Redox Reactions — NEET Notes

Classical idea of redox — how the definition grew

The word oxidation began as a description of what happens when something burns in air. Magnesium glows white in a flame and forms MgO; sulphur burns to SO₂; carbon burns to CO₂. The unifying feature was gain of oxygen, and reduction was its opposite — loss of oxygen, as when mercuric oxide decomposes back into mercury and O₂ on heating. This worked for the chemistry of the 18th and 19th centuries, but it left out everything that did not involve oxygen at all.

Chemists then broadened the definition. When CH₄ burns, hydrogen is replaced by oxygen in the product — so loss of hydrogen from a substance must also count as oxidation. Reactions like 2H₂S + O₂ → 2S + 2H₂O are oxidation of H₂S by exactly this criterion. The definition expanded further once reactions like Mg + Cl₂ → MgCl₂ came under scrutiny: magnesium is oxidised here despite no oxygen being involved, because a more electronegative element (chlorine) has been added. So the classical picture settled into a four-part rule.

Oxidation = gain of O or loss of H. Reduction = loss of O or gain of H.

The classical (pre-electron) definition

Extending the rule to electronegative and electropositive elements gives the fully general classical definition: oxidation is the addition of oxygen or any electronegative element, or the removal of hydrogen or any electropositive element. Reduction is the reverse. From the very first analyses it was clear that oxidation and reduction never happen alone — whenever one species loses oxygen, something else gains it. The two processes are inseparable, and the word redox was coined to capture this.

Redox as electron transfer

The classical picture breaks down when the product is not obviously ionic — when methane and chlorine make CCl₄, where exactly is the "loss of hydrogen"? The modern picture answers this cleanly. In 2Na + Cl₂ → 2NaCl, the product is genuinely Na⁺Cl⁻, so we can split the overall change into two half-reactions: 2Na → 2Na⁺ + 2e⁻, and Cl₂ + 2e⁻ → 2Cl⁻. Sodium loses electrons; chlorine gains them. The same logic, when applied to covalent compounds with the assumption that the more electronegative atom takes both bonding electrons, gives a consistent definition for every reaction.

In NCERT's competitive electron-transfer experiments, a strip of zinc placed in copper nitrate solution turns reddish (copper deposits on it) while the blue Cu²⁺ colour fades — zinc is oxidised, Cu²⁺ is reduced. Copper does not do the reverse in zinc sulphate solution. Copper, on the other hand, can be oxidised by Ag⁺. Stack these observations and an electron-releasing order emerges: Zn > Cu > Ag. This is the embryonic electrochemical series — the same ordering used in Class XII to predict cell EMF and metal-reactivity.

Oxidation number — the six rules

For covalent compounds, "loss of electrons" is fuzzy because the electrons are shared. The oxidation number (ON) of an atom is a bookkeeping device: we pretend that the bonding pair belongs entirely to the more electronegative atom and assign each atom the charge it would carry under that pretence. In CO₂, both C=O bonds give all four shared electrons to oxygen, so each O is −2 and C is +4. The assignment is artificial, but it is consistent and lets us track electron flow in any reaction — covalent or ionic. NCERT lists six rules; the NIOS supplement gives the same set with worked examples.

Two NCERT subtleties matter for NEET. First, the maximum positive oxidation state of a main-group element equals its group number for groups 1 and 2, or group number − 10 (long form) for groups 13–17. So across the third period the highest oxidation state rises 1 → 2 → 3 → 4 → 5 → 6 → 7 from Na through Cl. Second, when the same element appears in several environments inside one species — as in C₃O₂, Br₃O₈ or Na₂S₄O₆ — averaging gives a fractional oxidation number. The fraction is real on paper but not on the structure: it is the average of distinct integer states in different positions. NCERT calls this the paradox of fractional oxidation number.

The five types of redox reactions

NCERT classifies every redox reaction into four main types — combination, decomposition, displacement and disproportionation — and singles out displacement into a metal-vs-metal flavour and a non-metal-vs-non-metal flavour. A fifth type, comproportionation, is the formal reverse of disproportionation and is examined less often but worth recognising. The classification is exam-bait: NEET 2021 directly asked which of four reactions was a metal displacement, and the answer required ruling out a non-metal displacement and two decomposition reactions.

Disproportionation — the signature pattern

Disproportionation deserves a section to itself because NEET tests it almost every alternate year, and because the pattern is unmistakable once you see it. The requirement is exact: one element, present in a single intermediate oxidation state, simultaneously oxidised and reduced. So the element must be capable of at least three oxidation states — the starting one and two more, one higher and one lower. Hydrogen peroxide is the textbook case: oxygen in H₂O₂ is in the −1 state, and the reaction 2H₂O₂ → 2H₂O + O₂ produces oxygen in the −2 state (in H₂O) and in the 0 state (in O₂).

3 ClO⁻ → 2 Cl⁻ + ClO₃⁻  ·  same element, two destinies.

Disproportionation of hypochlorite — chlorine goes from +1 to −1 and +5

The chlorine oxoanions illustrate the rule beautifully. ClO⁻, ClO₂⁻ and ClO₃⁻ all disproportionate; ClO₄⁻ cannot, because chlorine is already at its maximum oxidation state of +7 and has nowhere higher to go. NCERT writes the three permitted disproportionations as: 3ClO⁻ → 2Cl⁻ + ClO₃⁻ · 6ClO₂⁻ → 4ClO₃⁻ + 2Cl⁻ · 4ClO₃⁻ → Cl⁻ + 3ClO₄⁻. Fluorine is the exception even among halogens: because it cannot exhibit a positive oxidation state, F₂ does not disproportionate in alkali. Instead it gives 2F₂ + 2OH⁻ → 2F⁻ + OF₂ + H₂O.

Balancing redox — oxidation number method

Two algorithms balance any redox reaction: the oxidation-number method, which uses the change in ON of the oxidised and reduced atoms; and the half-reaction (ion-electron) method, which splits the reaction into the two halves and balances each separately. NCERT teaches both, and NEET has tested both — the half-reaction method appears in NEET 2018 (MnO₄⁻ + C₂O₄²⁻) and again in NEET 2023 (Cr₂O₇²⁻ + SO₃²⁻). Knowing both is non-negotiable; preferring one is a matter of style.

The final balanced equation for the NEET 2023 problem is Cr₂O₇²⁻ + 3 SO₃²⁻ + 8 H⁺ → 2 Cr³⁺ + 3 SO₄²⁻ + 4 H₂O, giving coefficients a = 1, b = 3, c = 8. Notice how mechanical the algorithm is: at no point did we have to "see" anything clever. That is the whole point — the ON method is a procedure, not a puzzle.

Balancing redox — half-reaction (ion-electron) method

The half-reaction method does the same job by splitting the redox reaction into its two halves, balancing each half for atoms and charge separately, then adding them back together so the electrons cancel. This is the method working chemists actually use, because each half-reaction is a complete electrochemical statement that can be looked up in standard tables of electrode potentials. NCERT illustrates the method by balancing Fe²⁺ + Cr₂O₇²⁻ → Fe³⁺ + Cr³⁺ in acidic medium. The seven steps below adapt to any reaction in either medium.

Worked through carefully on Fe²⁺ + Cr₂O₇²⁻: the two halves are Fe²⁺ → Fe³⁺ + e⁻ (oxidation, 1-electron) and Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O (reduction, 6-electron). Multiply the oxidation half by 6 so both involve 6 electrons, then add: 6 Fe²⁺ + Cr₂O₇²⁻ + 14 H⁺ → 6 Fe³⁺ + 2 Cr³⁺ + 7 H₂O. The same engine works for MnO₄⁻ + C₂O₄²⁻ + H⁺ → Mn²⁺ + CO₂ + H₂O — split into 2 e⁻ (oxalate) and 5 e⁻ (permanganate) halves, multiply by 5 and 2, and recover the NEET 2018 coefficients 2 : 5 : 16 for MnO₄⁻ : C₂O₄²⁻ : H⁺.

Basic medium — the OH⁻ adjustment

For basic medium, NCERT prescribes a neat patch on top of the acidic-medium procedure: balance the equation exactly as you would in acid (using H⁺), then for every H⁺ on either side, add an equal number of OH⁻ ions to both sides. Wherever H⁺ and OH⁻ end up on the same side, combine them to H₂O. Wherever H₂O molecules now appear on both sides, cancel the redundant ones. The result is a clean basic-medium equation with OH⁻ instead of H⁺.

NCERT's worked example is permanganate oxidising iodide in basic solution: MnO₄⁻ + I⁻ → MnO₂ + I₂. The acidic-balanced halves are 2 I⁻ → I₂ + 2 e⁻ (oxidation) and MnO₄⁻ + 4 H⁺ + 3 e⁻ → MnO₂ + 2 H₂O (reduction). Multiply oxidation by 3 and reduction by 2, add 4 OH⁻ to each side to neutralise the 4 H⁺, and combine. The final basic equation is 6 I⁻ + 2 MnO₄⁻ + 4 H₂O → 3 I₂ + 2 MnO₂ + 8 OH⁻. NEET 2022 tested this exact pattern: in neutral or faintly alkaline medium KMnO₄ oxidises iodide all the way to iodate (+5), with Mn changing from +7 → +4 (MnO₂), not +2.

Redox titrations

Redox titrations measure the strength of an oxidising or reducing solution by reacting it stoichiometrically with a solution of the opposite kind, exactly as acid-base titrations do for H⁺ and OH⁻. The end point is signalled by a colour change, and NCERT lists three flavours of indicator system. Permanganate is its own indicator — the intense purple of MnO₄⁻ vanishes as it is reduced to colourless Mn²⁺, and the first persistent pink tinge after the equivalence point marks the end. Dichromate needs an external indicator like diphenylamine, which Cr₂O₇²⁻ oxidises to an intense blue just after the equivalence point. Iodimetric titration exploits the deep blue starch–iodine complex: iodine liberated by an oxidising agent is then titrated against standard thiosulphate, with starch added near the end point so that the blue colour disappears exactly when iodine is consumed.

Underlying every redox titration is the same conservation that drives balancing: moles of electrons gained equal moles of electrons lost. This is why a sharp stoichiometric ratio always exists between the two solutions — once you know which is oxidised, which is reduced, and by how many electrons, the entire titration becomes one equation.

Redox reactions and electrode processes

If the two halves of a redox reaction can be physically separated, the electrons forced to travel from one half to the other through an external wire become a current. NCERT motivates this by modifying the Zn + Cu²⁺ experiment: place a zinc rod in zinc sulphate solution in one beaker, a copper rod in copper sulphate in another, connect the two solutions with a salt bridge (a U-tube of KCl or NH₄NO₃ in agar), and connect the two metal rods externally with a wire and an ammeter. The resulting device is the Daniell cell — the prototype of every battery.

At the zinc rod (the anode), Zn → Zn²⁺ + 2 e⁻ — oxidation, electrons released. At the copper rod (the cathode), Cu²⁺ + 2 e⁻ → Cu — reduction, electrons consumed. Electrons travel through the wire from zinc to copper; ions migrate through the salt bridge to keep each beaker electrically neutral. The potential difference between the two electrodes — measured at unit concentration, 1 atm pressure, 298 K — is the cell's standard EMF. By convention, the standard hydrogen electrode is assigned E° = 0.00 V; every other electrode is measured against it.

E°cell = E°cathode − E°anode  ·  positive E°cell means the reaction is spontaneous.

Standard EMF — the test for spontaneity of any redox reaction

Standard reduction potentials let you predict any redox outcome on paper. NEET 2022 used this: given E° for MnO₄⁻/Mn²⁺ (+1.51 V) and O₂/H₂O (+1.23 V), will MnO₄⁻ liberate O₂ from water in acid? Treat MnO₄⁻/Mn²⁺ as the cathode (it has the higher reduction potential) and O₂/H₂O as the anode (lower; it gets reversed and acts as an oxidation, so it loses water → O₂). E°cell = 1.51 − 1.23 = +0.287 V. Positive → spontaneous → yes, MnO₄⁻ will oxidise water to O₂ in acid. The full chapter on these calculations is Class XII Electrochemistry.