Redox

Redox

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

A redox reaction couples oxidation (loss of electrons, rise in oxidation number) with reduction (gain of electrons, fall in oxidation number). Conservation requires total electron loss = total electron gain, expressed as

ΔO.N.(oxidation) × n(oxidant) = ΔO.N.(reduction) × n(reductant).

Assign oxidation numbers using the priority: free element → 0; F is always −1; O is −2 except in peroxides (−1) and OF₂ (+2); H is +1 except in metal hydrides (−1); the sum equals the overall charge.

E°cell = E°cathode − E°anode; spontaneous when E°cell > 0. For JEE Main, master n-factor of KMnO₄ (acidic = 5, neutral = 3, strong base = 1) and the Nernst equation: E = E° − (0.0591/n) log Q at 298 K.

---

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Oxidation Number Rules

Oxidation number (O.N.) is a bookkeeping charge assuming heterolytic splitting: the more electronegative atom takes the bonding electrons. Apply the IUPAC priority ladder above, and check using the sum rule (sum of O.N. = charge of the species). Example: in Cr₂O₇²⁻, let O.N. of Cr = x; then 2x + 7(−2) = −2 → x = +6.

Balancing Methods

Oxidation-number method: identify atoms changing O.N., find ΔO.N. per atom, cross-multiply to balance electron equivalents, then balance O (with H₂O) and H (with H⁺/OH⁻).

Ion-electron (half-reaction) method: split into two half-reactions, balance atoms other than O and H, then balance O using H₂O and H using H⁺ (acidic) or OH⁻ (basic — add H₂O on the H-deficient side and OH⁻ on the other). Charge is balanced using electrons — this is the most-skipped step in JEE.

n-Factor Shortcuts

Species	Medium	n-factor
KMnO₄	acidic	5
KMnO₄	neutral/faintly basic	3
KMnO₄	strongly basic	1
K₂Cr₂O₇	acidic	6
H₂O₂ → H₂O	(as oxidant)	2
H₂O₂ → O₂	(as reductant)	2
FeSO₄	acidic	1

Equivalent mass = Molar mass / n-factor — directly used in normality and titration problems.

Electrochemical Series

Standard potentials are tabulated as reduction potentials. The species with the higher E° is reduced (cathode); the one with lower E° is oxidised (anode). A positive E°cell signals a spontaneous galvanic reaction; a negative E°cell needs electrolysis.

---

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Disproportionation and Comproportionation

In disproportionation, the same element is both oxidised and reduced: 2H₂O₂ → 2H₂O + O₂ (O goes from −1 to −2 and 0); 2Cu⁺ → Cu²⁺ + Cu; Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O (cold, dilute alkali). The reverse process — two different O.N.s converging to one — is comproportionation, e.g., 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O where Mn ends at +2 from +7 and Fe ends at +3 from +2 (not strictly comproportionation, but a useful contrast).

Nernst and Thermodynamic Link

The Nernst equation is derived from ΔG = ΔG° + RT ln Q combined with ΔG = −nFE:

E = E° − (RT/nF) ln Q = E° − (0.0591/n) log Q at 298 K.

So ΔG° = −nFE°cell; a larger E°cell means a more negative ΔG° and a more exergonic reaction. JEE sometimes disguises electrochemistry as thermodynamics: if ΔG° < 0, E°cell > 0.

Common Pitfalls

• Treating O.N. as identical to valency (in H₂O₂, valency of O is 2, O.N. is −1).
• Forgetting the peroxide exception: in Na₂O₂, BaO₂, H₂O₂, O = −1.
• Using 0.0591 in the Nernst equation at temperatures other than 298 K — replace it with 0.0257·T/298 or 0.0591 T/298.
• Miscounting n-factor of K₂Cr₂O₇ as 2 (it’s 6: each Cr goes from +6 to +3, two Cr atoms).
• Assuming all metal + O₂ reactions are redox — they are, but noble metals (Au, Pt) need extremely harsh conditions, often tested as a trick option.

Exam Strategy

Redox contributes roughly 3% of JEE Main Chemistry — about 1 question in each shift, often an integer-type (matching n-factor to equivalent mass) or a numerical (E°cell, Nernst). High-yield drill: balance a full equation in basic medium (e.g., MnO₄⁻ + I⁻ → MnO₂ + IO₃⁻) and compute E°cell from a two-line standard-potential table.

Practice Prompts

1. Balance in basic medium: Cr(OH)₃ + H₂O₂ → CrO₄²⁻ and find n-factor of H₂O₂.
2. For Fe³⁺ + I⁻ → Fe²⁺ + I₂, given E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(I₂/I⁻) = +0.54 V, compute E°cell, ΔG°, and predict direction of spontaneity.

---

Content adapted based on your selected roadmap duration. Switch tiers using the selector above.