 2\text{PCl}_3 \quad (\text{P} = +3, \text{lower oxidation state})$

Conclusion: These three illustrations confirm that excess oxidising agent leads to higher oxidation state products, while excess reducing agent leads to lower oxidation state products.

7.12How do you count for the following observations?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?Show solution

(a) Use of alcoholic KMnO₄ for oxidation of toluene to benzoic acid:

Reason:

Acidic KMnO₄ is a very strong oxidising agent. It would not only oxidise the –CH₃ group of toluene to –COOH but would also further oxidise/degrade the benzene ring, giving unwanted products.
Alkaline KMnO₄ is also a strong oxidant and may cause over-oxidation.
Alcoholic (neutral) KMnO₄ is a milder oxidising agent. It selectively oxidises the methyl group (–CH₃) to carboxyl group (–COOH) without attacking the benzene ring.

Balanced redox equation:
$\text{C}_6\text{H}_5\text{CH}_3 + 2\text{KMnO}_4 \rightarrow \text{C}_6\text{H}_5\text{COOK} + 2\text{MnO}_2 + \text{KOH} + \text{H}_2\text{O}$

Or in ionic form:
$\text{C}_6\text{H}_5\text{CH}_3 + 2[\text{MnO}_4]^- \rightarrow \text{C}_6\text{H}_5\text{COO}^- + 2\text{MnO}_2 + \text{OH}^- + \text{H}_2\text{O}$

(b) HCl gas vs Br₂ vapour with concentrated H₂SO₄:

With chloride (Cl⁻):
$\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}\uparrow$

Concentrated H₂SO₄ is not a strong enough oxidising agent to oxidise HCl (since Cl⁻/Cl₂ has a high reduction potential, Cl⁻ is a weak reducing agent). So HCl is simply displaced as a gas — no redox, only acid-base reaction. Colourless pungent HCl gas is obtained.

With bromide (Br⁻):
$\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}$

HBr is a stronger reducing agent than HCl (Br⁻ is more easily oxidised than Cl⁻). Concentrated H₂SO₄ is strong enough to oxidise HBr:
$2\text{HBr} + \text{H}_2\text{SO}_4(\text{conc.}) \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}$

Br⁻ is oxidised to Br₂ (red-brown vapour), while H₂SO₄ is reduced to SO₂.

Conclusion: The difference in reducing power of Cl⁻ and Br⁻ accounts for the different observations. Br⁻ is a stronger reductant and gets oxidised by conc. H₂SO₄, while Cl⁻ is not.

7.13Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:
(a) 2AgBr(s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)
(b) HCHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH⁻(aq) → 2Ag(s) + HCOO⁻(aq) + 4NH₃(aq) + 2H₂O(l)
(c) HCHO(l) + 2Cu²⁺(aq) + 5OH⁻(aq) → Cu₂O(s) + HCOO⁻(aq) + 3H₂O(l)
(d) N₂H₄(l) + 2H₂O₂(l) → N₂(g) + 4H₂O(l)
(e) Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)Show solution

Method: Assign O.N. to all elements and identify changes.

(a) 2AgBr(s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)

C₆H₆O₂ = hydroquinone; C₆H₄O₂ = benzoquinone.

Ag in AgBr: +1 → Ag: 0 (reduced)
C in C₆H₆O₂: average O.N. increases (hydroquinone → quinone, oxidation)

	Substance	Role
Oxidised	C₆H₆O₂ (hydroquinone)	Reducing agent
Reduced	AgBr	Oxidising agent

(b) HCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → 2Ag + HCOO⁻ + 4NH₃ + 2H₂O

C in HCHO: O.N. = 0 (H=+1, O=−2: $x+1-2=0 \Rightarrow x=+1$ ... let me recalculate: HCHO: $x+2(+1)+(-2)=0 \Rightarrow x=0$ ). C in HCOO⁻: $x+1+2(-2)=-1 \Rightarrow x=+2$ . So C: 0 → +2 (oxidised).
Ag in [Ag(NH₃)₂]⁺: +1 → Ag: 0 (reduced)

	Substance	Role
Oxidised	HCHO (formaldehyde)	Reducing agent
Reduced	[Ag(NH₃)₂]⁺	Oxidising agent

(c) HCHO + 2Cu²⁺ + 5OH⁻ → Cu₂O + HCOO⁻ + 3H₂O

C in HCHO: 0 → C in HCOO⁻: +2 (oxidised)
Cu²⁺: +2 → Cu in Cu₂O: +1 (reduced)

	Substance	Role
Oxidised	HCHO	Reducing agent
Reduced	Cu²⁺	Oxidising agent

(d) N₂H₄(l) + 2H₂O₂(l) → N₂(g) + 4H₂O(l)

(Note: The question writes H₂O₈ which appears to be a typo; it should be H₂O₂.)

N in N₂H₄: −2 → N in N₂: 0 (oxidised)
O in H₂O₂: −1 → O in H₂O: −2 (reduced)

	Substance	Role
Oxidised	N₂H₄	Reducing agent
Reduced	H₂O₂	Oxidising agent

(e) Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

Pb: 0 → Pb in PbSO₄: +2 (oxidised)
Pb in PbO₂: +4 → Pb in PbSO₄: +2 (reduced)

	Substance	Role
Oxidised	Pb (metal)	Reducing agent
Reduced	PbO₂	Oxidising agent

This is a disproportionation reaction (Pb acts as both oxidant and reductant in different species).

7.14Consider the reactions:
2S₂O₃²⁻(aq) + I₂(s) → S₄O₆²⁻(aq) + 2I⁻(aq)
S₂O₃²⁻(aq) + 2Br₂(l) + 5H₂O(l) → 2SO₄²⁻(aq) + 4Br⁻(aq) + 10H⁺(aq)
Why does the same reductant, thiosulphate react differently with iodine and bromine?Show solution

Given: Thiosulphate (S₂O₃²⁻) reacts with I₂ to give tetrathionate (S₄O₆²⁻), but with Br₂ to give sulphate (SO₄²⁻).

Oxidation states of S:

In S₂O₃²⁻: O.N. of S = $x$ : $2x + 3(-2) = -2 \Rightarrow x = +2$
In S₄O₆²⁻: O.N. of S = $y$ : $4y + 6(-2) = -2 \Rightarrow y = +2.5$
In SO₄²⁻: O.N. of S = +6

With I₂:

S is oxidised from +2 to +2.5 (a small change of +0.5 per S atom)
I₂ is a mild oxidising agent (moderate reduction potential)
I₂ can only partially oxidise S₂O₃²⁻ to S₄O₆²⁻

With Br₂:

S is oxidised from +2 to +6 (a large change of +4 per S atom)
Br₂ is a stronger oxidising agent than I₂ (higher reduction potential: $E°_{\text{Br}_2/\text{Br}^-} = +1.09\text{ V}$ vs $E°_{\text{I}_2/\text{I}^-} = +0.54\text{ V}$ )
Br₂ completely oxidises S₂O₃²⁻ to SO₄²⁻

Conclusion: The difference in oxidising power of I₂ and Br₂ accounts for the different products. Br₂, being a stronger oxidant, oxidises thiosulphate completely to sulphate (+6), while the weaker oxidant I₂ only partially oxidises it to tetrathionate (+2.5).

7.15Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.Show solution

Fluorine is the best oxidant among halogens:

Fluorine has the highest reduction potential ( $E° = +2.87$ V), meaning it has the greatest tendency to get reduced (gain electrons). It can oxidise all other halide ions:

$\text{F}_2 + 2\text{Cl}^- \rightarrow 2\text{F}^- + \text{Cl}_2$
$\text{F}_2 + 2\text{Br}^- \rightarrow 2\text{F}^- + \text{Br}_2$
$\text{F}_2 + 2\text{I}^- \rightarrow 2\text{F}^- + \text{I}_2$

No other halogen can oxidise F⁻ to F₂. This confirms F₂ is the strongest oxidising agent among halogens.

Hydroiodic acid (HI) is the best reductant among hydrohalic acids:

I⁻ has the lowest reduction potential (most negative) among halide ions, meaning it is most easily oxidised. HI can reduce:

$2\text{HI} + \text{Cl}_2 \rightarrow 2\text{HCl} + \text{I}_2$
$2\text{HI} + \text{Br}_2 \rightarrow 2\text{HBr} + \text{I}_2$

HI can even reduce concentrated H₂SO₄:
$8\text{HI} + \text{H}_2\text{SO}_4 \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{H}_2\text{O}$

HCl and HBr cannot reduce H₂SO₄ to H₂S. This confirms HI is the strongest reducing agent among hydrohalic acids.

Reason: The reducing power of HX increases as the size of X increases (HF < HCl < HBr < HI) because the H–X bond becomes weaker with increasing size of X, making it easier to release electrons (H–I bond is weakest).

7.16Why does the following reaction occur?
XeO₆⁴⁻(aq) + 2F⁻(aq) + 6H⁺(aq) → XeO₃(g) + F₂(g) + 3H₂O(l)
What conclusion about the compound Na₂XeO₆ (of which XeO₆⁴⁻ is a part) can be drawn from the reaction?

Given reaction:
$\text{XeO}_6^{4-}(\text{aq}) + 2\text{F}^-(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow \text{XeO}_3(\text{g}) + \text{F}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l})$

Oxidation number analysis:

Xe in XeO₆⁴⁻: $x + 6(-2) = -4 \Rightarrow x = +8$
Xe in XeO₃: $x + 3(-2) = 0 \Rightarrow x = +6$
F in F⁻: −1 → F in F₂: 0

So:

Xe is reduced from +8 to +6
F⁻ is oxidised from −1 to 0

Why does this reaction occur?

Fluorine is the most electronegative element and F₂ has the highest reduction potential (+2.87 V). Normally, F₂ oxidises everything and F⁻ is not oxidised by ordinary oxidants. However, Xe in XeO₆⁴⁻ is in the unusually high oxidation state of +8, which is extremely unstable. The strong tendency of Xe to revert to a lower, more stable oxidation state (+6 in XeO₃) provides the driving force to oxidise even F⁻ to F₂.

Conclusion about Na₂XeO₆:

Since XeO₆⁴⁻ can oxidise F⁻ (the most difficult species to oxidise) to F₂, Na₂XeO₆ is an extremely strong oxidising agent. The Xe in +8 state is highly unstable and Na₂XeO₆ is one of the strongest oxidants known. This also confirms that Xe can exhibit an oxidation state of +8.

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7.17Consider the reactions:
(a) H₃PO₂(aq) + 4AgNO₃(aq) + 2H₂O(l) → H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)
(b) H₃PO₂(aq) + 2CuSO₄(aq) + 2H₂O(l) → H₃PO₄(aq) + 2Cu(s) + H₂SO₄(aq)
(c) C₆H₅CHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH⁻(aq) → C₆H₅COO⁻(aq) + 2Ag(s) + 4NH₃(aq) + 2H₂O(l)
(d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH⁻(aq) → No change observed.
What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions?

Analysis:

(a) and (b): H₃PO₂ (hypophosphorous acid, P = +1) is oxidised to H₃PO₄ (P = +5) by both AgNO₃ and CuSO₄. Both Ag⁺ and Cu²⁺ are reduced to their respective metals.

(c): Benzaldehyde (C₆H₅CHO) is oxidised to benzoate (C₆H₅COO⁻) by [Ag(NH₃)₂]⁺ (Tollens' reagent). Ag⁺ is reduced to Ag metal.

(d): Benzaldehyde does not react with Cu²⁺ under the given conditions.

Inference:

Ag⁺ is a stronger oxidising agent than Cu²⁺.
Ag⁺ can oxidise both H₃PO₂ and C₆H₅CHO.
Cu²⁺ can oxidise H₃PO₂ (a stronger reducing agent) but cannot oxidise C₆H₅CHO (a weaker reducing agent).

This means the reducing power order is: H₃PO₂ > C₆H₅CHO.
And the oxidising power order is: Ag⁺ > Cu²⁺.

In other words, Cu²⁺ is a weaker oxidant compared to Ag⁺, and it can only oxidise stronger reducing agents like H₃PO₂ but not weaker ones like benzaldehyde.

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7.18Balance the following redox reactions by ion-electron method:
(a) MnO₄⁻(aq) + I⁻(aq) → MnO₂(s) + I₂(s) (in basic medium)
(b) MnO₄⁻(aq) + SO₂(g) → Mn²⁺(aq) + HSO₄⁻(aq) (in acidic solution)
(c) H₂O₂(aq) + Fe²⁺(aq) → Fe³⁺(aq) + H₂O(l) (in acidic solution)
(d) Cr₂O₇²⁻ + SO₂(g) → Cr³⁺(aq) + SO₄²⁻(aq) (in acidic solution)

Ion-electron method steps: (1) Write half-reactions. (2) Balance atoms other than 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 equalise electrons. (7) Add half-reactions.

(a) MnO₄⁻ + I⁻ → MnO₂ + I₂ (basic medium)

Reduction half-reaction:
$\text{MnO}_4^- \rightarrow \text{MnO}_2$
Balance O: $\text{MnO}_4^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$
Balance H (basic: add OH⁻ to side needing H, add H₂O to other side):
$\text{MnO}_4^- + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{OH}^-$
Balance charge:
Left: $-1 + 0 = -1$ ; Right: $0 + 4(-1) = -4$
Add 3e⁻ to left:
$\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^- \quad \cdots(i)$

Oxidation half-reaction:
$2\text{I}^- \rightarrow \text{I}_2 + 2e^- \quad \cdots(ii)$

Multiply (i) by 2 and (ii) by 3:
$2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6e^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^-$
$6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-$

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

(b) MnO₄⁻ + SO₂ → Mn²⁺ + HSO₄⁻ (acidic medium)

Reduction half-reaction:
$\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$
Balance O: $\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$
Balance H: $\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$
Balance charge: Left = $-1+8 = +7$ ; Right = $+2$ . Add 5e⁻:
$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad \cdots(i)$

Oxidation half-reaction:
$\text{SO}_2 \rightarrow \text{HSO}_4^-$
Balance O: $\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{HSO}_4^-$
Balance H: $\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+$
Balance charge: Left = 0; Right = $-1+3 = +2$ . Add 2e⁻ to right:
$\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+ + 2e^- \quad \cdots(ii)$

Multiply (i) by 2 and (ii) by 5:
$2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$
$5\text{SO}_2 + 10\text{H}_2\text{O} \rightarrow 5\text{HSO}_4^- + 15\text{H}^+ + 10e^-$

Adding and simplifying (cancel 15H⁺ from both sides, cancel 8H₂O):
$2\text{MnO}_4^- + 16\text{H}^+ + 5\text{SO}_2 + 10\text{H}_2\text{O} \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 5\text{HSO}_4^- + 15\text{H}^+$
$\boxed{2\text{MnO}_4^- + \text{H}^+ + 5\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Mn}^{2+} + 5\text{HSO}_4^-}$

(c) H₂O₂ + Fe²⁺ → Fe³⁺ + H₂O (acidic medium)

Oxidation half-reaction:
$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \quad \cdots(i)$

Reduction half-reaction:
$\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O}$
Balance H: $\text{H}_2\text{O}_2 + 2\text{H}^+ \rightarrow 2\text{H}_2\text{O}$
Balance charge: Left = $0+2 = +2$ ; Right = 0. Add 2e⁻:
$\text{H}_2\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow 2\text{H}_2\text{O} \quad \cdots(ii)$

Multiply (i) by 2:
$2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2e^-$

Adding:
$\boxed{\text{H}_2\text{O}_2 + 2\text{Fe}^{2+} + 2\text{H}^+ \rightarrow 2\text{Fe}^{3+} + 2\text{H}_2\text{O}}$

(d) Cr₂O₇²⁻ + SO₂ → Cr³⁺ + SO₄²⁻ (acidic medium)

Reduction half-reaction:
$\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}$
Balance O: $\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
Balance H: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
Balance charge: Left = $-2+14 = +12$ ; Right = $+6$ . Add 6e⁻:
$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \quad \cdots(i)$

Oxidation half-reaction:
$\text{SO}_2 \rightarrow \text{SO}_4^{2-}$
Balance O: $\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-}$
Balance H: $\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+$
Balance charge: Left = 0; Right = $-2+4 = +2$ . Add 2e⁻:
$\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+ + 2e^- \quad \cdots(ii)$

Multiply (ii) by 3:
$3\text{SO}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{SO}_4^{2-} + 12\text{H}^+ + 6e^-$

Adding (i) and 3×(ii):
$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 3\text{SO}_2 + 6\text{H}_2\text{O} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{SO}_4^{2-} + 12\text{H}^+$

Simplify (cancel 12H⁺ and 6H₂O):
$\boxed{\text{Cr}_2\text{O}_7^{2-} + 2\text{H}^+ + 3\text{SO}_2 \rightarrow 2\text{Cr}^{3+} + 3\text{SO}_4^{2-} + \text{H}_2\text{O}}$

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7.19Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P₄(s) + OH⁻(aq) → PH₃(g) + HPO₂⁻(aq)
(b) N₂H₄(l) + ClO₃⁻(aq) → NO(g) + Cl⁻(g)
(c) Cl₂O₇(g) + H₂O₂(aq) → ClO₂⁻(aq) + O₂(g) + H⁺

(a) P₄ + OH⁻ → PH₃ + HPO₂⁻ (basic medium)

Oxidation number method:

P in P₄: 0
P in PH₃: −3 (reduced)
P in HPO₂⁻: +1 (oxidised)

This is a disproportionation reaction. P₄ is both oxidised and reduced.

Change per P atom: reduction = 3 (0→−3); oxidation = 1 (0→+1)
To balance: 1 P reduced for every 3 P oxidised? No — LCM of 3 and 1 is 3.
So 1 P → PH₃ (gains 3e⁻) and 3 P → 3HPO₂⁻ (each loses 1e⁻, total 3e⁻).

So from 1 P₄: 1 P → PH₃ and 3 P → 3HPO₂⁻. But P₄ has 4 P atoms, so we need to use P₄ in multiples.

Actually: ratio PH₃ : HPO₂⁻ = 1:3. From P₄ (4 P atoms): 1 PH₃ + 3 HPO₂⁻ per P₄.

$\text{P}_4 + \text{OH}^- \rightarrow \text{PH}_3 + 3\text{HPO}_2^-$

Balance H and O (basic medium, add H₂O and OH⁻):
Left H: from OH⁻. Right H: 3 (PH₃) + 3×1 (HPO₂⁻) = 6 H in products (not counting OH⁻ contribution).

Ion-electron method:

Reduction: $\text{P}_4 \rightarrow \text{PH}_3$
$\frac{1}{4}\text{P}_4 \rightarrow \text{PH}_3$
Balance H: $\frac{1}{4}\text{P}_4 + 3\text{H}_2\text{O} \rightarrow \text{PH}_3 + 3\text{OH}^-$
Balance charge: Left = 0; Right = −3. Add 3e⁻ to left:
$\frac{1}{4}\text{P}_4 + 3\text{H}_2\text{O} + 3e^- \rightarrow \text{PH}_3 + 3\text{OH}^- \quad \cdots(i)$
Multiply by 4: $\text{P}_4 + 12\text{H}_2\text{O} + 12e^- \rightarrow 4\text{PH}_3 + 12\text{OH}^-$

Oxidation: $\text{P}_4 \rightarrow \text{HPO}_2^-$
$\frac{1}{4}\text{P}_4 \rightarrow \text{HPO}_2^-$
Balance O: $\frac{1}{4}\text{P}_4 + 2\text{H}_2\text{O} \rightarrow \text{HPO}_2^- + ...$
Actually HPO₂⁻ has 2 O: $\frac{1}{4}\text{P}_4 + 2\text{OH}^- \rightarrow \text{HPO}_2^- + \text{H}_2\text{O}$ ?

Let me balance carefully:
$\frac{1}{4}\text{P}_4 \rightarrow \text{HPO}_2^-$
O: add 2H₂O to left: $\frac{1}{4}\text{P}_4 + 2\text{H}_2\text{O} \rightarrow \text{HPO}_2^- + 3\text{H}^+$
In basic medium, replace H⁺ with H₂O/OH⁻: add 3OH⁻ to both sides:
$\frac{1}{4}\text{P}_4 + 2\text{H}_2\text{O} + 3\text{OH}^- \rightarrow \text{HPO}_2^- + 3\text{H}_2\text{O}$
$\frac{1}{4}\text{P}_4 + 3\text{OH}^- \rightarrow \text{HPO}_2^- + \text{H}_2\text{O}$
Balance charge: Left = −3; Right = −1. Add 2e⁻ to right? No — oxidation, so electrons on right:
Left = −3; Right = −1 + electrons. For oxidation: $-3 = -1 - 2 \Rightarrow$ add 2e⁻ to right? That means left is more negative, so we need to remove electrons from left (oxidation means losing electrons):
Actually: charge left = −3; charge right = −1. To balance: $-3 = -1 - 2$ , so add 2e⁻ to right (product side for oxidation)?

Correct approach: oxidation half-reaction has electrons on the right:
$\frac{1}{4}\text{P}_4 + 3\text{OH}^- \rightarrow \text{HPO}_2^- + \text{H}_2\text{O} + e^- \quad \cdots(ii)$
Check: Left charge = −3; Right charge = −1 + 0 − 1 = −2. Not balanced.

Let me redo: $\frac{1}{4}\text{P}_4 + 3\text{OH}^- \rightarrow \text{HPO}_2^- + \text{H}_2\text{O}$
Left charge = −3; Right charge = −1. Difference = 2. So 2e⁻ on right:
$\frac{1}{4}\text{P}_4 + 3\text{OH}^- \rightarrow \text{HPO}_2^- + \text{H}_2\text{O} + 2e^-$
Left: −3; Right: −1 + 0 − 2 = −3. ✓

Multiply by 4: $\text{P}_4 + 12\text{OH}^- \rightarrow 4\text{HPO}_2^- + 4\text{H}_2\text{O} + 8e^- \quad \cdots(ii')$

To equalise electrons: multiply reduction (×4 already gives 12e⁻) and oxidation (×4 gives 8e⁻). LCM = 24.
Multiply reduction by 2: $2\text{P}_4 + 24\text{H}_2\text{O} + 24e^- \rightarrow 8\text{PH}_3 + 24\text{OH}^-$
Multiply oxidation by 3: $3\text{P}_4 + 36\text{OH}^- \rightarrow 12\text{HPO}_2^- + 12\text{H}_2\text{O} + 24e^-$

Adding:
$5\text{P}_4 + 24\text{H}_2\text{O} + 36\text{OH}^- \rightarrow 8\text{PH}_3 + 12\text{HPO}_2^- + 12\text{H}_2\text{O} + 24\text{OH}^-$

Simplify (cancel 12H₂O and 24OH⁻):
$\boxed{5\text{P}_4 + 12\text{H}_2\text{O} + 12\text{OH}^- \rightarrow 8\text{PH}_3 + 12\text{HPO}_2^-}$

Oxidising agent: P₄ (gets reduced to PH₃)
Reducing agent: P₄ (gets oxidised to HPO₂⁻)
(P₄ undergoes disproportionation)

(b) N₂H₄(l) + ClO₃⁻(aq) → NO(g) + Cl⁻(g) (basic medium)

Oxidation numbers:

N in N₂H₄: −2 → N in NO: +2 (oxidised, change = +4 per N, +8 per N₂H₄)
Cl in ClO₃⁻: +5 → Cl in Cl⁻: −1 (reduced, change = −6)

LCM of 8 and 6 = 24. So multiply N₂H₄ by 3 and ClO₃⁻ by 4.

Ion-electron method:

Oxidation: $\text{N}_2\text{H}_4 \rightarrow 2\text{NO}$
Balance O: $\text{N}_2\text{H}_4 + 2\text{H}_2\text{O} \rightarrow 2\text{NO} + ...$
Balance H: $\text{N}_2\text{H}_4 + 2\text{H}_2\text{O} \rightarrow 2\text{NO} + 8\text{H}^+$
Basic medium: add 8OH⁻: $\text{N}_2\text{H}_4 + 2\text{H}_2\text{O} + 8\text{OH}^- \rightarrow 2\text{NO} + 8\text{H}_2\text{O}$
$\text{N}_2\text{H}_4 + 8\text{OH}^- \rightarrow 2\text{NO} + 6\text{H}_2\text{O}$
Balance charge: Left = −8; Right = 0. Add 8e⁻ to right:
$\text{N}_2\text{H}_4 + 8\text{OH}^- \rightarrow 2\text{NO} + 6\text{H}_2\text{O} + 8e^- \quad \cdots(i)$
Check: Left = −8; Right = 0 + 0 − 8 = −8. ✓

Reduction: $\text{ClO}_3^- \rightarrow \text{Cl}^-$
Balance O: $\text{ClO}_3^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}$
Balance H: $\text{ClO}_3^- + 6\text{H}^+ \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}$
Basic: add 6OH⁻: $\text{ClO}_3^- + 6\text{H}_2\text{O} \rightarrow \text{Cl}^- + 3\text{H}_2\text{O} + 6\text{OH}^-$
$\text{ClO}_3^- + 3\text{H}_2\text{O} \rightarrow \text{Cl}^- + 6\text{OH}^-$
Balance charge: Left = −1; Right = −1 − 6 = −7. Add 6e⁻ to left:
$\text{ClO}_3^- + 3\text{H}_2\text{O} + 6e^- \rightarrow \text{Cl}^- + 6\text{OH}^- \quad \cdots(ii)$
Check: Left = −1 − 6 = −7; Right = −1 − 6 = −7. ✓

Multiply (i) by 3 and (ii) by 4:
$3\text{N}_2\text{H}_4 + 24\text{OH}^- \rightarrow 6\text{NO} + 18\text{H}_2\text{O} + 24e^-$
$4\text{ClO}_3^- + 12\text{H}_2\text{O} + 24e^- \rightarrow 4\text{Cl}^- + 24\text{OH}^-$

Adding (cancel 24OH⁻ and 12H₂O):
$\boxed{3\text{N}_2\text{H}_4 + 4\text{ClO}_3^- \rightarrow 6\text{NO} + 4\text{Cl}^- + 6\text{H}_2\text{O}}$

Oxidising agent: ClO₃⁻ (Cl reduced from +5 to −1)
Reducing agent: N₂H₄ (N oxidised from −2 to +2)

(c) Cl₂O₇(g) + H₂O₂(aq) → ClO₂⁻(aq) + O₂(g) + H⁺ (basic medium)

Oxidation numbers:

Cl in Cl₂O₇: +7 → Cl in ClO₂⁻: +3 (reduced, change = −4 per Cl, −8 per Cl₂O₇)
O in H₂O₂: −1 → O in O₂: 0 (oxidised, change = +1 per O, +2 per H₂O₂)

LCM of 8 and 2 = 8. So 1 Cl₂O₇ and 4 H₂O₂.

Ion-electron method:

Reduction: $\text{Cl}_2\text{O}_7 \rightarrow 2\text{ClO}_2^-$
Balance O: $\text{Cl}_2\text{O}_7 \rightarrow 2\text{ClO}_2^- + 3\text{H}_2\text{O}$ (7 O on left, 4 O in 2ClO₂⁻, so 3 extra O → 3H₂O)
Balance H: $\text{Cl}_2\text{O}_7 + 6\text{H}^+ \rightarrow 2\text{ClO}_2^- + 3\text{H}_2\text{O}$
Basic: add 6OH⁻: $\text{Cl}_2\text{O}_7 + 6\text{H}_2\text{O} \rightarrow 2\text{ClO}_2^- + 3\text{H}_2\text{O} + 6\text{OH}^-$
$\text{Cl}_2\text{O}_7 + 3\text{H}_2\text{O} \rightarrow 2\text{ClO}_2^- + 6\text{OH}^-$
Balance charge: Left = 0; Right = −2 − 6 = −8. Add 8e⁻ to left:
$\text{Cl}_2\text{O}_7 + 3\text{H}_2\text{O} + 8e^- \rightarrow 2\text{ClO}_2^- + 6\text{OH}^- \quad \cdots(i)$
Check: Left = −8; Right = −2 − 6 = −8. ✓

Oxidation: $\text{H}_2\text{O}_2 \rightarrow \text{O}_2$
Balance H: $\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+$
Basic: $\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O}$
Balance charge: Left = −2; Right = 0. Add 2e⁻ to right:
$\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 2e^- \quad \cdots(ii)$
Check: Left = −2; Right = 0 + 0 − 2 = −2. ✓

Multiply (ii) by 4:
$4\text{H}_2\text{O}_2 + 8\text{OH}^- \rightarrow 4\text{O}_2 + 8\text{H}_2\text{O} + 8e^-$

Adding (i) and 4×(ii):
$\text{Cl}_2\text{O}_7 + 3\text{H}_2\text{O} + 4\text{H}_2\text{O}_2 + 8\text{OH}^- \rightarrow 2\text{ClO}_2^- + 6\text{OH}^- + 4\text{O}_2 + 8\text{H}_2\text{O}$

Simplify (cancel 6OH⁻ and 3H₂O):
$\boxed{\text{Cl}_2\text{O}_7 + 4\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow 2\text{ClO}_2^- + 4\text{O}_2 + 5\text{H}_2\text{O}}$

Oxidising agent: Cl₂O₇ (Cl reduced from +7 to +3)
Reducing agent: H₂O₂ (O oxidised from −1 to 0)

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7.20What sorts of information can you draw from the following reaction?
(CN)₂(g) + 2OH⁻(aq) → CN⁻(aq) + CNO⁻(aq) + H₂O(l)

Given reaction:
$(\text{CN})_2(\text{g}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{CN}^-(\text{aq}) + \text{CNO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})$

Oxidation number analysis:

In (CN)₂: Let O.N. of C = $x$ , N = −3 (as in most cyanides):
$x + (-3) = 0 \Rightarrow x = +3$ (for each C in CN)

In CN⁻: $x + (-3) = -1 \Rightarrow x = +2$

In CNO⁻: $x + (-3) + (-2) = -1 \Rightarrow x = +4$

So:

C in (CN)₂: +3 → C in CN⁻: +2 (reduced by 1)
C in (CN)₂: +3 → C in CNO⁻: +4 (oxidised by 1)

Information drawn:

(CN)₂ undergoes disproportionation — it is simultaneously oxidised and reduced. This is similar to the behaviour of halogens (Cl₂, Br₂, I₂) with alkali, hence (CN)₂ is a pseudohalogen (behaves like a halogen).

The reaction is a redox reaction (disproportionation).

(CN)₂ behaves like a halogen (X₂ + 2OH⁻ → X⁻ + XO⁻ + H₂O), confirming its pseudohalogen character.

The products CN⁻ (cyanide) and CNO⁻ (cyanate) are analogous to X⁻ and XO⁻ formed by halogens.

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7.21The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂, and H⁺ ion. Write a balanced ionic equation for the reaction.

Given: Mn³⁺ disproportionates to Mn²⁺ and MnO₂.

Oxidation number changes:

Mn³⁺ → Mn²⁺: reduction (change = −1)
Mn³⁺ → MnO₂ (Mn = +4): oxidation (change = +1)

Since changes are equal (1 each), ratio of Mn²⁺ : MnO₂ = 1:1, so 2 Mn³⁺ are needed.

Half-reactions:

Reduction: $\text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+} \quad \cdots(i)$

Oxidation: $\text{Mn}^{3+} \rightarrow \text{MnO}_2$
Balance O: $\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^+$
Balance charge: Left = +3; Right = 0 + 4 = +4. Add 1e⁻ to right:
$\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^+ + e^- \quad \cdots(ii)$
Check: Left = +3; Right = 0 + 4 − 1 = +3. ✓

Adding (i) and (ii):
$\boxed{2\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{Mn}^{2+} + \text{MnO}_2 + 4\text{H}^+}$

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7.22Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor the positive oxidation state.

(a) Element exhibiting only negative oxidation state:

Fluorine (F)

Fluorine is the most electronegative element. It cannot exhibit a positive oxidation state. It only shows −1 (in compounds) or 0 (in F₂).

(b) Element exhibiting only positive oxidation state:

Caesium (Cs)

Cs is an alkali metal. It readily loses its one valence electron to form Cs⁺ (+1). It never gains electrons (too large, too low electronegativity) and hence never shows negative oxidation state.

(c) Element exhibiting both positive and negative oxidation states:

Iodine (I)

Iodine can show negative oxidation state (−1, e.g., HI, KI) as well as positive oxidation states (+1, +3, +5, +7, e.g., ICl, IF₃, IF₅, IF₇, HIO₃).

(d) Element exhibiting neither negative nor positive oxidation state:

Neon (Ne)

Neon is a noble gas with a completely filled electronic configuration. It does not form compounds under normal conditions and always has an oxidation state of 0.

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7.23Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

Given: Excess Cl₂ is removed by SO₂ in water.

Reaction: Cl₂ oxidises SO₂ to sulphate (SO₄²⁻) and is itself reduced to Cl⁻.

Oxidation number changes:

Cl in Cl₂: 0 → Cl⁻: −1 (reduction, gain of 1e⁻ per Cl, 2e⁻ per Cl₂)
S in SO₂: +4 → S in SO₄²⁻: +6 (oxidation, loss of 2e⁻)

Electrons gained = electrons lost, so 1:1 ratio.

Unbalanced: $\text{Cl}_2 + \text{SO}_2 + \text{H}_2\text{O} \rightarrow \text{HCl} + \text{H}_2\text{SO}_4$

Balanced equation:
$\boxed{\text{Cl}_2(\text{aq}) + \text{SO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{HCl}(\text{aq}) + \text{H}_2\text{SO}_4(\text{aq})}$

Verification:

Cl: 2 = 2 ✓; S: 1 = 1 ✓; O: 2+2 = 4 = 4 ✓; H: 4 = 2+2 = 4 ✓
Charge balance (ionic form): $\text{Cl}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{H}^+ + 2\text{Cl}^- + \text{SO}_4^{2-} + 2\text{H}^+$ ... ✓

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7.24Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non-metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.

Concept: Disproportionation requires an element to be in an intermediate oxidation state so that it can simultaneously be oxidised and reduced.

(a) Non-metals that can show disproportionation:

Non-metals that have intermediate oxidation states:

Phosphorus (P) — intermediate states: 0 (P₄), can disproportionate to PH₃ (−3) and H₃PO₂ (+1) or HPO₂⁻ (+1)
Chlorine (Cl) — Cl₂ (0) disproportionates in alkali: $\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}$
Sulphur (S) — can disproportionate in certain reactions
Fluorine (F) — cannot disproportionate (only −1 or 0)

Possible non-metals: P, Cl, Br, I, S (those with intermediate oxidation states)

(b) Three metals that can show disproportionation:

Metals with intermediate oxidation states:

Copper (Cu) — Cu⁺ (intermediate) disproportionates:

$2\text{Cu}^+ \rightarrow \text{Cu}^0 + \text{Cu}^{2+}$

Manganese (Mn) — Mn³⁺ disproportionates:

$2\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{Mn}^{2+} + \text{MnO}_2 + 4\text{H}^+$

Gold (Au) — Au⁺ disproportionates:

$3\text{Au}^+ \rightarrow 2\text{Au}^0 + \text{Au}^{3+}$

(Also: Mercury — Hg₂²⁺ can disproportionate; Iron — Fe²⁺ in some conditions)

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7.25In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen?

Given:

Mass of NH₃ = 10.00 g
Mass of O₂ = 20.00 g
Find: maximum mass of NO produced

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

Molar masses:

NH₃ = 14 + 3 = 17 g/mol
O₂ = 32 g/mol
NO = 14 + 16 = 30 g/mol

Moles of reactants:
$n(\text{NH}_3) = \frac{10.00}{17} = 0.588 \text{ mol}$
$n(\text{O}_2) = \frac{20.00}{32} = 0.625 \text{ mol}$

Identify limiting reagent:

From the equation, 4 mol NH₃ requires 5 mol O₂.
For 0.588 mol NH₃, O₂ required = $0.588 \times \frac{5}{4} = 0.735$ mol

But only 0.625 mol O₂ is available. So O₂ is the limiting reagent.

Moles of NO produced (based on O₂):
$n(\text{NO}) = n(\text{O}_2) \times \frac{4}{5} = 0.625 \times \frac{4}{5} = 0.500 \text{ mol}$

Mass of NO:
$m(\text{NO}) = 0.500 \times 30 = 15.00 \text{ g}$

$\boxed{\text{Maximum mass of NO} = 15.00 \text{ g}}$

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7.26Using the standard electrode potentials given in Table 8.1, predict if the reaction between the following is feasible:
(a) Fe³⁺(aq) and I⁻(aq)
(b) Ag⁺(aq) and Cu(s)
(c) Fe³⁺(aq) and Cu(s)
(d) Ag(s) and Fe³⁺(aq)
(e) Br₂(aq) and Fe²⁺(aq)

Concept: A redox reaction is feasible if $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} > 0$ .

Standard electrode potentials (from NCERT Table):

$E°(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77$ V
$E°(\text{I}_2/\text{I}^-) = +0.54$ V
$E°(\text{Ag}^+/\text{Ag}) = +0.80$ V
$E°(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V
$E°(\text{Fe}^{3+}/\text{Fe}) = -0.04$ V (or use Fe³⁺/Fe²⁺ = +0.77 V)
$E°(\text{Br}_2/\text{Br}^-) = +1.09$ V
$E°(\text{Fe}^{2+}/\text{Fe}) = -0.44$ V

(a) Fe³⁺(aq) and I⁻(aq):

Possible reaction: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$

Cathode (reduction): Fe³⁺ + e⁻ → Fe²⁺, $E° = +0.77$ V
Anode (oxidation): 2I⁻ → I₂ + 2e⁻, $E° = +0.54$ V

$E°_{\text{cell}} = 0.77 - 0.54 = +0.23 \text{ V} > 0$
Feasible. ✓

(b) Ag⁺(aq) and Cu(s):

Possible reaction: $2\text{Ag}^+ + \text{Cu} \rightarrow 2\text{Ag} + \text{Cu}^{2+}$

Cathode: Ag⁺ + e⁻ → Ag, $E° = +0.80$ V
Anode: Cu → Cu²⁺ + 2e⁻, $E° = +0.34$ V

$E°_{\text{cell}} = 0.80 - 0.34 = +0.46 \text{ V} > 0$
Feasible. ✓

(c) Fe³⁺(aq) and Cu(s):

Possible reaction: $2\text{Fe}^{3+} + \text{Cu} \rightarrow 2\text{Fe}^{2+} + \text{Cu}^{2+}$

Cathode: Fe³⁺ + e⁻ → Fe²⁺, $E° = +0.77$ V
Anode: Cu → Cu²⁺ + 2e⁻, $E° = +0.34$ V

$E°_{\text{cell}} = 0.77 - 0.34 = +0.43 \text{ V} > 0$
Feasible. ✓

(d) Ag(s) and Fe³⁺(aq):

Possible reaction: $\text{Ag} + \text{Fe}^{3+} \rightarrow \text{Ag}^+ + \text{Fe}^{2+}$

Cathode: Fe³⁺ + e⁻ → Fe²⁺, $E° = +0.77$ V
Anode: Ag → Ag⁺ + e⁻, $E° = +0.80$ V

$E°_{\text{cell}} = 0.77 - 0.80 = -0.03 \text{ V} < 0$
Not feasible. ✗

(e) Br₂(aq) and Fe²⁺(aq):

Possible reaction: $\text{Br}_2 + 2\text{Fe}^{2+} \rightarrow 2\text{Br}^- + 2\text{Fe}^{3+}$

Cathode: Br₂ + 2e⁻ → 2Br⁻, $E° = +1.09$ V
Anode: Fe²⁺ → Fe³⁺ + e⁻, $E° = +0.77$ V

$E°_{\text{cell}} = 1.09 - 0.77 = +0.32 \text{ V} > 0$
Feasible. ✓

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7.27Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes
(ii) An aqueous solution of AgNO₃ with platinum electrodes
(iii) A dilute solution of H₂SO₄ with platinum electrodes
(iv) An aqueous solution of CuCl₂ with platinum electrodes

Concept: During electrolysis, at cathode (reduction) the species with higher reduction potential is preferentially reduced; at anode (oxidation) the species with lower oxidation potential (or the electrode material if active) is oxidised.

(i) Aqueous AgNO₃ with silver electrodes:

Cathode: Ag⁺ ions are preferentially reduced (higher $E°$ than H₂O):

$\text{Ag}^+(\text{aq}) + e^- \rightarrow \text{Ag}(\text{s})$
Silver is deposited at cathode.

Anode: Silver electrode dissolves (active electrode):

$\text{Ag}(\text{s}) \rightarrow \text{Ag}^+(\text{aq}) + e^-$
Silver dissolves from anode.

Net result: Silver is transferred from anode to cathode (electroplating/refining).

(ii) Aqueous AgNO₃ with platinum electrodes:

Cathode: Ag⁺ is reduced:

$\text{Ag}^+(\text{aq}) + e^- \rightarrow \text{Ag}(\text{s})$
Silver is deposited.

Anode: Pt is inert. Water is oxidised:

$2\text{H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4e^-$
Oxygen gas is evolved at anode.

(iii) Dilute H₂SO₄ with platinum electrodes:

Cathode: H⁺ ions are reduced:

$2\text{H}^+(\text{aq}) + 2e^- \rightarrow \text{H}_2(\text{g})$
Hydrogen gas is evolved at cathode.

Anode: Water is oxidised (SO₄²⁻ is not discharged in dilute solution):

$2\text{H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4e^-$
Oxygen gas is evolved at anode.

Overall: Electrolysis of water.

(iv) Aqueous CuCl₂ with platinum electrodes:

Cathode: Cu²⁺ is preferentially reduced (higher $E°$ than H⁺):

$\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu}(\text{s})$
Copper is deposited at cathode.

Anode: Cl⁻ is preferentially oxidised (lower discharge potential than OH⁻/H₂O in concentrated solution):

$2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2e^-$
Chlorine gas is evolved at anode.

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7.28Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.

Concept: A metal higher in the activity series (more reactive, lower reduction potential) can displace a metal lower in the series from its salt solution.

Standard reduction potentials:

$E°(\text{Mg}^{2+}/\text{Mg}) = -2.37$ V
$E°(\text{Al}^{3+}/\text{Al}) = -1.66$ V
$E°(\text{Zn}^{2+}/\text{Zn}) = -0.76$ V
$E°(\text{Fe}^{2+}/\text{Fe}) = -0.44$ V
$E°(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V

Lower the reduction potential → greater the reducing power → higher in activity series → displaces metals below it.

Order of displacement (most reactive to least reactive):

$\boxed{\text{Mg} > \text{Al} > \text{Zn} > \text{Fe} > \text{Cu}}$

Mg can displace all others; Cu cannot displace any of the others from their salt solutions.

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7.29Given the standard electrode potentials,
K⁺/K = −2.93 V, Ag⁺/Ag = 0.80 V,
Hg²⁺/Hg = 0.79 V
Mg²⁺/Mg = −2.37 V, Cr³⁺/Cr = −0.74 V
Arrange these metals in their increasing order of reducing power.

Concept: Reducing power of a metal is its tendency to lose electrons (get oxidised). Lower the standard reduction potential (more negative), greater the reducing power.

Standard reduction potentials:

K⁺/K = −2.93 V (most negative → strongest reducing agent)
Mg²⁺/Mg = −2.37 V
Cr³⁺/Cr = −0.74 V
Hg²⁺/Hg = +0.79 V
Ag⁺/Ag = +0.80 V (most positive → weakest reducing agent)

Increasing order of reducing power (weakest to strongest):

$\boxed{\text{Ag} < \text{Hg} < \text{Cr} < \text{Mg} < \text{K}}$

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7.30Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) takes place. Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode.

Given reaction: $\text{Zn}(\text{s}) + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{Ag}(\text{s})$

Galvanic Cell Representation:

$\text{Zn}(\text{s}) \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Ag}^+(\text{aq}) \mid \text{Ag}(\text{s})$

Left half-cell: Zn electrode in ZnSO₄ solution (anode)
Right half-cell: Ag electrode in AgNO₃ solution (cathode)
Salt bridge connects the two half-cells

(i) Negatively charged electrode:

The Zn electrode (anode) is negatively charged. At the anode, Zn is oxidised and releases electrons, making it electron-rich (negative).

(ii) Carriers of current in the cell:

In the external circuit (wire): Current is carried by electrons flowing from Zn (anode) to Ag (cathode).
In the electrolyte solution: Current is carried by ions — cations (Zn²⁺, Ag⁺) and anions (SO₄²⁻, NO₃⁻).
In the salt bridge: Current is carried by ions (e.g., K⁺ and NO₃⁻ or K⁺ and Cl⁻).

(iii) Individual reactions at each electrode:

At Anode (Zn electrode) — Oxidation:
$\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$

At Cathode (Ag electrode) — Reduction:
$2\text{Ag}^+(\text{aq}) + 2e^- \rightarrow 2\text{Ag}(\text{s})$

Overall cell reaction:
$\text{Zn}(\text{s}) + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{Ag}(\text{s})$

EMF of cell:
$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = 0.80 - (-0.76) = +1.56 \text{ V}$

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Formula Sheet

Chapter Summary

Concept Maps

Study Plan

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