Redox

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Redox — Key Facts for JEE Advanced

Oxidation Number Rules:

• Free elements: oxidation number = 0
• O: −2 (except in peroxides where it is −1, and in OF₂ where it is +2)
• H: +1 (except in metal hydrides where it is −1)
• Monoatomic ion: oxidation number = charge
• Sum of oxidation numbers equals zero for neutral compounds; equals charge for polyatomic ions

Quick Assignment Method:

1. Assign O.N. to known elements first (use rules above)
2. Sum all O.N.s and solve for the unknown element
3. Check: Max possible O.N. for a main group element ≤ Group number (for metals ≤ Group number − 10)

Identifying Redox Reactions:

• Oxidation: O.N. increases (loss of electrons) — Reducing agent
• Reduction: O.N. decreases (gain of electrons) — Oxidizing agent
• Non-redox: O.N. stays same for all elements (precipitation, acid-base)

Balancing Redox Equations:

• Half-reaction method (acidic):

  1. Split into oxidation and reduction half-reactions
  2. Balance atoms other than O and H
  3. Balance O by adding H₂O, H by adding H⁺
  4. Balance charge with electrons (e⁻)
  5. Multiply halves so e⁻ cancel
  6. Add halves, simplify

• Half-reaction method (basic): Same as acidic, then add OH⁻ to neutralize H⁺ and balance

Common JEE Redox Pairs:

Oxidizing Agent	Reducing Product	Environment
KMnO₄	Mn²⁺ (acidic), MnO₂ (basic/neutral), MnO₄²⁻ (strongly basic)	Various
K₂Cr₂O₇	Cr³⁺	Acidic
H₂SO₄ (hot concentrated)	SO₂, S	Reducing acids

⚡ Exam Tip: In any redox problem, ALWAYS assign oxidation numbers first before deciding what type of reaction it is. Many JEE questions trap students who assume all reactions involving oxygen are redox.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Redox — Chemistry Study Guide

Oxidation Number (O.N.) in Depth:

The oxidation number is a formal concept used to track electrons in redox reactions. Unlike oxidation state, which is a real integer, O.N. is assigned by set rules and can be fractional (e.g., in Fe₃O₄ where average O.N. = +8/3).

Common oxidation number patterns for JEE:

• Carbon in organic compounds: −4 to +4 depending on substituents
• Halogens: −1 (in metal halides), +1, +3, +5, +7 (in oxyacids/oxyhalides)
• Phosphorus: −3, +3, +5
• Sulfur: −2, +2, +4, +6
• Nitrogen: −3, +1, +2, +3, +4, +5

Special cases to memorize for JEE:

• In FeS₂ (pyrite): Fe = +2, S = −1
• In MnO₄²⁻: Mn = +6
• In H₂S₂O₈ (Marshall’s acid): each S = +6, each O in peroxide bridge = −1
• In CrO₅: Cr = +6, each peroxide O = −1

Types of Redox Reactions:

1. Combination Reactions: A + B → AB (both O.N. change) Example: N₂ + O₂ → 2NO (O.N. of N changes from 0 to +2)

2. Decomposition Reactions: AB → A + B (requires energy input) Example: 2H₂O₂ → 2H₂O + O₂ (decomposition of H₂O₂)

3. Displacement Reactions: A + BC → AC + B (metal displaces metal, or halogen displaces halogen) Activity series: Li > K > Na > Ca > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Hg > Ag > Au Nonmetal displacement: F₂ > Cl₂ > Br₂ > I₂

4. Disproportionation (Self Redox): Same element undergoes both oxidation and reduction Example: 2Cu⁺ → Cu²⁺ + Cu⁰ Example: Cl₂ + 2OH⁻ → Cl⁻ + OCl⁻ + H₂O Condition: The intermediate oxidation state must be stable enough to exist

5. Comproportionation: Opposite of disproportionation: two species of the same element at different O.N. react to give a product at an intermediate O.N. Example: 2Fe³⁺ + Fe → 3Fe²⁺

Balancing by Oxidation Number Method:

1. Write skeleton equation
2. Assign O.N. to elements undergoing change
3. Find change in O.N. per atom × number of atoms
4. Cross-multiply: oxidizing agent and reducing agent coefficients
5. Balance remaining atoms by inspection

JEE-Important Balancing Examples:

Permanganate in acidic medium: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ + H₂O

Balancing: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O …(reduction, n-factor = 5) Fe²⁺ → Fe³⁺ + e⁻ …(oxidation, n-factor = 1)

Multiply Fe half by 5: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Iodometric titration balance: S₂O₃²⁻ + I₂ → I⁻ + S₄O₆²⁻

n-factor: S₂O₃²⁻ → S₄O₆²⁻: S goes from +2 to +2.5 (average)… actually: In S₂O₃²⁻: average S O.N. = +2; In S₄O₆²⁻: average S O.N. = +2.5 But actually the reaction is: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻ and I₂ + 2e⁻ → 2I⁻ So: 2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻

Equivalent Mass in Redox:

Equivalent mass of an oxidizing agent = Molar mass / n-factor (where n = number of electrons gained per formula unit)

Substance	Reaction Condition	n-factor	Reasoning
KMnO₄	Acidic	5	Mn⁷⁺ → Mn²⁺ (+5e⁻)
KMnO₄	Neutral/basic	3	Mn⁷⁺ → Mn⁴⁺ (+3e⁻)
K₂Cr₂O₇	Acidic	6	2Cr⁶⁺ → 2Cr³⁺ (+3e⁻ each)
H₂C₂O₄	Acidic	2	C³⁺ → C⁴⁺ (+2e⁻ per C₂O₄²⁻)
FeSO₄	Acidic	1	Fe²⁺ → Fe³⁺ (+1e⁻)
Na₂S₂O₃	Acidic	1	S₂O₃²⁻ → S₄O₆²⁻ (1 e⁻ per S₂O₃²⁻)

⚡ Exam Tip: Always identify the actual product in the given condition before calculating n-factor. KMnO₄ in alkaline medium has n=3, not 5. This is a common JEE trap.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Redox — Comprehensive Chemistry Notes

Advanced Oxidation Number Concepts:

Peroxide and Superoxide O.N.:

• In H₂O₂: O.N. of O = −1 (each O bonded to H)
• In Na₂O₂: Same, −1
• In BaO₂: Same, −1
• In KO₂ (superoxide): O.N. of O = −½ (average; one O is −1, one is 0 in formal terms, but averaged = −½)
• In CrO₅: Four peroxide O’s (−1 each) and one oxide O (−2), Cr = +6
• In Fe₃O₄: FeO·Fe₂O₃; Fe in FeO is +2, Fe in Fe₂O₃ is +3

O.N. in Complex Ions:

• In [Fe(CN)₆]⁴⁻: Fe = +2 (low spin, t₂g⁶)
• In [Fe(CN)₆]³⁻: Fe = +3 (low spin, t₂g⁵)
• In [Co(NH₃)₆]³⁺: Co = +3 (low spin, t₂g⁶, diamagnetic)
• In [Co(NH₃)₆]²⁺: Co = +2 (high spin, t₂g⁵e_g², paramagnetic)

Fractional Oxidation States:

When an element has fractional O.N., check if: (a) The compound is a mixed oxide (e.g., Fe₃O₄ = FeO·Fe₂O₃) (b) The element exists in multiple oxidation states simultaneously (c) The O.N. is an average (only meaningful for bookkeeping)

JEE Example: In Pb₃O₄ (red lead/minium): Pb in PbO = +2, Pb in PbO₂ = +4 Average = (2 + 4 + 4)/3 = 10/3 ≈ 3.33 Can also be viewed as 2PbO·PbO₂

N-Factor (Total Change in O.N.) in Complex Redox Systems:

For a compound that can act as both oxidizing and reducing agent (amphoteric redox):

• I₂ in碱性 medium undergoes disproportionation: I₂ + 2OH⁻ → I⁻ + OI⁻ + H₂O
• NO₂ in water: 2NO₂ + H₂O → HNO₂ + HNO₃ (disproportionation)

n-Factor for compounds with multiple atoms of the same element in different environments:

Example: N₂H₄ (hydrazine) N oxidation state: −2 In reaction: N₂H₄ → N₂ + 4H⁺ + 4e⁻ (each N goes from −2 to 0, change of 2 per N, total 4e⁻) n-factor = 4

Example: CuH (copper hydride) Cu = +1, H = −1 Reaction: 2CuH → Cu²⁺ + Cu + 2H⁻ + e⁻… wait Actually: Cu⁺ undergoes disproportionation: 2Cu⁺ → Cu²⁺ + Cu⁰ n-factor = 1 per Cu⁺ (one electron per Cu)

Equivalent Mass Derivation:

E = M/n where n = total electrons transferred per mole of substance

For oxidizing agents: E = M / (number of electrons gained per formula unit) For reducing agents: E = M / (number of electrons lost per formula unit) For salts undergoing redox: E = M / (electrons transferred per metal atom × number of metal atoms)

Volumetric Analysis — Redox Titrations:

1. Permanganometry (KMnO₄ titration): Indicator: KMnO₄ is self-indicating (pink in acidic, decolorizes when reduced) End point: Pink color disappears at endpoint Standardization: With oxalic acid (H₂C₂O₄·2H₂O) or sodium oxalate (Na₂C₂O₄)

Reaction in acidic medium: 2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 10CO₂ + 8H₂O

Calculation: meq of KMnO₄ = meq of oxalic acid N₁V₁ = N₂V₂

2. Dichrometry (K₂Cr₂O₇ titration): Indicator: Diphenylamine (blue-green color change) or N-phenylanthranilic acid Reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O n-factor = 6

3. Iodometry: Involves liberation of I₂ which is titrated with Na₂S₂O₃ I₂ + 2Na₂S₂O₃ → 2NaI + Na₂S₄O₆ Indicator: Starch solution (blue-black complex with I₂)

Steps: (a) Oxidizing agent + KI (excess) → I₂ liberated (b) I₂ titrated with Na₂S₂O₃ (c) Starch added at end point (when I₂ color fades to pale yellow)

⚡ Exam Tip: In iodometry, always add starch indicator only AFTER the I₂ color has faded to pale yellow (near endpoint), otherwise starch-I₂ complex is hard to dissociate and causes error.

Redox and Equivalent Mass in Organic Chemistry:

Equivalent mass of organic oxidizing/reducing agents:

• KMnO₄ (as oxidant in alkaline medium): E = M/3
• K₂Cr₂O₇ (as oxidant in acidic medium): E = M/6
• H₂SO₄ (as oxidant, hot concentrated): E = M/2 (reduces to SO₂)

For organic compounds undergoing oxidation:

• Ethanol (CH₃CH₂OH) → Acetic acid: n = 4e⁻ (C₂ goes from −2 to +2 avg, change of 4)
• Glucose (C₆H₁₂O₆) → CO₂: n = 24e⁻ (each C from 0 avg to +4, total change = 24e⁻ per molecule)

Biological Redox Systems:

Cellular respiration chain: NAD⁺ → NADH (biological reducing agent, NAD⁺ + H⁺ + 2e⁻ → NADH) FAD → FADH₂ O₂ is the final electron acceptor in aerobic respiration

Photosynthesis: 2H₂O → O₂ + 4H⁺ + 4e⁻ (water splitting, oxygen evolving complex)

Redox in Industrial Processes:

Thermite process: Fe₂O₃ + 2Al → 2Fe + Al₂O₃ (ΔH < 0, highly exothermic) Used in rail welding and metal extraction

Bessemer process: Conversion of FeO (impurity in steel) - actually removes carbon by oxidation: 2FeO + Si → 2Fe + SiO₂

Hydrogen-oxygen fuel cell: 2H₂ + O₂ → 2H₂O, E°cell = 1.23 V

JEE Advanced Problems — Advanced Level:

Problem type 1: Finding O.N. in complex compounds In CrO₂Cl₂ (chromyl chloride): Cr = +6, O = −2 (×2 = −4), Cl = −1 (×2 = −2) Sum = +6 − 4 − 2 = 0 ✓

Problem type 2: Balancing in basic medium Cl₂ + OH⁻ → Cl⁻ + ClO₃⁻ + H₂O

Half-reactions: 3Cl₂ → 2ClO₃⁻: Cl₂ + 6H₂O → 2ClO₃⁻ + 12H⁺ + 10e⁻ Cl₂ → 2Cl⁻: Cl₂ + 2e⁻ → 2Cl⁻

Multiply second by 5: 5Cl₂ + 10e⁻ → 10Cl⁻ Add: 6Cl₂ + 6H₂O → 2ClO₃⁻ + 10Cl⁻ + 12H⁺ Neutralize H⁺: 12H⁺ + 12OH⁻ → 12H₂O 6Cl₂ + 6H₂O + 12OH⁻ → 2ClO₃⁻ + 10Cl⁻ + 12H₂O Simplify: 3Cl₂ + 6OH⁻ → ClO₃⁻ + 5Cl⁻ + 3H₂O

Problem type 3: n-factor of compounds In Marshall’s acid (H₂S₂O₈): The structure is HO₃S-O-O-SO₃H with peroxide bridge Each S = +6 (as in sulfuric acid) Each oxygen in bridge (peroxide) = −1, all other oxygens = −2 n-factor calculation: The peroxide O’s each undergo reduction to −2 (change of 1 per O, 2 O’s total): n = 2

⚡ Exam Tip: When asked for n-factor of a salt like KMnO₄ in a specific condition, first identify the change in oxidation state of the element, multiply by number of atoms of that element, then divide molar mass by this number.

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