Redox Reactions: Oxidation Numbers, Balancing, Equivalent Mass and Titrimetric Analysis
Exam-oriented notes for NEET, JEE (Main & Advanced), IIT-JAM, BITSAT, GATE, CSIR-NET, TGT/PGT
Table of Contents
- 1. Why Redox Reactions Matter
- 2. Oxidation Number: The Bookkeeping Tool of Redox Chemistry
- 3. Oxidation and Reduction — Classical and Electronic Views
- 4. Types of Redox Reactions
- 5. Calculation of n-Factor
- 6. Balancing Redox Reactions
- 7. Equivalent Mass
- 8. Stoichiometry of Redox Reactions and Gram Equivalents
- 9. Volumetric Analysis
- 10. Types of Titrations
- 11. Applications of Redox Reactions
- 12. Common Pitfalls and Exam Tips
- 13. Practice MCQs (with Answers)
1. Why Redox Reactions Matter
Long before anyone spoke of electrons, chemists were already describing a very specific kind of chemical change: the combination of a substance with oxygen. Combustion, rusting, the calcination of ores — all of it fell under a single word, oxidation. The reverse process, stripping oxygen away from a metal oxide to recover the free metal, was called reduction. It was Antoine Lavoisier, working in 1789, who first showed that combustion is nothing mysterious — it is simply a reaction with the oxygen present in air, whether the fuel is wood, coal, or anything else that burns.
What took much longer to appreciate is that "reaction with oxygen" is only one face of something far more general. Whenever electrons move from one chemical species to another, you have the same underlying phenomenon, whether or not oxygen is anywhere in sight. These electron-transfer processes are what we now call oxidation–reduction reactions, or simply redox reactions — and this broader, electron-based understanding is what makes modern inorganic and analytical chemistry work. I want to build this chapter the way the concept itself grew historically: starting from oxidation numbers as a bookkeeping device, moving to the electron-transfer picture, and then putting the ideas to work in balancing equations and running titrations.
2. Oxidation Number: The Bookkeeping Tool of Redox Chemistry
2.1 Oxidation Number and Oxidation State
The oxidation number of an element tells you how many electrons that atom has effectively lost, gained, or shared relative to its free (uncombined) state. Tracking the change in oxidation number across a reaction is really the whole game: oxidants undergo a decrease in oxidation number (they have the higher oxidation number among a conjugate redox pair), while reductants undergo an increase (they carry the lower oxidation number in that pair).
For a genuinely ionic compound, the oxidation number is a real, physical charge. For a covalent compound, it is a fiction we impose deliberately — we pretend the molecule is ionic and hand every shared pair of electrons to whichever atom is more electronegative. It doesn't matter whether the compound is actually ionic at all. Strontium in SrF₂ and carbon in CO are both assigned +2, even though one compound is essentially ionic and the other is thoroughly covalent. That's the trick: oxidation number is a bookkeeping convention, not a claim about real charge distribution, except in the ionic case where the two happen to coincide.
Oxidation state is often used as a synonym, but there is a precise distinction worth knowing for exams: oxidation number is the total charge summed over all atoms of one kind in the formula, while oxidation state is the charge per atom. So oxidation state = oxidation number × (number of atoms of that element in the compound).
2.2 Stock Notation
When it is useful to specify oxidation state alongside an element's name, chemists write the oxidation number as a Roman numeral in parentheses — "iron(III)" means Fe is in the +3 state. In formulas, the numeral follows the metal's symbol: ferric chloride is Fe(III)Cl₃ and ferrous chloride is Fe(II)Cl₂. This lets you see immediately, for instance, that stannous chloride Sn(II)Cl₂ is the reduced form of stannic chloride Sn(IV)Cl₄. This convention — known as Stock notation — was introduced by the German chemist Alfred Stock.
2.3 Rules for Assigning Oxidation Numbers
Assigning oxidation numbers in covalent compounds isn't always obvious, since it's not always clear which atom is more electronegative just by looking. The following ordered rules resolve almost every case you'll meet:
- The oxidation number of any free (uncombined) element is zero — this holds regardless of how structurally complex the molecule is. Every atom in H₂, O₂, S₈, P₄, or in a piece of metallic Ca or K, has oxidation number 0.
- For a simple monoatomic ion, the oxidation number equals the ionic charge (Na⁺ is +1, Cl⁻ is −1). For a polyatomic ion, the oxidation number of the ion as a whole equals its net charge — for example, SO₄²⁻ is −2 overall and PO₄³⁻ is −3 overall.
- The algebraic sum of oxidation numbers of all atoms in a neutral molecule must be zero; in a polyatomic ion, the sum must equal the ion's charge.
- Fluorine is always −1 in its compounds. Other halogens are −1 in simple halides, but can show positive oxidation numbers in oxoacids and oxoanions.
- Hydrogen is usually +1, except when bonded to a less electronegative element (typically a metal), where it becomes −1, as in LiH.
- Oxygen is usually −2. Exceptions: in peroxides (H₂O₂, O₂²⁻) it is −1; in superoxides it is −1/2; and when bonded to the more electronegative fluorine (in OF₂ or O₂F₂) it turns positive, +2 and +1 respectively.
- In binary ionic compounds with metals, non-metals take the oxidation number equal to the charge of their anion — the phosphide ion P³⁻ in Mg₃P₂ is −3, for instance.
Whenever two rules conflict, the lower-numbered rule wins and the higher-numbered (conflicting) rule is set aside for that case.
Worked reasoning: why oxygen is −1 in H₂O₂
Applying Rule 6 to H₂O₂ would say oxygen is −2, but that conflicts with the sum rule once hydrogen (Rule 5, +1) is fixed. Since Rule 5 is lower-numbered, we keep it and let oxygen's value float: with H at +1 (two atoms) contributing +2, and total charge zero, the two oxygens must together contribute −2, so each oxygen is −1. This is exactly the exception Rule 6 already flags for O–O bonded species — the rules are self-consistent once you apply the priority correctly.
Worked reasoning: the azide ion in NaN₃
Sodium is fixed at +1 (Rule 2), and the compound is neutral, so the three nitrogen atoms together must sum to −1. Divided equally, each nitrogen carries an oxidation number of −1/3 — a clean illustration of how "average" oxidation number works out to a fraction even for something as simple-looking as an azide salt.
Solved Example: Structural oxidation numbers in peroxo-species
Na₂S₄O₆ (sodium tetrathionate): The structure is NaO₃S–S–S–SO₃Na. Only the two terminal sulphur atoms are bonded to oxygen and sodium; they carry +5 each. The two central sulphur atoms, joined only to each other by a pure S–S covalent bond, have oxidation number 0. (A naive formula-based calculation would instead give the misleading "average" of +2.5 for every sulphur — structural reasoning is essential here.)
H₂SO₅ (peroxomonosulphuric acid): Structure H–O–S(=O)(=O)–O–O–H. One O–O (peroxo) linkage exists, and peroxo oxygen is assigned −1 while the other oxygens are −2. With H at +1: +2 + (−6) + (−2) + x = 0 → x = +6.
H₂S₂O₈ (peroxodisulphuric/Marshall's acid): Two sulphurs, one peroxo linkage (−1 each for those two oxygens), six ordinary oxygens (−2 each). Working through the sum gives an oxidation state of +6 for each sulphur — the peroxo linkage does not change sulphur's oxidation number here because both sulphurs are structurally equivalent, unlike in tetrathionate.
CrO₅: Four of the five oxygens are joined in two peroxo (O–O) linkages, each contributing −1 per oxygen, while the fifth oxygen is doubly bonded to chromium (−2). Solving x + 4(−1) − 2 = 0 gives x = +6 for chromium — not the "+10" a blind formula approach might suggest.
2.4 Group-wise Trends in Oxidation States
A periodic table annotated with common oxidation numbers (as most textbooks show) reveals patterns that are genuinely useful to memorize rather than derive each time:
- Group 1 (ns¹) metals show a uniform +1 state; Group 2 (ns²) metals show a uniform +2 state.
- Group 13 (ns²np¹): +1 and +3.
- Group 14 (ns²np²): maximum +4, minimum −4; Sn and Pb additionally show a stable +2 (the "inert pair" effect, and this +2 state is metallic in character).
- Group 15 (ns²np³): maximum +5, minimum −3.
- Group 16 (ns²np⁴): maximum +6, minimum −2.
- Group 17 (ns²np⁵): maximum +7, minimum −1.
- Transition metals show a wide spread of oxidation states because (n−1)d electrons participate alongside the ns electrons.
Two more general rules worth internalizing: the most positive oxidation number an atom shows is usually its group number (except O and F); and the most negative oxidation number a non-metal shows is usually (group number − 8). The maximum oxidation number available to an element increases as you move across a period — this is exactly why, for example, Mn (Group 7) can reach +7 in KMnO₄, but Fe (Group 8) tops out lower in common chemistry.
2.5 Fractional Oxidation Numbers
A fractional oxidation number is always a signal that the same element exists in more than one distinct oxidation state within that structure, and the number reported is simply the arithmetic average. The textbook example is magnetite, Fe₃O₄: with oxygen fixed at −2, 3x + 4(−2) = 0 gives x = +8/3. Structurally, this is because two of the three iron atoms carry +3 and one carries +2 — average (2×3 + 2)/3 = 8/3. The average here describes non-equivalent atoms, not some strange "in-between" charge every iron atom shares equally.
2.6 Oxidation Number vs Valency
These two ideas get conflated constantly, so it's worth being precise. Valence electrons are the electrons in an atom's outermost shell, available for bonding. Valency is the number of such electrons an atom loses or gains (or shares) when it combines. The differences that matter for exams:
| Property | Oxidation number | Valency |
|---|---|---|
| Sign | Carries a sign (+ or −) | No sign |
| Can be fractional? | Yes | Never |
| Can be zero? | Yes, for any element | Only for noble gases |
A nice illustration: across CH₄, CH₃Cl, CH₂Cl₂, CHCl₃, and CCl₄, carbon's valency stays fixed at 4 throughout, but its oxidation number walks steadily from −4 up to +4 as each hydrogen is swapped for the more electronegative chlorine. This is a classic exam question precisely because it shows the two concepts diverging sharply for the same series of molecules.
3. Oxidation and Reduction — Classical and Electronic Views
3.1 The Classical Definitions
Before the electron-transfer picture took hold, chemists defined oxidation and reduction operationally, in terms of what atoms were gained or lost:
Reduction = removal of oxygen (or an electronegative element), OR addition of hydrogen (or an electropositive element).
Each half of these definitions is illustrated by a familiar reaction. Sulphur burning in oxygen, S + O₂ → SO₂, is oxidation by addition of oxygen. Methane's complete combustion, CH₄ + 2O₂ → CO₂ + 2H₂O, removes hydrogen from carbon while adding oxygen — both descriptions of the same oxidation. Sodium reacting with chlorine, 2Na + Cl₂ → 2NaCl, involves no oxygen or hydrogen at all, yet it is unmistakably an oxidation of sodium, because we've broadened the definition to include addition of any electronegative element (or, symmetrically, removal of an electropositive one). On the reduction side, CuO + H₂ → Cu + H�2O is removal of oxygen from copper, and C₂H₄ + H₂ → C₂H₆ is addition of hydrogen.
These paired definitions always march together. You cannot oxidize something without reducing something else in the same reaction — this is the conceptual seed of the electron-transfer picture that follows.
3.2 Oxidation and Reduction as Electron Transfer
The modern, unifying view: oxidation is loss of one or more electrons (with an accompanying increase in oxidation number), and reduction is gain of electrons (with a decrease in oxidation number). Any redox reaction can be split conceptually into two half-reactions — one showing the oxidation, one showing the reduction — even though both happen simultaneously in the actual chemistry.
For sodium burning in chlorine:
Cl₂ + 2e⁻ → 2Cl⁻ (reduction)
Overall: 2Na(s) + Cl₂(g) → 2Na⁺Cl⁻(s)
Notice that the electrons lost by sodium (two, since two Na atoms are oxidized) exactly match the electrons gained by chlorine (two, in reducing one Cl₂ molecule). This conservation is not a coincidence — electrons never appear as a leftover "product" in a balanced overall equation, precisely because whatever is lost by one species must be captured by another.
3.3 Oxidizing and Reducing Agents
An oxidizing agent accepts electrons from another substance, causing that substance to be oxidized — and in doing so, the oxidizing agent itself is reduced (its own valence/oxidation number falls). A reducing agent donates electrons, causing the other substance to be reduced, while it itself is oxidized. This pairing — oxidizing agent gets reduced, reducing agent gets oxidized — trips students up constantly in exams, so it is worth saying twice, in plain terms: the agent's own fate is the opposite of what it does to the other reactant.
When copper wire sits in an ammoniacal solution of Ag⁺, silver metal deposits as visible "whiskers" on the wire while the solution takes on a blue-green tint from the Cu(NH₃)₄²⁺ complex that forms. The net transfer, Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), makes Ag⁺ the oxidizing agent and Cu the reducing agent — copper gives up electrons, silver ions accept them.
3.4 Activity Series and Displacement Reactions
Dip a strip of zinc into copper sulphate solution and, after a while, a reddish deposit of metallic copper coats the zinc while the solution's blue colour fades as Zn²⁺ takes the place of Cu²⁺. The net ionic equation is simply Zn(s) + Cu²⁺(aq) → Cu(s) + Zn²⁺(aq), with sulphate as a spectator. At the atomic level, Cu²⁺ ions collide with the zinc surface, pick up electrons directly from zinc atoms, and deposit as neutral copper, while the zinc atoms that lost electrons enter solution as Zn²⁺.
This behavior generalizes into the activity series: a metal that is more easily oxidized ("more active") will displace, from its compounds, any metal that is less easily oxidized. Reading from least active (top) to most active (bottom): Au, Hg, Ag, Cu, H, Pb, Sn, Co, Cd, Fe, Cr, Zn, Mn, Al, Mg, Na, Ca, Sr, Ba, K, Rb, Cs.
- Any metal will be displaced from its salt solution by any metal positioned below it in the series.
- Metals below hydrogen can displace H₂ from acids containing H⁺; metals above hydrogen cannot. Metals at the very bottom (K, Na, Ca) are such powerful reducing agents that they can even reduce the hydrogen in water itself: 2Na(s) + 2H₂O → 2NaOH(aq) + H₂(g).
- For metals below hydrogen, greater ease of oxidation correlates with a faster reaction rate against H⁺ — this is why Mg reacts with dilute HCl briskly while Fe reacts sluggishly.
Halogens have their own activity series as oxidizing agents, running the opposite direction: F₂ is the strongest oxidant and reactivity falls steadily down the group to I₂. Fluorine is so aggressive that it will even displace oxygen from water (2F₂ + 2H₂O → 4HF + O₂), which is why fluorine displacement reactions on other halides are never carried out in aqueous medium. Chlorine, however, can cleanly displace bromine and iodine from aqueous solution, and bromine can displace iodine — this ordering underlies the classic "layer test" used to identify bromide and iodide ions using CCl₄, where Br₂ imparts a reddish-brown colour to the organic layer and I₂ a violet one.
4. Types of Redox Reactions
Four broad patterns cover essentially every redox reaction you'll encounter in this syllabus.
(1) Combination reactions
Two or more reactants combine into a single product, A + B → AB. For this to be redox, at least one of A or B must be in the elemental form. Combustion of fuels is the classic case: C(s) + O₂(g) → CO₂(g); CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). Magnesium burning in oxygen (2Mg + O₂ → 2MgO) and lithium reacting with nitrogen (6Li + N₂ → 2Li₃N) are similarly combination-type redox reactions.
(2) Decomposition reactions
The reverse pattern, AB → A + B, is redox only when at least one product emerges in elemental form. Decomposition of potassium chlorate (2KClO₃ → 2KCl + 3O₂) and hydrogen peroxide (2H₂O₂ → 2H₂O + O₂) qualify; decomposition of CaCO₃ into CaO and CO₂ does not, since no element appears free in the products.
(3) Displacement reactions
One element replaces another within a compound, AB + C → AC + B, requiring C to be more reactive than B. These split into metal-displacement (a more active metal displaces a less active one — Cu(s) + 2AgNO₃ → 2Ag(s) + Cu(NO₃)₂) and non-metal-displacement, most commonly hydrogen displacement from water, steam, or acid (Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)), where magnesium reacts fastest and iron slowest among the common laboratory metals.
(4) Disproportionation reactions
This is the pattern students find most conceptually interesting: one and the same element, starting in a single intermediate oxidation state, is simultaneously oxidized and reduced, ending up in two different oxidation states among the products. The essential prerequisite is that the element must be capable of existing in at least three oxidation states, with the reactant sitting in the middle one. Chlorine gas reacting with cold NaOH illustrates this perfectly:
Chlorine starts at 0 and ends up at −1 (in NaCl, reduced) and +1 (in NaOCl, oxidized) simultaneously. Bromine and iodine behave the same way with alkali. Fluorine is the odd one out: because it is the most electronegative element and can only ever show −1 in its compounds, it simply cannot disproportionate — with alkali it instead gives 2F₂ + 2NaOH → 2NaF + OF₂ + H₂O, an entirely different reaction pathway.
5. Calculation of n-Factor
The n-factor (also called the valence factor) of an oxidizing or reducing agent is the number of electrons gained or lost, respectively, per formula unit in a given reaction. It is the single most important number for equivalent-mass and titration calculations, and there are several distinct patterns to recognize:
- One atom changes oxidation state, one product: for MnO₄⁻ → Mn²⁺, n = (+7) − (+2) = 5.
- One atom changes state but splits into two products at the same new oxidation state: for Cr₂O₇²⁻ → 2Cr³⁺, n = 2×[(+6) − (+3)] = 6.
- One atom changes state and forms two products at different oxidation states (mixed oxidation-only or reduction-only outcome): for 3MnO₄⁻ → 2Mn²⁺ + Mn⁶⁺ (as MnO₄²⁻), total electrons for 3 mol MnO₄⁻ = 2(7−2) + 1(7−6) = 11, so n-factor per mole of MnO₄⁻ = 11/3.
- One atom's oxidation state changes in one product but stays the same as the reactant in another product: in K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 3Cl₂ + 7H₂O, of the 14 Cl atoms from HCl, 6 are oxidized to Cl₂ (0 state) while 8 remain as Cl⁻ in KCl/CrCl₃. Only the 6 oxidized chlorines contribute electrons, so n-factor of HCl here = 6/14 = 3/7.
- Two or more atoms in the same species change oxidation state (same direction): for FeC₂O₄ → Fe³⁺ + 2CO₂, n-factor = [1×(3−2)] + [2×(4−3)] = 1 + 2 = 3.
- Two atoms in the same species change state in opposite directions (one oxidized, one reduced) — here you compute the change for whichever atom is being oxidized or whichever is being reduced (they must come out equal by conservation): for (NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4H₂O, considering the reduction of Cr, n-factor = (+6−+3)×2 = 6.
- Disproportionation, where oxidant and reductant are the same species: for 2H₂O₂ → 2H₂O + O₂, the oxidation half gives n = 2×0 − (−1)×2 = 2, and the reduction half gives n = (−2)×2 − (−1)×2 = 2. Both halves agree — the n-factor of H₂O₂ (whether you view it via oxidation or reduction) is 2.
Solved Example: n-factor of HCl with two different oxidants
With KMnO₄: 2KMnO₄ + 16HCl → 2MnCl₂ + 5Cl₂ + 2KCl + 8H₂O. Here 2 mol KMnO₄ ≡ 16 mol HCl overall, but the redox-active portion is governed by the half reaction MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, so 1/5 mol KMnO₄ ≡ 1 mol HCl (electron-equivalent basis). Combining the stoichiometric ratio (1 mol KMnO₄ : 8 mol HCl) with the equivalence relation gives n-factor of HCl = 5/8.
With K₂Cr₂O₇: K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 3Cl₂ + 7H₂O. The half-reaction Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O gives K₂Cr₂O₇ an n-factor of 6. Since 1 mol K₂Cr₂O₇ ≡ 14 mol HCl stoichiometrically, matching gram-equivalents gives n-factor of HCl = 6/14 = 3/7 — matching the direct reasoning in point 4 above.
6. Balancing Redox Reactions
6.1 Half-Reaction (Ion–Electron) Method
This method is powerful precisely because it tells you automatically how H⁺, OH⁻, and H₂O participate — you don't need to guess that in advance. The steps, illustrated with Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ in acidic solution:
- Split into half-reactions: Cr₂O₇²⁻(aq) → Cr³⁺(aq); Fe²⁺(aq) → Fe³⁺(aq).
- Balance atoms other than H and O: two chromiums on the left need a coefficient of 2 on the right: Cr₂O₇²⁻ → 2Cr³⁺.
- Balance oxygen using H₂O: add 7H₂O to the deficient side: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O.
- Balance hydrogen using H⁺: add 14H⁺ to the left: 14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O.
- Balance charge using electrons: net charge on the left is (+14) + (−2) = +12; on the right, 2×(+3) = +6. The difference, 6, is added as electrons to the more positive side: 6e⁻ + 14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O. For iron: Fe²⁺ → Fe³⁺ + e⁻.
- Equalize electrons and add: the chromium half needs 6 electrons but iron's half only supplies 1, so multiply the iron half-reaction by 6, then add: 14H⁺ + Cr₂O₇²⁻ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺.
- Cancel anything common to both sides — here the six electrons cancel exactly, giving the final balanced equation.
For basic solution, the trick is to first balance as if the medium were acidic (Steps 1–7 above), then convert using three additional steps: (8) add the same number of OH⁻ to both sides as there are H⁺; (9) combine each H⁺ + OH⁻ pair into H₂O; (10) cancel any H₂O that appears on both sides. Applying this to SO₃²⁻(aq) + MnO₄⁻(aq) → SO₄²⁻(aq) + MnO₂(s) in basic medium eventually yields the clean result H₂O + 3SO₃²⁻ + 2MnO₄⁻ → 3SO₄²⁻ + 2MnO₂(s) + 2OH⁻(aq).
6.2 Oxidation Number Method
This method focuses directly on the atoms whose oxidation states change, using the following steps (illustrated for HNO₃(aq) + H₂S(aq) → NO(g) + S(s) + H₂O(l)):
- Write correct formulas for every reactant and product.
- Identify the atoms whose oxidation states have changed: N goes from +5 to +2 (a decrease of 3), and S goes from −2 to 0 (an increase of 2).
- Make sure the number of atoms of the changing elements is balanced on both sides.
- Draw a "bridge" connecting the same atoms whose states changed, marking the electron change per atom.
- Multiply each species by whatever integer makes the total electrons lost equal the total electrons gained — here, ×2 for the nitrogen species and ×3 for the sulphur species (since 3 and 2 share a common multiple of 6): 2HNO₃ + 3H₂S → 2NO + 3S + (H₂O, unbalanced yet).
- Balance everything else by inspection — 8 hydrogens on the left require 4 H₂O on the right, and checking confirms the oxygens balance too: 2HNO₃ + 3H₂S → 2NO + 3S + 4H₂O.
6.3 The Fastest Method (n-Factor Cross-Multiplication)
For quick balancing under exam time pressure, this shortcut is genuinely fast once you're comfortable with n-factors. Applied to MnO₄⁻ + Cl⁻ → Cl₂ + Mn²⁺:
- Write n-factors above the species on one side — say the left side. MnO₄⁻ has n = 5 (going to Mn²⁺); Cl⁻ has n = 1 (going to ½Cl₂, i.e., losing one electron per Cl).
- Cross-multiply the n-factors as new coefficients (don't touch this side's original coefficients otherwise): 1 MnO₄⁻ + 5 Cl⁻ → Cl₂ + Mn²⁺.
- Balance the oxidized/reduced elements on the product side: 1 MnO₄⁻ + 5 Cl⁻ → (5/2)Cl₂ + 1 Mn²⁺.
- Balance oxygen with H₂O: 1 MnO₄⁻ + 5 Cl⁻ → (5/2)Cl₂ + 1 Mn²⁺ + 4H₂O.
- Balance hydrogen with H⁺: 8H⁺ + 1 MnO₄⁻ + 5 Cl⁻ → (5/2)Cl₂ + 1 Mn²⁺ + 4H₂O.
This equation, balanced for acidic medium, requires no separate charge-balancing step — charge automatically comes out balanced. To convert to basic medium, repeat Steps 1–5, then add OH⁻ = H⁺ on both sides and cancel water exactly as in Section 6.1, arriving at 4H₂O + 1MnO₄⁻ + 5Cl⁻ → (5/2)Cl₂ + 1Mn²⁺ + 8OH⁻.
7. Equivalent Mass
The equivalent mass of a substance is the number of parts by mass of it that combines with, or displaces, 1.008 parts of hydrogen, 8 parts of oxygen, 35.5 parts of chlorine, or 108 parts of silver. It relates to atomic (or molecular) mass through:
Three commonly needed definitions:
- Equivalent mass of an acid = molecular mass ÷ basicity (number of replaceable H⁺ per molecule).
- Equivalent mass of a base = molecular mass ÷ acidity (number of replaceable OH⁻ per molecule).
- Equivalent mass of an oxidizing/reducing agent = molecular mass ÷ number of electrons gained or lost per molecule (i.e., its n-factor).
Several experimental methods exist for determining equivalent mass, all leaning on the definitions above: the hydrogen-displacement method (Eq. mass = mass of metal × 1.008 ÷ mass of H₂ liberated), the oxide-formation method (Eq. mass = mass of metal × 8 ÷ mass of oxygen combined), the chloride-formation method (Eq. mass = mass of metal × 35.5 ÷ mass of chlorine combined), the metal-to-metal displacement method (equivalent masses are in the same ratio as the reacting masses, m₁/m₂ = E₁/E₂), and the double decomposition method (for AB + CD → AD + CB, equivalent mass of the compound AB equals the sum of equivalent masses of its constituent radicals A and B).
- Medium-dependent: KMnO₄ has n = 5 in acidic medium (MnO₄⁻ → Mn²⁺) but also n = 5 in alkaline medium in some contexts (MnO₄⁻ → MnO₂ would be n=3, so always check the actual half-reaction given). Na₂S₂O₃ has n = 1 in acidic medium (→ S₄O₆²⁻) but n = 8 in alkaline medium (→ SO₄²⁻), giving very different equivalent weights (158 vs 19.75).
- Dilution-dependent: Concentrated HNO₃ acts with n = 1 (→ NO₂, equivalent weight 63) while dilute HNO₃ acts with n = 3 (→ NO, equivalent weight 21) — a case where a smaller n-factor (dilute acid) is nonetheless found experimentally, from electrode-potential data, to correspond to a genuinely different (not necessarily "weaker") oxidizing pathway.
- Disproportionation: equivalent weight = (sum of the two "half" molecular weights contributed by oxidation and reduction) ÷ (number of electrons transferred). If the electron count is the same on both sides (as in H₂O₂ → H₂O + O₂, n = 2 either way, giving Eq. wt. = 68/2 = 34), the formula simplifies neatly. If the electron counts differ between the two half-reactions (as for Br₂ + OH⁻ → Br⁻ + BrO₃⁻ + H₂O, where oxidation loses 10e⁻ but reduction gains only 2e⁻ per Br₂, requiring the reduction half to be multiplied by 5), you must use the weighted total: Eq. wt. = 479.4/10 = 47.94.
8. Stoichiometry of Redox Reactions and Gram Equivalents
The central working relationship for redox stoichiometry, once you have a balanced equation with stoichiometric coefficients n₁ (reductant) and n₂ (oxidant), is the molarity equation:
This single equation lets you solve for any one unknown (molarity or volume of either reactant) once the other three quantities are known — it is the redox analogue of the simple mole-ratio calculation, but expressed through gram-equivalents rather than raw moles, which is why n₁ and n₂ appear explicitly.
The underlying principle, for a reaction A (reductant) + B (oxidant) → C + D, is that
or equivalently, milliequivalents of oxidizing agent = milliequivalents of reducing agent, at the point of exact stoichiometric reaction. This "milliequivalents are always equal" idea is the single most-used relationship in every titration calculation that follows in this chapter.
9. Volumetric Analysis
Volumetric (titrimetric) analysis is the quantitative determination of an unknown concentration by measuring the exact volume of a standard solution (known concentration) needed to react completely with a measured volume of the unknown. The process of adding titrant until the reaction is judged complete is titration.
Volumetric analysis rests on reactions of four broad types — precipitation (forms a sparingly soluble salt), complexometric (forms a soluble complex between metal ion and ligand), neutralization (acid–base), and redox (oxidant + reductant) — and on the Law of Chemical Equivalence: at the end-point, the number of gram-equivalents of the two reacting substances becomes equal, expressible as the familiar normality equation:
Key vocabulary
| Term | Meaning |
|---|---|
| Titrant | Standard solution of accurately known strength |
| Titrate | Solution whose concentration is to be found |
| Equivalence point | Theoretical point where reaction is exactly complete |
| End-point | Practically observed point (colour change, etc.) near the equivalence point |
| Indicator | Substance signalling the end-point by a visible change; must be sensitive with a sharp colour change, and its working pH range should coincide with the titration's equivalence-point region |
Common indicator classes: acid–base indicators (colour changes with pH), redox indicators (oxidized and reduced forms differ in colour), precipitation indicators (form a coloured precipitate at the end-point), and complexometric indicators (change colour on complexing with a metal). The three most familiar acid–base indicators and their working ranges:
| Indicator | pH range | Acidic colour | Basic colour |
|---|---|---|---|
| Phenolphthalein | 8.3–10.0 | Colourless | Pink |
| Methyl orange | 3.1–4.4 | Pink/Red | Yellow/Orange |
| Methyl red | 4.2–6.3 | Red | Yellow |
Standard solutions come in two kinds. A primary standard can be prepared directly, by dissolving an accurately weighed pure amount and making up to volume — it should be highly pure, non-hygroscopic, stable to oven-drying, and readily soluble (examples: oxalic acid, AgNO₃, anhydrous Na₂CO₃). A secondary standard (NaOH, KMnO₄, HCl, and similarly reactive or hygroscopic substances) cannot be weighed out accurately in this way, so its exact concentration is determined afterward by titrating it against a primary standard — a process called standardization.
10. Types of Titrations
10.1 Redox Titrations
A redox titration pairs a reducing agent against an oxidizing agent — for example, MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. The three most important reagent systems are worth knowing in detail.
Potassium permanganate (KMnO₄) titrations
In acidic medium: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (n = 5). H₂SO₄ is the acidifying agent of choice — HCl is avoided because KMnO₄ oxidizes chloride to Cl₂ gas, and HNO₃ is avoided because it is itself a strong oxidant and would interfere by oxidizing the analyte independently. In the classic estimation of oxalic acid, 2KMnO₄ + 3H₂SO₄ + 5H₂C₂O₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 10CO₂, no external indicator is needed: KMnO₄ is its own self-indicator, going from intense purple to colourless as it's reduced, with the first persistent faint pink marking the end-point. Its major drawback is that it is a secondary standard and must be standardized before use.
Potassium dichromate (K₂Cr₂O₇) titrations
K₂Cr₂O₇ oxidizes only in acidic medium: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (n = 6). Unlike KMnO₄, dichromate does not oxidize HCl at room temperature, so either H₂SO₄ or HCl can be used for acidification. In the classic estimation of Fe(II) in Mohr's salt, diphenylamine is used as an internal indicator, turning from colourless to intense blue once all Fe²⁺ is consumed and the first excess of dichromate appears. K₂Cr₂O₇ is a weaker oxidant than KMnO₄, but its advantage is that it is a genuine primary standard.
Iodine titrations — iodometry vs iodimetry
Iodine is a comparatively mild oxidizing agent, acting via I₂ + 2e⁻ → 2I⁻. Starch is the standard indicator, turning deep blue with even traces of free iodine. Two distinct strategies exist, and distinguishing them is a favourite exam question:
- Iodometry: an unknown oxidizing agent first liberates I₂ from excess iodide, and this liberated iodine is then titrated against a standard reductant (typically Na₂S₂O₃, "hypo"). The classic case is estimating Cu²⁺: 2Cu²⁺ + 4I⁻ → 2CuI + I₂ (Cu²⁺ is reduced to Cu⁺, iodide is oxidized to I₂), followed by I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. The amount of thiosulphate consumed is stoichiometrically tied back to the original Cu²⁺.
- Iodimetry: iodine itself (as a standard solution) is titrated directly against a reducing agent — used, for example, to determine the strength of ferrous ions, sulphite, arsenite, or thiosulphate solutions.
Redox indicator types
Three categories, based on how the end-point is signalled: self-indicators (the titrant's own colour change, as with KMnO₄); internal indicators (added directly to the titration flask, like diphenylamine with K₂Cr₂O₇); and external indicators (spotted on a separate tile and tested against withdrawn drops of the reaction mixture — for example, K₃Fe(CN)₆ in the titration of K₂Cr₂O₇ against Mohr's salt, which stops giving its blue colour with Fe²⁺ once all the ferrous ion is consumed).
10.2 Complexometric Titrations and Water Hardness
Complexometric titrations rely on formation of a soluble complex between a metal ion and a complexing agent (ligand), most commonly used for estimating Ca²⁺ and Mg²⁺ with EDTA. The indicator (typically Eriochrome Black T, EBT) must form a complex with the metal ion that is less stable than the metal–EDTA complex, so that as EDTA is added it progressively displaces the metal from the EBT complex, releasing free indicator (which changes colour) right at the equivalence point.
Hardness of water — its capacity to precipitate soap — arises from dissolved Ca²⁺ and Mg²⁺. Temporary hardness comes from bicarbonates and is removable by boiling, since the bicarbonates decompose to insoluble carbonates/hydroxides: Ca(HCO₃)₂ → CaCO₃↓ + H₂O + CO₂. Permanent hardness comes from chlorides and sulphates (MgCl₂, MgSO₄, CaCl₂, CaSO₄, and salts of Fe/Al) and survives boiling — it requires chemical softening agents instead. Hardness is conventionally expressed as ppm (equivalent parts CaCO₃ per 10⁶ parts water by weight; also equal to mg L⁻¹), Degree Clarke (per 70,000 parts water), or Degree French (per 10⁵ parts water), related by 1 ppm = 1 mg L⁻¹ = 0.1°Fr = 0.07°Cl.
Clarke's method removes temporary hardness using lime: Ca(OH)₂ + Ca(HCO₃)₂ → 2CaCO₃↓ + 2H₂O, and similarly for magnesium bicarbonate.
10.3 Precipitation (Argentometric) Titrations
Precipitation titrations rely on formation of a sparingly soluble salt — occurring once the ionic product exceeds the solubility product — and the precipitate must form quickly and be genuinely insoluble for a sharp end-point. The most useful class involves silver nitrate and is called argentometric titration, with two named methods:
- Mohr's method (for chloride estimation): AgNO₃ is titrated against halide in neutral medium using K₂CrO₄ indicator. AgCl (white) precipitates first; once chloride is exhausted, the first excess of Ag⁺ forms brick-red Ag₂CrO₄, marking the end-point. Neutral medium is essential — Ag₂CrO₄ dissolves in acid, and AgNO₃ itself precipitates as AgOH in base. This method cannot be used for iodide, since the iodide precipitate would mask the indicator's colour change.
- Volhard's method (for silver estimation): AgNO₃ is titrated against thiocyanate in dilute HNO₃ using ferric ion as indicator. AgSCN (white) precipitates until Ag⁺ is exhausted, then the first excess of SCN⁻ forms blood-red Fe(SCN)₃. This method extends to bromide estimation but not chloride, since AgCl precipitate redissolves ("disappears") on stirring under these conditions.
10.4 Neutralization Titrations and Double Titration
Acid–base titrations (also called neutralization titrations) involve H⁺ and OH⁻ combining to form water. The pH at the equivalence point depends on the relative strengths of the acid and base involved — it need not be exactly 7. A plot of pH against volume of titrant added (the pH titration curve) is the tool for choosing an appropriate indicator: the indicator's working range should coincide with the steep, near-vertical portion of the curve near the equivalence point.
| Titration type | Approx. equivalence pH | Suitable indicator(s) |
|---|---|---|
| Strong acid – strong base | ~4 to 10 | Phenolphthalein, methyl orange, or methyl red |
| Weak acid – strong base | ~7.5 to 10 | Phenolphthalein |
| Strong acid – weak base | ~4 to 6.5 | Methyl orange or methyl red |
| Weak acid – weak base | ~6.5 to 7.5, no sharp jump | No simple indicator works well |
A special and heavily examined case is the titration of a mixture of two carbonate/bicarbonate/hydroxide bases using two indicators — double titration. With phenolphthalein, only NaOH is fully neutralized and Na₂CO₃ is only half-neutralized (to NaHCO₃), because phenolphthalein changes colour exactly when the weakly-basic NaHCO₃ stage is reached: Na₂CO₃ + HCl → NaHCO₃ + NaCl. Adding methyl orange to the same, already-titrated solution and continuing the titration then neutralizes all remaining base (the NaHCO₃, both original and just-formed) down to carbonic acid. If a mL of acid is used up to the phenolphthalein end-point and a further b mL to the methyl orange end-point, then acid used by Na₂CO₃ alone = 2b, and acid used by NaOH alone = (a − b). The analogous case for a Na₂CO₃/NaHCO₃ mixture gives acid used by Na₂CO₃ = 2a, and acid used by NaHCO₃ alone = (b − a).
10.5 Back Titration
Back titration is the go-to strategy whenever an impure solid Z cannot be titrated directly — typically because its impurities don't react cleanly, or because the reaction with a direct titrant is too slow or ill-defined. A known excess of a reagent X (of known but excess quantity) is added to react completely with Z; then the leftover, unreacted X is titrated against a second standard reagent Y. Subtracting the amount of X consumed by Y from the total X added gives exactly the amount of X that reacted with Z, from which the mass — and hence purity — of Z is calculated:
where N₁V₁ is total milliequivalents of X added, N₂V₂ is milliequivalents of X back-titrated by Y, and W is the mass of impure sample taken.
11. Applications of Redox Reactions
Volume strength of H₂O₂
H₂O₂ concentration is conventionally expressed as "x volume," meaning 1 volume of solution liberates x volumes of O₂ gas (at STP) on complete decomposition: 2H₂O₂ → 2H₂O + O₂. Since 68 g of H₂O₂ (2 mol) liberates 22,400 mL of O₂ at STP, a short derivation connects volume strength directly to normality, molarity, and strength in g/L — relationships that appear constantly in numericals:
- % strength (by mass) = (17/56) × Volume strength
- Volume strength = (56/17) × % strength
- Volume strength = 11.2 × Molarity
- Volume strength = 5.6 × Normality
- Normality = Volume strength ÷ 5.6
- Molarity = Volume strength ÷ 11.2
- Strength (g L⁻¹) = Volume strength × 68/22.4 = Volume strength × 34/11.2
- Strength (g L⁻¹) = Normality × Equivalent weight = Molarity × Molecular mass
Percentage labeling of oleum
Oleum ("fuming sulphuric acid") is SO₃ dissolved in H₂SO₄. On adding water, the free SO₃ converts to additional H₂SO₄ (SO₃ + H₂O → H₂SO₄), which is why the mass of the diluted sample increases. "% Labeling" reports the total mass of H₂SO₄ (initial plus that generated from SO₃) obtainable from 100 g of oleum after full dilution — a labeling above 100% signals genuine free SO₃ content, and this is a very common numerical-problem setup (e.g., "109% oleum" contains 40% free SO₃ and 60% H₂SO₄ by mass, worked out from the stoichiometry of SO₃ + H₂O → H₂SO₄).
Broader applications
- Industrial manufacture: Chlorine gas, NaOH, and KOH are produced industrially by electrolysis, an inherently redox-based process.
- Metallurgy: Metal oxides are reduced to free metals by suitable reducing agents — electrolytically for Al₂O₃, or in a blast furnace for Fe₂O₃. Metal-displacement redox reactions underlie the thermite reaction (2Al + Fe₂O₃ → 2Fe + Al₂O₃) used to produce molten iron, and the Kroll process (2Mg + TiCl₄ → Ti + 2MgCl₂) used to extract titanium.
- Photosynthesis: 6CO₂ + 6H₂O → (chlorophyll, sunlight) → C₆H₁₂O₆ + 6O₂ is a redox process in which CO₂ is reduced to carbohydrate while water is oxidized to O₂.
- Combustion of fuels: Hydrocarbon fuels burn to CO₂ and H₂O with large energy release; the biological oxidation of glucose in living cells proceeds by the analogous redox pathway.
- Batteries and electrochemical cells: Electrode processes are fundamentally redox reactions between metal electrodes and electrolyte salts.
- Quantitative estimation: The entire apparatus of redox titrimetry described above (Section 10) is itself a direct application of redox chemistry to analytical measurement.
12. Common Pitfalls and Exam Tips
- Don't assume every atom of the same element in a compound shares the identical oxidation state — always check the structure for species like Na₂S₄O₆, H₂SO₅, H₂S₂O₈, and CrO₅ before applying the simple sum rule.
- KMnO₄'s n-factor is not a fixed "5" universally — it depends on the actual reduction product (Mn²⁺ in acid gives n=5; MnO₂ in neutral/faintly alkaline medium gives n=3; MnO₄²⁻ in strongly alkaline medium gives n=1). Always check what the question specifies as the product.
- Concentrated versus dilute HNO₃ genuinely change the n-factor (and hence equivalent weight) because the reduction product itself changes (NO₂ vs NO) — this is decided by experimental electrode-potential data, not simply by "dilution weakens the oxidant" reasoning.
- Fluorine never disproportionates — it is the one halogen that cannot, since −1 is its only accessible negative state and it cannot be oxidized further by anything in ordinary aqueous chemistry.
- Mohr's method requires strictly neutral medium; Volhard's method cannot be used directly for chloride estimation. These medium/scope restrictions are frequently tested as standalone objective questions.
- In double titrations, always work out which base reacts fully with phenolphthalein-stage acid and which only half-reacts — a very common source of calculation errors under time pressure.
13. Practice MCQs (with Answers)
The following 35 questions test understanding of the concepts above rather than repeating textbook wording verbatim. Try each before checking the answer.
1. The oxidation number of chromium in CrO₅ is:
(a) +5 (b) +6 (c) +8 (d) +10Answer: (b) — four oxygens form two peroxo linkages (−1 each), one oxygen is doubly bonded (−2); x + 4(−1) − 2 = 0 → x = +6.
2. Which of the following elements CANNOT undergo disproportionation in alkaline medium?
(a) Cl₂ (b) Br₂ (c) I₂ (d) F₂Answer: (d) — fluorine, being the most electronegative element, only ever shows −1 and cannot be simultaneously oxidized.
3. In the reaction of Zn with CuSO₄ solution, the spectator ion is:
(a) Zn²⁺ (b) Cu²⁺ (c) SO₄²⁻ (d) NoneAnswer: (c) — sulphate takes no part in the electron transfer; only Zn/Zn²⁺ and Cu²⁺/Cu are involved.
4. The n-factor of K₂Cr₂O₇ as an oxidizing agent in acidic medium is:
(a) 1 (b) 2 (c) 3 (d) 6Answer: (d) — from Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
5. Which acidifying agent should NOT be used with KMnO₄ titrations, and why?
(a) H₂SO₄, too weak (b) HCl, gets oxidized to Cl₂ (c) HNO₃, too dilute (d) CH₃COOH, insolubleAnswer: (b) — HCl is oxidized by KMnO₄ to Cl₂ gas, introducing error; HNO₃ is avoided because it is itself oxidizing.
6. A metal oxide MO has M in oxidation state +2. If M is oxidized further, which activity-series position best predicts easy further oxidation?
(a) A metal high in the activity series (near Au) (b) A metal low in the activity series (near Cs) (c) Position is irrelevant (d) Only noble gases show thisAnswer: (b) — metals lower in the activity series are more easily oxidized (more "active").
7. Which of the following pairs correctly matches indicator with its titration?
(a) Diphenylamine — K₂Cr₂O₇ vs Fe²⁺ (b) Starch — acid-base titration (c) Phenolphthalein — iodometric titration (d) EBT — argentometric titrationAnswer: (a) — diphenylamine is the standard internal indicator for dichromate–ferrous titrations.
8. In Mohr's method for chloride estimation, the medium must be:
(a) Strongly acidic (b) Strongly basic (c) Neutral (d) Any pHAnswer: (c) — Ag₂CrO₄ dissolves in acid and Ag⁺ precipitates as AgOH in base.
9. The oxidation number of S in the two distinct sulphur environments of Na₂S₄O₆ are:
(a) +5 and +5 (b) +5 and 0 (c) +6 and −2 (d) +2.5 and +2.5Answer: (b) — the two terminal sulphurs (bonded to O) are +5; the two central S–S bonded sulphurs are 0.
10. Volume strength of an H₂O₂ sample is 22.4. Its molarity is:
(a) 1 M (b) 2 M (c) 5.6 M (d) 11.2 MAnswer: (b) — Molarity = Volume strength ÷ 11.2 = 22.4/11.2 = 2 M.
11. Which of these is an example of a decomposition reaction that is NOT a redox reaction?
(a) 2KClO₃ → 2KCl + 3O₂ (b) CaCO₃ → CaO + CO₂ (c) 2H₂O₂ → 2H₂O + O₂ (d) 2HgO → 2Hg + O₂Answer: (b) — no element is present in free/elemental form among CaO and CO₂, so no electron transfer occurs.
12. What is the n-factor of FeC₂O₄ when oxidized fully to Fe³⁺ and CO₂?
(a) 1 (b) 2 (c) 3 (d) 5Answer: (c) — Fe: (3−2)=1, plus 2C: 2×(4−3)=2; total = 3.
13. Which halogen can displace oxygen from water?
(a) Cl₂ (b) Br₂ (c) I₂ (d) F₂Answer: (d) — 2F₂ + 2H₂O → 4HF + O₂, unique to fluorine's exceptional oxidizing power.
14. In back titration, which quantity is subtracted to find the moles of impure analyte Z that reacted?
(a) Milliequivalents of Y from milliequivalents of X initially added (b) Milliequivalents of Z from milliequivalents of X (c) Molarity of X from molarity of Y (d) Volume of Y from volume of XAnswer: (a) — remaining (unreacted) X = total X added minus X consumed by Y (titrated back); that remainder equals the X that reacted with Z.
15. Which oxidation state is impossible for oxygen in ordinary compounds (excluding OF₂/O₂F₂ and peroxides/superoxides)?
(a) −2 (b) −1 (c) +2 (d) 0 (in O₂)Answer: (c) — oxygen is positive only when bonded to fluorine; otherwise −2 (normal), −1 (peroxide), −1/2 (superoxide), or 0 (free element).
16. Diphenylamine indicator changes from colourless to:
(a) Pink (b) Blue (c) Yellow (d) Brick-redAnswer: (b)
17. Temporary hardness of water is due to:
(a) Chlorides of Ca and Mg (b) Sulphates of Ca and Mg (c) Bicarbonates of Ca and Mg (d) Carbonates of Na Answer: (c)18. 1 ppm of hardness equals:
(a) 1 mg CaCO₃ per litre (b) 1 g CaCO₃ per litre (c) 0.1°Fr equivalent, wrongly stated as 10 mg/L (d) 1°ClAnswer: (a) — 1 ppm = 1 mg L⁻¹ = 0.1°Fr = 0.07°Cl.
19. Which titration uses EBT as the indicator?
(a) Mohr's method (b) Volhard's method (c) EDTA hardness titration (d) Iodometric titrationAnswer: (c)
20. What is the correct order of increasing oxidation number for nitrogen: NH₄⁺, N₂, NO, NO₃⁻?
(a) NH₄⁺ < N₂ < NO < NO₃⁻ (b) N₂ < NH₄⁺ < NO < NO₃⁻ (c) NO₃⁻ < NO < N₂ < NH₄⁺ (d) NH₄⁺ < NO₃⁻ < N₂ < NOAnswer: (a) — oxidation states: NH₄⁺ (−3), N₂ (0), NO (+2), NO₃⁻ (+5).
21. Volhard's method cannot be reliably used for the direct estimation of:
(a) Ag⁺ (b) Br⁻ (c) SCN⁻ (d) Cl⁻Answer: (d) — AgCl precipitate redissolves on stirring under Volhard conditions, giving unreliable end-points.
22. Which is a primary standard suitable for direct preparation of a standard solution?
(a) NaOH (b) KMnO₄ (c) Anhydrous Na₂CO₃ (d) HCl (conc.)Answer: (c) — NaOH is hygroscopic, KMnO₄ and HCl are secondary standards requiring standardization.
23. For H₂O₂ acting simultaneously as oxidant/reductant in disproportionation (2H₂O₂ → 2H₂O + O₂), the n-factor is:
(a) 1 (b) 2 (c) 3 (d) 4Answer: (b)
24. In the titration of a Na₂CO₃/NaHCO₃ mixture, if a mL acid is used to the phenolphthalein end-point and b mL total to the methyl orange end-point, the acid used by NaHCO₃ alone is:
(a) a (b) 2a (c) b − a (d) a + bAnswer: (c)
25. Which statement about oxidation number vs valency is FALSE?
(a) Oxidation number can be fractional; valency cannot (b) Oxidation number can be zero for any element; valency is zero only for noble gases (c) Oxidation number always carries a sign; valency never does (d) Oxidation number and valency are always numerically identicalAnswer: (d) — they coincide only in simple ionic compounds; in general they differ, as the CH₄→CCl₄ series shows.
26. The reduction half-reaction for acidified KMnO₄ acting as an oxidant is:
(a) MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O (b) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (c) MnO₄⁻ + e⁻ → MnO₄²⁻ (d) 2MnO₄⁻ + 2e⁻ → Mn₂O₇Answer: (b)
27. A sample of oleum is labeled "118%." This means:
(a) 100 g oleum gives 118 g pure H₂SO₄ on complete dilution (b) The sample contains 118% SO₃ by mass (c) The density of the oleum is 1.18 g/mL (d) The sample is 18% impureAnswer: (a)
28. Which of the following is the correct relationship between normality and molarity of an acid/base?
(a) Normality = Molarity ÷ n (b) Normality = Molarity × n (c) Normality = Molarity + n (d) Normality and Molarity are always equalAnswer: (b) — where n is the n-factor (basicity/acidity/valence factor as relevant).
29. Which metal, when placed in a solution of Cu²⁺, will NOT show any displacement reaction?
(a) Zn (b) Fe (c) Ag (d) AlAnswer: (c) — Ag lies above Cu in the activity series and cannot displace it.
30. Fractional oxidation number of Fe in Fe₃O₄ is best explained by:
(a) Fe existing as a genuinely fractional-charge species (b) Averaging of two Fe³⁺ and one Fe²⁺ per formula unit (c) An error in classical bonding theory (d) Covalent bonding between Fe and O onlyAnswer: (b)
31. In the reaction Cl₂ + 2NaOH → NaCl + NaOCl + H₂O, which best describes chlorine's role?
(a) Only oxidized (b) Only reduced (c) Simultaneously oxidized and reduced (disproportionation) (d) Neither oxidized nor reducedAnswer: (c)
32. What volume (mL) of 0.1 N KMnO₄ is needed to exactly oxidize 25 mL of 0.2 N FeSO₄ solution (acidic medium)?
(a) 12.5 mL (b) 25 mL (c) 50 mL (d) 5 mLAnswer: (c) — N₁V₁ = N₂V₂ (milliequivalents equal): 0.1 × V = 0.2 × 25 → V = 50 mL.
33. Which best explains why concentrated and dilute HNO₃ have different equivalent weights as oxidizing agents?
(a) Concentration never affects equivalent weight (b) They form different reduction products (NO₂ vs NO), changing the n-factor (c) Dilution changes the molecular formula of HNO₃ (d) Only temperature, not concentration, affects equivalent weightAnswer: (b)
34. The equivalent weight of an oxidizing agent undergoing disproportionation with equal electrons transferred in both half-reactions is calculated as:
(a) Molecular weight ÷ (sum of electrons in both halves) (b) (Sum of molecular weights contributed by both halves) ÷ (number of electrons transferred) (c) Molecular weight × n (d) Molecular weight ÷ 2n alwaysAnswer: (b)
35. Which of these correctly pairs a reaction type with an example?
(a) Combination — 2KClO₃ → 2KCl + 3O₂ (b) Decomposition — Zn + 2HCl → ZnCl₂ + H₂ (c) Displacement — Cu + 2AgNO₃ → 2Ag + Cu(NO₃)₂ (d) Disproportionation — 2Mg + O₂ → 2MgOAnswer: (c) — (a) is actually decomposition, (b) is actually displacement, (d) is actually combination.
