Redox Reactions

Long before electrons were understood, chemists defined oxidation as the addition of oxygen (or a more electronegative element) to a substance, or the removal of hydrogen from it. Reduction was the exact opposite — addition of hydrogen, or removal of oxygen. For example, when magnesium burns in air, 2Mg + O₂ → 2MgO, magnesium is oxidised because oxygen is added to it. In the reverse sense, when CuO is heated with hydrogen, CuO + H₂ → Cu + H₂O, copper oxide is reduced because it loses oxygen.

This classical picture works only for simple cases and breaks down for reactions where no oxygen or hydrogen is involved at all — for instance 2Na + Cl₂ → 2NaCl clearly involves oxidation and reduction, but neither oxygen nor hydrogen appears anywhere. This is exactly the gap that the electron-transfer definition (next section) was built to close. You should still know the classical definitions because they are how the terms 'oxidation' and 'reduction' are first defined in NCERT and are often asked as a two-mark definition-with-example question.

A reaction in which one reactant is oxidised and another is simultaneously reduced is called a redox (reduction–oxidation) reaction. Oxidation and reduction can never occur alone — every redox reaction is really two half-reactions happening together.

The modern and more general definition: oxidation is the loss of one or more electrons by an atom, ion or molecule; reduction is the gain of one or more electrons. Consider 2Na(s) + Cl₂(g) → 2NaCl(s). Sodium loses one electron to become Na⁺ (oxidation) and chlorine gains one electron to become Cl⁻ (reduction). Because electrons lost by one species must be gained by another, oxidation and reduction always occur together — hence the combined name 'redox'.

The species that gets oxidised (loses electrons) is called the reducing agent or reductant, because it causes reduction of the other species while itself getting oxidised. The species that gets reduced (gains electrons) is called the oxidising agent or oxidant, because it causes oxidation of the other species while itself getting reduced. In Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s), zinc loses two electrons (Zn → Zn²⁺ + 2e⁻, oxidation, Zn is the reducing agent) and Cu²⁺ gains two electrons (Cu²⁺ + 2e⁻ → Cu, reduction, Cu²⁺ is the oxidising agent).

This electron-transfer picture is completely general — it applies even when no oxygen or hydrogen is involved — and it is the definition examiners expect when they say 'in terms of electron transfer'. Remember the memory device: OIL RIG — Oxidation Is Loss, Reduction Is Gain (of electrons).

Oxidation number (O.N.) is a formal charge assigned to an atom in a molecule or ion, assuming every bond is fully ionic and the shared electron pair goes entirely to the more electronegative atom. It lets us track electron transfer even in covalent molecules where no full ionic charge actually exists. An increase in oxidation number during a reaction signals oxidation; a decrease signals reduction.

The rules NCERT gives, in the order you should apply them:

• The oxidation number of an atom in its free (uncombined) element is zero — e.g. O.N. of Na in Na(s), of O in O₂, of P in P₄ is 0.
• For a monoatomic ion, the oxidation number equals the charge on the ion — O.N. of Na⁺ is +1, of Cl⁻ is −1.
• Combined oxygen is usually −2 (CO₂, H₂O). Exceptions: in peroxides (H₂O₂, Na₂O₂) it is −1; in superoxides (KO₂) it is −½; in OF₂ it is +2 and in O₂F₂ it is +1, because F is more electronegative than O.
• Combined hydrogen is usually +1 (H₂O, NH₃). Exception: in ionic/metal hydrides (NaH, CaH₂) it is −1, because H is more electronegative than the metal.
• Fluorine is always −1 in every compound (it is the most electronegative element).
• Alkali metals (Li, Na, K…) are always +1; alkaline earth metals (Be, Mg, Ca…) are always +2 in their compounds.
• The algebraic sum of oxidation numbers of all atoms in a neutral molecule is zero; in a polyatomic ion it equals the net charge on the ion.

Worked example: find O.N. of Mn in KMnO₄. Let O.N. of Mn = x. K is +1, each O is −2 (4 oxygens). Sum = 0: (+1) + x + 4(−2) = 0 ⇒ x = +7. So Mn is in the +7 state — the highest oxidation state manganese shows, which is why KMnO₄ is such a strong oxidising agent.

NCERT groups redox reactions into four familiar reaction types, viewed now through the oxidation-number lens.

Combination reactions — two or more elements/compounds combine to form one product. When at least one reactant is a free element, the reaction is necessarily redox: C(s) + O₂(g) → CO₂(g) (C goes 0 → +4; O goes 0 → −2). Not every combination reaction is redox — CaO + CO₂ → CaCO₃ has no change in any oxidation number, so it is not a redox reaction.

Decomposition reactions — one compound breaks into two or more simpler substances; the reverse of combination. 2KClO₃(s) → 2KCl(s) + 3O₂(g): Cl goes +5 → −1 (reduced) and O goes −2 → 0 (oxidised). 2H₂O₂(aq) → 2H₂O(l) + O₂(g) is also decomposition and redox. Like combination reactions, not all decompositions are redox — CaCO₃ → CaO + CO₂ is not, since no atom changes oxidation state.

Displacement reactions — an atom or ion in a compound is replaced by an atom or ion of another element. Metal displacement: CuSO₄(aq) + Zn(s) → Cu(s) + ZnSO₄(aq), where a more reactive metal displaces a less reactive one from its salt solution. Non-metal displacement (mostly hydrogen or halogen displacement): Cl₂(g) + 2KBr(aq) → 2KCl(aq) + Br₂(l), where a more reactive halogen displaces a less reactive one — this is only possible because Cl₂ is a stronger oxidising agent than Br₂.

Disproportionation reactions — a special class of redox reaction in which the same element, in one intermediate oxidation state, is simultaneously oxidised and reduced. Classic example: 2H₂O₂(aq) → 2H₂O(l) + O₂(g); oxygen in H₂O₂ is at −1, and it goes to −2 in H₂O (reduced) and to 0 in O₂ (oxidised) at the same time. Another NCERT favourite: P₄(s) + 3NaOH(aq) + 3H₂O(l) → PH₃(g) + 3NaH₂PO₂(aq), where P (O.N. = 0) disproportionates to −3 (in PH₃) and +1 (in NaH₂PO₂).

This method balances an equation by making the total increase in oxidation number equal to the total decrease. The steps (as NCERT sets them out):

1. Write the skeletal (unbalanced) equation showing all reactants and products.
2. Assign oxidation numbers to every atom and identify the atoms whose oxidation number changes.
3. Calculate the increase or decrease in oxidation number per atom, then multiply by the number of such atoms in the formula to get the total change for that species.
4. Multiply the formulas of the oxidising and reducing agents by suitable numbers so that the total increase in O.N. equals the total decrease.
5. Balance all the atoms other than H and O.
6. Finally balance O by adding H₂O molecules, and balance H by adding H⁺ ions (acidic medium) — if the medium is basic, add an equal number of OH⁻ ions to both sides at the end to neutralise the H⁺.

Worked example: KMnO₄ + KI + H₂SO₄ → MnSO₄ + I₂ + K₂SO₄ + H₂O (acidic medium).

Mn goes from +7 (in KMnO₄) to +2 (in MnSO₄): a decrease of 5 per Mn atom. I goes from −1 (in KI) to 0 (in I₂): an increase of 1 per I atom, and since I₂ needs 2 iodine atoms, we need 2 KI to supply 2 electrons lost, matched against 5 electrons gained per Mn — so we need 2 KMnO₄ and 10 KI to make the total increase (10 × 1 = 10) equal the total decrease (2 × 5 = 10). Balancing the rest atom by atom and then O and H gives:

2KMnO₄ + 10KI + 8H₂SO₄ → 2MnSO₄ + 5I₂ + 6K₂SO₄ + 8H₂O

This method, developed by Jorgensen, splits the redox equation into two half-reactions — one for oxidation, one for reduction — balances each separately, then adds them so electrons cancel. Steps: (1) write the skeletal ionic equation; (2) split it into an oxidation half and a reduction half; (3) balance all atoms except O and H in each half; (4) balance O by adding H₂O to the side short of oxygen; (5) balance H by adding H⁺ to the side short of hydrogen (this gives the acidic-medium version); (6) balance charge on each half by adding electrons; (7) multiply the two half-reactions by numbers that make the electrons lost equal the electrons gained, then add and cancel; (8) for a reaction in basic medium, add OH⁻ equal to the number of H⁺ on both sides, combine H⁺ and OH⁻ into H₂O, and cancel water molecules common to both sides.

Acidic-medium example (Fe²⁺ reducing MnO₄⁻): Oxidation half: Fe²⁺ → Fe³⁺ + e⁻. Reduction half: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Multiply the oxidation half by 5 and add: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. (Charge check: LHS = −1+8+10 = +17; RHS = +2+15 = +17. Balanced.)

Basic-medium example (MnO₄⁻ oxidising I⁻ to IO₃⁻, itself reduced to MnO₂): Reduction half: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. Oxidation half: I⁻ + 6OH⁻ → IO₃⁻ + 3H₂O + 6e⁻. Multiply the reduction half by 2 (to get 6 electrons on each side) and add, then cancel common H₂O and OH⁻: 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻. (Oxygen check: LHS 8+1=9, RHS 4+3+2=9 — balanced; charge check: LHS −3, RHS −3 — balanced.)

Redox titrations use the fact that oxidising and reducing agents react in a fixed, calculable electron ratio, exactly as acid–base titrations use the mole ratio of H⁺ and OH⁻. The two workhorse oxidants in the school lab are potassium permanganate (KMnO₄) and potassium dichromate (K₂Cr₂O₇).

KMnO₄ is used in acidic medium (usually acidified with dilute H₂SO₄, never HCl, because HCl itself gets oxidised to Cl₂ and would consume extra KMnO₄) to estimate reducing agents such as Fe²⁺ (from Mohr's salt or ferrous sulphate) or oxalate ions. Its half-reaction, MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, shows Mn changes from purple MnO₄⁻ to almost colourless Mn²⁺. Because of this sharp colour change, KMnO₄ acts as its own indicator — the appearance of a permanent faint pink colour marks the end point, with no separate indicator needed.

K₂Cr₂O₇ is used similarly in acidic medium: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Unlike KMnO₄, it needs an external indicator (such as diphenylamine) because the orange-to-green colour change is not sharp enough to see clearly on its own.

The general idea to remember for a titration question: at the equivalence point, moles of electrons lost by the reducing agent = moles of electrons gained by the oxidising agent, and you use each half-reaction's electron count (5 for MnO₄⁻→Mn²⁺, 6 for Cr₂O₇²⁻→2Cr³⁺) to convert between moles of titrant and moles of analyte.

Oxidation number is a bookkeeping device, not a measure of real charge, and it has real limits that NCERT explicitly flags. Fractional oxidation numbers can arise when we average over multiple non-equivalent atoms of the same element in one formula — for example the O.N. of carbon in C₃O₂ works out to +4/3, and in Fe₃O₄ the average O.N. of Fe is +8/3 (in reality Fe₃O₄ contains two Fe³⁺ and one Fe²⁺, whose average is +8/3, but no single iron atom actually carries a fractional charge). Such fractional values are perfectly valid for balancing equations but should not be read as the literal charge on any one atom.

The oxidation number concept also cannot always distinguish between compounds that are genuinely different in structure but happen to give the same average O.N., and it does not by itself explain bonding character (ionic vs covalent) — it is purely a formal, calculated quantity assuming complete electron transfer, which is an approximation for covalent bonds. Despite these limitations, oxidation number remains extremely useful for identifying which atoms are oxidised or reduced and for balancing equations systematically, which is why it is taught before the more rigorous ion-electron method.

When a strip of zinc is dipped directly into copper sulphate solution, Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) occurs with electrons transferring directly at the metal surface, and the chemical energy is lost as heat. If instead the two half-reactions are physically separated — zinc in a ZnSO₄ solution, copper in a CuSO₄ solution, the two beakers connected by a salt bridge and the two metal strips connected by a wire through a voltmeter — the same net reaction occurs, but now the electrons must travel through the external wire to get from Zn to Cu²⁺. This device is a galvanic (voltaic) cell, and it converts chemical energy directly into electrical energy. The zinc electrode, where oxidation occurs, is the anode (negative terminal in a galvanic cell); the copper electrode, where reduction occurs, is the cathode (positive terminal). The salt bridge maintains electrical neutrality in both half-cells by allowing ion flow as the reaction proceeds.

Each electrode has a tendency to lose or gain electrons, measured (relative to a chosen reference) as its standard electrode potential. The reference is the standard hydrogen electrode (SHE), arbitrarily assigned a standard reduction potential of exactly 0.00 V, against which every other electrode's potential is measured under standard conditions (1 bar pressure, 1 M concentration, 298 K). Arranging elements in order of increasing standard reduction potential gives the electrochemical series: strong reducing agents (Li, K, Ca, Na… with large negative potentials) sit at the top, and strong oxidising agents (F₂, with a large positive potential) sit at the bottom. A species higher in the series (more negative potential) can reduce — and will displace — a species lower in the series from its salt solution, which is exactly why Zn displaces Cu²⁺ but Cu cannot displace Zn²⁺.