Chapters

Everything NCERT Class 11 Unit 7 (Redox Reactions) says that can be asked, plus what JEE Main adds beyond the book (marked Beyond NCERT): the n-factor, equivalents, normality and the titration arithmetic built on them. The short note stays for the last week; this is for a proper revision pass.

Formula sheet

Every formula in the chapter on one screen, for the last pass. Details and examples are in the sections below.

Quantity	Formula	Watch
Oxidation	loss of e⁻ = rise in oxidation number	The reductant is oxidised
Reduction	gain of e⁻ = fall in oxidation number	The oxidant is reduced
Oxidation numbers	Σ = 0 (molecule) or = charge (ion)	O −2, H +1, F −1, with exceptions
Balancing	e⁻ lost = e⁻ gained	Then charge (H⁺ or OH⁻), then H₂O
n-factor (redox)	change in ON × atoms changing, per formula unit	Beyond NCERT
n-factor (acid / base)	replaceable H⁺ / OH⁻ per formula unit	H₃PO₃ 2, H₃PO₂ 1
Equivalent mass	molar mass / n-factor
Equivalents	mass / equivalent mass = moles × n
Normality	N = M × n-factor	Equivalents per litre
Law of equivalence	N₁V₁ = N₂V₂; equivalents of oxidant = of reductant	Same as e⁻ lost = gained
H₂O₂ volume strength	"x volume" = 11.2 × M = 5.6 × N	At STP (22.4 L mol⁻¹)
Feasibility	higher E° couple oxidises a lower one	From the E° table

0 · What JEE Main asks from this chapter

59 questions in JEE Main 2019–2026, about 1 every second paper, mostly easy. Half are numerical-answer questions. Grouped by type (counts are approximate):

Type	Share	Typical question
Redox titrations: n-factor, equivalents, molarity, % purity	~32%	20 mL 0.02 M K₂Cr₂O₇ titrates 10 mL Fe²⁺: [Fe²⁺]
Disproportionation and types of redox reaction	~25%	Which species can't disproportionate: ClO⁻, ClO₂⁻, ClO₃⁻, ClO₄⁻
Balancing: coefficients, H⁺ or OH⁻ count, electrons	~20%	2MnO₄⁻ + bC₂O₄²⁻ + cH⁺ → …: find c
Oxidation numbers (average and real)	~12%	Oxidation states of S in polythionic acids
Oxidant or reductant; indicators; E°	~11%	Which can't act as an oxidising agent: N³⁻

Titration questions use the n-factors of KMnO₄, K₂Cr₂O₇, Fe²⁺, oxalate, H₂O₂ and thiosulphate again and again. Know section 8's table by heart.

1 · Classical idea: oxygen and hydrogen

The original meaning of oxidation was adding oxygen (the air is ~20% O₂, which is why most elements occur as oxides): 2Mg + O₂ → 2MgO; S + O₂ → SO₂; CH₄ + 2O₂ → CO₂ + 2H₂O. It was widened step by step:

Oxidation	Reduction
adding oxygen or another electronegative element	removing oxygen or an electronegative element
removing hydrogen or an electropositive element	adding hydrogen or an electropositive element

• Removing hydrogen: 2H₂S + O₂ → 2S + 2H₂O (H₂S oxidised).
• Adding electronegative elements: Mg + F₂ → MgF₂; Mg + Cl₂ → MgCl₂; Mg + S → MgS.
• Removing an electropositive element: 2K₄[Fe(CN)₆] + H₂O₂ → 2K₃[Fe(CN)₆] + 2KOH (ferrocyanide → ferricyanide, K removed).
• Reductions: 2HgO → 2Hg + O₂ (O removed); 2FeCl₃ + H₂ → 2FeCl₂ + 2HCl (Cl removed); CH₂=CH₂ + H₂ → CH₃–CH₃ (H added); 2HgCl₂ + SnCl₂ → Hg₂Cl₂ + SnCl₄ (Hg added; SnCl₂ is oxidised at the same time).
• Oxidation and reduction always happen together, hence "redox".

Example (NCERT Problem 7.1) · H₂S + Cl₂ → 2HCl + S: H₂S oxidised (H removed from S), Cl₂ reduced (H added). 3Fe₃O₄ + 8Al → 9Fe + 4Al₂O₃: Al oxidised, Fe₃O₄ reduced. 2Na + H₂ → 2NaH: only electronegativity tells you Na is oxidised, H reduced, which leads to the electron idea.

2 · Redox as electron transfer

2Na + Cl₂ → 2NaCl is really 2Na → 2Na⁺ + 2e⁻ and Cl₂ + 2e⁻ → 2Cl⁻. Each step is a half reaction; their sum is the overall reaction. Same for 4Na + O₂ → 2Na₂O, 2Na + S → Na₂S and 2Na + H₂ → 2Na⁺H⁻.

• Oxidation: loss of electrons. Reduction: gain of electrons.
• Oxidising agent (oxidant): electron acceptor; it gets reduced.
• Reducing agent (reductant): electron donor; it gets oxidised.

Competitive electron transfer (NCERT's experiments):

• Zn strip in Cu(NO₃)₂: the strip turns reddish with copper and the blue colour fades. Zn + Cu²⁺ → Zn²⁺ + Cu. (Zn²⁺ is confirmed as white ZnS with H₂S after adding ammonia.)
• Cu strip in ZnSO₄: nothing. Even H₂S, an extremely sensitive test (black CuS), finds no Cu²⁺. So the equilibrium lies far towards Zn²⁺ + Cu.
• Cu in AgNO₃: the solution turns blue; Cu + 2Ag⁺ → Cu²⁺ + 2Ag, far to the right.
• Co in NiSO₄: Co + Ni²⁺ ⇌ Co²⁺ + Ni, with both sides present in moderate amounts.
• Electron-releasing tendency: Zn > Cu > Ag. Ranking metals this way gives the activity (electrochemical) series, and the basis of galvanic cells.

3 · Oxidation number

In covalent compounds electrons shift rather than transfer (2H₂ + O₂ → 2H₂O; H₂ + Cl₂ → 2HCl; CH₄ + 4Cl₂ → CCl₄ + 4HCl). For book-keeping, assume the bonding pair belongs entirely to the more electronegative atom. The charge an atom would then carry is its oxidation number (used interchangeably with oxidation state).

Rules (NCERT):

1. An element in the free state: 0 (H₂, O₂, Cl₂, O₃, P₄, S₈, Na, Mg, Al).
2. A monatomic ion: its charge (Na⁺ +1, Mg²⁺ +2, Fe³⁺ +3, Cl⁻ −1, O²⁻ −2). Alkali metals are +1 and alkaline earth metals +2 in all compounds; Al is +3.
3. Oxygen is −2, except:
   • peroxides (H₂O₂, Na₂O₂, O–O bond): −1;
   • superoxides (KO₂, RbO₂): −½;
   • bonded to fluorine: OF₂ +2, O₂F₂ +1.
4. Hydrogen is +1, except in binary metal hydrides (LiH, NaH, CaH₂): −1.
5. Fluorine is −1 in all compounds. Cl, Br and I are −1 as halides but positive with oxygen (oxoacids, oxoanions).
6. The sum is 0 for a compound and equals the charge for an ion (in CO₃²⁻, C + 3(−2) = −2, so C = +4).

• If an element appears more than once (Na₂S₂O₃, Cr₂O₇²⁻), the rules give the average.
• Metals have positive oxidation numbers; non-metals positive or negative. Transition metals usually show several positive states.
• Highest oxidation number of a representative element = group number (groups 1, 2), group number − 10 (groups 13–17). It rises across a period:

Group	Compound	Highest ON
1	NaCl	+1
2	MgSO₄	+2
13	AlF₃	+3
14	SiCl₄	+4
15	P₄O₁₀	+5
16	SF₆	+6
17	HClO₄	+7

Stock notation: the metal's oxidation number in Roman numerals after its symbol. Aurous and auric chloride: Au(I)Cl, Au(III)Cl₃. Stannous and stannic chloride: Sn(II)Cl₂, Sn(IV)Cl₄. Hg₂(I)Cl₂ is the reduced form of Hg(II)Cl₂.

Example (NCERT Problem 7.3) · HAu(III)Cl₄, Tl₂(I)O, Fe(II)O, Fe₂(III)O₃, Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O₂.

Redefined with oxidation numbers:

• Oxidation: increase in oxidation number. Reduction: decrease.
• Oxidant: raises another element's oxidation number. Reductant: lowers it.
• Redox reaction: any reaction in which oxidation numbers change.

Example (NCERT Problem 7.4) · 2Cu₂O + Cu₂S → 6Cu + SO₂. Cu: +1 → 0 (reduced); S: −2 → +4 (oxidised). Cu(I) is the oxidant; S of Cu₂S is the reductant.

Example (NCERT exercise 7.1) · P in NaH₂PO₄ +5; S in NaHSO₄ +6; P in H₄P₂O₇ +5; Mn in K₂MnO₄ +6; O in CaO₂ −1; B in NaBH₄ +3 (H is −1); S in H₂S₂O₇ +6; S in KAl(SO₄)₂·12H₂O +6.

Example (2020, 2023) · K is +1 in K₂O, K₂O₂ and KO₂ (only oxygen changes). Fe(CO)₅ 0, VO²⁺ +4, WO₃ +6: sum 10.

Example (2021) · N in NO₃⁻ (+5) > NO₂ (+4) > NO (+2) > N₂O (+1).

Fractional oxidation numbers are averages (NCERT's box). Structures show whole-number states on different atoms:

Species	Average	Real
C₃O₂ (O=C=C=C=O)	+4/3	+2, 0, +2
Br₃O₈	+16/3	+6, +4, +6
S₄O₆²⁻ (tetrathionate)	+2.5	+5, 0, 0, +5
Fe₃O₄ (FeO·Fe₂O₃)	+8/3	+2, +3, +3

Mn₃O₄ and Pb₃O₄ (2PbO·PbO₂) are mixed oxides too. O₂⁺ and O₂⁻ really are +½ and −½.

Structures you need for "real" oxidation numbers · Beyond NCERT

• Peroxide links lower the count: H₂SO₅ (one O–O): S = +6, not +8. CrO₅ (two O–O): Cr = +6. H₂S₂O₈: S = +6.
• Polythionic acids H₂SₓO₆ (x = 3–5): the two end S are +5, the middle S are 0 (2021).
• Thiosulphate S₂O₃²⁻: average +2. By structure the two S differ (central S +4 and terminal S 0 when the S–S bond is split evenly; +6 and −2 in other books). Use the average for n-factors.
• Organic carbon: CH₃CH₂OH, CH₃ −3 and CH₂OH −1 (average −2); CH₃COOH, CH₃ −3 and COOH +3 (average 0) (NCERT exercise 7.2).
• KI₃ = K⁺[I–I←I⁻]: average −⅓; the I₃⁻ ion has one I⁻ and an I₂ unit.

4 · Types of redox reaction

1. Combination (A + B → C): redox only if A or B (or both) is an element. All combustions with O₂ are redox.

• C + O₂ → CO₂; 3Mg + N₂ → Mg₃N₂ (N −3); CH₄ + 2O₂ → CO₂ + 2H₂O (H unchanged).

2. Decomposition: a compound breaks into parts, at least one an element.

• 2H₂O → 2H₂ + O₂; 2NaH → 2Na + H₂; 2KClO₃ → 2KCl + 3O₂ (K unchanged).
• Not every decomposition is redox: CaCO₃ → CaO + CO₂ changes no oxidation number.

3. Displacement (X + YZ → XZ + Y):

• Metal displacement (used in metallurgy): CuSO₄ + Zn → Cu + ZnSO₄; V₂O₅ + 5Ca → 2V + 5CaO; TiCl₄ + 2Mg → Ti + 2MgCl₂; Cr₂O₃ + 2Al → Al₂O₃ + 2Cr. The displacing metal is the better reductant.
• Hydrogen displacement:

Metal	Reacts with	Example
Alkali metals; Ca, Sr, Ba	cold water	2Na + 2H₂O → 2NaOH + H₂
Mg, Fe	steam	2Fe + 3H₂O → Fe₂O₃ + 3H₂
Zn, Mg, Fe; also Cd, Sn (no reaction with steam)	acids	Zn + 2HCl → ZnCl₂ + H₂
Ag, Au (occur native)	not even HCl	—

The rate of H₂ from acid follows reactivity: fastest with Mg, slowest with Fe. Zn, Mg and Fe with HCl are the laboratory preparations of H₂.

• Halogen displacement: oxidising power falls from F₂ to I₂. Each halogen displaces the ones below it from their halides.
  • Cl₂ + 2KBr → 2KCl + Br₂; Cl₂ + 2KI → 2KCl + I₂; Br₂ + 2I⁻ → 2Br⁻ + I₂.
  • The layer test for Br⁻ and I⁻: Br₂ and I₂ colour a CCl₄ layer.
  • F₂ is so strong it displaces O₂ from water: 2H₂O + 2F₂ → 4HF + O₂. So F₂ displacements aren't done in water.
  • Halogens are recovered from halides by oxidation (2X⁻ → X₂ + 2e⁻). F⁻ can't be oxidised chemically; F₂ is made only electrolytically.

4. Disproportionation: one element in one oxidation state is oxidised and reduced at the same time. It needs an element with at least three oxidation states, sitting in an intermediate one.

• 2H₂O₂ → 2H₂O + O₂ (O: −1 → −2 and 0).
• P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻ (P: 0 → −3 and +1).
• S₈ + 12OH⁻ → 4S²⁻ + 2S₂O₃²⁻ + 6H₂O (S: 0 → −2 and +2). (2021: a = 12.)
• Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O (Cl: 0 → +1 and −1): household bleach; ClO⁻ oxidises coloured stains. Br₂ and I₂ behave the same.
• Fluorine doesn't disproportionate: it has no positive state. 2F₂ + 2OH⁻ → 2F⁻ + OF₂ + H₂O (and some O₂ from water).
• 2NO₂ + 2OH⁻ → NO₂⁻ + NO₃⁻ + H₂O (N: +4 → +3 and +5).
• 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O; 2Cu⁺ → Cu²⁺ + Cu; 2Mn³⁺ + 2H₂O → Mn²⁺ + MnO₂ + 4H⁺ (NCERT exercise 7.21).

An element in its highest or lowest state can't disproportionate: ClO₄⁻ (+7), MnO₄⁻ (+7), Cr₂O₇²⁻ (+6), BrO₄⁻, F₂, F⁻, K⁺.

Example (NCERT Problem 7.5) · ClO₄⁻ doesn't disproportionate (+7, the highest). The others do: 3ClO⁻ → 2Cl⁻ + ClO₃⁻; 6ClO₂⁻ → 4ClO₃⁻ + 2Cl⁻; 4ClO₃⁻ → Cl⁻ + 3ClO₄⁻.

Example (NCERT Problem 7.6, asked in 2024) · N₂ + O₂ → 2NO: combination. 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂: decomposition. NaH + H₂O → NaOH + H₂: displacement (hydride displaces H from water). 2NO₂ + 2OH⁻ → NO₂⁻ + NO₃⁻ + H₂O: disproportionation.

Example (2024) · Of H₂O₂, ClO₃⁻, P₄, Cl₂, Ag, Cu⁺, F₂, NO₂, K⁺, those that can disproportionate: H₂O₂, ClO₃⁻, P₄, Cl₂, Cu⁺, NO₂ = 6.

Example (2024) · 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂ is not disproportionation: Mn is reduced and O oxidised (two elements). 2MnO₄⁻ + 3Mn²⁺ + 2H₂O → 5MnO₂ + 4H⁺ is the reverse, comproportionation: +7 and +2 meet at +4. A half reaction (MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O) is neither.

Comproportionation · Beyond NCERT · Two states of one element give an intermediate one: IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O; 2H₂S + SO₂ → 3S + 2H₂O.

Example (NCERT Problem 7.7) · Pb₃O₄ = 2PbO·PbO₂. With HCl: 2PbO + 4HCl → 2PbCl₂ + 2H₂O (acid–base) and PbO₂ + 4HCl → PbCl₂ + Cl₂ + 2H₂O (redox: Pb(IV) oxidises Cl⁻), overall Pb₃O₄ + 8HCl → 3PbCl₂ + Cl₂ + 4H₂O. With HNO₃, only the PbO reacts (HNO₃ is itself an oxidant and PbO₂ is passive to it): Pb₃O₄ + 4HNO₃ → 2Pb(NO₃)₂ + PbO₂ + 2H₂O.

Can it act as oxidant, reductant or both?

• Element in its highest state: only an oxidant (HNO₃, KMnO₄, K₂Cr₂O₇, conc. H₂SO₄). O₃ acts only as an oxidant too.
• Lowest state: only a reductant (N³⁻, S²⁻, I⁻, H⁻). N³⁻ can't be an oxidising agent (2024).
• Intermediate state: both (SO₂ S +4, H₂O₂ O −1, HNO₂ N +3, H₂SO₃). H₃PO₄ (P +5) can't act as both (2020). (NCERT exercise 7.8: SO₂ and H₂O₂ both ways; O₃ and HNO₃ only oxidants.)

5 · Balancing redox equations

(a) Oxidation number method (NCERT's steps):

1. Write correct formulas for every reactant and product.
2. Assign oxidation numbers; find the atoms that change.
3. Find the increase and decrease per atom and per formula unit; multiply so that increase = decrease. (If two things are reduced and nothing oxidised, a formula or an ON is wrong.)
4. Balance charge: add H⁺ (acidic) or OH⁻ (basic) on the side that needs it.
5. Balance H with H₂O; check O.

Example (NCERT Problem 7.8) · Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻ (acidic). Cr +6 → +3: fall of 3 per Cr, 6 per Cr₂O₇²⁻. S +4 → +6: rise of 2. So 3SO₃²⁻ per Cr₂O₇²⁻. Charge: left −2 − 6 = −8, right +6 − 6 = 0, add 8H⁺ on the left; then 4H₂O on the right. Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O.

Example (NCERT Problem 7.9) · MnO₄⁻ + Br⁻ → MnO₂ + BrO₃⁻ (basic). Mn falls by 3, Br rises by 6, so 2MnO₄⁻ per Br⁻. Charge: left −3, right −1, add 2OH⁻ on the right; then one H₂O on the left. 2MnO₄⁻ + Br⁻ + H₂O → 2MnO₂ + BrO₃⁻ + 2OH⁻.

(b) Half-reaction (ion–electron) method:

1. Write the unbalanced ionic equation.
2. Split into oxidation and reduction halves.
3. Balance atoms other than O and H.
4. Acidic medium: balance O with H₂O, then H with H⁺.
5. Balance charge with electrons; multiply the halves so the electrons are equal.
6. Add; cancel electrons (and anything common).
7. Check atoms and charge.

Basic medium: balance as if acidic, then add as many OH⁻ as there are H⁺ to both sides; combine H⁺ + OH⁻ into H₂O and cancel water.

Example (NCERT) · Fe²⁺ + Cr₂O₇²⁻ (acidic). Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Oxidation: Fe²⁺ → Fe³⁺ + e⁻, times 6. 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O. (2023: X + Y + Z = 14 + 2 + 7 = 23.)

Example (NCERT Problem 7.10, asked in 2024) · MnO₄⁻ + I⁻ → MnO₂ + I₂ (basic). Reduction: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O; add 4OH⁻: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. Oxidation: 2I⁻ → I₂ + 2e⁻. Times 2 and 3: 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻ (z = 8).

Half reactions worth knowing:

Half reaction	e⁻
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O (acidic)	5
MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O (or + 2H₂O → MnO₂ + 4OH⁻, neutral/weakly basic)	3
MnO₄⁻ → MnO₄²⁻ (strongly basic)	1
Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O	6
C₂O₄²⁻ → 2CO₂	2 (1 per CO₂)
Fe²⁺ → Fe³⁺	1
H₂O₂ → O₂ + 2H⁺ (as reductant)	2
H₂O₂ + 2H⁺ → 2H₂O (as oxidant)	2
2S₂O₃²⁻ → S₄O₆²⁻	2 (1 per S₂O₃²⁻)
2I⁻ → I₂	2
IO₃⁻ + 6H⁺ → ½I₂ + 3H₂O	5

Example (2021, 2024) · 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O: c = 16; H⁺ per MnO₄⁻ = 8; electrons per CO₂ = 1 (2019, 2022).

Example (2023) · 2IO₃⁻ + xI⁻ + 12H⁺ → 6I₂ + 6H₂O: I balance 2 + x = 12, x = 10.

6 · Redox titrations

Like acid–base titrations, but an indicator signals the end of the redox reaction.

1. Self-indicator: MnO₄⁻ is intensely purple. After the last Fe²⁺ or C₂O₄²⁻ is oxidised, the first lasting pink appears at MnO₄⁻ around 10⁻⁶ mol L⁻¹, so the overshoot is tiny.
2. External redox indicator: Cr₂O₇²⁻ is not a self-indicator. It oxidises diphenylamine just after the equivalence point, giving an intense blue.
3. Iodometric: for oxidants that oxidise I⁻, such as Cu²⁺: 2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂. The liberated I₂ is titrated with thiosulphate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. Starch gives an intense blue with I₂ that vanishes when the iodine is used up. (I₂ stays dissolved in KI as KI₃.)

• Equivalence point: oxidant and reductant have reacted in exactly the stoichiometric ratio.
• Permanganate titrations use H₂SO₄, not HCl: MnO₄⁻ would oxidise Cl⁻ to Cl₂ and use up extra titrant (2022).
• Redox indicators respond to the redox potential, acid–base indicators to pH (2023: both statements, swapped, were false).

Primary standards · Beyond NCERT (2025) · Pure, stable, non-hygroscopic, high molar mass: oxalic acid, K₂Cr₂O₇, Mohr's salt FeSO₄(NH₄)₂SO₄·6H₂O, borax. Not primary standards: NaOH (absorbs moisture and CO₂), KMnO₄, Na₂Cr₂O₇ (hygroscopic), FeSO₄·7H₂O (air-oxidised).

7 · Limitations, electrode processes and the E° series

Limitation of oxidation numbers: the idea keeps evolving. Today oxidation is seen as a decrease in electron density around an atom, and reduction as an increase.

Electrode processes: dip Zn in CuSO₄ and electrons pass directly (heat is released). Separate the two:

• Zn rod in ZnSO₄ in one beaker, Cu rod in CuSO₄ in another.
• Redox couple: the oxidised and reduced forms of one species in a half reaction, written oxidised/reduced: Zn²⁺/Zn, Cu²⁺/Cu.
• A salt bridge (U-tube of KCl or NH₄NO₃ set in agar jelly) connects the solutions without mixing them; ions move through it.
• A wire (with an ammeter and switch) joins the rods. This is the Daniell cell. Switch on: electrons flow from Zn to Cu through the wire, and current flows the opposite way.
• Current flows because there is a potential difference between the electrodes. Each electrode's potential is its electrode potential.
• Standard electrode potential E°: every species at unit concentration (gases at 1 atm), 298 K. The hydrogen electrode is 0.00 V by convention.
• Negative E°: the couple is a stronger reducing agent than H⁺/H₂. Positive E°: weaker reducing agent (stronger oxidant).

Read it as a ladder of reductions (oxidised form + ne⁻ → reduced form). The oxidised form of a higher couple oxidises the reduced form of any lower couple. F₂ is the strongest oxidant listed; Li the strongest reductant.

Rest of NCERT's Table 7.1 (V): Co³⁺/Co²⁺ +1.81; H₂O₂/H₂O +1.78; Au³⁺/Au +1.40; O₂/H₂O +1.23; MnO₂/Mn²⁺ +1.23; NO₃⁻/NO +0.97; Hg²⁺/Hg₂²⁺ +0.92; O₂/H₂O₂ +0.68; Cu⁺/Cu +0.52; AgCl/Ag +0.22; AgBr/Ag +0.10; Pb²⁺/Pb −0.13; Sn²⁺/Sn −0.14; Ni²⁺/Ni −0.25; Cr³⁺/Cr −0.74; H₂O/H₂, OH⁻ −0.83; Al³⁺/Al −1.66; Na⁺/Na −2.71; Ca²⁺/Ca −2.87; K⁺/K −2.93.

Example (NCERT exercise 7.26) · Feasible: Fe³⁺ + I⁻ (0.77 > 0.54); Ag⁺ + Cu (0.80 > 0.34); Fe³⁺ + Cu (0.77 > 0.34); Br₂ + Fe²⁺ (1.09 > 0.77). Not feasible: Ag + Fe³⁺ (Ag⁺/Ag 0.80 is above Fe³⁺/Fe²⁺ 0.77).

Example (NCERT exercise 7.28) · Order in which metals displace each other from their salts: Mg > Al > Zn > Fe > Cu.

Example (2024, NCERT exercise 7.14) · Thiosulphate gives S₄O₆²⁻ with I₂ but SO₄²⁻ with Br₂: bromine is the stronger oxidant and takes S to +6; iodine only to +2.5.

8 · n-factor, equivalents and normality · Beyond NCERT

NCERT has dropped equivalents, but JEE titration questions are built on them. They are the same idea as balancing: electrons lost = electrons gained.

Law of equivalence: in any reaction, the equivalents of every reactant that takes part are equal. For a titration:

n-factor:

Substance	n-factor	Examples
Acid	replaceable H⁺ per molecule (basicity)	HCl 1, H₂SO₄ 2, H₃PO₄ 3, H₃PO₃ 2, H₃PO₂ 1, H₃BO₃ 1, CH₃COOH 1
Base	replaceable OH⁻ (acidity)	NaOH 1, Ca(OH)₂ 2, Al(OH)₃ 3
Salt (no redox)	total charge of cations (or anions)	Na₂CO₃ 2, Al₂(SO₄)₃ 6
Redox	change in ON × atoms changing, per formula unit	see below

Redox n-factors (memorise):

Reagent	Changes to	n
KMnO₄, acidic	Mn²⁺	5
KMnO₄, neutral or weakly basic	MnO₂	3
KMnO₄, strongly basic	MnO₄²⁻	1
K₂Cr₂O₇, acidic	2Cr³⁺	6
Fe²⁺ (FeSO₄, Mohr's salt)	Fe³⁺	1
H₂C₂O₄ or C₂O₄²⁻	2CO₂	2
FeC₂O₄	Fe³⁺ + 2CO₂	3
Fe₂(C₂O₄)₃	only oxalate oxidised	6
H₂O₂ (either way)	O₂ or H₂O	2
Na₂S₂O₃	S₄O₆²⁻	1
Na₂S₂O₃ (with Br₂/Cl₂)	SO₄²⁻	8
I₂	I⁻	2
KIO₃ (with I⁻, acid)	I₂	5
MnO₂	Mn²⁺	2
Cu²⁺ (with I⁻)	Cu⁺ (Cu₂I₂)	1

• In a disproportionation, n for the substance = n₁n₂/(n₁ + n₂), where n₁ and n₂ are the n-factors of the two changes.
• Oxalic acid dihydrate H₂C₂O₄·2H₂O = 126 g mol⁻¹, equivalent mass 63 (as acid and as reductant). KMnO₄ 158, equivalent mass 31.6 in acid. K₂Cr₂O₇ 294, equivalent mass 49. Mohr's salt 392, equivalent mass 392.

Volume strength of H₂O₂: "x volume" means 1 L of the solution gives x L of O₂ at STP. 2H₂O₂ → 2H₂O + O₂: 2 mol H₂O₂ give 22.4 L, so

(With NCERT's 1 bar STP, 22.7 L, use 11.35 × M; questions say which.)

The method for any titration:

1. Write each species' change; get its n-factor (or balance the equation).
2. Equivalents (mmol × n) of titrant = equivalents of the analyte.
3. Several steps (iodometry, back titration): follow the electrons through; intermediates cancel.

Example (2022) · 20 mL 0.02 M K₂Cr₂O₇ (n = 6) titrates 10 mL Fe²⁺ (n = 1): 20 × 0.02 × 6 = 10 × M, M = 0.24.

Example (2021) · 10.0 mL 0.05 M KMnO₄ (acid, n = 5) oxidises 10.0 mL oxalic acid (n = 2): 10 × 0.05 × 5 = 10 × M × 2, M = 0.125. Strength = 0.125 × 126 = 15.75 g L⁻¹.

Example (2024) · 2 mL KMnO₄ for 20 mL 2 M oxalic acid: 2 × M × 5 = 20 × 2 × 2, M = 8.

Example (2022) · 0.01 M KMnO₄ from a 50 mL burette into 20.0 mL 0.05 M Mohr's salt: 1 mmol Fe²⁺ needs 0.2 mmol KMnO₄ = 20 mL. 30 mL left in the burette.

Example (2019, 2026) · Oxidise 1 mol each of FeC₂O₄ (3), Fe₂(C₂O₄)₃ (6; Fe³⁺ can't be oxidised further), FeSO₄ (1), Fe₂(SO₄)₃ (0) with acidic KMnO₄: 10 mol e⁻ / 5 = 2 mol KMnO₄.

Example (2020) · 0.2 g impure H₂O₂ reacts with 0.316 g KMnO₄ (0.002 mol × 5 = 0.01 eq). H₂O₂ (n = 2) = 0.005 mol = 0.17 g: 85% pure.

Example (2021) · CrO₄²⁻ (Cr +6 → +3 in Cr(OH)₄⁻, n = 3) oxidises S₂O₃²⁻ to SO₄²⁻ (2 S from +2 to +6, n = 8). 40 × 0.25 × 8 = 80 meq; V = 80 / (0.154 × 3) = 173 mL.

Example (2022) · Iodometry: 2.0 g sample with MnO₂ + HCl → Cl₂; Cl₂ + 2KI → I₂; I₂ needs 60.0 mL 0.1 M thiosulphate = 6 mmol. Electrons: 6 meq → MnO₂ (n = 2) = 3 mmol = 0.261 g → 13%.

Example (2022) · 20 mL 0.02 M hypo titrates the I₂ from 10 mL CuSO₄ with excess KI: 0.4 mmol S₂O₃²⁻ = 0.4 mmol Cu²⁺ (both n = 1). [Cu²⁺] = 0.04 M.

Example (2026) · 500 mL 0.2 M KMnO₄ (100 mmol) + 500 mL 1.2 M KI, basic: MnO₄⁻ → MnO₂ (n = 3), 300 meq. I⁻ (600 mmol) is in excess, so I₂ = 150 mmol, needing 300 mmol thiosulphate = 3 L of 0.1 M.

Example (2026) · 1 mol Cl₂ into 2 L of cold 2 M KOH (4 mol): Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. Cl⁻ 1 mol, ClO⁻ 1 mol, OH⁻ left 2 mol: 0.5, 0.5, 1 M.

Example (2023) · KMnO₄ with Mohr's salt in dilute H₂SO₄: 2KMnO₄ + 10FeSO₄(NH₄)₂SO₄·6H₂O + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 10(NH₄)₂SO₄ + 68H₂O. Water: 8 from the reaction + 60 of crystallisation = 68.

Double-indicator titration of NaOH + Na₂CO₃ · Phenolphthalein end point: all NaOH + half the Na₂CO₃ (CO₃²⁻ → HCO₃⁻). Then methyl orange: the other half (HCO₃⁻ → CO₂). So the second reading alone gives half the carbonate. (2021: 1.5 mL of N/10 HCl after phenolphthalein → 0.15 mmol Na₂CO₃ = 15.9 mg in 0.4 g = 4%.)

9 · Traps

Trap	What goes wrong	Do this
Oxidant vs oxidised	Calling the oxidant "oxidised"	The oxidant is reduced; the reductant is oxidised
Peroxide O–O	S = +8 in H₂SO₅, Cr = +10 in CrO₅	Count O–O as −1 each: S +6, Cr +6
Average ON	Taking +2.5 as real in S₄O₆²⁻	Real: +5, 0, 0, +5
KMnO₄'s n	Using 5 in neutral or basic medium	5 acid, 3 neutral/weak base, 1 strong base
FeC₂O₄	n = 1 (only Fe)	Fe²⁺ and C₂O₄²⁻ both oxidised: n = 3
Fe₂(C₂O₄)₃	Counting Fe³⁺	Fe³⁺ can't be oxidised: n = 6
H₃PO₃, H₃PO₂	Basicity 3	P–H bonds aren't acidic: 2 and 1
Basic-medium balancing	Leaving H⁺ in the answer	Add OH⁻ to both sides, combine to water
Disproportionation	Two elements change (2KMnO₄ → …)	One element, one state, both up and down
Highest state	ClO₄⁻, MnO₄⁻ "can disproportionate"	Highest or lowest state can't
HCl with KMnO₄	Titrating in HCl	Use H₂SO₄: MnO₄⁻ oxidises Cl⁻
Volume strength	Using 22.4 with moles of H₂O₂ directly	2H₂O₂ → O₂: vol strength = 11.2 × M
Hydrates	90 g mol⁻¹ for oxalic acid crystals	H₂C₂O₄·2H₂O = 126
E° comparison	Reading the table as oxidation potentials	Higher reduction E° = stronger oxidant

10 · Numbers worth knowing

Compound	M (g mol⁻¹)	Equivalent mass
KMnO₄ (acid)	158	31.6
K₂Cr₂O₇	294	49
H₂C₂O₄·2H₂O	126	63
FeSO₄·7H₂O	278	278
Mohr's salt	392	392
FeC₂O₄	144	48
H₂O₂	34	17
Na₂S₂O₃·5H₂O	248	248
MnO₂	87	43.5

• E°: F₂ +2.87, MnO₄⁻/Mn²⁺ +1.51, Cl₂ +1.36, Cr₂O₇²⁻ +1.33, Br₂ +1.09, Ag⁺ +0.80, Fe³⁺/Fe²⁺ +0.77, I₂ +0.54, Cu²⁺ +0.34, Zn −0.76, Li −3.05 V.
• 1 volume H₂O₂ ≈ 1/11.2 M; "20 volume" ≈ 1.79 M ≈ 6.1% w/v.

11 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• Where redox matters: burning fuels, electrochemical extraction of reactive metals and non-metals, manufacture of caustic soda, dry and wet batteries, corrosion; lately the hydrogen economy (liquid hydrogen as fuel) and the ozone hole.
• The unit's motto: "Where there is oxidation, there is always reduction. Chemistry is essentially a study of redox systems."
• NCERT exercise 7.9: writing 6CO₂ + 12H₂O → C₆H₁₂O₆ + 6H₂O + 6O₂ shows that photosynthetic O₂ comes from water; isotopically labelled H₂¹⁸O traces the path.
• NCERT exercise 7.25: 10.00 g NH₃ and 20.00 g O₂ in 4NH₃ + 5O₂ → 4NO + 6H₂O. O₂ (0.625 mol) limits: NO = 0.500 mol = 15.0 g.
• NCERT exercise 7.23: excess chlorine in water is removed with SO₂: Cl₂ + SO₂ + 2H₂O → 2Cl⁻ + SO₄²⁻ + 4H⁺.