Chapter 7 · Redox Reactions

7.2 Types of Redox Reactions and Balancing Redox Equations

Combination, decomposition, displacement and disproportionation; two reliable ways to balance any redox equation — the oxidation-number method and the half-reaction (ion–electron) method in acidic and basic solution; and redox titrations with permanganate, dichromate and iodine.

Household bleach is made by bubbling chlorine through cold sodium hydroxide solution. In that one reaction chlorine is both oxidised and reduced: half of the chlorine atoms become chloride ions and half become hypochlorite ions, the active bleaching agent. A water-testing laboratory finds the iron content of a sample by adding purple potassium permanganate from a burette until the first permanent pink tinge appears — no indicator needed. To use reactions like these you must be able to recognise what kind of redox change is happening and, above all, write a correctly balanced equation: the permanganate reaction needs exactly five iron(II) ions for every $\mathrm{MnO_{4}^{-}}$, and an unbalanced equation would give a result five times too large or too small.

Simple trial and error, which works for $\mathrm{2H_{2}+O_{2}\rightarrow 2H_{2}O}$, fails for equations such as $\mathrm{MnO_{4}^{-}+C_{2}O_{4}^{2-}+H^{+}\rightarrow Mn^{2+}+CO_{2}+H_{2}O}$. This section first sorts redox reactions into four types, then develops two systematic balancing methods built on the fact proved in Section 7.1 — that electrons lost must equal electrons gained — and finally puts them to work in redox titrations.

Types of Redox Reactions

📘 Definition 7.6 — Combination, decomposition and displacement reactions
  • In a combination reaction two or more substances combine to form a single product, $\mathrm{A+B\rightarrow C}$. It is a redox reaction when at least one reactant is an element: $\mathrm{C(s)+O_{2}(g)\rightarrow CO_{2}(g)}$, $\mathrm{3Mg(s)+N_{2}(g)\xrightarrow{\Delta}Mg_{3}N_{2}(s)}$.
  • In a decomposition reaction one compound breaks down into two or more substances, $\mathrm{C\rightarrow A+B}$. It is a redox reaction when at least one product is an element: $\mathrm{2H_{2}O(l)\xrightarrow{\text{electrolysis}}2H_{2}(g)+O_{2}(g)}$, $\mathrm{2KClO_{3}(s)\xrightarrow{\Delta}2KCl(s)+3O_{2}(g)}$. The decomposition of calcium carbonate into CaO and $\mathrm{CO_{2}}$ is not a redox reaction.
  • In a displacement reaction an atom or ion in a compound is replaced by an atom or ion of another element, $\mathrm{X+YZ\rightarrow XZ+Y}$. In metal displacement a more reactive metal replaces a less reactive one; in non-metal displacement hydrogen, oxygen or a halogen is displaced.

Metal displacements are the competitive electron transfer reactions of Section 7.1, and they are used to extract metals: a reactive metal reduces the oxide or chloride of a less reactive one.

$$\mathrm{TiCl_{4}(l)+2Mg(s)\xrightarrow{\Delta}Ti(s)+2MgCl_{2}(s)}$$ $$\mathrm{Cr_{2}O_{3}(s)+2Al(s)\xrightarrow{\Delta}Al_{2}O_{3}(s)+2Cr(s)}$$

The second, a thermite-type reaction, releases so much heat that the chromium is produced molten. Among non-metal displacements, the most familiar is the displacement of hydrogen. The alkali metals and calcium displace it from cold water, magnesium from hot water, iron only from steam, and metals such as zinc and iron from dilute acids; metals below hydrogen in the activity series (copper, silver, gold) do not displace it at all:

$$\mathrm{2Na(s)+2H_{2}O(l)\rightarrow 2NaOH(aq)+H_{2}(g)}$$ $$\mathrm{3Fe(s)+4H_{2}O(g)\xrightarrow{\Delta}Fe_{3}O_{4}(s)+4H_{2}(g)}$$

The halogens displace each other. Their oxidising power falls down the group, $\mathrm{F_{2}\gt Cl_{2}\gt Br_{2}\gt I_{2}}$, so a halogen displaces any halogen below it from a solution of its halide ions. Chlorine water added to potassium bromide gives orange bromine, and to potassium iodide gives iodine, which colours a layer of carbon tetrachloride violet — a laboratory test for $\mathrm{Br^{-}}$ and $\mathrm{I^{-}}$:

$$\mathrm{Cl_{2}(aq)+2Br^{-}(aq)\rightarrow 2Cl^{-}(aq)+Br_{2}(aq)}$$

Fluorine is so strong an oxidising agent that it displaces oxygen from water, $\mathrm{2F_{2}(g)+2H_{2}O(l)\rightarrow 4HF(aq)+O_{2}(g)}$; for this reason its displacement of other halogens cannot be carried out in aqueous solution.

The fourth type is the most interesting, because one element plays both parts.

📘 Definition 7.7 — Disproportionation and comproportionation
A disproportionation reaction is a redox reaction in which an element in one oxidation state is simultaneously oxidised and reduced: part of it goes to a higher and part to a lower oxidation number. The reacting substance must therefore contain an element that can exist in at least three oxidation states, and it must be in an intermediate one. Examples (oxidation numbers of the disproportionating element in brackets): $$\mathrm{2H_{2}O_{2}(aq)\rightarrow 2H_{2}O(l)+O_{2}(g)}\qquad (-1\rightarrow-2,\ 0)$$ $$\mathrm{Cl_{2}(g)+2OH^{-}(aq)\rightarrow Cl^{-}(aq)+ClO^{-}(aq)+H_{2}O(l)}\qquad (0\rightarrow-1,\ +1)$$ $$\mathrm{P_{4}(s)+3OH^{-}(aq)+3H_{2}O(l)\rightarrow PH_{3}(g)+3H_{2}PO_{2}^{-}(aq)}\qquad (0\rightarrow-3,\ +1)$$ The reverse process, in which an element in two different oxidation states forms a single intermediate oxidation state, is called comproportionation: $\mathrm{2H_{2}S(g)+SO_{2}(g)\rightarrow 3S(s)+2H_{2}O(l)}$ (S: $-2$ and $+4\rightarrow0$).
⚠️ Which species can disproportionate?
An element in its highest or lowest oxidation state cannot disproportionate: $\mathrm{ClO_{4}^{-}}$ (Cl $+7$) can only be reduced and $\mathrm{Cl^{-}}$ ($-1$) only oxidised. Fluorine never disproportionates either, because it has only two oxidation states, $0$ and $-1$; with alkali it gives $\mathrm{2F_{2}(g)+2OH^{-}(aq)\rightarrow}$ $\mathrm{2F^{-}(aq)+OF_{2}(g)+H_{2}O(l)}$, in which oxygen is oxidised. Chlorine, bromine and iodine (states from $-1$ to $+7$) disproportionate in alkali, and so do $\mathrm{ClO^{-}}$, $\mathrm{ClO_{2}^{-}}$ and $\mathrm{ClO_{3}^{-}}$, but not $\mathrm{ClO_{4}^{-}}$.
✏️ Example 1 — Classifying redox reactions
Assign oxidation numbers and classify each reaction as combination, decomposition, displacement (metal or non-metal) or disproportionation. Name any reaction that is not redox. (a) $\mathrm{2Pb(NO_{3})_{2}(s)\xrightarrow{\Delta}2PbO(s)+4NO_{2}(g)+O_{2}(g)}$ (b) $\mathrm{V_{2}O_{5}(s)+5Ca(s)\xrightarrow{\Delta}2V(s)+5CaO(s)}$ (c) $\mathrm{3HNO_{2}(aq)\rightarrow HNO_{3}(aq)+2NO(g)+H_{2}O(l)}$ (d) $\mathrm{S(s)+O_{2}(g)\rightarrow SO_{2}(g)}$ (e) $\mathrm{Mg(s)+2H_{2}O(l)\xrightarrow{\Delta}Mg(OH)_{2}(s)+H_{2}(g)}$ (f) $\mathrm{SO_{3}(g)+H_{2}O(l)\rightarrow H_{2}SO_{4}(l)}$
(a) N $+5\rightarrow+4$ (reduced), O $-2\rightarrow0$ in $\mathrm{O_{2}}$ (oxidised), Pb stays $+2$. One compound gives three substances, one an element: redox decomposition.
(b) V $+5\rightarrow0$, Ca $0\rightarrow+2$: calcium replaces vanadium, a metal displacement (used to make vanadium metal).
(c) N in $\mathrm{HNO_{2}}$ is $+3$; it goes to $+5$ in $\mathrm{HNO_{3}}$ and to $+2$ in NO: disproportionation. Check: one N rises by 2, two N fall by 1 each ✓.
(d) S $0\rightarrow+4$, O $0\rightarrow-2$: combination.
(e) Mg $0\rightarrow+2$, H $+1\rightarrow0$: magnesium displaces hydrogen from (hot) water — non-metal displacement.
(f) S $+6$, O $-2$, H $+1$ on both sides: a combination, but not redox.
✏️ Example 2 — Who can disproportionate?
(a) Which of these can undergo disproportionation: $\mathrm{ClO_{4}^{-}}$, $\mathrm{ClO_{3}^{-}}$, $\mathrm{F_{2}}$, $\mathrm{MnO_{4}^{2-}}$, $\mathrm{S^{2-}}$? Give reasons. (b) Manganate(VI) disproportionates in acid: $\mathrm{3MnO_{4}^{2-}(aq)+4H^{+}(aq)\rightarrow}$ $\mathrm{2MnO_{4}^{-}(aq)+MnO_{2}(s)+2H_{2}O(l)}$. Identify the changes and show that the electrons balance. (c) In hot alkali, hypochlorite disproportionates to chloride and chlorate: $\mathrm{3ClO^{-}(aq)\rightarrow 2Cl^{-}(aq)+ClO_{3}^{-}(aq)}$. Check the equation.
(a) $\mathrm{ClO_{3}^{-}}$ (Cl $+5$, between $-1$ and $+7$) and $\mathrm{MnO_{4}^{2-}}$ (Mn $+6$, between $+2$ and $+7$) can. $\mathrm{ClO_{4}^{-}}$ (highest, $+7$) and $\mathrm{S^{2-}}$ (lowest, $-2$) cannot; $\mathrm{F_{2}}$ cannot because fluorine has no positive oxidation state.
(b) Mn $+6\rightarrow+7$ in two $\mathrm{MnO_{4}^{-}}$ (oxidation, $2\times1=2$ electrons lost) and $+6\rightarrow+4$ in one $\mathrm{MnO_{2}}$ (reduction, 2 electrons gained) ✓. Charge: $3(-2)+4=-2$ on the left, $2(-1)=-2$ on the right ✓.
(c) Cl $+1\rightarrow-1$ for two Cl (4 electrons gained) and $+1\rightarrow+5$ for one Cl (4 lost) ✓. Atoms: 3 Cl, 3 O on each side; charge $-3$ on each side ✓.

Balancing Redox Equations: the Oxidation-Number Method

Every balanced redox equation must satisfy three conditions: the atoms of each element balance, the total charge is the same on both sides, and the total increase in oxidation number equals the total decrease. The oxidation-number method uses the third condition first to fix the coefficients of the species that are oxidised and reduced, then uses the other two to finish.

🧭 Problem-solving strategy — the oxidation-number method
  1. Write the skeleton equation with the correct formulas of the reactants and products.
  2. Assign oxidation numbers and find the elements whose oxidation numbers change. If needed, balance the atoms of these elements first.
  3. Find the total increase (per formula unit of the reducing agent) and the total decrease (per formula unit of the oxidising agent).
  4. Multiply by the smallest factors that make the total increase equal the total decrease; this fixes the coefficients of the redox species.
  5. For an ionic equation, balance the charge by adding $\mathrm{H^{+}}$ ions (acidic solution) or $\mathrm{OH^{-}}$ ions (basic solution) to the appropriate side.
  6. Balance hydrogen by adding $\mathrm{H_{2}O}$ molecules, then check oxygen: if the oxygen atoms balance, the equation is correct.
✏️ Example 3 — The oxidation-number method
Balance (a) $\mathrm{Cr_{2}O_{7}^{2-}(aq)+SO_{3}^{2-}(aq)\rightarrow}$ $\mathrm{Cr^{3+}(aq)+SO_{4}^{2-}(aq)}$ in acidic solution; (b) the reaction of copper with dilute nitric acid, $\mathrm{Cu(s)+HNO_{3}(aq)\rightarrow}$ $\mathrm{Cu(NO_{3})_{2}(aq)+NO(g)+H_{2}O(l)}$; (c) $\mathrm{MnO_{4}^{-}(aq)+I^{-}(aq)\rightarrow}$ $\mathrm{MnO_{2}(s)+IO_{3}^{-}(aq)}$ in basic solution.
(a) changes Cr: $+6\rightarrow+3$, a decrease of 3 per Cr, so $2\times3=6$ per $\mathrm{Cr_{2}O_{7}^{2-}}$ (which gives $2\mathrm{Cr^{3+}}$). S: $+4\rightarrow+6$, an increase of 2. Equalise: $1\times6=3\times2$, so $\mathrm{Cr_{2}O_{7}^{2-}+3SO_{3}^{2-}\rightarrow 2Cr^{3+}+3SO_{4}^{2-}}$.
(a) finish Charge: left $-2-6=-8$, right $+6-6=0$; add $8\mathrm{H^{+}}$ on the left. Hydrogen: $8$ on the left, so $4\mathrm{H_{2}O}$ on the right: $$\mathrm{Cr_{2}O_{7}^{2-}(aq)+3SO_{3}^{2-}(aq)+8H^{+}(aq)\rightarrow}$$ $$\mathrm{2Cr^{3+}(aq)+3SO_{4}^{2-}(aq)+4H_{2}O(l)}$$ Check O: $7+9=16$ on the left, $12+4=16$ on the right ✓.
(b) changes Cu: $0\rightarrow+2$, increase 2. N: $+5\rightarrow+2$ (in NO), decrease 3. Equalise: $3\times2=2\times3$, so $3\mathrm{Cu}$ and $2\mathrm{HNO_{3}}$ acting as oxidant give $3\mathrm{Cu(NO_{3})_{2}}$ and $2\mathrm{NO}$.
(b) finish $3\mathrm{Cu(NO_{3})_{2}}$ contains six nitrate groups whose N does not change; they come from six more $\mathrm{HNO_{3}}$ acting only as an acid. Total $8\mathrm{HNO_{3}}$, i.e. 8 H, so $4\mathrm{H_{2}O}$: $$\mathrm{3Cu(s)+8HNO_{3}(aq)\rightarrow 3Cu(NO_{3})_{2}(aq)+2NO(g)+4H_{2}O(l)}$$ Check O: $24=18+2+4$ ✓.
(c) changes Mn: $+7\rightarrow+4$, decrease 3. I: $-1\rightarrow+5$, increase 6. Equalise: $2\times3=1\times6$: $\mathrm{2MnO_{4}^{-}+I^{-}\rightarrow 2MnO_{2}+IO_{3}^{-}}$.
(c) finish Charge: left $-3$, right $-1$. In basic solution add $2\mathrm{OH^{-}}$ on the right (making it $-3$). Hydrogen: 2 on the right, so $1\mathrm{H_{2}O}$ on the left: $$\mathrm{2MnO_{4}^{-}(aq)+I^{-}(aq)+H_{2}O(l)\rightarrow 2MnO_{2}(s)+IO_{3}^{-}(aq)+2OH^{-}(aq)}$$ Check O: $8+1=9$; $4+3+2=9$ ✓.
Interactive 1 — Oxidation-Number Method, Step by Step

Choose a skeleton equation and step through the oxidation-number method with the slider, or press ▶ Play. The lab finds the oxidation numbers of the elements that change (red = increase, blue = decrease), equalises the total increase and decrease, adds $\mathrm{H^{+}}$ or $\mathrm{OH^{-}}$ to balance the charge and $\mathrm{H_{2}O}$ to balance hydrogen, and finally checks every element and the charge.

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Balancing Redox Equations: the Half-Reaction Method

The second method follows the physical picture of Section 7.1: the reaction is split into an oxidation half-reaction and a reduction half-reaction, each is balanced on its own, and the two are then added so that the electrons cancel. It needs no oxidation numbers at all (though they are a useful check), and it is the method used for electrode reactions in Section 7.3.

🧭 Problem-solving strategy — the half-reaction (ion–electron) method
  1. Write the skeleton ionic equation and split it into an oxidation half and a reduction half.
  2. In each half, balance the atoms other than O and H.
  3. Balance O by adding $\mathrm{H_{2}O}$ to the side short of oxygen.
  4. Balance H by adding $\mathrm{H^{+}}$ to the side short of hydrogen.
  5. Basic solution only: add as many $\mathrm{OH^{-}}$ ions to both sides as there are $\mathrm{H^{+}}$ ions; combine $\mathrm{H^{+}+OH^{-}}$ into $\mathrm{H_{2}O}$ and cancel water that appears on both sides.
  6. Balance the charge by adding electrons, $\mathrm{e^{-}}$, to the more positive side.
  7. Multiply the halves so that both contain the same number of electrons, add them, and cancel everything that appears on both sides.
  8. Check that every element and the total charge balance.
✏️ Example 4 — Half-reactions in acidic solution
Balance in acidic solution: (a) $\mathrm{MnO_{4}^{-}(aq)+C_{2}O_{4}^{2-}(aq)\rightarrow}$ $\mathrm{Mn^{2+}(aq)+CO_{2}(g)}$ — the reaction used to standardise permanganate with sodium oxalate; (b) $\mathrm{Cr_{2}O_{7}^{2-}(aq)+Fe^{2+}(aq)\rightarrow}$ $\mathrm{Cr^{3+}(aq)+Fe^{3+}(aq)}$.
(a) reduction $\mathrm{MnO_{4}^{-}\rightarrow Mn^{2+}}$; Mn balanced. O: add $4\mathrm{H_{2}O}$ on the right. H: add $8\mathrm{H^{+}}$ on the left. Charge: left $-1+8=+7$, right $+2$; add $5\mathrm{e^{-}}$ on the left: $$\mathrm{MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{2+}+4H_{2}O}$$
(a) oxidation $\mathrm{C_{2}O_{4}^{2-}\rightarrow 2CO_{2}}$ (balance C). O: $4=4$. Charge: left $-2$, right $0$; add $2\mathrm{e^{-}}$ on the right: $\mathrm{C_{2}O_{4}^{2-}\rightarrow 2CO_{2}+2e^{-}}$.
(a) combine LCM of 5 and 2 is 10: reduction $\times2$, oxidation $\times5$: $$\mathrm{2MnO_{4}^{-}(aq)+5C_{2}O_{4}^{2-}(aq)+16H^{+}(aq)\rightarrow}$$ $$\mathrm{2Mn^{2+}(aq)+10CO_{2}(g)+8H_{2}O(l)}$$ Check: Mn 2, C 10, O $8+20=28=20+8$, H 16; charge $-2-10+16=+4=2(+2)$ ✓.
(b) Reduction: $\mathrm{Cr_{2}O_{7}^{2-}\rightarrow 2Cr^{3+}}$; $+7\mathrm{H_{2}O}$ right; $+14\mathrm{H^{+}}$ left; charge $-2+14=+12$ vs $+6$, so $+6\mathrm{e^{-}}$ left. Oxidation: $\mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^{-}}$, multiplied by 6: $$\mathrm{Cr_{2}O_{7}^{2-}(aq)+6Fe^{2+}(aq)+14H^{+}(aq)\rightarrow}$$ $$\mathrm{2Cr^{3+}(aq)+6Fe^{3+}(aq)+7H_{2}O(l)}$$ Check: charge $-2+12+14=+24$ and $6+18=+24$ ✓.
✏️ Example 5 — Half-reactions in basic solution
Balance in basic solution: (a) $\mathrm{Cr(OH)_{3}(s)+ClO^{-}(aq)\rightarrow}$ $\mathrm{CrO_{4}^{2-}(aq)+Cl^{-}(aq)}$; (b) the disproportionation of chlorine in hot alkali, $\mathrm{Cl_{2}(g)\rightarrow Cl^{-}(aq)+ClO_{3}^{-}(aq)}$.
(a) oxidation $\mathrm{Cr(OH)_{3}\rightarrow CrO_{4}^{2-}}$. O: add $1\mathrm{H_{2}O}$ on the left. H: left has $3+2=5$, so $5\mathrm{H^{+}}$ on the right. Basic: add $5\mathrm{OH^{-}}$ to both sides; on the right $5\mathrm{H^{+}}+5\mathrm{OH^{-}}=5\mathrm{H_{2}O}$; cancel one $\mathrm{H_{2}O}$: $\mathrm{Cr(OH)_{3}+5OH^{-}\rightarrow CrO_{4}^{2-}+4H_{2}O}$. Charge: $-5$ vs $-2$, so $3\mathrm{e^{-}}$ on the right: $$\mathrm{Cr(OH)_{3}+5OH^{-}\rightarrow CrO_{4}^{2-}+4H_{2}O+3e^{-}}$$
(a) reduction $\mathrm{ClO^{-}\rightarrow Cl^{-}}$: $+\mathrm{H_{2}O}$ right, $+2\mathrm{H^{+}}$ left; basic: $+2\mathrm{OH^{-}}$ both sides, giving $\mathrm{ClO^{-}+2H_{2}O\rightarrow Cl^{-}+H_{2}O+2OH^{-}}$, i.e. $\mathrm{ClO^{-}+H_{2}O\rightarrow Cl^{-}+2OH^{-}}$. Charge: $-1$ vs $-3$; add $2\mathrm{e^{-}}$ on the left: $\mathrm{ClO^{-}+H_{2}O+2e^{-}\rightarrow Cl^{-}+2OH^{-}}$.
(a) combine Oxidation $\times2$, reduction $\times3$ (6 electrons); cancel $3\mathrm{H_{2}O}$ and $6\mathrm{OH^{-}}$: $$\mathrm{2Cr(OH)_{3}(s)+3ClO^{-}(aq)+4OH^{-}(aq)\rightarrow}$$ $$\mathrm{2CrO_{4}^{2-}(aq)+3Cl^{-}(aq)+5H_{2}O(l)}$$ Check: Cr 2; Cl 3; O $6+3+4=13=8+5$; H $6+4=10=10$; charge $-7=-4-3$ ✓.
(b) Reduction: $\mathrm{Cl_{2}+2e^{-}\rightarrow 2Cl^{-}}$. Oxidation: $\mathrm{Cl_{2}\rightarrow 2ClO_{3}^{-}}$; $+6\mathrm{H_{2}O}$ left, $+12\mathrm{H^{+}}$ right; basic: $+12\mathrm{OH^{-}}$ both sides and cancel water: $\mathrm{Cl_{2}+12OH^{-}\rightarrow 2ClO_{3}^{-}+6H_{2}O+10e^{-}}$. Reduction $\times5$ and add: $\mathrm{6Cl_{2}+12OH^{-}\rightarrow 10Cl^{-}+2ClO_{3}^{-}+6H_{2}O}$; divide by 2: $$\mathrm{3Cl_{2}(g)+6OH^{-}(aq)\rightarrow 5Cl^{-}(aq)+ClO_{3}^{-}(aq)+3H_{2}O(l)}$$ Check: Cl 6; O 6; H 6; charge $-6=-5-1$ ✓. Five of the six Cl atoms are reduced and one is oxidised.
⚠️ Common mistakes in balancing redox equations
  • Leaving $\mathrm{H^{+}}$ in the final equation for a reaction in basic solution (or $\mathrm{OH^{-}}$ for one in acid). Neither ion exists in appreciable amount in the other medium.
  • Writing $\mathrm{O^{2-}}$, O atoms or $\mathrm{O_{2}}$ to balance oxygen. Oxygen is balanced only with $\mathrm{H_{2}O}$ (and $\mathrm{OH^{-}}$ in base), unless $\mathrm{O_{2}}$ really is a reactant or product.
  • Putting the electrons on the wrong side: in a reduction half they are a reactant, in an oxidation half a product.
  • Forgetting to check the charge. An equation can balance in every atom and still be wrong: $\mathrm{Fe^{3+}+Cu\rightarrow Fe^{2+}+Cu^{2+}}$ is unbalanced in charge; correct is $\mathrm{2Fe^{3+}+Cu\rightarrow 2Fe^{2+}+Cu^{2+}}$.
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — when two elements are oxidised at once

Not required by the CBSE syllabus. Balancing redox equations is listed by JEE Main and NEET, and JEE Advanced lists balanced equations and redox calculations; this is a harder case of the same method.

Some reductants contain two elements that are both oxidised. Treat the whole formula unit as one: in arsenic(III) sulphide, $\mathrm{As_{2}S_{3}}$, oxidised by nitrate to $\mathrm{H_{3}AsO_{4}}$, $\mathrm{SO_{4}^{2-}}$ and NO, each As rises $+3\rightarrow+5$ and each S rises $-2\rightarrow+6$, a total increase of $2(2)+3(8)=28$ per $\mathrm{As_{2}S_{3}}$; each N falls by 3. The LCM of 28 and 3 is 84, so $3\mathrm{As_{2}S_{3}}$ react with $28\mathrm{NO_{3}^{-}}$. Finishing with $\mathrm{H^{+}}$ and $\mathrm{H_{2}O}$: $$\mathrm{3As_{2}S_{3}(s)+28NO_{3}^{-}(aq)+10H^{+}(aq)+4H_{2}O(l)\rightarrow}$$ $$\mathrm{6H_{3}AsO_{4}(aq)+9SO_{4}^{2-}(aq)+28NO(g)}$$ Check: charge $-28+10=-18=9(-2)$; O $84+4=88=24+36+28$ ✓.
Interactive 2 — Half-Reaction Balancer (Acidic and Basic Solution)

Choose a reaction and step through the half-reaction method with the slider, or press ▶ Play. The two halves are balanced side by side — key atoms, then O with $\mathrm{H_{2}O}$, H with $\mathrm{H^{+}}$ (and, in basic solution, $\mathrm{OH^{-}}$), then charge with electrons — and finally multiplied, added and simplified. The bars at the bottom compare each element and the charge on the two sides of the current overall equation; green means balanced.

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Redox Titrations and the Limits of Oxidation Numbers

A balanced redox equation is a recipe for chemical analysis. If a solution of an oxidising agent of known concentration is added from a burette to a measured volume of a reducing agent, the volume needed to reach the equivalence point — the point at which the two have reacted in exactly the ratio of the equation — gives the amount of the reducing agent. The only difficulty is seeing the equivalence point.

📘 Definition 7.8 — Redox titration and redox indicators
A redox titration is a titration in which the titrant and the analyte react by electron transfer. The end point is shown in one of three ways:
  • Self-indicator. Potassium permanganate is intensely purple, while its reduction product in acid, $\mathrm{Mn^{2+}}$, is almost colourless. As long as reductant remains, each drop of $\mathrm{MnO_{4}^{-}}$ is decolourised; the first excess drop gives a permanent pale pink colour. The acid used is dilute $\mathrm{H_{2}SO_{4}}$, not HCl, which would itself be oxidised to chlorine.
  • Redox indicator. The colour change of dichromate (orange $\mathrm{Cr_{2}O_{7}^{2-}}$ to green $\mathrm{Cr^{3+}}$) is too gradual to fix the end point, so an indicator such as diphenylamine is added; it is oxidised by the first excess of dichromate to an intensely blue product.
  • Iodine and starch. Iodine forms a deep blue complex with starch, visible at very low concentration. In iodometric titrations an oxidising agent (for example $\mathrm{Cu^{2+}}$) liberates iodine from excess iodide, and the iodine is then titrated with sodium thiosulphate until the blue colour disappears.

The reactions of an iodometric estimation of copper are

$$\mathrm{2Cu^{2+}(aq)+4I^{-}(aq)\rightarrow 2CuI(s)+I_{2}(aq)}$$ $$\mathrm{I_{2}(aq)+2S_{2}O_{3}^{2-}(aq)\rightarrow 2I^{-}(aq)+S_{4}O_{6}^{2-}(aq)}$$

In the second reaction sulphur rises from an average of $+2$ in thiosulphate to $+2.5$ in tetrathionate (Section 7.1), so each thiosulphate ion loses one electron and two of them reduce one $\mathrm{I_{2}}$.

📐 Rule 7.4 — Stoichiometry of a redox titration
If the balanced equation is $\;a\,\mathrm{Ox}+b\,\mathrm{Red}\rightarrow\text{products}$, then at the equivalence point $$\frac{n_{\mathrm{Ox}}}{a}=\frac{n_{\mathrm{Red}}}{b}\qquad\text{i.e.}\qquad \frac{M_{\mathrm{Ox}}V_{\mathrm{Ox}}}{a}=\frac{M_{\mathrm{Red}}V_{\mathrm{Red}}}{b}$$ with molarities $M$ in $\mathrm{mol\,L^{-1}}$ and both volumes in the same unit. Equivalently, the moles of electrons accepted by the oxidising agent equal the moles of electrons given by the reducing agent. For $\mathrm{MnO_{4}^{-}}$ and $\mathrm{Fe^{2+}}$, $a=1$ and $b=5$.
✏️ Example 6 — Two redox titrations
(a) $25.0\ \mathrm{mL}$ of an iron(II) sulphate solution, acidified with dilute sulphuric acid, needs $20.0\ \mathrm{mL}$ of $0.0200\ \mathrm{mol\,L^{-1}}$ $\mathrm{KMnO_{4}}$ to reach the end point. Find the molarity of $\mathrm{Fe^{2+}}$ and the mass of iron per litre ($M(\mathrm{Fe})=55.85\ \mathrm{g\,mol^{-1}}$). (b) A solution containing $\mathrm{Cu^{2+}}$ is treated with excess KI. The iodine liberated needs $24.0\ \mathrm{mL}$ of $0.100\ \mathrm{mol\,L^{-1}}$ $\mathrm{Na_{2}S_{2}O_{3}}$ (starch indicator). Find the mass of copper in the solution ($M(\mathrm{Cu})=63.55\ \mathrm{g\,mol^{-1}}$).
(a) equation $\mathrm{MnO_{4}^{-}(aq)+5Fe^{2+}(aq)+8H^{+}(aq)\rightarrow}$ $\mathrm{Mn^{2+}(aq)+5Fe^{3+}(aq)+4H_{2}O(l)}$, so $a=1$, $b=5$.
(a) amounts $n(\mathrm{MnO_{4}^{-}})=0.0200\times0.0200=4.00\times10^{-4}\ \mathrm{mol}$; $n(\mathrm{Fe^{2+}})=5\times4.00\times10^{-4}=2.00\times10^{-3}\ \mathrm{mol}$.
(a) result $M(\mathrm{Fe^{2+}})=\dfrac{2.00\times10^{-3}}{0.0250}=0.0800\ \mathrm{mol\,L^{-1}}$; mass of iron $=0.0800\times55.85=4.47\ \mathrm{g\,L^{-1}}$.
(b) From the two equations, $2\mathrm{Cu^{2+}}\equiv\mathrm{I_{2}}\equiv2\mathrm{S_{2}O_{3}^{2-}}$, so $n(\mathrm{Cu^{2+}})=n(\mathrm{S_{2}O_{3}^{2-}})=0.100\times0.0240=2.40\times10^{-3}\ \mathrm{mol}$. Mass $=2.40\times10^{-3}\times63.55=0.153\ \mathrm{g}$.
Check Electron count in (a): $4.00\times10^{-4}\ \mathrm{mol}$ $\mathrm{MnO_{4}^{-}}$ accept $5\times4.00\times10^{-4}=2.00\times10^{-3}\ \mathrm{mol}$ of electrons, one from each $\mathrm{Fe^{2+}}$ ✓.
⚠️ Limitations of the oxidation-number concept
Oxidation numbers are a bookkeeping device, not a measurement. They assume that every bond is ionic, which is far from true for covalent molecules: carbon in $\mathrm{CH_{4}}$ is "$-4$" and in $\mathrm{CCl_{4}}$ "$+4$", yet neither carbon atom carries anything like that charge. They depend on the structure, not just the formula, so formula-based values can be fractional ($\mathrm{Fe_{3}O_{4}}$, $\mathrm{S_{4}O_{6}^{2-}}$) or impossible ($\mathrm{CrO_{5}}$). And in many organic and biochemical reactions the real change is a shift of electron density, not a transfer. Modern chemistry therefore describes redox change as a change in electron density around an atom, which the oxidation number tracks only approximately — but the oxidation number remains the quickest reliable way to recognise and balance redox reactions.

Exercises

I
Types of redox reactions.
  • 1 Give one example each of a redox combination, a redox decomposition and a decomposition that is not redox.
  • 2 Classify: (a) $\mathrm{2NaH(s)\xrightarrow{\Delta}2Na(s)+H_{2}(g)}$ (b) $\mathrm{Zn(s)+2HCl(aq)\rightarrow ZnCl_{2}(aq)+H_{2}(g)}$ (c) $\mathrm{2Cu^{+}(aq)\rightarrow Cu^{2+}(aq)+Cu(s)}$ (d) $\mathrm{N_{2}(g)+O_{2}(g)\rightarrow 2NO(g)}$.
  • 3 Which of the halogens can displace bromine from potassium bromide solution? Write one ionic equation.
  • 4 Why is fluorine not used to displace chlorine from aqueous sodium chloride?
  • 5 State the condition for a species to undergo disproportionation, and say which of $\mathrm{H_{2}O_{2}}$, $\mathrm{SO_{4}^{2-}}$, $\mathrm{NO_{2}}$, $\mathrm{I^{-}}$ can do so.
  • 6 In $\mathrm{2NO_{2}(g)+H_{2}O(l)\rightarrow}$ $\mathrm{HNO_{3}(aq)+HNO_{2}(aq)}$, give the oxidation-number changes and name the type of reaction.
  • 7 In $\mathrm{IO_{3}^{-}(aq)+5I^{-}(aq)+6H^{+}(aq)\rightarrow}$ $\mathrm{3I_{2}(aq)+3H_{2}O(l)}$, which type of redox reaction is this? Verify the electron balance.
1 Combination: $\mathrm{2Mg+O_{2}\rightarrow 2MgO}$. Redox decomposition: $\mathrm{2HgO\rightarrow 2Hg+O_{2}}$. Not redox: $\mathrm{CaCO_{3}\rightarrow CaO+CO_{2}}$.
2 (a) Redox decomposition (H $-1\rightarrow0$, Na $+1\rightarrow0$). (b) Displacement of hydrogen (Zn $0\rightarrow+2$, H $+1\rightarrow0$). (c) Disproportionation (Cu $+1\rightarrow+2$ and $0$). (d) Combination (N $0\rightarrow+2$, O $0\rightarrow-2$).
3 Fluorine and chlorine (above bromine in the group). $\mathrm{Cl_{2}(aq)+2Br^{-}(aq)\rightarrow 2Cl^{-}(aq)+Br_{2}(aq)}$.
4 Fluorine oxidises the water itself ($\mathrm{2F_{2}+2H_{2}O\rightarrow 4HF+O_{2}}$) before it can oxidise chloride ions.
5 It must contain an element in an intermediate oxidation state, with at least one higher and one lower state available. $\mathrm{H_{2}O_{2}}$ (O $-1$) and $\mathrm{NO_{2}}$ (N $+4$) can; $\mathrm{SO_{4}^{2-}}$ (S $+6$, highest) and $\mathrm{I^{-}}$ ($-1$, lowest) cannot.
6 N: $+4\rightarrow+5$ (in $\mathrm{HNO_{3}}$) and $+4\rightarrow+3$ (in $\mathrm{HNO_{2}}$): disproportionation.
7 Comproportionation: I $+5$ (one atom) and $-1$ (five atoms) both go to $0$. Electrons: one I gains 5; five I lose 1 each ✓.
II
The oxidation-number method. Balance, showing the oxidation-number changes.
  • 1 $\mathrm{Fe_{2}O_{3}(s)+C(s)\rightarrow Fe(s)+CO(g)}$
  • 2 $\mathrm{MnO_{4}^{-}(aq)+Fe^{2+}(aq)\rightarrow}$ $\mathrm{Mn^{2+}(aq)+Fe^{3+}(aq)}$ (acidic)
  • 3 $\mathrm{Cr_{2}O_{7}^{2-}(aq)+H_{2}S(g)\rightarrow Cr^{3+}(aq)+S(s)}$ (acidic)
  • 4 $\mathrm{Zn(s)+NO_{3}^{-}(aq)\rightarrow Zn^{2+}(aq)+NH_{4}^{+}(aq)}$ (acidic)
  • 5 $\mathrm{MnO_{4}^{-}(aq)+SO_{3}^{2-}(aq)\rightarrow}$ $\mathrm{MnO_{2}(s)+SO_{4}^{2-}(aq)}$ (basic)
  • 6 $\mathrm{P_{4}(s)+OH^{-}(aq)\rightarrow}$ $\mathrm{PH_{3}(g)+H_{2}PO_{2}^{-}(aq)}$ (basic)
  • 7 Concentrated nitric acid with copper gives $\mathrm{NO_{2}}$: balance $\mathrm{Cu(s)+HNO_{3}(aq)\rightarrow}$ $\mathrm{Cu(NO_{3})_{2}(aq)+NO_{2}(g)+H_{2}O(l)}$.
  • 8 In part 7, what fraction of the nitric acid acts as an oxidising agent?
1 Fe $+3\rightarrow0$ (decrease $2\times3=6$ per $\mathrm{Fe_{2}O_{3}}$); C $0\rightarrow+2$ (increase 2). $\mathrm{Fe_{2}O_{3}(s)+3C(s)\rightarrow 2Fe(s)+3CO(g)}$.
2 Mn: decrease 5; Fe: increase 1. $\mathrm{MnO_{4}^{-}+5Fe^{2+}+8H^{+}\rightarrow Mn^{2+}+5Fe^{3+}+4H_{2}O}$.
3 Cr: decrease 6 per $\mathrm{Cr_{2}O_{7}^{2-}}$; S: $-2\rightarrow0$, increase 2. $\mathrm{Cr_{2}O_{7}^{2-}+3H_{2}S+8H^{+}\rightarrow 2Cr^{3+}+3S+7H_{2}O}$.
4 Zn: increase 2; N: $+5\rightarrow-3$, decrease 8. $\mathrm{4Zn+NO_{3}^{-}+10H^{+}\rightarrow 4Zn^{2+}+NH_{4}^{+}+3H_{2}O}$.
5 Mn: decrease 3; S: increase 2. $\mathrm{2MnO_{4}^{-}+3SO_{3}^{2-}+H_{2}O\rightarrow 2MnO_{2}+3SO_{4}^{2-}+2OH^{-}}$.
6 P: $0\rightarrow-3$ (decrease 3) and $0\rightarrow+1$ (increase 1), so one P reduced for every three oxidised. $\mathrm{P_{4}+3OH^{-}+3H_{2}O\rightarrow PH_{3}+3H_{2}PO_{2}^{-}}$.
7 Cu: increase 2; N: $+5\rightarrow+4$, decrease 1; so $2\mathrm{NO_{2}}$ per Cu, plus 2 nitrate in $\mathrm{Cu(NO_{3})_{2}}$. $\mathrm{Cu+4HNO_{3}\rightarrow Cu(NO_{3})_{2}+2NO_{2}+2H_{2}O}$.
8 Two of the four $\mathrm{HNO_{3}}$ are reduced: one half.
III
The half-reaction method.
  • 1 Balance the half-reaction $\mathrm{NO_{3}^{-}\rightarrow NO}$ in acidic solution.
  • 2 Balance the half-reaction $\mathrm{SO_{2}\rightarrow SO_{4}^{2-}}$ in acidic solution.
  • 3 Using the half-reaction of part 2, balance $\mathrm{MnO_{4}^{-}(aq)+SO_{2}(g)\rightarrow}$ $\mathrm{Mn^{2+}(aq)+SO_{4}^{2-}(aq)}$ in acidic solution.
  • 4 Balance $\mathrm{H_{2}O_{2}(aq)+I^{-}(aq)\rightarrow H_{2}O(l)+I_{2}(aq)}$ in acidic solution.
  • 5 Balance $\mathrm{MnO_{4}^{-}(aq)+H_{2}O_{2}(aq)\rightarrow}$ $\mathrm{Mn^{2+}(aq)+O_{2}(g)}$ in acidic solution. Is $\mathrm{H_{2}O_{2}}$ the oxidising or the reducing agent here?
  • 6 Balance $\mathrm{MnO_{4}^{-}(aq)+I^{-}(aq)\rightarrow MnO_{2}(s)+I_{2}(aq)}$ in basic solution.
  • 7 Balance $\mathrm{ClO_{2}(g)+H_{2}O_{2}(aq)\rightarrow}$ $\mathrm{ClO_{2}^{-}(aq)+O_{2}(g)}$ in basic solution.
  • 8 Balance $\mathrm{Al(s)+NO_{3}^{-}(aq)\rightarrow}$ $\mathrm{[Al(OH)_{4}]^{-}(aq)+NH_{3}(g)}$ in basic solution.
1 $\mathrm{NO_{3}^{-}+4H^{+}+3e^{-}\rightarrow NO+2H_{2}O}$.
2 $\mathrm{SO_{2}+2H_{2}O\rightarrow SO_{4}^{2-}+4H^{+}+2e^{-}}$.
3 Reduction ($\mathrm{MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{2+}+4H_{2}O}$) $\times2$, oxidation $\times5$, then cancel $16\mathrm{H^{+}}$ and $8\mathrm{H_{2}O}$: $\mathrm{2MnO_{4}^{-}+5SO_{2}+2H_{2}O\rightarrow 2Mn^{2+}+5SO_{4}^{2-}+4H^{+}}$. Charge $-2=+4-10+4$ ✓.
4 $\mathrm{H_{2}O_{2}+2H^{+}+2e^{-}\rightarrow 2H_{2}O}$ and $\mathrm{2I^{-}\rightarrow I_{2}+2e^{-}}$: $\mathrm{H_{2}O_{2}+2I^{-}+2H^{+}\rightarrow 2H_{2}O+I_{2}}$.
5 $\mathrm{H_{2}O_{2}\rightarrow O_{2}+2H^{+}+2e^{-}}$ ($\times5$) with the permanganate half ($\times2$): $\mathrm{2MnO_{4}^{-}+5H_{2}O_{2}+6H^{+}\rightarrow 2Mn^{2+}+5O_{2}+8H_{2}O}$. $\mathrm{H_{2}O_{2}}$ is the reducing agent (O $-1\rightarrow0$).
6 $\mathrm{MnO_{4}^{-}+2H_{2}O+3e^{-}\rightarrow MnO_{2}+4OH^{-}}$ ($\times2$) and $\mathrm{2I^{-}\rightarrow I_{2}+2e^{-}}$ ($\times3$): $\mathrm{2MnO_{4}^{-}+6I^{-}+4H_{2}O\rightarrow 2MnO_{2}+3I_{2}+8OH^{-}}$.
7 $\mathrm{ClO_{2}+e^{-}\rightarrow ClO_{2}^{-}}$ ($\times2$) and $\mathrm{H_{2}O_{2}+2OH^{-}\rightarrow O_{2}+2H_{2}O+2e^{-}}$: $\mathrm{2ClO_{2}+H_{2}O_{2}+2OH^{-}\rightarrow 2ClO_{2}^{-}+O_{2}+2H_{2}O}$.
8 $\mathrm{Al+4OH^{-}\rightarrow [Al(OH)_{4}]^{-}+3e^{-}}$ ($\times8$) and $\mathrm{NO_{3}^{-}+6H_{2}O+8e^{-}\rightarrow NH_{3}+9OH^{-}}$ ($\times3$): $\mathrm{8Al+3NO_{3}^{-}+5OH^{-}+18H_{2}O\rightarrow}$ $\mathrm{8[Al(OH)_{4}]^{-}+3NH_{3}}$. Check: O $32=32$, H $41=41$, charge $-8=-8$ ✓.
IV
Redox titrations. Use $M(\mathrm{Na_{2}C_{2}O_{4}})=134.0$, $M(\mathrm{Fe})=55.85$, $M(\mathrm{C_{6}H_{8}O_{6}})=176.1\ \mathrm{g\,mol^{-1}}$.
  • 1 Why is no indicator added in permanganate titrations? What colour marks the end point?
  • 2 Why is a permanganate titration acidified with dilute sulphuric acid rather than hydrochloric acid?
  • 3 $0.250\ \mathrm{g}$ of sodium oxalate, dissolved in dilute acid, needs $37.3\ \mathrm{mL}$ of a $\mathrm{KMnO_{4}}$ solution (Example 4(a)). Find the molarity of the permanganate.
  • 4 $20.0\ \mathrm{mL}$ of a hydrogen peroxide solution needs $18.0\ \mathrm{mL}$ of $0.0200\ \mathrm{mol\,L^{-1}}$ $\mathrm{KMnO_{4}}$ in acid (part III.5). Find the molarity of $\mathrm{H_{2}O_{2}}$.
  • 5 A $0.500\ \mathrm{g}$ sample of iron ore is dissolved and all its iron reduced to $\mathrm{Fe^{2+}}$. Titration needs $24.5\ \mathrm{mL}$ of $0.0170\ \mathrm{mol\,L^{-1}}$ $\mathrm{K_{2}Cr_{2}O_{7}}$. Find the percentage of iron in the ore.
  • 6 Name the indicator used in part 5 and explain why one is needed.
  • 7 Applied. Vitamin C (ascorbic acid) is titrated directly with iodine: $\mathrm{C_{6}H_{8}O_{6}(aq)+I_{2}(aq)\rightarrow}$ $\mathrm{C_{6}H_{6}O_{6}(aq)+2I^{-}(aq)+2H^{+}(aq)}$. A tablet labelled "250 mg vitamin C" needs $27.3\ \mathrm{mL}$ of $0.0500\ \mathrm{mol\,L^{-1}}$ $\mathrm{I_{2}}$ with starch. Find the mass of vitamin C and compare with the label. What colour change marks the end point?
1 $\mathrm{MnO_{4}^{-}}$ is intensely purple and $\mathrm{Mn^{2+}}$ nearly colourless, so permanganate is its own indicator; the end point is the first permanent pale pink.
2 $\mathrm{MnO_{4}^{-}}$ oxidises $\mathrm{Cl^{-}}$ to $\mathrm{Cl_{2}}$, so some permanganate would be used up by the acid and the titre would be too large. Sulphate is not oxidised (S is already $+6$).
3 $n(\mathrm{C_{2}O_{4}^{2-}})=0.250/134.0=1.866\times10^{-3}\ \mathrm{mol}$; $n(\mathrm{MnO_{4}^{-}})=\tfrac25\times1.866\times10^{-3}=7.46\times10^{-4}\ \mathrm{mol}$; $M=7.46\times10^{-4}/0.0373=0.0200\ \mathrm{mol\,L^{-1}}$.
4 $n(\mathrm{MnO_{4}^{-}})=3.60\times10^{-4}\ \mathrm{mol}$; $n(\mathrm{H_{2}O_{2}})=\tfrac52\times3.60\times10^{-4}=9.00\times10^{-4}\ \mathrm{mol}$; $M=9.00\times10^{-4}/0.0200=0.0450\ \mathrm{mol\,L^{-1}}$.
5 $n(\mathrm{Cr_{2}O_{7}^{2-}})=0.0170\times0.0245=4.165\times10^{-4}\ \mathrm{mol}$; $n(\mathrm{Fe^{2+}})=6\times$ that $=2.499\times10^{-3}\ \mathrm{mol}$; mass $=0.1396\ \mathrm{g}$; percentage $=27.9\ \%$.
6 Diphenylamine. The orange-to-green change of dichromate is gradual and masked by the green $\mathrm{Cr^{3+}}$ formed, so the end point cannot be seen without an indicator that gives a sharp, intense colour (blue).
7 $n(\mathrm{I_{2}})=0.0500\times0.0273=1.365\times10^{-3}\ \mathrm{mol}=n(\text{vitamin C})$; mass $=1.365\times10^{-3}\times176.1=0.240\ \mathrm{g}=240\ \mathrm{mg}$, i.e. $96\ \%$ of the label value. End point: the first permanent blue colour (excess iodine with starch).
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET Which of the following is a disproportionation reaction? (A) $\mathrm{2KClO_{3}\rightarrow 2KCl+3O_{2}}$ (B) $\mathrm{2H_{2}O_{2}\rightarrow 2H_{2}O+O_{2}}$ (C) $\mathrm{Zn+CuSO_{4}\rightarrow ZnSO_{4}+Cu}$ (D) $\mathrm{2H_{2}S+SO_{2}\rightarrow 3S+2H_{2}O}$

Idea One element, starting in one oxidation state, must go both up and down. In (B) O goes from $-1$ to $-2$ (in $\mathrm{H_{2}O}$) and to $0$ (in $\mathrm{O_{2}}$). (A) is a redox decomposition in which two different elements (Cl and O) change; (C) is a displacement; (D) is comproportionation.
Answer (B).

E2. JEE Main Numerical answer. For the reaction $a\,\mathrm{MnO_{4}^{-}}+b\,\mathrm{C_{2}O_{4}^{2-}}+c\,\mathrm{H^{+}}\rightarrow$ $x\,\mathrm{Mn^{2+}}+y\,\mathrm{CO_{2}}+z\,\mathrm{H_{2}O}$ balanced with the smallest whole numbers, the value of $a+b+c$ is ____.

Balance Mn gains 5 electrons, each $\mathrm{C_{2}O_{4}^{2-}}$ loses 2: ratio $2:5$. Charge: $-2-10+c=+4$ gives $c=16$ (Example 4).
Answer $2+5+16=$ 23.

E3. Board · 3 marks Balance by the ion–electron method in basic medium: $\mathrm{MnO_{4}^{-}(aq)+I^{-}(aq)\rightarrow}$ $\mathrm{MnO_{2}(s)+I_{2}(s)}$.

1 mark Oxidation: $\mathrm{2I^{-}\rightarrow I_{2}+2e^{-}}$. Reduction: $\mathrm{MnO_{4}^{-}\rightarrow MnO_{2}}$; $+2\mathrm{H_{2}O}$ right, $+4\mathrm{H^{+}}$ left, $+3\mathrm{e^{-}}$ left.
1 mark Basic medium: add $4\mathrm{OH^{-}}$ to both sides and simplify: $\mathrm{MnO_{4}^{-}+2H_{2}O+3e^{-}\rightarrow MnO_{2}+4OH^{-}}$.
1 mark Reduction $\times2$, oxidation $\times3$ (6 electrons): $\mathrm{2MnO_{4}^{-}(aq)+6I^{-}(aq)+4H_{2}O(l)\rightarrow}$ $\mathrm{2MnO_{2}(s)+3I_{2}(s)+8OH^{-}(aq)}$; charge $-8=-8$ ✓.

E4. JEE Main The number of moles of $\mathrm{KMnO_{4}}$ that react completely with one mole of iron(II) oxalate, $\mathrm{FeC_{2}O_{4}}$, in acidic solution is (A) $0.2$ (B) $0.6$ (C) $1.0$ (D) $1.67$

Electrons Both ions of $\mathrm{FeC_{2}O_{4}}$ are oxidised: $\mathrm{Fe^{2+}\rightarrow Fe^{3+}}$ gives 1 electron and $\mathrm{C_{2}O_{4}^{2-}\rightarrow 2CO_{2}}$ gives 2, so 3 mol of electrons per mole.
Permanganate Each $\mathrm{MnO_{4}^{-}}$ accepts 5 electrons: $3/5=0.6\ \mathrm{mol}$. (A) counts only the iron; (D) inverts the ratio.
Answer (B).

E5. JEE Advanced One or more options correct. For $\mathrm{3Cl_{2}(g)+6OH^{-}(aq)\rightarrow}$ $\mathrm{5Cl^{-}(aq)+ClO_{3}^{-}(aq)+3H_{2}O(l)}$: (A) chlorine acts as both oxidising and reducing agent; (B) the ratio of chlorine atoms reduced to those oxidised is $5:1$; (C) the oxidation number of Cl in the oxidised product is $+5$; (D) it is a comproportionation reaction.

(A), (B) True: five Cl go $0\rightarrow-1$ (5 electrons gained) and one goes $0\rightarrow+5$ (5 lost).
(C) True: $x+3(-2)=-1$, $x=+5$.
(D) False: one oxidation state splits into two — this is disproportionation, the reverse of comproportionation.
Answer (A), (B), (C).

E6. Board · Case study A quality-control laboratory checks iron tablets (iron(II) sulphate). One tablet is dissolved in dilute sulphuric acid and titrated with $0.0100\ \mathrm{mol\,L^{-1}}$ $\mathrm{KMnO_{4}}$; the mean titre is $12.5\ \mathrm{mL}$. (i) Write the balanced ionic equation for the titration reaction. (2 marks) (ii) Calculate the amount of $\mathrm{Fe^{2+}}$ in the tablet. (1 mark) (iii) Find the mass of iron per tablet in mg ($M(\mathrm{Fe})=55.85\ \mathrm{g\,mol^{-1}}$) and say how the end point is detected. (1 mark)

(i) $\mathrm{MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\rightarrow}$ $\mathrm{Mn^{2+}(aq)+4H_{2}O(l)}$ and $\mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^{-}}$ ($\times5$): $\mathrm{MnO_{4}^{-}(aq)+5Fe^{2+}(aq)+8H^{+}(aq)\rightarrow}$ $\mathrm{Mn^{2+}(aq)+5Fe^{3+}(aq)+4H_{2}O(l)}$.
(ii) $n(\mathrm{MnO_{4}^{-}})=0.0100\times0.0125=1.25\times10^{-4}\ \mathrm{mol}$; $n(\mathrm{Fe^{2+}})=5\times1.25\times10^{-4}=6.25\times10^{-4}\ \mathrm{mol}$.
(iii) $6.25\times10^{-4}\times55.85=0.0349\ \mathrm{g}=34.9\ \mathrm{mg}$. Permanganate is its own indicator: the end point is the first permanent pale pink colour.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How much oxygen is dissolved in a river, and how can a chain of three redox reactions measure it to a fraction of a milligram per litre? Design and analyse a Winkler titration for a river near you.

Why it matters. Fish need at least about $5\ \mathrm{mg}$ of dissolved oxygen per litre; below about $2\ \mathrm{mg\,L^{-1}}$ water becomes "hypoxic" and fish kills follow. Pollution-control boards in India, the US Environmental Protection Agency and the EU Water Framework Directive all use dissolved oxygen (DO) and biochemical oxygen demand (BOD) as key indicators of river health — the Ganga, Yamuna, Thames and Rhine are all monitored this way. The Winkler method, devised in 1888, is still the reference against which electronic oxygen probes are calibrated.

What to do.

  1. The sample is "fixed" with $\mathrm{Mn^{2+}}$ in alkali: $\mathrm{Mn^{2+}+OH^{-}+O_{2}\rightarrow MnO(OH)_{2}}$ (brown). Balance it, and give the oxidation-number change of Mn and of O.
  2. On acidifying with excess iodide, $\mathrm{MnO(OH)_{2}+I^{-}+H^{+}\rightarrow}$ $\mathrm{Mn^{2+}+I_{2}+H_{2}O}$. Balance it by the half-reaction method.
  3. The iodine is titrated with thiosulphate (starch indicator). Combine the three equations to show that $1\ \mathrm{mol}\ \mathrm{O_{2}}\equiv4\ \mathrm{mol}\ \mathrm{S_{2}O_{3}^{2-}}$.
  4. A $200\ \mathrm{mL}$ sample needs $10.5\ \mathrm{mL}$ of $0.0100\ \mathrm{mol\,L^{-1}}$ $\mathrm{Na_{2}S_{2}O_{3}}$. Find the DO in $\mathrm{mg\,L^{-1}}$ and judge the river's health.

Tools: the half-reaction method, Rule 7.4, a water-quality standard from a pollution-control board, a spreadsheet.

Hand in: The three balanced equations, the overall mole ratio, your DO calculation and a 150-word note on why samples must be fixed at the riverside rather than carried back to the laboratory.

Going further. The solubility of oxygen in water falls from about $11\ \mathrm{mg\,L^{-1}}$ at $10\ ^{\circ}\mathrm{C}$ to about $7.5\ \mathrm{mg\,L^{-1}}$ at $30\ ^{\circ}\mathrm{C}$. Why are summer fish kills more common, even without extra pollution?

Instructor note: $\mathrm{2Mn^{2+}+4OH^{-}+O_{2}\rightarrow 2MnO(OH)_{2}}$ (Mn $+2\rightarrow+4$, O $0\rightarrow-2$); $\mathrm{MnO(OH)_{2}+2I^{-}+4H^{+}\rightarrow Mn^{2+}+I_{2}+3H_{2}O}$; $\mathrm{I_{2}+2S_{2}O_{3}^{2-}\rightarrow 2I^{-}+S_{4}O_{6}^{2-}}$. So $1\ \mathrm{O_{2}}\equiv2\ \mathrm{MnO(OH)_{2}}\equiv2\ \mathrm{I_{2}}\equiv4\ \mathrm{S_{2}O_{3}^{2-}}$. Sample: $n(\mathrm{S_{2}O_{3}^{2-}})=1.05\times10^{-4}\ \mathrm{mol}$, $n(\mathrm{O_{2}})=2.625\times10^{-5}\ \mathrm{mol}=0.840\ \mathrm{mg}$ in $0.200\ \mathrm{L}$, DO $=4.20\ \mathrm{mg\,L^{-1}}$ — below the $5\ \mathrm{mg\,L^{-1}}$ needed by most fish: a stressed river. Fixing traps the oxygen as a solid before it can exchange with air or be used by microbes. Going further: warm water holds less oxygen while warm fish and microbes use more.

🧪 Self-check quiz

0/5 correct

Q1. Which reaction is a disproportionation?

Q2. In the balanced half-reaction for $\mathrm{MnO_{4}^{-}\rightarrow Mn^{2+}}$ in acidic solution, the numbers of $\mathrm{H^{+}}$ and $\mathrm{e^{-}}$ are

Q3. In $\mathrm{3Cu+8HNO_{3}\rightarrow 3Cu(NO_{3})_{2}+2NO+4H_{2}O}$, how many of the eight $\mathrm{HNO_{3}}$ molecules act as an oxidising agent?

Q4. The end point of a titration of $\mathrm{Fe^{2+}}$ with acidified $\mathrm{KMnO_{4}}$ is detected by

Q5. In basic solution, the balanced equation $\mathrm{2MnO_{4}^{-}+I^{-}+H_{2}O\rightarrow}$ $\mathrm{2MnO_{2}+IO_{3}^{-}+2OH^{-}}$ shows that the change in oxidation number of iodine is

← 7.1 Oxidation, Reduction and Oxidation Number7.3 Redox Reactions and Electrode Processes →