Redox Reactions: Class 11 Chemistry · Chapter 7

Chemistry Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 22 Oct 2026 · Updated 22 Oct 2026

85 min
1 simulation
37 practice questions
27 MCQs

What you will learn in Redox Reactions

Recognise redox reactions through the oxygen–hydrogen, electron-transfer and oxidation-number views, and explain why oxidation and reduction always occur together
Name the oxidising and reducing agents in a reaction and compare their strengths from displacement reactions and standard electrode potentials
Assign oxidation numbers with a consistent set of rules, interpret fractional (average) values and use Stock notation
Classify redox reactions as combination, decomposition, displacement or disproportionation
Balance redox equations in acidic or basic solution by the oxidation-number method and the half-reaction method
Explain how redox titrations show their end points and use their stoichiometry in calculations
Describe a redox couple, the working of a Daniell cell and the meaning of a standard electrode potential

Original companion text aligned to NCERT Chemistry Class 11, Chapter 7, pp. 235–255. Not affiliated with or endorsed by NCERT.

Strike a matchstick, watch an iron gate rust through the monsoon, leave a cut apple on the plate or let a phone battery run down. In every case electrons are moving from one substance to another. Changes in which one substance is oxidised while another is reduced are called redox reactions (अपचयोपचय अभिक्रियाएँ), a word stitched together from reduction and oxidation.

Redox chemistry quietly runs much of daily life. Burning LPG, coal and petrol releases energy by oxidation. Aluminium and sodium are won from their compounds by electrolysis, a redox change driven by electricity, and caustic soda and chlorine come from brine the same way. Your cells oxidise glucose to stay alive, green leaves reduce carbon dioxide, and bleaching, corrosion and every battery are redox at heart. Even debates about hydrogen as a clean fuel, or about the ozone layer, turn on redox reactions.

Chemists did not reach today's meaning in one step. Oxidation first meant gaining oxygen, then losing electrons, and finally a rise in oxidation number. We follow that path, then use the idea to sort redox reactions into families, balance their equations, run titrations and build a simple cell.

7.1 Classical idea of redox reactions: oxidation and reduction

The oldest meaning of oxidation is the plainest: a substance combines with oxygen. Heat a copper wire in a flame and it blackens; burn spirit in a lamp and it becomes carbon dioxide and water:

\begin{aligned} 2 C u + O_{2} \overset{}{\to} 2 C u O \\ C_{2} H_{5} O H + 3 O_{2} \\ \overset{}{\to} 2 C O_{2} + 3 H_{2} O \end{aligned}

Oxygen's appetite for metals also explains the rocks we mine. Over geological time, metals near the surface combined with the oxygen of the air (about 21 % by volume), which is why ores such as haematite ( $F e_{2} O_{3}$ ) and bauxite (mostly hydrated $A l_{2} O_{3}$ ) are oxides.

A second clue came from reactions that strip hydrogen away. Ethanol vapour passed over hot copper loses hydrogen and becomes ethanal, $C_{2} H_{5} O H \overset{}{\to} C H_{3} C H O + H_{2} .$ Chemists began to count removal of hydrogen as oxidation too, and burning fuels fit the pattern, since their hydrogen ends up attached to oxygen.

Next, chemists noticed that other electronegative elements behave like oxygen. Aluminium burning in chlorine, $2 A l + 3 C l_{2} \overset{}{\to} 2 A l C l_{3},$ or zinc heated with sulphur, $Z n + S \overset{}{\to} Z n S,$ oxidises the metal just as surely as oxygen would. Taking away an electropositive partner counts too: when chlorine water turns potassium iodide into iodine, $2 K I + C l_{2} \overset{}{\to} 2 K C l + I_{2},$ the iodide loses its potassium and is oxidised.

Reduction grew as the mirror image. It began as removing oxygen, the old furnace meaning of 'reducing' an ore to its metal, as when roasted zinc ore is heated with coke: $Z n O + C \overset{}{\to} Z n + C O$ (Fig. 7.1). Today it also covers removing an electronegative element, adding hydrogen, as in $C H_{3} C H O + H_{2} \overset{}{\to} C_{2} H_{5} O H,$ or adding an electropositive element.

Compare

Oxidation and reduction, classical view

Oxidation

adds oxygen or another electronegative element
removes hydrogen or an electropositive element (a metal)

Reduction

removes oxygen or another electronegative element
adds hydrogen or an electropositive element

3/3Zinc oxide has been reduced and carbon oxidised, in the same event.

Fig. 7.1One transfer, two names: the oxide is reduced while the carbon is oxidised.

Three circles stand for zinc, oxygen and carbon. Step 1: oxygen is bonded to zinc as zinc oxide, with a separate carbon atom on the right. Step 2: a dashed arrow shows the oxygen atom moving across to the carbon. Step 3: zinc is left alone with a tag saying lost oxygen equals reduced, and carbon monoxide carries a tag saying gained oxygen equals oxidised.

Run through any of these reactions again and the same pattern appears: whatever one substance gives up, another takes. In $2 F e C l_{3} + S n C l_{2} \overset{}{\to} 2 F e C l_{2} + S n C l_{4},$ iron(III) chloride loses chlorine (reduced) exactly as tin(II) chloride gains it (oxidised). Oxidation never happens without reduction, which is why the pair is named as one: redox.

Example 7.1NCERT Problem 7.1 · p. 236

Spotting oxidation and reduction the classical way

In each reaction, say which substance is oxidised and which is reduced, using only the oxygen, hydrogen and electronegativity ideas.

(i) $P b O + C O \overset{}{\to} P b + C O_{2}$ (ii) $2 H_{2} S + S O_{2} \overset{}{\to} 3 S + 2 H_{2} O$ (iii) $6 L i + N_{2} \overset{}{\to} 2 L i_{3} N$

Try it yourself first. Then reveal the solution one step at a time.

Part (iii) needed a judgement about electronegativity, and for many reactions that judgement is awkward. A sharper test comes from asking where the electrons go.

7.2 Redox reactions in terms of electron transfer reactions

The classical rules work, but they ask you to remember which partner is 'oxygen-like'. The products tell a deeper story. Calcium chloride, magnesium oxide and lithium nitride are ionic solids, better written $C a^{2 +} (C l -)_{2},$ $M g^{2 +} O^{2 -}$ and $(L i^{+})_{3} N^{3 -} .$ To make them, metal atoms must hand electrons to non-metal atoms.

We can write each change as two half reactions (अर्ध अभिक्रियाएँ), one that releases electrons and one that takes them up (Fig. 7.2). For calcium burning in chlorine:

Two half reactions

\begin{aligned} C a \overset{}{\to} C a^{2 +} + 2 e^{-} \\ C l_{2} + 2 e^{-} \overset{}{\to} 2 C l^{-} \end{aligned}

Adding the two lines cancels the electrons and leaves $C a + C l_{2} \overset{}{\to} C a C l_{2} .$ The half that loses electrons is the oxidation; the half that gains them is the reduction. Calcium, which gives electrons away, reduces chlorine, so calcium is the reducing agent. Chlorine, which takes them, oxidises calcium and is the oxidising agent. The electron picture explains every classical case: oxygen, chlorine and sulphur are electron-hungry, and metals give electrons readily.

3/3Calcium is left as $C a^{2 +},$ oxidised; the chlorines are now $C l^{-},$ reduced.

Fig. 7.2Calcium loses two electrons (oxidation); each chlorine atom gains one (reduction).

A calcium atom sits between two chlorine atoms, each drawn with a dashed outer shell. Step 1: calcium shows two outer electrons and each chlorine shows one empty place. Step 2: curved arrows carry one electron from calcium to each chlorine. Step 3: calcium has shrunk into a calcium ion labelled oxidised, minus two electrons, and each chlorine has grown into a chloride ion labelled reduced, plus one electron.

Definition

Oxidation

Loss of electrons by an atom, ion or molecule. The species that loses electrons is the reducing agent.

Example $Z n \overset{}{\to} Z n^{2 +} + 2 e^{-}$

Definition

Reduction

Gain of electrons by an atom, ion or molecule. The species that gains electrons is the oxidising agent.

Example $C u^{2 +} + 2 e^{-} \overset{}{\to} C u$

Compare

Two kinds of reagent

Oxidising agent (oxidant)

accepts electrons
is reduced itself
examples: $O_{2},$ $C l_{2},$ $M n O_{4}^{-}$

Reducing agent (reductant)

donates electrons
is oxidised itself
examples: Mg, Zn, $H_{2},$ C

A memory aid many students use is OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons). The naming of the agents trips people more often: the reducing agent is the one that gets oxidised.

Example 7.2NCERT Problem 7.2 · p. 237

Proving a redox change with half reactions

Lithium reacts with nitrogen even at room temperature: $6 L i + N_{2} \overset{}{\to} 2 L i_{3} N .$ Show that this is a redox reaction and name the reducing agent.

Try it yourself first. Then reveal the solution one step at a time.

7.2.1 Competitive electron transfer reactions

If two substances compete for electrons, which one wins? A simple bench test answers it. Try it before reading on.

Try this

Which metal gives up its electrons?

You’ll need

zinc granules
bare copper wire, cleaned with sandpaper
1 M copper sulphate
1 M zinc sulphate
0.1 M silver nitrate
three test tubes in a stand

Predict first

In which tubes do you expect a new coating on the metal?

A onlyA and CB and Call three

Pour 5 mL of each solution into three labelled test tubes: copper sulphate (A), zinc sulphate (B) and dilute silver nitrate (C).
Drop a clean zinc granule into A. Push a cleaned coil of copper wire into B and another into C.
Watch for five minutes, then again after an hour. Note any coating on the metal and any colour change in the solution.

What did you see?

Tube	Prediction	What happened
A: Zn in $C u S O_{4}$
B: Cu in $Z n S O_{4}$
C: Cu in $A g N O_{3}$

Your entries are saved automatically.

In tube A zinc atoms hand electrons straight to copper ions touching the metal surface (Fig. 7.3). Zinc atoms become zinc ions that wander off into the solution, while copper ions turn into copper atoms that stay on the strip:

\begin{aligned} Z n (s) + C u^{2 +} (a q) \\ \overset{}{\to} Z n^{2 +} (a q) + C u (s) \end{aligned}

(7.15)

3/4The zinc atom leaves as $Z n^{2 +};$ a copper atom takes its place on the surface.

Fig. 7.3Zinc atoms give electrons to copper ions at the metal surface.NCERT Fig. 7.1 · p. 238

A close-up of the zinc surface as two rows of grey zinc atoms under a blue solution of copper ions; a legend on the right names zinc atoms, copper ions, copper atoms and zinc ions. Step 1: copper ions move freely in the solution. Step 2: one copper ion touches the surface and an arrow shows two electrons passing from a zinc atom to it. Step 3: that zinc atom has become a dashed zinc ion drifting up into the solution, and a brown copper atom sits in its place. Step 4: three surface sites are now copper, two more zinc ions are in solution and the blue colour has faded.

Can the reverse happen? In tube B nothing visible occurs, and even the most sensitive chemical tests find essentially no copper ions in the solution. The equilibrium for reaction (7.15) lies overwhelmingly to the right: its equilibrium constant at 298 K is about $10^{37},$ a figure you will learn to calculate in Class 12.

Copper in turn passes electrons to silver ions, as tube C showed: $C u + 2 A g^{+} \overset{}{\to} C u^{2 +} + 2 A g .$ Here too the products are hugely favoured. Not every pair is so one-sided, though. Tin metal in lead(II) nitrate solution reacts only partly, $S n + P b^{2 +} \overset{}{⇌} S n^{2 +} + P b,$ and at equilibrium both kinds of ion remain in comparable amounts ( $K \approx 2$ ), because tin and lead hold their electrons almost equally tightly.

Ranking metals by how readily they release electrons gives an activity series, or electrochemical series; our three tubes already show $Z n > C u > A g .$ Arranging the contest so that the electrons travel through a wire instead of straight across a surface gives a galvanic cell, which turns chemical energy into electrical energy (Section 7.4, and in detail in Class 12).

InteractiveDisplacement bench

Buttons choose one of six metal strips (magnesium, zinc, iron, nickel, copper, silver) and one of seven solutions (the sulphates of magnesium, zinc, iron, nickel and copper, silver nitrate, dilute hydrochloric acid). You first predict whether anything happens, then dip the strip: a coating of the displaced metal or hydrogen bubbles appear and the solution colour changes when a reaction goes. A panel gives the balanced ionic equation, the two standard electrode potentials, the cell voltage and an estimate of the equilibrium constant, keeps a score of your predictions and lists the metals tested in order of reducing power.

Fig. 7.4Pick a metal and a solution, predict, then press Dip the strip to see whether the metal is coated, what the equation is and how the E° values decide it.

7.3 Oxidation number

Electron transfer is easy to see when the product is ionic. Now think of hydrogen burning, $2 H_{2} + O_{2} \overset{}{\to} 2 H_{2} O .$ Water is a covalent molecule, so no electron has fully left hydrogen. Yet the shared pairs in the O–H bonds sit closer to oxygen, the more electronegative atom: hydrogen has partly given up electron density and oxygen has partly gained it. Electron density has moved, but no electron has fully changed hands. The same holds for $H_{2} + B r_{2} \overset{}{\to} 2 H B r$ and for nitrogen in $N_{2} + 3 H_{2} \overset{}{\to} 2 N H_{3} .$

To follow such partial shifts with whole numbers, chemists use oxidation numbers. The rule behind them is a deliberate exaggeration: in every bond, pretend the shared electrons belong entirely to the more electronegative atom. The charge each atom would then carry is its oxidation number. Water gives H = +1 and O = −2; hydrogen bromide gives H = +1 and Br = −1; ammonia gives N = −3 and H = +1. The pretence is harmless, and it describes redox changes very neatly.

Definition

Oxidation number

The charge an atom would have in a substance if every shared electron pair were handed wholly to the more electronegative atom of its bond (pairs between identical atoms are split equally).

ExampleIn $H_{2} O$ oxygen is −2 and each hydrogen +1.

4/4H has lost 1 electron (+1), O has gained 2 (−2) and Cl has lost 1 (+1); the sum is zero.

Fig. 7.5Give each shared pair to the more electronegative atom, then count.

The molecule H–O–Cl drawn with bonding pairs as black dots between atoms and lone pairs as grey dots; electronegativities 2.2, 3.4 and 3.2 are written below. Step 1: both bonding pairs sit midway. Step 2: the H–O pair slides onto oxygen. Step 3: the O–Cl pair also slides onto oxygen, because oxygen is more electronegative than chlorine. Step 4: badges show +1 for hydrogen, −2 for oxygen and +1 for chlorine, adding to zero.

Drawing a structure every time would be slow, so we use a short list of rules that give the same answers (Fig. 7.5 shows why chlorine can end up positive).

Free elements are zero. Every atom in Fe, $N_{2},$ $B r_{2},$ $O_{3},$ $P_{4}$ or $S_{8}$ has oxidation number 0.
Everything adds up to the overall charge: zero for a neutral compound, the ion's charge for an ion. This is the rule you use to find the unknown one.
Some values never change. Fluorine is always −1; group 1 metals are +1, group 2 metals +2 and aluminium +3 in all their compounds; a monatomic ion such as $S^{2 -}$ carries its own charge.
Hydrogen is +1 when bonded to non-metals, but −1 in binary metal hydrides such as KH and $M g H_{2} .$
Oxygen is −2 except where it bonds to itself or to fluorine: −1 in peroxides ( $H_{2} O_{2},$ $B a O_{2}$ ), $- \frac{1}{2}$ in superoxides such as $C s O_{2},$ +2 in $O F_{2}$ and +1 in $O_{2} F_{2} .$
Chlorine, bromine and iodine are −1 in halides, but positive when bonded to oxygen or a lighter halogen: Cl is +7 in $H C l O_{4}$ and +3 in $C l F_{3} .$

For example, in $S O_{4}^{2 -}$ rule 2 gives $x + 4 (- 2) = - 2,$ so sulphur is +6.

When a formula contains several atoms of one element, the rules return their average value. Usually that is also each atom's value, but not always, as the box at the end of this section shows.

Line up the period 3 elements and a pattern appears (Fig. 7.6). The highest oxidation number climbs one step per group, from +1 for sodium to +7 for chlorine, while the non-metals also reach down to negative values, from −4 for silicon to −1 for chlorine. In general a main-group element tops out at its group number in groups 1 and 2, and at (group number − 10) in groups 13 to 17. Metals never go negative in their compounds, non-metals can go either way, and transition metals usually show several positive states: iron +2 and +3, manganese anything from +2 to +7.

2/3Non-metals also show negative values, as in $M g_{2} S i$ (−4), $N a_{3} P$ (−3), $N a_{2} S$ (−2) and NaCl (−1).

Fig. 7.6Highest and lowest oxidation numbers across period 3.NCERT p. 241

Bars for Na, Mg, Al, Si, P, S and Cl, with their group numbers underneath. Step 1: solid bars above zero show the highest oxidation numbers, +1 for sodium rising one step at a time to +7 for chlorine. Step 2: dashed bars below zero show the lowest values of the non-metals: −4 for silicon, −3 for phosphorus, −2 for sulphur and −1 for chlorine. Step 3: brackets beside silicon, phosphorus, sulphur and chlorine show that the highest and lowest values are always 8 apart.

Oxidation state is used as a synonym: sulphur in $S O_{3}$ is in the +6 oxidation state. For metals that show more than one state, the German chemist Alfred Stock proposed in 1919 putting the oxidation number into the name; in the form adopted a few years later it is a Roman numeral in brackets after the metal. This is Stock notation. 'Ferrous chloride' becomes iron(II) chloride, $F e C l_{2},$ and 'ferric chloride' iron(III) chloride, $F e C l_{3};$ cuprous and cupric oxides become copper(I) oxide, $C u_{2} O,$ and copper(II) oxide, CuO. The numeral tells you at once which form is more oxidised.

Example 7.3NCERT Problem 7.3 · p. 241

Naming compounds in Stock notation

Give the Stock name of each compound: $C u_{2} S,$ $P b O_{2},$ $V_{2} O_{5},$ $M n_{2} O_{7},$ $C o C l_{2},$ $K_{2} P t C l_{6} .$

Try it yourself first. Then reveal the solution one step at a time.

Oxidation numbers turn every redox word into something you can check with arithmetic. The table sets the two views side by side.

Table 7.1Redox terms in the electron view and the oxidation-number view

	Electron view	Oxidation-number view
Oxidation	electrons lost	number goes up
Reduction	electrons gained	number goes down
Oxidising agent	takes electrons; is reduced	its own number falls
Reducing agent	gives electrons; is oxidised	its own number rises
Redox reaction	electrons transferred	some numbers change

Example 7.4NCERT Problem 7.4 · p. 241

Finding the oxidant and the reductant

Black copper(II) oxide heated in ammonia gas turns into shiny copper: $3 C u O + 2 N H_{3} \overset{}{\to} 3 C u + N_{2} + 3 H_{2} O .$ Show that this is redox and name the oxidant and the reductant.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

What is the oxidation number of chromium in

K_{2} C r O_{4}

Show answer

Answer 6

2 (+ 1) + x + 4 (- 2) = 0

gives

x = + 6.

Did you know?NCERT Box: fractional oxidation number · p. 245

When the oxidation number is a fraction

Rules sometimes return a fraction. In magnetite, $F e_{3} O_{4},$ iron comes out at $+ \frac{8}{3};$ in the azide ion, $N_{3}^{-},$ nitrogen averages $- \frac{1}{3};$ in propane, $C_{3} H_{8},$ carbon averages $- \frac{8}{3} .$ No atom can lose a third of an electron, so what is going on?

A fraction is a signal that atoms of the same element sit in different environments. Magnetite is really $F e O \cdot F e_{2} O_{3}$ : one iron(II) and two iron(III) per formula (Fig. 7.7). The azide ion is a line of three nitrogens; giving each bond's electrons out by the rules makes the end atoms −1 and the middle one +1, which averages to $- \frac{1}{3} .$ In propane the two end carbons are −3 and the middle one −2. Red lead, $P b_{3} O_{4},$ is similarly two lead(II) and one lead(IV). Only the structure reveals the real values; a fractional oxidation number is an average.

2/2One iron(II) and two iron(III): the average of +2, +3 and +3 is $+ \frac{8}{3} .$

Fig. 7.7The fraction is an average of whole-number states.

Three circles labelled Fe stand for the iron atoms in one Fe3O4 unit. Step 1: all three look alike and each is labelled plus eight thirds, the value the rules give. Step 2: the first circle turns dashed and is labelled +2 under a bracket marked FeO part, while the other two are labelled +3 under a bracket marked Fe2O3 part; a line shows that two plus three plus three over three equals eight thirds.

7.3.1 Types of redox reactions

Oxidation numbers let us sort redox reactions into four families. In each case, the test for 'redox' is simply whether any oxidation number changes.

Combination reactions

Two substances joining into one, $A + B \to C,$ is redox whenever at least one of them is a free element, because an element's zero oxidation number must change when it bonds to something different. All burning in oxygen falls in this family. Aluminium foil's protective skin forms by $4 A l + 3 O_{2} \overset{}{\to} 2 A l_{2} O_{3}$ (Al 0 → +3, O 0 → −2). Propane in LPG burns as $C_{3} H_{8} + 5 O_{2}$ $\overset{}{\to} 3 C O_{2} + 4 H_{2} O$ : carbon rises from an average of $- \frac{8}{3}$ to +4 and oxygen falls to −2. Combining two compounds need not be redox at all: in $C a O + C O_{2} \overset{}{\to} C a C O_{3}$ no oxidation number changes.

Decomposition reactions

When one compound falls apart into simpler substances, look at the products: if a free element appears among them, some oxidation number must have changed, and the decomposition is redox. Heating saltpetre, $2 K N O_{3} \overset{}{\to} 2 K N O_{2} + O_{2},$ lowers nitrogen from +5 to +3 and raises some oxygen from −2 to 0. Older car airbags inflated by $2 N a N_{3} \overset{}{\to} 2 N a + 3 N_{2},$ where sodium falls from +1 to 0 and nitrogen rises from $- \frac{1}{3}$ to 0. But plenty of decompositions are not redox: $M g C O_{3} \overset{}{\to} M g O + C O_{2}$ changes no oxidation number.

Displacement reactions

In a displacement a free element trades places with an element in a compound: $X + Y Z \to X Z + Y .$ The incoming element may be a metal or a non-metal.

Metal displacement. A more reactive metal pushes a less reactive one out of its compound, which makes these reactions useful for extracting metals. The thermite reaction, $F e_{2} O_{3} + 2 A l \overset{}{\to} A l_{2} O_{3} + 2 F e,$ gives white-hot molten iron and is used to weld railway rails. Titanium can be won from its chloride with sodium, $T i C l_{4} + 4 N a \overset{}{\to} T i + 4 N a C l,$ and an iron nail in copper sulphate solution gets a reddish-brown coat of copper. The metal that ends up as ions is always the stronger reducing agent of the pair.

Non-metal displacement: hydrogen. How much coaxing a metal needs before it releases hydrogen is a direct measure of its reducing power (Table 7.2). With dilute acids, the bubbling is fastest for magnesium, slower for zinc and slower still for iron, the same order as their activity. Zinc with dilute sulphuric acid, $Z n + H_{2} S O_{4}$ $\overset{}{\to} Z n S O_{4} + H_{2},$ is the usual laboratory source.

Table 7.2Where metals get their hydrogen from (most reactive first)

Hydrogen source	Metals	Example
cold water	group 1 metals; Ca, Sr, Ba	$2 K + 2 H_{2} O$ $\overset{}{\to} 2 K O H + H_{2}$
steam	Mg, Zn, Fe	$3 F e + 4 H_{2} O (g)$ $\overset{}{\to} F e_{3} O_{4} + 4 H_{2}$
dilute acid only	Ni, Sn (slowly)	$S n + 2 H C l$ $\overset{}{\to} S n C l_{2} + H_{2}$
none of these	Cu, Ag, Au	no hydrogen, even with dilute HCl

Non-metal displacement: halogens. A favourite laboratory test shows this family at work. Add chlorine water to a halide solution, shake with a little non-polar solvent, and watch the solvent layer: orange-brown means bromine was set free, violet means iodine. Chlorine has taken electrons from the halide ions, as in $C l_{2} + 2 B r^{-} \overset{}{\to} 2 C l^{-} + B r_{2} .$ This layer test works because oxidising power falls down group 17: chlorine frees bromine and iodine, bromine frees only iodine, and iodine frees neither. Fluorine, strongest of all, would attack the water itself and release oxygen, so it is never used this way. Industry recovers bromine from sea water and brines by the same chlorine reaction.

Disproportionation reactions

In disproportionation one element, starting in a single oxidation state, is oxidised and reduced in the same reaction. It is possible only for an element with three or more accessible oxidation states, and only when it starts on a middle one. A bottle of hydrogen peroxide kept for cleaning wounds slowly loses strength this way: $2 H_{2} O_{2} \overset{}{\to} 2 H_{2} O + O_{2} .$ The oxygen of the peroxide starts at −1; some goes up to 0 in $O_{2}$ and some down to −2 in water.

Chlorine shows the idea clearly on an oxidation-state ladder (Fig. 7.8). In cold dilute alkali, $C l_{2} + 2 O H^{-}$ $\overset{}{\to} C l^{-} + C l O^{-} + H_{2} O$ : half the chlorine atoms drop to −1 and half rise to +1. The hypochlorite ion is the active part of household bleach, oxidising coloured stains to colourless products. In hot alkali the hypochlorite itself disproportionates further to chloride and chlorate ( $C l O_{3}^{-},$ +5). Bromine and iodine behave similarly, as do phosphorus and sulphur in hot alkali, and copper(I) in water: $2 C u^{+} \overset{}{\to} C u + C u^{2 +} .$

Fluorine is the exception among the halogens. Nothing can pull electrons away from it, so it has no positive oxidation states at all and no higher rung to climb to. With alkali it is simply reduced to fluoride, and the oxygen of the hydroxide ion is oxidised instead, giving oxygen difluoride, $O F_{2} .$

2/4 $C l_{2}$ (0) splits into $C l O^{-}$ (+1) and $C l^{-}$ (−1): the bleach reaction.

Fig. 7.8Disproportionation needs a middle rung.

A vertical ladder of chlorine oxidation states from −1 at the bottom to +7 at the top, with chloride, chlorine, hypochlorite, chlorite, chlorate and perchlorate written beside their rungs. Step 1: the ladder alone. Step 2: chlorine at 0 is highlighted, with a solid arrow up to hypochlorite at +1 marked oxidised and a dashed arrow down to chloride at −1 marked reduced, labelled cold alkali. Step 3: hypochlorite at +1 is highlighted, with arrows up to chlorate at +5 and down to chloride, labelled hot alkali. Step 4: perchlorate at the top and chloride at the bottom are boxed with notes that they can only go down or only go up.

Example 7.5NCERT Problem 7.5 · p. 244

Which species can disproportionate?

Which of $N H_{3},$ $H N O_{2},$ $N O_{3}^{-}$ and $H_{2} O_{2}$ cannot disproportionate? Write an equation for each that can.

Try it yourself first. Then reveal the solution one step at a time.

Example 7.6NCERT Problem 7.6 · p. 244

Classifying redox reactions

Name the family of each redox reaction: (a) $2 M g + C O_{2} \overset{}{\to} 2 M g O + C$ (b) $2 A g_{2} O \overset{}{\to} 4 A g + O_{2}$ (c) $2 F e + 3 C l_{2} \overset{}{\to} 2 F e C l_{3}$ (d) $4 K C l O_{3} \overset{}{\to} 3 K C l O_{4} + K C l$ (gentle heating)

Try it yourself first. Then reveal the solution one step at a time.

Example 7.7NCERT Problem 7.7 · p. 246

Same acid, different behaviour

Hausmannite, $M n_{3} O_{4},$ and magnetite, $F e_{3} O_{4},$ are both mixed oxides of a metal in its +2 and +3 states. Warmed with concentrated hydrochloric acid, one releases greenish chlorine gas and the other gives only a solution of chlorides. Which is which, and why?

Try it yourself first. Then reveal the solution one step at a time.

7.3.2 Balancing of redox reactions

Redox equations often look hard to balance because several species change at once. Two methods tame them. The oxidation number method tracks the rise and fall of oxidation numbers; the half-reaction method (also called the ion–electron method) balances the oxidation and reduction parts separately. Both rest on one principle: the electrons lost by the reducing agent must equal the electrons gained by the oxidising agent (Fig. 7.9). Use whichever you find clearer.

2/3Five oxalates and two permanganates: 10 electrons each way.

Fig. 7.9Electrons lost must equal electrons gained before the halves are added.

A beam balance with an oxidation half on the left (oxalate to carbon dioxide) and a reduction half on the right (permanganate to manganese(II)). Step 1: the left pan holds 2 electron counters and the right pan 5, and the beam tilts down on the right. Step 2: after multiplying the halves by 5 and 2, both pans hold 10 counters and the beam is level. Step 3: the pans are empty because the electrons cancel, and the balanced overall equation appears below.

(a) Oxidation number method

Step by step

The oxidation number method, one step at a time

Step 1 of 5

Write the skeleton

Use correct formulas for everything that reacts and forms. Example, in acid: $M n O_{4}^{-} + F e^{2 +}$ $\overset{}{\to} M n^{2 +} + F e^{3 +} .$

1 / 5

Example 7.8NCERT Problem 7.8 · p. 246

Oxidation number method in acid: the breathalyser

Older roadside breath testers used acidified orange dichromate, which turns green as it oxidises ethanol to acetic acid. Balance $C r_{2} O_{7}^{2 -} + C_{2} H_{5} O H$ $\overset{}{\to} C r^{3 +} + C H_{3} C O O H$ in acidic solution.

Try it yourself first. Then reveal the solution one step at a time.

Example 7.9NCERT Problem 7.9 · p. 247

Oxidation number method in base

In qualitative analysis, green chromium(III) hydroxide is oxidised to yellow chromate by hydrogen peroxide in alkali. Balance $C r (O H)_{3} + H_{2} O_{2}$ $\overset{}{\to} C r O_{4}^{2 -} + H_{2} O$ in basic solution.

Try it yourself first. Then reveal the solution one step at a time.

(b) Half-reaction method

This method treats the oxidation and the reduction as two separate equations, balances each one fully, and only then combines them. In acidic solution:

Split the skeleton ionic equation into an oxidation half and a reduction half.
In each half, fix the 'key' atoms first: everything except O and H.
Fix oxygen with water molecules, then hydrogen with $H^{+}$ ions.
Fix the charge with electrons, which always go on the more positive side.
Scale the halves to the same electron count, add them, cancel what appears on both sides, and check atoms and charge.

A basic solution needs one extra move. Free $H^{+}$ ions cannot survive among hydroxide ions, so after balancing as if in acid, neutralise them: put as many $O H^{-}$ on both sides as there are $H^{+},$ let each $H^{+}$ / $O H^{-}$ pair become a water molecule, and strike out water that appears on both sides. Here is the method on the reaction behind permanganate titrations of oxalate; reveal it one line at a time:

Permanganate oxidises oxalate

\begin{aligned} 2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} \\ + 16 H^{+} \overset{}{\to} 2 M n^{2 +} \\ + 10 C O_{2} + 8 H_{2} O \end{aligned}

InteractiveHalf-reaction balancer

A menu offers six redox reactions in acidic or basic solution, such as permanganate with iron(II), dichromate with iodide, copper with nitrate and permanganate with sulphite. Next and Back buttons walk through the half-reaction method one step at a time: the skeleton, the two halves with their oxidation numbers, balancing other atoms, oxygen with water, hydrogen with hydrogen ions, the switch to hydroxide ions in base, electrons, equalising and adding the halves, and a final table checking every element and the charge.

Fig. 7.10Choose a reaction and press Next step to balance it by the half-reaction method, including the switch to basic solution.

Example 7.10NCERT Problem 7.10 · p. 248

Half reactions in basic solution

Zinc dust in hot sodium hydroxide reduces nitrate ions to ammonia, a laboratory test for nitrate. Balance $Z n + N O_{3}^{-}$ $\overset{}{\to} Z n O_{2}^{2 -} + N H_{3}$ in basic solution by the half-reaction method.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Put the half-reaction steps (acidic solution) in order.

Add H₂O to balance O
Balance atoms except O and H
Add electrons to balance charge
Equalise electrons and add
Add H⁺ to balance H
Split into two halves

Show answer

Answer

Split into two halves
Balance atoms except O and H
Add H₂O to balance O
Add H⁺ to balance H
Add electrons to balance charge
Equalise electrons and add

Atoms other than O and H first, then O with water, H with hydrogen ions, and only then the charge with electrons.

7.3.3 Redox reactions as the basis for titrations

A redox titration measures the amount of an oxidant or a reductant, much as an acid–base titration measures an acid or a base. The difference lies in how you see the end point: instead of a pH indicator you need something that responds to the redox change itself. Three approaches are common.

A self-coloured reagent. Potassium permanganate is intensely purple, but the manganese(II) ion it becomes in acid is almost colourless. While any reductant such as iron(II) or oxalate remains, each drop is decolourised at once. When the reductant is used up, the next drop leaves a faint pink that stays (Fig. 7.11). So little permanganate is needed to tint the solution, about $10^{- 6} m o l L^{- 1},$ that the end point you see lies almost exactly at the equivalence point, the moment when the oxidant added matches the reductant in their reacting mole ratio. Permanganate is its own indicator.

3/3One drop in excess leaves a pale pink that stays.

Fig. 7.11Permanganate signals its own end point.

A burette filled with purple potassium permanganate solution stands over a conical flask of almost colourless iron(II) solution. Step 1: a drop lands and makes a small purple patch that disappears at once. Step 2: near the end point the purple patch is larger and lingers before it fades. Step 3: the whole solution in the flask stays pale pink, marking the end point.

An added redox indicator. Potassium dichromate does change from orange to the green of chromium(III), but too gradually to judge the end point. So an indicator such as diphenylamine is added: the first slight excess of dichromate oxidises it to a deep violet.

Iodine and starch (iodometry). Many oxidants are measured indirectly. Excess potassium iodide is added, and the oxidant turns an equivalent amount of iodide into iodine, which stays dissolved as the triiodide ion, $I_{3}^{-} .$ The iodine is then titrated with sodium thiosulphate, $I_{2} + 2 S_{2} O_{3}^{2 -}$ $\overset{}{\to} 2 I^{-} + S_{4} O_{6}^{2 -} .$ A little starch added near the end gives the deep blue starch–iodine colour, which vanishes sharply when the last iodine is used. Copper(II) is estimated this way, through $2 C u^{2 +} + 4 I^{-}$ $\overset{}{\to} 2 C u I + I_{2},$ and so is the 'available chlorine' of bleaching powder or liquid bleach, through $C l O^{-} + 2 I^{-} + 2 H^{+}$ $\overset{}{\to} C l^{-} + I_{2} + H_{2} O .$

Try this

Vitamin C against iodine

You’ll need

povidone-iodine antiseptic (the brown liquid in a first-aid kit)
a vitamin C tablet, crushed
water in which rice was washed or boiled (it contains starch)
two clear glasses and a spoon

Predict first

What will happen to the dark blue colour when vitamin C is stirred in?

it disappearsit deepensit turns pinknothing changes

Half fill two clear glasses with water and stir 5 drops of the antiseptic into each: a pale yellow-brown colour.
Add a spoonful of rice water to each glass and stir. Note the colour.
Stir the crushed vitamin C tablet into the second glass only, and watch for a minute.

Colours you saw

Glass	With rice water	After vitamin C
1		(none added)
2

Your entries are saved automatically.

ExampleNCERT p. 249

Iron in a tablet by permanganate titration

An iron–folic acid tablet is dissolved in dilute sulphuric acid so that all its iron is $F e^{2 +} .$ The solution needs 10.8 mL of 0.0200 M $K M n O_{4}$ to reach a permanent pale pink. How much iron did the tablet contain? (Fe = 55.85 g mol⁻¹)

Try it yourself first. Then reveal the solution one step at a time.

7.3.4 Limitations of concept of oxidation number

Oxidation numbers are a tool, not a picture of what electrons really do. They rest on a convenient fiction, the complete transfer of bonding electrons, so they say nothing about how far electrons really move. They become uncertain when two bonded atoms have almost the same electronegativity, and, as the box on fractional values showed, they can hide atoms in different states. Chemists now often describe oxidation more generally as a decrease in electron density around an atom, and reduction as an increase, a view that also covers changes in which electrons are only partly shifted.

7.4 Redox reactions and electrode processes

Drop a zinc rod into copper sulphate solution and the reaction of Section 7.2.1 runs at the metal surface: electrons pass straight from zinc atoms to copper ions, and the energy released only warms the solution a little. To collect that energy as electricity instead, zinc must not meet the copper ions at all. The electrons have to be sent the long way round, through a wire.

So we keep the partners apart: a zinc rod in zinc sulphate solution in one beaker, a copper rod in copper sulphate solution in another. Each beaker now holds one pair, a metal touching its own ions. Such a pair is a redox couple. It is written with the oxidised form first and a slash, standing for the boundary, before the reduced form: $Z n^{2 +}$ /Zn and $C u^{2 +}$ /Cu.

Definition

Redox couple

A pair made of a species in its oxidised form and the same species in its reduced form, linked by one half reaction; written oxidised/reduced.

Example $C u^{2 +}$ /Cu; or $F e^{3 +}$ / $F e^{2 +}$ when both forms are ions.

Now join the rods by a wire through a voltmeter, and join the solutions with a salt bridge (लवण सेतु): an inverted U-tube packed with a jelly of potassium nitrate or potassium chloride set in agar, which lets ions through but keeps the solutions apart. This arrangement is the Daniell cell (Fig. 7.12). Without the bridge nothing happens; with it, the voltmeter reads about 1.1 V and three things go on together:

At the zinc rod, zinc atoms are oxidised, $Z n \overset{}{\to} Z n^{2 +} + 2 e^{-} .$ This rod is the negative electrode, the anode.
The electrons travel through the external wire to the copper rod, where copper ions are reduced, $C u^{2 +} + 2 e^{-} \overset{}{\to} C u .$ This rod is the positive electrode, the cathode. Conventional current flows the other way, from copper to zinc.
Inside the cell, ions carry the charge. Negative ions drift through the bridge towards the zinc side, where $Z n^{2 +}$ is building up, and positive ions towards the copper side, where $C u^{2 +}$ is being used up. Both solutions stay neutral, so the reaction can continue.

3/4Electrons flow through the wire from zinc to copper; ions move through the bridge.

Fig. 7.12Sending the electrons through a wire turns the zinc–copper reaction into a source of electricity.NCERT Fig. 7.3 · p. 250

Two beakers: on the left a zinc rod in colourless zinc sulphate solution, on the right a copper rod in blue copper sulphate solution, joined by a wire through a voltmeter. Step 1: no salt bridge and the meter reads 0.00 volts. Step 2: a U-shaped salt bridge joins the solutions and the meter reads 1.10 volts. Step 3: arrows show electrons moving through the wire from zinc to copper, current in the opposite direction, anions moving through the bridge towards the zinc side and cations towards the copper side; the zinc is labelled anode, negative, oxidation and the copper cathode, positive, reduction. Step 4: after running, the zinc rod is thinner, the copper rod thicker and the blue solution paler.

SimulationOutside the cell: electrons in the wire

A simple circuit with a 1.10 volt cell standing for the Daniell cell, a switch, an ammeter and an 11 ohm resistor labelled bulb. Pressing Play shows electrons leaving the negative terminal, passing through the bulb and returning to the positive terminal, with the current shown on the ammeter. Sliders change the cell voltage and the resistance, and the switch opens or closes the circuit.

Fig. 7.13A 1.10 V Daniell cell drives an 11 Ω bulb. Press Play to watch the electrons leave the negative terminal.

Try this

Press Play and follow the electrons: they leave the negative terminal (the zinc anode), pass through the bulb and enter the positive terminal (the copper cathode).
Read the ammeter: 0.10 A. That is 0.10 C each second, so about 3 × 10¹⁷ zinc atoms dissolve at the anode every second.
Raise the bulb's resistance to 22 Ω: the current halves, so the zinc dissolves half as fast. Open the switch and the reaction stops, just as when the salt bridge is lifted out.

A voltage between the rods means each electrode has its own electrode potential, the potential difference between the metal and its solution. Because it depends on concentration and temperature, values are compared under standard conditions: every dissolved species at unit concentration (1 mol L⁻¹), any gas at 1 bar, and 298 K. The potential is then the standard electrode potential, $E^{\circ} .$ A single electrode potential cannot be measured on its own, so all are measured against one reference, the standard hydrogen electrode, whose $E^{\circ}$ is fixed by convention at 0.00 V.

Tables list every couple as a reduction, oxidised form + electrons → reduced form (Table 7.3). A large positive $E^{\circ}$ means the oxidised form grabs electrons eagerly, so it is a strong oxidising agent, better than $H^{+} .$ A large negative $E^{\circ}$ means the reduced form parts with electrons readily, so it is a better reducing agent than hydrogen gas. Fluorine heads the list and lithium ends it. For a cell, the voltage under standard conditions is the difference of the two potentials; for the Daniell cell, $0.34 - (- 0.76) = 1.10 V,$ a result you will use often in Class 12.

Standard cell voltage

\begin{aligned} E_{cell}^{\circ} & = E_{cathode}^{\circ} \\ - E_{anode}^{\circ} \end{aligned}

where

$E_{cathode}^{\circ}$: standard potential of the couple that is reduced
$E_{anode}^{\circ}$: standard potential of the couple that is oxidised

Table 7.3Standard electrode potentials at 298 K (strongest oxidising agent at the top, strongest reducing agent at the bottom; ions in water)

Reduction half reaction	$E^{\circ}$ / V
$F_{2} + 2 e^{-} \overset{}{\to} 2 F^{-}$	+2.87
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \overset{}{\to} M n^{2 +} + 4 H_{2} O$	+1.51
$C l_{2} + 2 e^{-} \overset{}{\to} 2 C l^{-}$	+1.36
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \overset{}{\to} 2 C r^{3 +} + 7 H_{2} O$	+1.33
$O_{2} + 4 H^{+} + 4 e^{-} \overset{}{\to} 2 H_{2} O$	+1.23
$B r_{2} + 2 e^{-} \overset{}{\to} 2 B r^{-}$	+1.09
$A g^{+} + e^{-} \overset{}{\to} A g$	+0.80
$F e^{3 +} + e^{-} \overset{}{\to} F e^{2 +}$	+0.77
$I_{2} + 2 e^{-} \overset{}{\to} 2 I^{-}$	+0.54
$C u^{2 +} + 2 e^{-} \overset{}{\to} C u$	+0.34
$2 H^{+} + 2 e^{-} \overset{}{\to} H_{2}$	0.00
$P b^{2 +} + 2 e^{-} \overset{}{\to} P b$	−0.13
$S n^{2 +} + 2 e^{-} \overset{}{\to} S n$	−0.14
$N i^{2 +} + 2 e^{-} \overset{}{\to} N i$	−0.25
$F e^{2 +} + 2 e^{-} \overset{}{\to} F e$	−0.44
$Z n^{2 +} + 2 e^{-} \overset{}{\to} Z n$	−0.76
$2 H_{2} O + 2 e^{-} \overset{}{\to} H_{2} + 2 O H^{-}$	−0.83
$A l^{3 +} + 3 e^{-} \overset{}{\to} A l$	−1.66
$M g^{2 +} + 2 e^{-} \overset{}{\to} M g$	−2.37
$N a^{+} + e^{-} \overset{}{\to} N a$	−2.71
$K^{+} + e^{-} \overset{}{\to} K$	−2.93
$L i^{+} + e^{-} \overset{}{\to} L i$	−3.04

2/3 $C u^{2 +}$ (higher) oxidises Zn (lower): $Z n + C u^{2 +} \overset{}{\to} Z n^{2 +} + C u$ goes.

Fig. 7.14An oxidised form reacts with any reduced form below it on the ladder.

A vertical ladder, not to scale, with five couples: fluorine/fluoride at +2.87 volts, copper(II)/copper at +0.34, hydrogen ion/hydrogen at 0.00 in a box, zinc(II)/zinc at −0.76 and lithium ion/lithium at −3.04, with oxidised forms in the left column and reduced forms in the right. Arrows on the right show that oxidising power increases upwards and reducing power downwards. Step 2: copper(II) ion and zinc metal are highlighted and a solid arrow links them: they react. Step 3: zinc ion and copper metal are highlighted and a dashed arrow with a cross links them: no reaction.

ExampleNCERT Table 7.1 · p. 251

Using standard potentials to predict a reaction

Under standard conditions, (a) can acidified permanganate oxidise chloride ions? (b) can iron(III) ions oxidise bromide ions? Use the table.

Try it yourself first. Then reveal the solution one step at a time.

Recap

Redox Reactions: key points and summary

Redox reactions pair an oxidation with a reduction. Classically, oxidation is gain of oxygen or loss of hydrogen, and reduction the reverse (Section 7.1).
In electron terms, oxidation is loss of electrons and reduction is gain; the reducing agent is oxidised and the oxidising agent is reduced (Section 7.2).
Metals compete for electrons: zinc reduces $C u^{2 +}$ and copper reduces $A g^{+},$ never the reverse, which gives an activity series (Section 7.2.1).
Oxidation numbers come from treating every bond as ionic; simple rules give them, and the sum equals the overall charge (Section 7.3).
A fractional oxidation number is an average over atoms in different real states, such as Fe(II) and Fe(III) in $F e_{3} O_{4}$ (the box on fractions).
Stock notation shows a metal's oxidation number as a Roman numeral, as in iron(III) chloride (Example 7.3).
Redox reactions fall into four families: combination, decomposition, displacement and disproportionation (Section 7.3.1).
Disproportionation needs an element in an intermediate state, which is why fluorine never shows it (Fig. 7.8).
Equations are balanced by equalising electrons lost and gained, either through oxidation numbers or through half reactions, with $H^{+}$ / $H_{2} O$ in acid and $O H^{-}$ / $H_{2} O$ in base (Section 7.3.2).
Redox titrations see their end point through permanganate's own colour, a redox indicator such as diphenylamine, or the starch–iodine colour (Section 7.3.3).
Oxidation numbers are a book-keeping device; oxidation is more generally a decrease in electron density (Section 7.3.4).
In a Daniell cell the couples $Z n^{2 +}$ /Zn and $C u^{2 +}$ /Cu are joined by a wire and a salt bridge; standard electrode potentials, measured against the hydrogen electrode, rank oxidising and reducing agents (Section 7.4).

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In your NCERT textbook	Practice question here
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Quick check: MCQs on Redox Reactions

7.1 Classical idea of redox reactions: oxidation and reduction3 questions
7.2.1 Competitive electron transfer reactions4 questions
7.3 Oxidation number4 questions
Disproportionation reactions4 questions
(b) Half-reaction method4 questions
7.3.4 Limitations of concept of oxidation number4 questions
7.4 Redox reactions and electrode processes4 questions

Glossary · 15

Key terms (English–Hindi)

displacement reactionविस्थापन अभिक्रियाvisthapan abhikriya: A reaction in which an element pushes another element out of its compound: X + YZ → XZ + Y.First seen in the chapter →
disproportionationअसमानुपातनasamanupatan: A redox reaction in which one element, starting in a single oxidation state, is both oxidised and reduced.First seen in the chapter →

electrochemical series: Couples listed in order of standard electrode potential, ranking oxidising and reducing agents.First seen in the chapter →

half reactionअर्ध अभिक्रियाardh abhikriya: The oxidation part or the reduction part of a redox reaction written on its own, with the electrons shown.First seen in the chapter →

oxidationऑक्सीकरणoxikaran: Loss of electrons by a species, seen as a rise in oxidation number (classically, gain of oxygen or loss of hydrogen).
oxidation numberऑक्सीकरण संख्याoxikaran sankhya: The charge an atom would carry if every shared electron pair belonged wholly to the more electronegative atom.
oxidising agentऑक्सीकारकoxikarak: A species that accepts electrons and so oxidises another; it is itself reduced.First seen in the chapter →

redox couple: The oxidised and reduced forms of one species in a half reaction, written oxidised/reduced, such as $Z n^{2 +}$ / $Z n .$ First seen in the chapter →
redox reactionअपचयोपचय अभिक्रियाapachayopachay abhikriya: A reaction in which one species is oxidised while another is reduced, so oxidation numbers change.First seen in the chapter →
redox titration: A titration between an oxidant and a reductant whose end point is seen as a colour change.First seen in the chapter →
reducing agentअपचायकapachayak: A species that donates electrons and so reduces another; it is itself oxidised.First seen in the chapter →
reductionअपचयनapachayan: Gain of electrons by a species, seen as a fall in oxidation number (classically, loss of oxygen or gain of hydrogen).

salt bridgeलवण सेतुlavan setu: A tube of electrolyte gel joining two half-cells; ions move through it, keeping both solutions neutral without mixing them.First seen in the chapter →
standard electrode potentialमानक इलेक्ट्रोड विभवmaanak electrode vibhav: The reduction potential of a half-cell at unit concentrations, 1 bar and 298 K, measured against the hydrogen electrode (0.00 V).First seen in the chapter →
Stock notation: Writing a metal's oxidation number as a Roman numeral in brackets, as in iron(III) chloride.First seen in the chapter →

Redox Reactions: Class 11 Chemistry · Chapter 7

What you will learn in Redox Reactions

7.1 Classical idea of redox reactions: oxidation and reduction

7.2 Redox reactions in terms of electron transfer reactions

7.2.1 Competitive electron transfer reactions

7.3 Oxidation number

7.3.1 Types of redox reactions

Combination reactions

Decomposition reactions

Displacement reactions

Disproportionation reactions

7.3.2 Balancing of redox reactions

(a) Oxidation number method

(b) Half-reaction method

7.3.3 Redox reactions as the basis for titrations

7.3.4 Limitations of concept of oxidation number

7.4 Redox reactions and electrode processes

Redox Reactions: key points and summary

Practice questions with answers (NCERT “Exercises”)

Find your NCERT exercise question

Quick check: MCQs on Redox Reactions

Key terms (English–Hindi)

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