Class 11 · Chemistry · Chapter 07

Redox Reactions

Every rusting nail, every battery, every breath you take runs on the same currency: electrons changing hands. Learn to track that transfer and you can predict almost any reaction's direction.

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Chapter Snapshot

The chapter, at a glance

3
Concepts of redox — classical, electronic & oxidation number
2
Balancing methods: oxidation-number & ion-electron (half-reaction)
6–7
Marks typically weighted in board examinations
1
Gateway chapter into Electrochemistry (Class 12)
Why This Chapter Matters

The bridge chapter examiners love

  • Direct foundation for Electrochemistry, where redox reactions literally generate electricity.
  • Balancing equations is a guaranteed 3–5 mark question nearly every year — a skill, not a memorised fact.
  • Oxidation number rules reappear in Coordination Compounds and p-block chemistry in Class 12.
  • Conceptually testable as assertion-reason and match-the-column items, since the logic is rule-based.
Half-reaction check
OxidationZn → Zn²⁺ + 2e⁻
ReductionCu²⁺ + 2e⁻ → Cu
OverallZn + Cu²⁺ → Zn²⁺ + Cu
Key Concept Highlights

Six ideas driving every electron transfer

Classical

Oxidation & Reduction

Gain/loss of oxygen or hydrogen — the original, textbook definition.

Electronic

Electron Transfer View

Oxidation is loss of electrons (OIL); reduction is gain of electrons (RIG).

Numeric

Oxidation Number Rules

Assigning charges to atoms to track change even without visible electron transfer.

Agents

Oxidising & Reducing Agents

The species that gets reduced oxidises others, and vice versa — never confuse the two.

Method

Balancing by Half-Reactions

Splitting a redox equation into oxidation and reduction halves, then combining electrons.

Application

Redox Titrations

Using known concentrations and stoichiometry to determine an unknown's strength.

Important Formulae & Reactions

Reference cell — the working equations

O.N. sum = 0 (neutral)

Sum of oxidation numbers in a neutral compound is zero

O.N. sum = charge (ion)

For a polyatomic ion, oxidation numbers sum to its net charge

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Classic acidic-medium reduction half-reaction (KMnO₄ titrations)

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Dichromate reduction half-reaction in acidic medium

n-factor = electrons gained/lost

Used to relate equivalents in redox titration calculations

E°cell = E°cathode − E°anode

Standard cell potential from standard reduction potentials

What You Will Learn

Your circuit through the chapter

1 · Classical vs electronic concept of redox

Moving from oxygen/hydrogen transfer to electron transfer as the true definition.

2 · Oxidation number rules

Assigning and calculating oxidation states across compounds and ions confidently.

3 · Identifying oxidising & reducing agents

Recognising which species drives a reaction forward by being reduced or oxidised.

4 · Balancing redox equations

Both the oxidation-number method and the ion-electron (half-reaction) method.

5 · Redox reactions in acidic & basic media

Adjusting H⁺/OH⁻ and H₂O to balance charge and atoms under each condition.

6 · Applications & titrations

Connecting redox chemistry to real titration calculations and everyday reactions like corrosion.

Chapter Resources

Jump straight to what you need

📖Full NotesComplete theory 🎯MCQs PracticeTopic-wise quiz 🗂️PYQ BankPast-year questions ⚡True / FalseQuick-fire check ✍️ExercisesNCERT back-questions
Exam Strategy & Preparation Tips

How to actually score well here

01

Practise the ion-electron method until you can balance a redox equation without hesitation — it's the most-asked question type.

02

Build a quick-reference table of oxidation number rules — exceptions like peroxides and OF₂ are common trap questions.

03

Always label oxidising and reducing agents on the equation itself — examiners give marks for shown identification.

04

Practise both acidic and basic medium balancing — most students only prepare one and lose marks on the other.

05

Revisit KMnO₄ and K₂Cr₂O₇ half-reactions — they recur across titration numericals every year.

06

Use assertion-reason PYQs to sharpen conceptual clarity between oxidation state and actual charge.

Chapter 7 · CBSE · Class XI
🔄

Oxidation and Reduction

Redox Reactions NCERT Class 11 Chemistry Class 11 Chemistry Chapter 7 Oxidation Reduction Redox Process Oxidation Number Oxidation State Electronic Concept of Redox Electron Transfer Half Reactions Oxidising Agent Reducing Agent Oxidant Reductant Stock Notation Oxidation Number Method Half Reaction Method Balancing Redox Reactions Balancing Chemical Equations Acidic Medium Basic Medium Combination Reactions Decomposition Reactions Displacement Reactions Metal Displacement Non Metal Displacement Disproportionation Reactions Redox Titrations Redox Indicators Self Indicator Potassium Permanganate Potassium Dichromate Limitations of Oxidation Number Electron Density Concept Redox Couple Electrode Process Electrode Potential Standard Electrode Potential Standard Hydrogen Electrode Daniell Cell Salt Bridge Galvanic Cell Electrochemical Cell Cell Potential Anode Cathode Electrochemistry Basics CBSE Chemistry JEE Chemistry NEET Chemistry
📋
Table of Contents
14 sections
💡 Concept
🏛️ Historical Note
Historical Development of Oxidation
Initially, chemists observed that many substances combined readily with oxygen during burning or heating. Therefore, oxidation was originally defined as the addition of oxygen to a substance.

Examples \[\mathrm{2Mg(s)+O_2(g)\rightarrow2MgO(s)}\] \[\mathrm{S(s)+O_2(g)\rightarrow SO_2(g)}\] Since both magnesium and sulphur combine with oxygen, these reactions were considered oxidation reactions.
📘 Definition
Modern Definition of Oxidation
📘 Definition
Reduction
🤔 Did You Know?
Why Oxidation and Reduction Always Occur Together
Electrons cannot simply disappear from one substance. Whenever one species loses electrons, another species must accept them. Therefore oxidation and reduction always occur simultaneously.

This simultaneous occurrence is called a Redox Reaction.

Think of electron transfer like passing a ball between two players. One player can lose the ball only if another player catches it.
ℹ️ Meaning of Redox Reaction
The word Redox is formed by combining:
  • Red → Reduction
  • Ox → Oxidation

Thus every redox reaction contains both oxidation and reduction occurring simultaneously.
🔍 Electronic Interpretation
According to the electronic concept:
  • Oxidation = Loss of electrons
  • Reduction = Gain of electrons
Example
\[\mathrm{Zn\rightarrow Zn^{2+}+2e^{-}}\] Zinc loses two electrons. Hence zinc is oxidised.
\[\mathrm{Cu^{2+}+2e^{-}\rightarrow Cu}\] Copper ions gain two electrons. Hence copper ions are reduced.

Overall reaction: \[\mathrm{Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu}\]
📊 Comparison between Oxidation and Reduction
Oxidation Reaction Reduction Reaction
Oxidation Reduction
Addition of oxygen Removal of oxygen
Addition of electronegative element Removal of electronegative element
Removal of hydrogen Addition of hydrogen
Removal of electropositive element Addition of electropositive element
Loss of electrons Gain of electrons
Increase in oxidation number Decrease in oxidation number
🛠️ Applications of Oxidation and Reduction
  • Respiration is a biological oxidation process.
  • Photosynthesis involves reduction of carbon dioxide.
  • Rusting of iron is oxidation.
  • Combustion of fuels is oxidation.
  • Extraction of metals from ores involves reduction.
  • Rechargeable batteries operate through reversible redox reactions.
  • Bleaching agents work through oxidation.
  • Water purification often uses oxidation by chlorine or ozone.
🔤 Quick Memory Trick
✏️ Example
Solved Example
Identify oxidation and reduction in the following reaction. \[\mathrm{Zn+CuSO_4\rightarrow ZnSO_4+Cu}\]
Check electron transfer.
  1. 1
    Write ionic equation.
  2. 2
    Identify electron loss.
  3. 3
    Identify electron gain.
\[\mathrm{Zn\rightarrow Zn^{2+}+2e^{-}}\quad\color{gold}\text{→ Oxidation}\] \[\mathrm{Cu^{2+}+2e^{-}\rightarrow Cu}\quad\color{gold}\text{→ Reduction}\] Hence zinc acts as the reducing agent whereas copper(II) ions act as the oxidising agent.
Why is the following reaction considered oxidation? \[\mathrm{4Fe+3O_2\rightarrow2Fe_2O_3}\]
  • Iron combines with oxygen.
  • Iron loses electrons to oxygen.
  • The oxidation number of iron increases from 0 to +3.
Therefore the reaction satisfies all three definitions of oxidation.
Why is hydrogenation considered reduction? \[\mathrm{CH_2=CH_2+H_2\rightarrow CH_3CH_3}\]
Ethene gains hydrogen atoms. According to the classical definition, addition of hydrogen represents reduction.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming oxygen must always be present in oxidation.
  • Ignoring hydrogen addition or removal.
  • Forgetting that oxidation and reduction always occur together.
  • Confusing oxidising agent with reducing agent.
  • Not writing balanced half-reactions while explaining electron transfer.
📋 CBSE Case Study (HOTS)

A silver ornament gradually becomes black after being exposed to air for several months. The black coating mainly consists of silver sulphide.

Questions

  1. Is silver oxidised or reduced?
  2. Name the process responsible for the black coating.
  3. Which definition of oxidation is applicable here?
  4. Suggest one method to prevent this process.

Answer

  1. Silver undergoes oxidation.
  2. Tarnishing (a type of corrosion).
  3. Addition of an electronegative element (sulphur).
  4. Store in airtight containers or apply protective coating.
🌟 Why This Topic is Important for Board Examinations
  • Forms the basis of oxidation number method taught later in the chapter.
  • Frequently asked in one-mark and three-mark conceptual questions.
  • Essential for electrochemistry in Class XII.
  • Required to understand corrosion, metallurgy, batteries and chemical energetics.
  • Directly linked with identifying oxidising and reducing agents.
🔄

Redox Reactions in Terms of Electron Transfer

📋
Table of Contents
16 sections
📘 Definition
💡 Electronic Concept of Redox Reactions
✏️ Examples of Electron Transfer Reactions
Example 1
\[\mathrm{2Na(s)+Cl_2(g)\rightarrow2NaCl(s)}\]
Example 2
\[\mathrm{4Na(s)+O_2(g)\rightarrow2Na_2O(s)}\]
Example 3
\[\mathrm{2Na(s)+S(s)\rightarrow Na_2S(s)}\]
In all three reactions:
  • Sodium loses electrons.
  • Sodium is oxidised.
  • Chlorine, oxygen and sulphur gain electrons.
  • These elements are reduced.
Thus, these are all examples of redox reactions because oxidation and reduction occur together.
🤔 Did You Know?
How Electron Transfer Occurs
Consider the formation of sodium chloride.

Each sodium atom has one valence electron. It can attain a stable noble gas configuration by losing this electron.

Each chlorine atom requires one electron to complete its octet.

Therefore, sodium transfers its electron to chlorine.

The electron transfer can be represented as:

\[ \mathrm{Na\rightarrow Na^{+}+e^{-}} \] \[ \mathrm{Cl+e^{-}\rightarrow Cl^{-}} \] Since chlorine exists as a diatomic molecule, the balanced electron-transfer equations become: \[ \mathrm{2Na(s)\rightarrow2Na^{+}+2e^{-}} \] \[ \mathrm{Cl_2(g)+2e^{-}\rightarrow2Cl^{-}} \]
📘 Half Reactions
🗒️ Principle Of Electron Balance
One of the most important rules in redox chemistry is:

Total electrons lost = Total electrons gained
This is a direct consequence of the Law of Conservation of Charge.
Electrons cannot be created or destroyed during a chemical reaction.
📘 Important Definitions Based on Electronic Theory
🔎 Identifying Oxidising and Reducing Agents
🗺️ Roadmap for Solving Electron Transfer Questions

  1. Identify the reacting species.

  2. Determine which species loses electrons.

  3. Determine which species gains electrons.

  4. Write oxidation and reduction half reactions separately.

  5. Balance the electrons.

  6. Add both half reactions.

  7. Cancel common electrons.

  8. Write the final balanced reaction

⚖️ Comparison of Classical and Electronic Concepts
Classical Concept Electronic Concept
Addition of oxygen Loss of electrons
Removal of hydrogen Loss of electrons
Removal of oxygen Gain of electrons
Addition of hydrogen Gain of electrons
Applicable only to some reactions Applicable to almost all redox reactions
⚠️ Limitation of the Electronic Concept
The electronic concept explains ionic reactions very effectively. However, it cannot directly explain many covalent reactions where electrons are shared rather than completely transferred.

To explain such reactions, the concept of Oxidation Number is used, which is discussed later in this chapter.
✏️ Example
Solved Example
Identify oxidation, reduction, oxidising agent and reducing agent in the following reaction: \[\mathrm{2Na+Cl_2\rightarrow2NaCl}\]
Electron transfer determines oxidation and reduction.
Oxidation: \[ \mathrm{2Na\rightarrow2Na^{+}+2e^{-}} \] Reduction: \[ \mathrm{Cl_2+2e^{-}\rightarrow2Cl^{-}} \]
  • Sodium is oxidised.
  • Chlorine is reduced.
  • Sodium is the reducing agent.
  • Chlorine is the oxidising agent.
Identify the oxidising agent and reducing agent. \[\mathrm{Mg+Cu^{2+}\rightarrow Mg^{2+}+Cu}\]
\[ \mathrm{Mg\rightarrow Mg^{2+}+2e^{-}} \] \[ \mathrm{Cu^{2+}+2e^{-}\rightarrow Cu} \] Therefore,
  • Magnesium loses electrons and acts as the reducing agent.
  • Copper ions gain electrons and act as the oxidising agent.
⚡ Exam Tip
❌ Common Mistakes
  • Confusing oxidising agent with reducing agent.
  • Forgetting to balance the number of electrons.
  • Writing electrons in the final overall equation.
  • Assuming oxygen must always be present in redox reactions.
  • Ignoring charge balance while writing half reactions.
📋 Case Study
CBSE Case Study (HOTS)

During the extraction of copper, iron is added to copper sulphate solution. A reddish-brown deposit appears and the blue colour gradually fades.

Questions

  1. Write the balanced reaction.
  2. Which species is oxidised?
  3. Which species is reduced?
  4. Identify the oxidising agent.
  5. Identify the reducing agent.

Answer

\[ Fe+CuSO_4\rightarrow FeSO_4+Cu \]
  • Iron is oxidised.
  • Copper ions are reduced.
  • \(Cu^{2+}\) is the oxidising agent.
  • Iron is the reducing agent.
🌟 Significance
Importance for Board Examinations
  • This topic forms the conceptual basis of oxidation number and balancing redox equations.
  • Half reactions are extensively used in electrochemistry and galvanic cells.
  • CBSE frequently asks students to identify oxidising agents, reducing agents and electron transfer.
  • Understanding electron transfer is essential for Class XII Electrochemistry.
  • This concept is frequently tested in JEE Main, NEET and CUET examinations.
🔄

Oxidation Number (Oxidation State)

📋
Table of Contents
15 sections
🗺️ Overview
The concept of Oxidation Number (O.N.), also called the Oxidation State, is one of the most important concepts in inorganic chemistry. It provides a convenient method to identify oxidation and reduction even in reactions where electrons are not transferred completely, such as covalent compounds.

Oxidation number is extensively used in balancing redox equations, predicting oxidising and reducing agents, studying electrochemistry, coordination chemistry, metallurgy and many reactions of organic chemistry.

📘 Definition
🤔 Did You Know?
Why is Oxidation Number Important?
  • Identifies oxidation and reduction in complex reactions.
  • Helps balance redox equations using oxidation number method.
  • Determines oxidising and reducing agents.
  • Predicts chemical behaviour of elements.
  • Explains variable valency of transition elements.
  • Forms the basis of electrochemistry and coordination chemistry.
  • Frequently asked in CBSE Board, NEET, JEE Main and CUET examinations.
⚖️ Difference Between Oxidation Number and Valency
Oxidation Number Valency
May be positive, negative or zero. Always positive.
May be fractional in some compounds. Never fractional.
Depends upon the compound. Represents combining capacity.
Can vary in different compounds. Usually fixed for many main-group elements.
Used in redox reactions. Used in writing chemical formulae.
📜 Rules for Calculating Oxidation Number
📜 Rules
The following rules should always be applied in the given order while determining oxidation numbers.
Rule 1. Oxidation Number of Free Elements is Zero
Any element in its free, elemental or uncombined state has an oxidation number of zero.

Examples \[H_2,\ O_2,\ O_3,\ Cl_2,\ N_2,\ P_4,\ S_8,\ Na,\ Mg,\ Fe,\ Cu\] All these have \[\boxed{Oxidation\ Number=0}\] This rule applies irrespective of the number of atoms present in the molecule.
Rule 2. Oxidation Number of Monoatomic Ions
For ions consisting of only one atom, oxidation number equals the ionic charge.
Ion Oxidation Number
\(Na^+\) \(+1\)
\(Mg^{2+}\) \(+2\)
\(Al^{3+}\) \(+3\)
\(Cl^-\) \(-1\)
\(O^{2-}\) \(-2\)
\(S^{2-}\) \(-2\)
Rule 3. Oxidation Number of Oxygen
Oxygen usually has an oxidation number of \[\boxed{-2}\] However, there are important exceptions.
Compound Oxidation Number of Oxygen
Most compounds \(-2\)
Peroxides (\(H_2O_2,\ Na_2O_2\)) \(-1\)
Superoxides (\(KO_2,\ RbO_2\)) \(-\frac12\)
\(OF_2\) \(+2\)
\(O_2F_2\) \(+1\)
The positive oxidation number of oxygen in fluorides occurs because fluorine is the most electronegative element.
Rule 4. Oxidation Number of Hydrogen
Hydrogen generally possesses an oxidation number of \[\boxed{+1}\] Exception:
When hydrogen is bonded to highly electropositive metals in binary compounds (metal hydrides), its oxidation number becomes \[\boxed{-1}\] Examples \[LiH,\ NaH,\ KH,\ CaH_2\] When hydrogen is bonded to highly electropositive metals in binary compounds (metal hydrides), its oxidation number becomes
Rule 5. Oxidation Number of Halogens
Fluorine always has an oxidation number of \[-1\] Other halogens usually possess an oxidation number of \[-1\] However, chlorine, bromine and iodine may exhibit positive oxidation numbers when combined with oxygen or fluorine. Examples
Compound Oxidation Number
HCl Cl = -1
NaBr Br = -1
HClO Cl = +1
HClO3 Cl = +5
HClO4 Cl = +7
Rule 6. Sum of Oxidation Numbers
Neutral Compound
 The algebraic sum of oxidation numbers of all atoms in a neutral compound is always zero. \[\mathrm{H_2SO_4}\] \[2(+1)+S+4(-2)=0\] \[S=+6\] Polyatomic Ion
 The algebraic sum of oxidation numbers equals the charge on the ion.

Carbonate Ion
 \[\mathrm{CO_3^{2-}}\] \[\mathrm{C+3(-2)=-2}\] \[C=+4\]
Sulphate Ion
 \[\mathrm{SO_4^{2-}}\] \[S+4(-2)=-2\] \[S=+6\]
Nitrate Ion
 \[\mathrm{NO_3^-}\] \[N+3(-2)=-1\] \[N=+5\]
ℹ️ General Trends of Oxidation Number
  • Most metallic elements exhibit positive oxidation numbers.
  • Most non-metals generally exhibit negative oxidation numbers when combined with metals.
  • The same element may have different oxidation numbers in different compounds.
  • Transition elements show variable oxidation states due to participation of both ns and (n−1)d electrons.
🔎 Variable Oxidation States of Transition Elements
📎 Highest Oxidation Number in the Periodic Table
For representative (main-group) elements:
  • Groups 1 and 2 generally exhibit a maximum oxidation number equal to their group number.
  • For Groups 13 to 17, the highest oxidation number is approximately equal to Group Number − 10.
Group 1 2 13 14 15 16 17
Maximum Oxidation Number +1 +2 +3 +4 +5 +6 +7
Across the third period, the highest oxidation number increases progressively: \[\mathrm{Na(+1)\rightarrow Mg(+2)\rightarrow Al(+3)\rightarrow Si(+4)\rightarrow P(+5)\rightarrow S(+6)\rightarrow Cl(+7)}\]
🔄 Stepwise Method to Calculate Oxidation Number
  • 1
    Write the chemical formula.
  • 2
    Assign known oxidation numbers.
  • 3
    Let the unknown oxidation number be \(x\).
  • 4
    Use the algebraic sum rule.
  • 5
    Solve for \(x\).
✏️ Example
Solved Example
Calculate the oxidation number of sulphur in sulphuric acid.
\[2(+1)+x+4(-2)=0\] \[2+x-8=0\] \[x=+6\] Therefore, \[\boxed{S=+6}\]
Find the oxidation number of nitrogen in nitric acid.
\[(+1)+x+3(-2)=0\] \[1+x-6=0\] \[x=+5\]
Calculate the oxidation number of chromium in \(Cr_2O_7^{2-}\).
\[2x+7(-2)=-2\] \[2x-14=-2\] \[2x=12\] \[x=+6\]
🧠 Oxidation Number Memory Chart
⚡ Exam Tip
❌ Common Mistakes
  • Confusing oxidation number with valency.
  • Assigning oxygen as -2 in peroxides.
  • Assigning hydrogen as +1 in metal hydrides.
  • Forgetting that fluorine is always -1.
  • Ignoring the charge while calculating oxidation number of polyatomic ions.
  • Assuming transition metals have only one oxidation state.
📋 CBSE Case Study (HOTS)

Potassium permanganate is widely used as a disinfectant and oxidising agent in water treatment.

Questions

  1. Find the oxidation number of manganese in \(KMnO_4\).
  2. Why is potassium permanganate a strong oxidising agent?
  3. Name another compound in which manganese exhibits a lower oxidation state.

Solution

\[ (+1)+x+4(-2)=0 \] \[ 1+x-8=0 \] \[ x=+7 \]

The very high oxidation state (+7) makes manganese capable of accepting electrons easily, making \(KMnO_4\) a powerful oxidising agent.

🌟 Significance
Importance for Board Examinations
  • This is one of the highest-weightage concepts in the Redox Reactions chapter.
  • Direct questions are asked on oxidation number calculations.
  • Forms the basis of oxidation number method for balancing redox equations.
  • Essential for Electrochemistry, Coordination Compounds and d-Block Elements in Class XII.
  • Frequently tested in CBSE Board, NEET, JEE Main and CUET.
🔄

Stock Notation and Redox Reactions Based on Oxidation Number

📋
Table of Contents
14 sections
🗺️ Overview
After understanding the concept of oxidation number, chemists needed a systematic method to represent the oxidation state of elements, especially metals that exhibit variable oxidation states. For this purpose, the Stock Notation was introduced by the German chemist Alfred Stock. Today, Stock notation is the internationally accepted system recommended by IUPAC for naming compounds containing metals with more than one oxidation state.
📘 Definition
🌟 Why is Stock Notation Required?
Many transition metals form more than one stable ion. For example, iron forms both \(Fe^{2+}\) and \(Fe^{3+}\), while copper forms \(Cu^+\) and \(Cu^{2+}\).

Without mentioning the oxidation state, the name of the compound becomes incomplete because different oxidation states produce compounds with different chemical properties.

Stock notation clearly specifies the oxidation state of the metal, thereby eliminating confusion.
📜 Rules for Writing Stock Notation
📜 Rules
  1. Determine the oxidation number of the metal.
  2. Write the name or symbol of the metal.
  3. Write the oxidation number in Roman numerals inside parentheses immediately after the metal.
  4. Do not include the sign (+ or −) inside the parentheses.
Examples
  • \(Fe^{2+}\) → Iron(II)
  • \(Fe^{3+}\) → Iron(III)
  • \(Cu^{+}\) → Copper(I)
  • \(Cu^{2+}\) → Copper(II)
✏️ Examples of Stock Notation
Traditional Name Formula Stock Notation
Ferrous chloride \(FeCl_2\) Iron(II) chloride
Ferric chloride \(FeCl_3\) Iron(III) chloride
Cuprous oxide \(Cu_2O\) Copper(I) oxide
Cupric oxide \(CuO\) Copper(II) oxide
Aurous chloride \(AuCl\) Gold(I) chloride
Auric chloride \(AuCl_3\) Gold(III) chloride
Mercurous chloride \(Hg_2Cl_2\) Mercury(I) chloride
Mercuric chloride \(HgCl_2\) Mercury(II) chloride
Stannous chloride \(SnCl_2\) Tin(II) chloride
Stannic chloride \(SnCl_4\) Tin(IV) chloride
⚖️ Traditional Nomenclature versus Stock Notation
Traditional System Stock System
Ferrous chloride Iron(II) chloride
Ferric chloride Iron(III) chloride
Cuprous oxide Copper(I) oxide
Cupric oxide Copper(II) oxide
Aurous chloride Gold(I) chloride
Auric chloride Gold(III) chloride

Modern IUPAC nomenclature prefers the Stock system because it is more systematic and universally accepted.
✅ Advantages of Stock Notation
  • Provides the exact oxidation state of the metal.
  • Removes ambiguity from compound names.
  • Applicable to all transition elements.
  • Accepted by IUPAC.
  • Essential for writing correct chemical names in board examinations.
📌 Redox Reactions Based on Oxidation Number
📝 Summary of Redox Definitions
✏️ Example
Solved Example
Identify oxidation, reduction, oxidising agent and reducing agent in the following reaction. \[\mathrm{Fe+CuSO_4\rightarrow FeSO_4+Cu}\]
Species Oxidation Number Change Conclusion
Fe \(0\rightarrow+2\) Oxidation
\(Cu^{2+}\) \(+2\rightarrow0\) Reduction
Therefore,
  • Iron is the reducing agent.
  • Copper(II) ion is the oxidising agent.
Write the Stock notation for the following compounds.
  • \(\mathrm{FeCl_3\)}
  • \(\mathrm{Cu_2O\)}
  • \(\mathrm{AuCl_3\)}
Compound Stock Name
\(FeCl_3\) Iron(III) chloride
\(Cu_2O\) Copper(I) oxide
\(AuCl_3\) Gold(III) chloride
⚡ Exam Tip
❌ Common Mistakes
  • Writing oxidation number using Arabic numerals instead of Roman numerals.
  • Writing signs such as +2 or +3 inside parentheses.
  • Confusing oxidising agent with oxidation.
  • Confusing reducing agent with reduction.
  • Assuming oxidation number and valency are always identical.
📋 CBSE Case Study (HOTS)

Copper utensils gradually develop a green coating when exposed to moist air. During cleaning, dilute acid removes the coating.

Questions

  1. Which oxidation state of copper is commonly present in the green coating?
  2. Write the Stock name of \(CuO\).
  3. Is copper oxidised or reduced during formation of the coating?
  4. Identify the reducing agent if copper oxide is converted back into copper by hydrogen gas.

Answer

  • Copper generally exists in the +2 oxidation state.
  • Copper(II) oxide.
  • Copper undergoes oxidation.
  • Hydrogen acts as the reducing agent.
🌟 Importance for Board Examinations
  • Stock notation is a favourite topic in CBSE competency-based questions.
  • Students are often asked to identify oxidising and reducing agents using oxidation numbers.
  • The concept directly supports balancing redox equations by the oxidation number method.
  • Essential for understanding d-block chemistry, coordination compounds and electrochemistry in higher classes.
  • Frequently tested in CBSE, JEE Main, NEET and CUET examinations.
🔄

Types of Redox Reactions

📋
Table of Contents
13 sections
🗺️ Overview
A Redox Reaction is a chemical reaction in which the oxidation numbers of one or more elements change due to the transfer of electrons. One species undergoes oxidation while another simultaneously undergoes reduction.

Redox reactions are among the most important reactions in chemistry because they govern combustion, corrosion, respiration, photosynthesis, metallurgy, electrochemistry, batteries, bleaching, industrial synthesis and numerous biological processes.

The easiest way to identify a redox reaction is to compare the oxidation numbers of the elements before and after the reaction. If the oxidation number of at least one element increases and that of another decreases, the reaction is a redox reaction.
Classification of Redox Reactions
Redox reactions are broadly classified into the following categories.
  1. Combination (Synthesis) Reactions
  2. Decomposition Reactions
  3. Displacement Reactions
    • Metal Displacement Reactions
    • Non-Metal Displacement Reactions
  4. Disproportionation Reactions
📘 Combination (Synthesis) Redox Reactions
🛠️ Application
Applications of Combination Redox Reactions
  • Burning of fuels.
  • Formation of metal oxides.
  • Manufacture of ammonia.
  • Photosynthesis.
  • Industrial oxidation processes.
📘 Definition
Decomposition Redox Reactions
📎 Important Note
Every decomposition reaction is not a redox reaction.
Example
\[\mathrm{CaCO_3\rightarrow CaO+CO_2}] Oxidation numbers remain unchanged. Therefore, this is not a redox reaction.
📘 Displacement Redox Reactions
📘 Definition
In a displacement reaction, one element replaces another element from its compound.
General equation
\[\mathrm{X+YZ\rightarrow XZ+Y}\] Displacement reactions occur because one element is more reactive than the other.
Metal Displacement Reactions
A more reactive metal displaces a less reactive metal from its salt solution or oxide.
These reactions are widely used in metallurgy for extraction of metals from ores.
Example 1
\[\mathrm{CuSO_4(aq)+Zn(s)\rightarrow Cu(s)+ZnSO_4(aq)}\]
  • Zn: \(0\rightarrow+2\)
  • Cu: \(+2\rightarrow0\)
Zinc acts as the reducing agent.
Example 2
\[\mathrm{V_2O_5(s)+5Ca(s)\rightarrow2V(s)+5CaO(s)}\]
  • Ca is oxidised.
  • Vanadium is reduced.
Importance in Metallurgy
  • Extraction of chromium.
  • Extraction of vanadium.
  • Thermite process.
  • Recovery of metals from ores.
Non-Metal Displacement Reactions
A more reactive non-metal displaces a less reactive non-metal from its compound.
Hydrogen displacement and halogen displacement are the most common examples.
Hydrogen Displacement
Highly reactive metals such as alkali metals and alkaline earth metals react with water to liberate hydrogen gas.
Example 1
\[\mathrm{2Na(s)+2H_2O(l)\rightarrow2NaOH(aq)+H_2(g)}\]
Example 2
\[\mathrm{Ca(s)+2H_2O(l)\rightarrow Ca(OH)_2(aq)+H_2(g)}\] Less reactive metals react only with steam.
Example 3
\[\mathrm{Mg(s)+2H_2O(g)\rightarrow Mg(OH)_2(s)+H_2(g)}\]
Example 4
\[\mathrm{3Fe(s)+4H_2O(g)\rightarrow Fe_3O_4(s)+4H_2(g)}\] (Note: The balanced reaction with steam is shown above.)
Hydrogen Displacement from Acids
Many metals that do not react with cold water readily react with dilute acids.
Example
\[\mathrm{Zn+2HCl\rightarrow ZnCl_2+H_2}\] Here,
  • Zn is oxidised.
  • \(H^+\) ions are reduced.
Halogen Displacement
The oxidising power of halogens decreases down Group 17.
Order of oxidising power \[F_2>Cl_2>Br_2>I_2\]
Example
\[\mathrm{Cl_2+2KBr\rightarrow2KCl+Br_2}\] Chlorine displaces bromine because chlorine is the stronger oxidising agent.
📘 Disproportionation Reactions
⚖️ Comparison of Different Types of Redox Reactions
Type Main Feature Example
Combination Two reactants form one product. \(2Mg+O_2\rightarrow2MgO\)
Decomposition One compound breaks into simpler substances. \(2KClO_3\rightarrow2KCl+3O_2\)
Metal Displacement More reactive metal replaces less reactive metal. \(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)
Non-Metal Displacement More reactive non-metal replaces another non-metal. \(Cl_2+2KBr\rightarrow2KCl+Br_2\)
Disproportionation Same element is oxidised and reduced simultaneously. \(2H_2O_2\rightarrow2H_2O+O_2\)
✏️ Example
Solved Example
Classify the following reactions.
  1. \(\mathrm{Zn+CuSO_4\rightarrow ZnSO_4+Cu}\)
  2. \(\mathrm{2KClO_3\rightarrow2KCl+3O_2}\)
  3. \(\mathrm{2H_2O_2\rightarrow2H_2O+O_2}\)
Reaction Type
\(Zn+CuSO_4\) Metal displacement reaction
\(2KClO_3\) Decomposition redox reaction
\(2H_2O_2\) Disproportionation reaction
⚡ Exam Tip
❌ Common Mistakes
  • Assuming every combination reaction is a redox reaction.
  • Ignoring oxidation number changes before classification.
  • Confusing decomposition with disproportionation.
  • Using an incorrect reactivity order of metals or halogens.
  • Not identifying which species is oxidised and which is reduced.
📋 CBSE Case Study (HOTS)

A chemistry student performs the following experiments:

  1. Burning magnesium ribbon in air.
  2. Adding zinc to copper sulphate solution.
  3. Heating potassium chlorate.
  4. Allowing hydrogen peroxide to decompose.

Questions

  1. Identify the type of redox reaction in each case.
  2. Which reaction is a disproportionation reaction?
  3. Which reaction is widely used in metallurgy?
  4. Which reaction represents combustion?

Answers

  • Magnesium burning → Combination reaction.
  • Zinc + Copper sulphate → Metal displacement reaction.
  • Potassium chlorate decomposition → Decomposition redox reaction.
  • Hydrogen peroxide decomposition → Disproportionation reaction.
🌟 Significance
Importance for Board Examinations
  • Classification of redox reactions is frequently asked in competency-based and assertion-reason questions.
  • Students are expected to identify oxidation number changes along with the reaction type.
  • Displacement reactions are directly linked with the reactivity series and metallurgy.
  • Disproportionation reactions are important for higher studies in electrochemistry and inorganic chemistry.
  • This topic is regularly tested in CBSE, JEE Main, NEET and CUET examinations.
🔄

Balancing Redox Reactions

📋
Table of Contents
15 sections
🗺️ Overview
Balancing a redox reaction means ensuring that the number of atoms of each element and the total charge are equal on both sides of the chemical equation. Since redox reactions involve transfer of electrons, balancing requires both mass conservation and charge conservation.

Balanced redox equations are essential in electrochemistry, analytical chemistry, metallurgy, environmental chemistry and quantitative chemical calculations.

CBSE, JEE Main, NEET and CUET examinations frequently test students on balancing redox equations because it combines concepts of oxidation number, electron transfer and ionic reactions.
📜 Methods of Balancing Redox Reactions
📜 Rules
There are two standard methods for balancing redox reactions.
  1. Oxidation Number Method
  2. Half-Reaction (Ion-Electron) Method
Both methods produce the same balanced equation. The choice depends on the nature of the reaction and personal preference.
⚖️ Comparison of the Two Methods
Oxidation Number Method Half-Reaction Method
Uses oxidation number changes. Uses electron transfer explicitly.
Suitable for simple reactions. Best for ionic reactions.
Faster for board examinations. Widely used in electrochemistry.
Does not require writing half reactions. Requires oxidation and reduction half reactions.
📌 Method 1. Oxidation Number Method
✏️ Example
Solved Example
Balance \(\mathrm{Fe^{2+}+Cr_2O_7^{2-}\rightarrow Fe^{3+}+Cr^{3+}}\]) in acidic medium.
Step 1. Assign Oxidation Numbers
  • \(\mathrm{Fe:+2\rightarrow+3\})
  • \(\mathrm{Cr:+6\rightarrow+3}\)
Step 2. Calculate Electron Change
Each iron atom loses \[1e^-\] Each chromium atom gains \[3e^-\] Since there are two chromium atoms, \[2\times3=6e^-\] Hence six iron ions are required. \[\mathrm{6Fe^{2+}\rightarrow6Fe^{3+}}\]
Step 3. Balance Oxygen
Add seven water molecules. \[\mathrm{Cr_2O_7^{2-}\rightarrow2Cr^{3+}+7H_2O}\]
Step 4. Balance Hydrogen
\[\mathrm{Cr_2O_7^{2-}+14H^+\rightarrow2Cr^{3+}+7H_2O}\]
Final Balanced Equation
\[\mathrm{6Fe^{2+}+Cr_2O_7^{2-}+14H^+\rightarrow 6Fe^{3+}+2Cr^{3+}+7H_2O}\]
📌 Method 2. Half-Reaction (Ion-Electron) Method
✏️ Example
Solved Example
Balance \(\mathrm{Fe^{2+}+Cr_2O_7^{2-}\rightarrow Fe^{3+}+Cr^{3+}}\]) by Half Reaction Method
Step 1. Write the Ionic Equation
\[\mathrm{Fe^{2+}+Cr_2O_7^{2-}\rightarrow Fe^{3+}+Cr^{3+}}\]
Step 2. Separate into Half Reactions
Oxidation Half \[\mathrm{Fe^{2+}\rightarrow Fe^{3+}}\] Reduction Half \[\mathrm{Cr_2O_7^{2-}\rightarrow Cr^{3+}}\]
Step 3. Balance Atoms Except Oxygen and Hydrogen
\[\mathrm{Cr_2O_7^{2-}\rightarrow2Cr^{3+}}\]
Step 4. Balance Oxygen
\[\mathrm{Cr_2O_7^{2-}\rightarrow 2Cr^{3+}+7H_2O}\]
Step 5. Balance Hydrogen
\[\mathrm{Cr_2O_7^{2-}+14H^+\rightarrow 2Cr^{3+}+7H_2O}\]
Step 6. Balance Charges
Oxidation Half \[\mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-}\] Reduction Half \[\mathrm{Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2O}\]
Step 7. Equalise Electrons
Multiply oxidation half by 6. \[\mathrm{6Fe^{2+}\rightarrow 6Fe^{3+}+6e^-}\]
Step 8. Add Both Half Reactions
\[\mathrm{6Fe^{2+}+Cr_2O_7^{2-}+14H^+\rightarrow 6Fe^{3+}+2Cr^{3+}+7H_2O}\] Electrons cancel automatically because six electrons are produced and six are consumed.
🗒️ Balancing Redox Reactions in Basic Medium
When the reaction occurs in a basic solution, follow these additional steps after balancing the reaction in acidic medium.
  1. Add equal numbers of \(OH^-\) ions to both sides to neutralise every \(H^+\).
  2. Combine \(H^+\) and \(OH^-\) to form water.
  3. Cancel extra water molecules appearing on both sides.
Remember \[\mathrm{H^++OH^-\rightarrow H_2O}\]
📜 Important Rules to Remember
📜 Rules
  • Balance atoms before balancing charges.
  • Never add electrons in the final balanced equation.
  • Electrons appear only in half reactions.
  • Always balance oxygen before hydrogen.
  • Use \(H^+\) only in acidic medium.
  • Use \(OH^-\) only in basic medium.
  • Check both atom balance and charge balance at the end.
⚖️ Acidic vs Basic Medium
Acidic Medium Basic Medium
Use \(H^+\) Use \(OH^-\)
Add water for oxygen balance. Add water first, then neutralise using \(OH^-\).
Common in permanganate and dichromate reactions. Common in alkaline oxidising agents.
🔤 Memory Trick
⚡ Exam Tip
❌ Common Mistakes
  • Changing chemical formulae while balancing.
  • Balancing oxygen before oxidation number changes.
  • Using \(OH^-\) in acidic medium.
  • Leaving electrons in the final equation.
  • Ignoring charge balance.
  • Balancing atoms but forgetting total ionic charge.
  • Not cancelling common water molecules.
📋 CBSE Case Study (HOTS)

A laboratory technician uses acidified potassium dichromate solution to estimate the amount of iron(II) ions present in a water sample. During the reaction, iron(II) ions are oxidised to iron(III) ions while dichromate ions are reduced to chromium(III) ions.

Questions

  1. Which species undergoes oxidation?
  2. Which species undergoes reduction?
  3. Why is acidic medium necessary?
  4. Which balancing method is most suitable for this reaction?

Answers

  • \(Fe^{2+}\) undergoes oxidation.
  • \(Cr_2O_7^{2-}\) undergoes reduction.
  • \(H^+\) ions are required to balance hydrogen and oxygen atoms.
  • The Half-Reaction (Ion-Electron) Method is the most systematic approach.
🌟 Importance for Board Examinations
  • Balancing redox reactions is one of the highest-scoring topics in this chapter.
  • Frequently appears as competency-based, assertion-reason and long-answer questions.
  • Forms the conceptual foundation of electrochemistry in Class XII.
  • Extensively used in analytical chemistry, titrations and industrial oxidation-reduction processes.
  • Regularly tested in CBSE Board, JEE Main, NEET and CUET.
🔄

Redox Reactions as the Basis of Titrations

📋
Table of Contents
15 sections
🗺️ Overview
One of the most important practical applications of redox reactions is in volumetric analysis, where they are used to determine the concentration (strength) of an unknown solution. Such analytical procedures are known as Redox Titrations.

Just as acid-base titrations are based on the neutralisation reaction between an acid and a base, redox titrations are based on oxidation-reduction reactions involving the transfer of electrons.

These titrations are extensively used in chemistry laboratories, pharmaceutical industries, food analysis, environmental monitoring, water quality testing and metallurgy because they provide accurate and reliable quantitative analysis.
⚖️ Comparison Between Acid-Base and Redox Titrations
Acid-Base Titration Redox Titration
Based on neutralisation reaction. Based on oxidation-reduction reaction.
Transfer of protons (\(H^+\)). Transfer of electrons.
Uses acid-base indicators. Uses redox indicators or self-indicators.
End point detected by pH change. End point detected by oxidation state change.
Example: HCl vs NaOH. Example: \(\mathrm{KMnO_4}\) vs \(\mathrm{Fe^{2+}}\).
🗒️ Principle Of Redox Titration
A redox titration is based on the complete transfer of electrons between an oxidising agent and a reducing agent.

During the titration,
  • the oxidising agent gains electrons and gets reduced,
  • the reducing agent loses electrons and gets oxidised.
At the equivalence point, the total number of electrons lost by the reducing agent is exactly equal to the total number of electrons gained by the oxidising agent.

Mathematically, \[\boxed{\text{Electrons Lost}=\text{Electrons Gained}}\]
Important Terminology
Term Meaning
Titrant The solution of known concentration taken in the burette.
Analyte (Titrate) The solution of unknown concentration taken in the conical flask.
End Point The stage at which the indicator changes colour.
Equivalence Point The point where oxidant and reductant react exactly in stoichiometric proportion.
Redox Indicator An indicator that changes colour due to change in oxidation state.
📌 Redox Indicators
ℹ️ Common Redox Indicators
Indicator Oxidised Colour Reduced Colour
Diphenylamine Violet Colourless
N-Phenylanthranilic acid Red-violet Colourless
Ferroin Pale blue Red
Methylene Blue Blue Colourless
ℹ️ Information
Self Indicator
Some oxidising agents possess intense colours and therefore act as their own indicators. Such substances are called Self Indicators.

The best example is potassium permanganate.. Before the end point,
  • The purple colour of \(KMnO_4\) disappears because it is reduced to colourless \(Mn^{2+}\).
After the end point,
  • A single excess drop of \(KMnO_4\) produces a permanent light pink colour.
Hence, no external indicator is required in permanganate titrations.
📌 Common Types of Redox Titrations
✏️ Example
Permanganate Titration
A standard solution of potassium permanganate is used to determine the concentration of iron(II) ions in acidic medium.

Balanced reaction
\[\mathrm{MnO_4^-+5Fe^{2+}+8H^+\rightarrow Mn^{2+}+5Fe^{3+}+4H_2O}\]
Observation
  • Before the end point, the purple colour disappears.
  • At the end point, a permanent pale pink colour appears.
🛠️ Applications of Redox Titrations
  • Determination of iron content in ores.
  • Analysis of drinking water.
  • Estimation of dissolved oxygen.
  • Determination of hydrogen peroxide concentration.
  • Vitamin C estimation in fruits and medicines.
  • Quality control in pharmaceutical industries.
  • Analysis of bleaching powder.
  • Determination of oxidising and reducing impurities.
✏️ Example
Solved Example
Why does potassium permanganate not require an external indicator during titration?
A coloured reagent that indicates the end point by itself is called a self indicator.
Potassium permanganate is deep purple due to the presence of \(\mathrm{MnO_4^-}\) ions.
During titration, it is reduced to nearly colourless \(\mathrm{Mn^{2+}}\) ions in acidic medium.
Once all the reducing agent has reacted, the next drop of \(\mathrm{KMnO_4}\) remains unreacted and produces a permanent faint pink colour, indicating the end point.
Therefore, potassium permanganate acts as a self indicator.
📎 Basic Calculation Principle
In every redox titration, \[\boxed{\text{Total electrons lost}=\text{Total electrons gained}}\] This relationship forms the basis of concentration calculations in volumetric analysis.
For simple titration calculations, \[\mathrm{n_1M_1V_1=n_2M_2V_2}\] where,
  • \(\mathrm{M}\) = Molarity
  • \(\mathrm{V}\) = Volume
  • \(\mathrm{n}\) = Number of electrons exchanged (or stoichiometric factor)
⚡ Exam Tip
❌ Common Mistakes
  • Confusing acid-base indicators with redox indicators.
  • Assuming every coloured solution acts as a self indicator.
  • Using permanganate titration in neutral or basic medium without considering side reactions.
  • Ignoring the number of electrons exchanged during calculations.
  • Writing an unbalanced redox equation before calculating concentration.
📋 Case Study
CBSE Case Study (HOTS)

A chemist determines the concentration of iron(II) sulphate solution using acidified potassium permanganate. During titration, the purple colour disappears continuously and finally a faint permanent pink colour is observed.

Questions

  1. Why is sulphuric acid added during the titration?
  2. Why is potassium permanganate called a self indicator?
  3. Which species acts as the oxidising agent?
  4. What indicates the end point?

Answers

  1. Acidic medium is required for complete reduction of permanganate ions to \(Mn^{2+}\).
  2. Its own colour change indicates the end point without any external indicator.
  3. \(KMnO_4\) acts as the oxidising agent.
  4. The appearance of a permanent pale pink colour.
🌟 Importance for Board Examinations
  • Redox titrations are one of the most important practical applications of this chapter.
  • Questions are commonly asked on self indicators, redox indicators and the principle of electron transfer.
  • This topic forms the foundation for volumetric analysis in undergraduate chemistry.
  • Frequently tested in CBSE Board practical examinations, JEE Main, NEET and CUET.
🔄

Limitations of the Oxidation Number Concept

📋
Table of Contents
11 sections
🗺️ Overview
The concept of oxidation number is one of the most useful tools for identifying oxidation and reduction reactions. It successfully explains most ionic and covalent redox reactions by assigning formal charges to atoms. However, oxidation number is a formal or hypothetical quantity and does not always represent the actual electronic environment around an atom.

With the advancement of modern chemistry and quantum mechanics, chemists realised that many reactions cannot be explained completely by changes in oxidation numbers alone. Consequently, modern chemistry interprets oxidation and reduction in terms of changes in electron density rather than simply changes in oxidation state.
📌 Modern Concept of Oxidation and Reduction
🤔 Did You Know?
Why is Oxidation Number Only a Formal Quantity?
Oxidation number assumes that every covalent bond is completely ionic, meaning that the shared electron pair belongs entirely to the more electronegative atom.

In reality, electrons in covalent bonds are shared, not completely transferred. Therefore, oxidation numbers are imaginary values assigned for convenience rather than actual charges on atoms.
Example
\[\mathrm{H-Cl}\] In reality, the bonding electrons are shared between hydrogen and chlorine. However, while assigning oxidation numbers, both bonding electrons are assumed to belong entirely to chlorine because chlorine is more electronegative.
Major Limitations of Oxidation Number Concept
1. Oxidation Number is Hypothetical
The oxidation number is an assigned value and does not represent the actual charge present on atoms in covalent compounds. Example \[\mathrm{CH_4}\] Carbon is assigned an oxidation number of \(-4\), but carbon does not actually possess a charge of \(4-\).
2. Does Not Describe Electron Sharing Properly
In covalent compounds, electrons are shared between atoms. Oxidation number ignores this sharing and assumes complete transfer of electrons.

Therefore, it cannot accurately describe the true nature of covalent bonding.
3. Limited Use in Organometallic Compounds
Many organometallic compounds possess highly delocalised electrons. Assigning oxidation numbers in such compounds is often difficult or sometimes impossible. Example
  • Ferrocene
  • Metal carbonyl complexes
Their bonding is better explained using molecular orbital theory than oxidation numbers.
4. Cannot Explain Delocalised Electron Systems
Compounds containing resonance or delocalised electrons cannot always be represented accurately using oxidation numbers. Examples
  • Benzene
  • Graphite
  • Metallic bonding
5. Does Not Measure Actual Electron Density
Two atoms having identical oxidation numbers may possess different electron densities because of differences in electronegativity and molecular environment.

Hence oxidation number alone cannot fully explain chemical reactivity.
⚖️ Comparison of Classical and Modern Concepts
Oxidation Number Concept Modern Electron Density Concept
Based on formal oxidation state. Based on actual electron distribution.
Assumes complete electron transfer. Considers partial electron transfer and sharing.
Suitable for most ionic compounds. Applicable to ionic and covalent compounds.
Simple calculations. Requires advanced understanding of bonding.
Used extensively in school chemistry. Used mainly in higher chemistry.
✏️ Example
Illustrative Examples
Example 1

Reaction

\[ \mathrm{Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu} \]

Explanation

  • Oxidation number method: Zinc changes from 0 to +2.
  • Modern concept: Electron density decreases around zinc because it loses electrons.
Example 2

Reaction

\[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \]
  • Oxidation number decreases from +2 to 0.
  • Electron density around copper increases.
🌟 Significance
Significance of the Modern Concept
  • Explains covalent redox reactions more accurately.
  • Applicable to coordination compounds.
  • Useful in organometallic chemistry.
  • Explains catalytic mechanisms.
  • Forms the basis of molecular orbital theory.
  • Helps explain reaction mechanisms in organic chemistry.
✏️ Example
Solved Example
Why is oxidation number called a formal quantity?
Oxidation number is assigned assuming complete transfer of bonding electrons.
In covalent compounds, electrons are shared rather than completely transferred. While calculating oxidation number, all bonding electrons are assumed to belong to the more electronegative atom. Therefore, oxidation number does not represent the actual charge on an atom but only a formal value used for identifying oxidation and reduction.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming oxidation number equals actual charge in every compound.
  • Believing oxidation number explains all chemical reactions.
  • Ignoring electron sharing in covalent compounds.
  • Confusing oxidation number with valency.
  • Assuming electron density and oxidation number always change by the same magnitude.
📋 CBSE Case Study (HOTS)

A student states that carbon in methane possesses an actual charge of \(4-\) because its oxidation number is \(-4\).

Questions

  1. Is the statement correct?
  2. Why is oxidation number called a formal quantity?
  3. How does the modern concept explain oxidation?
  4. Which concept is more suitable for explaining covalent compounds?

Answers

  1. No. Carbon does not actually carry a charge of \(4-\).
  2. It assumes complete transfer of shared electrons to the more electronegative atom.
  3. Oxidation is explained as a decrease in electron density around an atom.
  4. The electron-density concept provides a more realistic explanation.
🌟 Importance for Board Examinations
🔄

Redox Reactions and Electrode Processes

📋
Table of Contents
18 sections
🗺️ Overview
Redox reactions are not only responsible for ordinary chemical reactions but also form the basis of electrochemical cells, batteries, corrosion, electroplating and electrolysis. In an electrochemical cell, oxidation and reduction occur at different locations called electrodes, and electrons travel through an external circuit to produce electric current.

The Daniell Cell is the simplest and most important electrochemical cell used to understand how chemical energy is converted into electrical energy. It is one of the most frequently asked topics in CBSE Board examinations and forms the foundation of Class XII Electrochemistry.
📘 Redox Couple
✏️ Examples of Redox Couples
Redox Couple Half Reaction
\(Zn^{2+}/Zn\) \(Zn^{2+}+2e^-\rightleftharpoons Zn\)
\(Cu^{2+}/Cu\) \(Cu^{2+}+2e^-\rightleftharpoons Cu\)
\(Fe^{3+}/Fe^{2+}\) \(Fe^{3+}+e^-\rightleftharpoons Fe^{2+}\)
\(Ag^+/Ag\) \(Ag^++e^-\rightleftharpoons Ag\)
By convention, the oxidised form is always written before the reduced form.
📌 Note
Daniell Cell
📘 Definition
A Daniell Cell is a galvanic (voltaic) cell in which zinc is oxidised and copper(II) ions are reduced, thereby converting chemical energy into electrical energy.

It consists of two half-cells connected by a salt bridge and an external metallic wire.
📐 Construction
A Daniell cell consists of:
  • A zinc rod dipped in zinc sulphate solution.
  • A copper rod dipped in copper sulphate solution.
  • A salt bridge connecting the two solutions.
  • A metallic wire connecting the two electrodes.
  • A switch and an ammeter to measure current.
Cell representation
\[\mathrm{\boxed{Zn(s)\;|\;Zn^{2+}(aq)\;||\;Cu^{2+}(aq)\;|\;Cu(s)}}\]
🎨 SVG Diagram
Daniel Cell
DANIELL CELL STANDARD GALVANIC CELL DEMONSTRATION Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) OVERALL REDOX REACTION e⁻ e⁻ 1.10 V VOLTMETER Zn Zn²⁺ Zn²⁺ Zn²⁺ Zn²⁺ SO₄²⁻ SO₄²⁻ Zn ANODE (–) Oxidation Half-Cell 1.0 M ZnSO₄ Colorless Electrolyte Cu Cu²⁺ Cu²⁺ Cu²⁺ Cu²⁺ SO₄²⁻ SO₄²⁻ Cu CATHODE (+) Reduction Half-Cell 1.0 M CuSO₄ Blue Electrolyte Cl⁻ Cl⁻ K⁺ K⁺ SALT BRIDGE (KCl) Maintains Charge Neutrality Cl⁻ Anions Flow to Anode Neutralizes excess Zn²⁺ K⁺ Cations Flow to Cathode Neutralizes excess SO₄²⁻ CELL POTENTIAL E° = 1.10 V ANODE HALF-CELL (OXIDATION) Electrode: Zinc (Zn) • Electrolyte: Zinc Sulfate (ZnSO₄) Zn(s) → Zn²⁺(aq) + 2e⁻ Zinc atoms release 2 electrons to the circuit and dissolve as ions. (-) Zn CATHODE HALF-CELL (REDUCTION) Electrode: Copper (Cu) • Electrolyte: Copper Sulfate (CuSO₄) Cu²⁺(aq) + 2e⁻ → Cu(s) Copper ions gain 2 electrons from the circuit and deposit as metal. (+) Cu
📘 Definition
Salt Bridge
📌 Note
📘 Definition
Electrode Potential
📘 Definition
Standard Electrode Potential (\(E^\Theta\))
📘 Definition
Standard Hydrogen Electrode (SHE)
🌟 Significance
Value of \(E^\Theta\) Meaning
Positive Greater tendency to undergo reduction.
Negative Greater tendency to undergo oxidation.
✏️ Example
Redox Couple \(E^\Theta\) (V)
\(Zn^{2+}/Zn\) -0.76
\(Cu^{2+}/Cu\) +0.34
Since zinc has a more negative standard electrode potential than copper, zinc acts as the reducing agent and copper ions act as the oxidising agent.
🔍 Interpretation of Standard Electrode Potential
  • A highly negative \(E^\Theta\) value indicates a strong reducing agent.
  • A highly positive \(E^\Theta\) value indicates a strong oxidising agent.
  • The greater the difference in electrode potentials, the larger is the cell voltage.
🗒️ Cell Potential
The voltage produced by an electrochemical cell is called the Cell Potential or Electromotive Force (EMF).

For a galvanic cell, \[ \boxed{E^\Theta_{cell}=E^\Theta_{cathode}-E^\Theta_{anode}} \] For the Daniell cell, \[ E^\Theta_{cell}=0.34-(-0.76) \] \[ \boxed{E^\Theta_{cell}=1.10\;V} \] Since the cell potential is positive, the reaction is spontaneous.
✏️ Example
Solved Example
Why does zinc act as the anode in a Daniell cell?
The electrode having the more negative standard electrode potential loses electrons more readily.
Standard electrode potentials are \[ E^\Theta_{Zn^{2+}/Zn}=-0.76\;V \] \[ E^\Theta_{Cu^{2+}/Cu}=+0.34\;V \] Since zinc has the more negative electrode potential, it loses electrons more easily. Therefore, oxidation occurs at the zinc electrode, making it the anode.
⚡ Exam Tip
❌ Common Mistakes
  • Confusing electron flow with conventional current.
  • Writing the cell notation in the wrong order.
  • Assuming the salt bridge transfers electrons instead of ions.
  • Confusing electrode potential with cell potential.
  • Assuming positive electrode potential means oxidation occurs.
📋 CBSE Case Study (HOTS)

A Daniell cell is prepared using zinc and copper electrodes. A student observes that the zinc electrode gradually becomes thinner while the copper electrode becomes thicker.

Questions

  1. Which electrode acts as the anode?
  2. Which electrode acts as the cathode?
  3. In which direction do electrons flow?
  4. What is the function of the salt bridge?
  5. Why is the standard electrode potential of zinc negative?

Answers

  1. Zinc electrode.
  2. Copper electrode.
  3. From zinc to copper through the external wire.
  4. To maintain electrical neutrality and complete the circuit by allowing ion migration.
  5. Because zinc readily loses electrons and is a stronger reducing agent than the Standard Hydrogen Electrode.
🌟 Importance for Board Examinations
· Updated

NCERT · Class XI Chemistry · Chapter 7

Redox Reactions

Electron transfer · oxidation numbers · balancing methods · titrations — a complete, self-contained learning engine

Five Core Concepts

Every idea in this chapter, and every question in the Practice tab, is organised under one of these five pillars.

C1

Classical & Electronic View of Redox

Oxidation and reduction were first defined through oxygen/hydrogen transfer, then generalised to electron transfer — the definition that actually explains every redox reaction, including those with no oxygen at all.

Classical (oxygen-based)

  • Oxidation — addition of oxygen / electronegative element, or removal of hydrogen / electropositive element.
  • Reduction — addition of hydrogen / electropositive element, or removal of oxygen / electronegative element.

Electronic (modern)

  • Oxidation — loss of one or more electrons by a species, increasing its oxidation number.
  • Reduction — gain of one or more electrons by a species, decreasing its oxidation number.
Why "redox"? Oxidation never happens alone. Every electron lost by one species must be gained by another — the two half-processes are chemically inseparable, hence reduction + oxidation = redox.
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Zn loses 2 e⁻ (oxidised, 0 → +2); Cu²⁺ gains 2 e⁻ (reduced, +2 → 0). One process cannot occur without the other.

C2

Oxidation Number

Oxidation number (O.N.) is the imaginary charge an atom would carry if every bond to a different element were treated as fully ionic. It lets us track electron transfer even in covalent molecules.

R1Free elements (uncombined, any allotrope) have O.N. = 0. e.g. Na, O₂, P₄, S₈.
R2Monoatomic ion O.N. = its charge. e.g. Mg²⁺ is +2, Cl⁻ is −1.
R3Sum of O.N. of all atoms in a neutral species = 0; in a polyatomic ion = the ion's charge.
R4Alkali metals (Gp 1) are always +1; alkaline earths (Gp 2) always +2 in compounds.
R5Hydrogen is +1 with non-metals, but −1 in ionic hydrides (NaH, CaH₂).
R6Oxygen is −2 usually; −1 in peroxides (H₂O₂); −½ in superoxides (KO₂); +2 only in OF₂.
R7Fluorine is always −1 (most electronegative element — never positive).
R8Other halogens are −1 unless bonded to O or F, in which case they take positive values solved algebraically.
Redox in O.N. language: Oxidation = increase in O.N. Reduction = decrease in O.N. This single re-statement is what makes the Oxidation-Number balancing method possible.
C3

Types of Redox Reactions

Redox reactions are classified by what the reactants and products look like — not by a different chemical mechanism.

Combination

Two or more elements/compounds unite; at least one reactant is a free element.

2Mg + O₂ → 2MgO

Decomposition

A single compound breaks into two or more simpler products, at least one a free element.

2KClO₃ → 2KCl + 3O₂

Displacement

An atom/ion in a compound is replaced by another (metal or non-metal).

Zn + CuSO₄ → ZnSO₄ + Cu

Disproportionation

The same element in one species is simultaneously oxidised and reduced, giving two different products.

Cl₂ + 2NaOH → NaCl + NaOCl + H₂O
Comproportionation (the reverse idea) — two different oxidation states of the same element combine to give one intermediate oxidation state, e.g. NH₄NO₃ → N₂O + 2H₂O, where N(−3) and N(+5) meet at N(+1).
C4

Balancing Redox Equations

Two systematic methods guarantee a correctly balanced equation without guesswork — both rely on the fact that electrons lost must equal electrons gained.

Oxidation Number Method

  1. Assign O.N. to every atom; spot the atoms that change.
  2. Compute total increase and total decrease in O.N. per formula unit.
  3. Multiply formulas by suitable factors so total increase = total decrease.
  4. Balance all other atoms, then balance O by H₂O and H by H⁺ (acidic) or OH⁻ (basic).

Ion-Electron (Half-Reaction) Method

  1. Split into oxidation and reduction half-reactions.
  2. Balance atoms other than O and H in each half.
  3. Balance O with H₂O, then H with H⁺ (acidic medium).
  4. Balance charge by adding electrons; equalise electrons and add the halves.
  5. For basic medium, add OH⁻ equal to left-over H⁺ on both sides to convert H⁺ → H₂O.
C5

Redox Titrations & Applications

Because electron transfer is quantitative, redox reactions form the basis of volumetric titrations used to determine unknown concentrations.

  • Law of chemical equivalence: at the equivalence point, milliequivalents of oxidant = milliequivalents of reductant: N₁V₁ = N₂V₂.
  • n-factor (equivalence factor) = number of electrons gained/lost by one formula unit of the species in that specific reaction — it can change with medium (e.g. KMnO₄ has n = 5 in acidic, 3 in neutral/faintly alkaline, 1 in strongly alkaline medium).
  • Self-indicating titrant: KMnO₄ is intensely purple as Mn(+7) and colourless as Mn²⁺(+2) — the end point needs no external indicator.
  • Iodometric/iodimetric titrations use the I₂/I⁻ couple with starch as indicator (deep blue colour with I₂).
  • Everyday redox: corrosion (rusting), combustion, respiration, photosynthesis, electroplating and batteries are all redox processes.

Step-by-Step AI Solver

A rule-based reasoning engine — runs entirely in your browser, no API key or server calls needed.

Compute the Oxidation Number of Any Atom

Pick a compound/ion from the library, or type a custom formula. The engine parses the formula, applies the eight assignment rules, and solves algebraically for the atom you choose.

Identify the Oxidising Agent, Reducing Agent & Electron Transfer

Choose a reaction — the engine tracks the oxidation-number change of every relevant atom and names both agents with full reasoning.

Classify the Redox Reaction Type

Combination, decomposition, displacement, disproportionation or comproportionation — with the reasoning spelled out.

Balance by the Oxidation Number Method

Full worked balancing, step by step, exactly as you'd write it in an exam.

Balance by the Ion-Electron (Half-Reaction) Method

Covers both acidic and basic medium, including the extra neutralisation step for basic medium.

Formula & Reference Vault

Everything you need on a single scrollable reference — grouped by use.

Oxidation Number — Assignment Rules

SpeciesO.N.
Free element0
Monoatomic ion= ionic charge
Group 1 metal (compound)+1
Group 2 metal (compound)+2
H (with non-metal)+1
H (ionic hydride, e.g. NaH)−1
O (usual)−2
O (peroxide, H₂O₂)−1
O (superoxide, KO₂)−½
O (in OF₂)+2
F−1 (always)

Sum Rule

Σ (O.N. × number of atoms) = overall charge on species

Neutral molecule → sum = 0. Polyatomic ion → sum = ionic charge. This single equation is what the Oxidation Number Finder tool solves for the unknown atom.

Law of Chemical Equivalence (Titrations)

Milliequivalents of oxidant = Milliequivalents of reductant
N₁V₁ = N₂V₂

where N = normality = Molarity × n-factor, and V = volume used.

n-factor (Equivalence Factor)

n-factor = number of electrons lost or gained by 1 formula unit in that reaction
SpeciesMediumn-factor
KMnO₄Acidic5
KMnO₄Neutral / faintly alkaline3
KMnO₄Strongly alkaline1
K₂Cr₂O₄Acidic6
H₂O₂as oxidant2
H₂O₂as reductant2
Oxalic acid H₂C₂O₄—2
Fe²⁺ → Fe³⁺—1

Oxidation Number Method — Procedure Card

  1. Write the skeletal equation and assign O.N. to all atoms.
  2. Identify the atom oxidised and the atom reduced; note Δ(O.N.) per atom.
  3. Multiply by the atomicity per formula unit if more than one such atom is present.
  4. Cross-multiply the two species so that total electrons lost = total electrons gained.
  5. Balance remaining atoms by inspection.
  6. Balance O using H₂O; balance H using H⁺ (acidic) or OH⁻ (basic).
  7. Verify: atom count and net charge equal on both sides.

Ion-Electron Method — Procedure Card

  1. Split the skeletal ionic equation into two half-reactions.
  2. Balance atoms other than O, H in each half.
  3. Balance O atoms by adding H₂O.
  4. Balance H atoms by adding H⁺ (acidic medium).
  5. Balance charge on each half by adding electrons (e⁻) to the more positive side.
  6. Multiply each half so electrons lost = electrons gained, then add.
  7. Cancel common terms (H₂O, H⁺, e⁻) on both sides.
  8. Basic medium only: add OH⁻ equal to remaining H⁺ on both sides → H⁺ + OH⁻ → H₂O, then cancel excess water.

Quick Half-Reaction Templates

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O acidic reduction
Cr₂O₄₄²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O acidic reduction
MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ basic reduction

Ticks & Tips

Fast recall devices and exam-hall shortcuts.

🎯

OIL RIG

Oxidation Is Loss, Reduction Is Gain (of electrons). The fastest way to never confuse the two directions.

🔧

LEO says GER

Lose Electrons = Oxidation; Gain Electrons = Reduction. Use whichever mnemonic sticks — both encode the same rule.

⚖

Fix the "usual suspects" first

Before solving for an unknown O.N., lock in H = +1, O = −2, Group 1 = +1, Group 2 = +2, F = −1. Only the exceptions (peroxide, hydride, OF₂) need special attention — everything else is one algebra step.

👀

Spot disproportionation instantly

If the same element appears in exactly two different products with two different oxidation numbers (one higher, one lower than in the reactant), it's disproportionation — no need to track every other atom.

⚖

Charge balance is a free checker

After balancing any ionic equation, add up the net charge on the reactant side and the product side separately. If they don't match, you have an arithmetic slip — this catches ~80% of balancing errors instantly.

📊

Acidic vs basic medium — add, don't guess

Always balance first as if in acidic medium (H⁺/H₂O). Only at the very last step, if the question says basic/alkaline, add OH⁻ to both sides to mop up the H⁺. Trying to use OH⁻ from the start causes more errors than it prevents.

✅

KMnO₄'s n-factor depends on medium

Remember the "5-3-1 ladder": acidic → 5, neutral/faintly alkaline → 3, strongly alkaline → 1. This single fact resolves most titration n-factor questions involving permanganate.

🔄

Average oxidation number for organic/complex species

When an element appears to have a "fractional" O.N. (e.g. C in C₃O₄²⁻ is +2.67), it's an average across non-equivalent atoms — perfectly valid, and a strong clue that individual atoms differ (as confirmed by structure).

Common Mistakes

The errors examiners see most often — flagged so you can avoid them.

×

Confusing "oxidised" with "oxidising agent"

The slip: calling the species that gets oxidised the "oxidising agent."

The fix: the substance that is itself reduced is the oxidising agent (it oxidises something else, at the cost of being reduced itself). Read the sentence twice — the agent name is the opposite of what happens to it.

×

Using O = −2 blindly in peroxides/superoxides

The slip: assigning Na in Na₂O₂ as +4 by forcing O = −2.

The fix: recognise the peroxide ion (O₂²⁻, each O = −1) or superoxide ion (O₂⁻, each O = −½) before applying the sum rule — Na stays +1 as always.

×

Forgetting to balance charge in ionic equations

The slip: balancing only atoms and stopping, leaving unequal net charge on both sides.

The fix: after atom-balancing every half-reaction, always add electrons to balance charge — this step is not optional and is exactly where most half-reaction attempts go wrong.

×

Mixing H⁺ and OH⁻ in the same balanced equation

The slip: ending up with both H⁺ and OH⁻ in a "final" answer for a basic-medium question.

The fix: basic medium equations should contain OH⁻ and H₂O only — never H⁺. If H⁺ is still present, the neutralisation step (add OH⁻ to both sides) was skipped or done incompletely.

×

Wrong n-factor for H₂O₂

The slip: assuming H₂O₂ always has n-factor = 1.

The fix: H₂O₂ has n-factor 2 both as an oxidant (O⁻¹ → O⁻², reduced) and as a reductant (O⁻¹ → O², oxidised, releasing O₂) — check which role it plays in the given reaction, but the number is 2 either way here.

×

Treating disproportionation as ordinary combination/decomposition

The slip: classifying Cl₂ + 2NaOH → NaCl + NaOCl + H₂O as "combination" because two reactants combine.

The fix: check the oxidation states of the same element (Cl here) across products first. If one element splits into two different oxidation states, it's disproportionation regardless of how many reactants were involved.

×

Ignoring atomicity when computing total electron change

The slip: in Cr₂O₄₄²⁻ → Cr³⁺, using "change per Cr = 3" without multiplying by 2 (two Cr atoms per formula unit), giving 3 instead of 6 electrons.

The fix: always multiply Δ(O.N.) by the number of that atom present in the formula unit before cross-balancing electrons.

Concept-Building Practice Questions

Original questions, organised by the same five concepts — every solution is fully worked, step by step. Click a question to reveal its solution.

C1 · Classical & Electronic View of Redox

Step 1 — Check the classical (oxygen/hydrogen) definitions. Neither oxygen nor hydrogen appears anywhere in this reaction, so neither the oxygen-transfer nor hydrogen-transfer definition of oxidation/reduction can be applied here at all.

Step 2 — Apply the electron-transfer definition. Na(0) → Na⁺ loses one electron per atom (oxidation); Cl(0) in Cl₂ → Cl⁻ gains one electron per atom (reduction).

Step 3 — Conclusion. This confirms the electron-transfer definition is the general, mechanism-independent definition — it works even when no oxygen or hydrogen is present, which is exactly why it replaced the classical definitions.

Step 1 — O.N. of carbon. C goes from 0 (free element) to +4 in CO₂. This is a loss of 4 electrons — oxidation, correctly identified.

Step 2 — O.N. of oxygen. O goes from 0 (free element, O₂) to −2 in CO₂. Each O atom gains 2 electrons — this is reduction, not "doing nothing."

Step 3 — Correction. The statement is incorrect: O₂ is the oxidising agent (it gets reduced while oxidising carbon), and C is the reducing agent. Redox always has both halves occurring together — there is no reaction where only one half-process happens.

Step 1 — Assign O.N. to every atom. H in HCl = +1; Cl in HCl = −1; Na in NaOH = +1; O in NaOH = −2; H in NaOH = +1. On the product side: Na in NaCl = +1; Cl in NaCl = −1; H and O in H₂O = +1 and −2.

Step 2 — Compare reactant and product O.N.s. Every single atom (H, Cl, Na, O) has the exact same oxidation number before and after the reaction.

Step 3 — Conclusion. No electron transfer has occurred anywhere, so this is not a redox reaction — it is a simple acid-base neutralisation (proton transfer), a different category entirely.

C2 · Oxidation Number

Step 1 — Fix known atoms. Na = +1 (Group 1) × 2 = +2. O = −2 × 6 = −12.

Step 2 — Set up the sum equation (neutral compound, sum = 0): 2(+1) + 4x + 6(−2) = 0 ⇒ 2 + 4x − 12 = 0 ⇒ 4x = 10 ⇒ x = +2.5 (average O.N. of S).

Step 3 — Why "average" makes sense. Tetrathionate's structure, −O₃S−S−S−SO₃−, has two terminal S atoms at +5 each and two central S atoms at 0 each; the mean is (5+5+0+0)/4 = 2.5 — consistent with the algebraic answer.

CrO₄²⁻: O = −2 × 4 = −8. Sum = ionic charge = −2. So x + (−8) = −2 ⇒ x = +6.

Cr₂O₄₄²⁻: O = −2 × 7 = −14. Sum = −2. So 2x − 14 = −2 ⇒ 2x = 12 ⇒ x = +6.

Conclusion. Chromium is +6 in both — dichromate is simply two chromate-like units sharing one bridging oxygen; the oxidation state of chromium is unaffected by whether it is monomeric or dimeric.

KO₂: K = +1. Sum = 0 (neutral). So 1 + 2x = 0 ⇒ x = −½. This half-integer value reflects the superoxide ion O₂⁻ where one unpaired electron is shared over two equivalent oxygen atoms.

K₂O: K = +1 × 2 = +2. Sum = 0. So 2 + x = 0 ⇒ x = −2 (ordinary oxide ion, O²⁻).

Conclusion. Both are valid oxidation states of oxygen; the fractional value in KO₂ is a real and expected result of applying the sum rule to the superoxide ion, not an error.

Step 1 — Split into the two ions. NH₄NO₃ is really NH₄⁺ and NO₃⁻ combined; each nitrogen must be solved within its own ion, not averaged together (they are not equivalent atoms).

NH₄⁺: H = +1 × 4 = +4. Sum = +1. x + 4 = +1 ⇒ x = −3.

NO₃⁻: O = −2 × 3 = −6. Sum = −1. x − 6 = −1 ⇒ x = +5.

Conclusion. One N is −3 (in NH₄⁺) and the other is +5 (in NO₃⁻) — this is exactly why thermal decomposition of NH₄NO₃ to N₂O (N = +1) is a comproportionation, the two extremes meeting at an intermediate state.

C3 · Types of Redox Reactions

Step 1 — O.N. of Mn in each species. MnO₄²⁻ (manganate): Mn = +6. MnO₄⁻ (permanganate): Mn = +7. MnO₂: Mn = +4.

Step 2 — Track the change. Mn(+6) splits into Mn(+7) (oxidised, one manganate loses an electron) and Mn(+4) (reduced, another manganate gains electrons) — the same starting element, same starting oxidation state, giving two different products.

Step 3 — Classification: disproportionation of manganate ion in acidic medium.

2AgBr → 2Ag + Br₂: a single compound (AgBr) breaks down into two elements (Ag and Br₂) with no second reactant — this is decomposition (redox: Ag +1→0 reduced, Br −1→0 oxidised).

2Al + Fe₂O₃ → Al₂O₃ + 2Fe: here a free element (Al) reacts with a compound (Fe₂O₃), and one atom (Fe) in the compound is replaced by another (Al) — this is displacement (redox: Al 0→+3 oxidised, Fe +3→0 reduced).

Distinguishing rule: decomposition has exactly one reactant; displacement always has an element reacting with a different compound and swapping places with one of its atoms.

Step 1 — O.N. of P. In P₄ (free element), P = 0. In PH₃, P = −3 (H is +1 with non-metal, sum = 0 ⇒ P = −3). In NaH₂PO₂ (sodium hypophosphite), P = +1.

Step 2 — Track the split. P(0) becomes both P(−3) (reduced) and P(+1) (oxidised) — same starting element and oxidation state, splitting into two different products.

Step 3 — Classification: disproportionation of elemental phosphorus in alkaline medium — directly analogous to the Cl₂/NaOH case but with phosphorus instead of chlorine.

C4 · Balancing Redox Equations

Step 1 — O.N. changes. I in KI: −1 → 0 in I₂ (oxidation, loses 1 e⁻ per I). O in H₂O₂: −1 → −2 in H₂O (reduction, each of the 2 O atoms gains 1 e⁻, so 2 e⁻ per H₂O₂).

Step 2 — Equalise electrons. 1 electron lost per I vs 2 electrons gained per H₂O₂. Take 2 KI (2 electrons lost) for every 1 H₂O₂ (2 electrons gained).

Step 3 — Place coefficients: 2KI + H₂SO₄ + H₂O₂ → K₂SO₄ + I₂ + 2H₂O. (Using K₂SO₄ instead of KHSO₄ keeps K and S balanced cleanly with 2 K.)

Step 4 — Check remaining atoms. K: 2 = 2 ✓. S: 1 = 1 ✓. H: 2(H₂SO₄)+2(H₂O₂) = 4 = 2×2 (H₂O) ✓. O: 4(SO₄) + 2(H₂O₂) = 6 = 4(SO₄)+2(H₂O) ✓.

Balanced equation: 2KI + H₂SO₄ + H₂O₂ → K₂SO₄ + I₂ + 2H₂O

Step 1 — O.N. changes. Cu: 0 → +2 in Cu(NO₃)₂ (oxidation, loses 2 e⁻). N: +5 in HNO₃ → +2 in NO (reduction, gains 3 e⁻ — only the N that becomes NO undergoes reduction; the N that stays as NO₃⁻ in Cu(NO₃)₂ is a spectator, its O.N. unchanged).

Step 2 — Equalise electrons. LCM of 2 and 3 is 6 ⇒ take 3 Cu (6 e⁻ lost) and 2 NO (6 e⁻ gained).

Step 3 — Place coefficients: 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O. (8 HNO₃ needed: 6 N end up as NO₃⁻ spectator in 3Cu(NO₃)₂, plus 2 N reduced to NO = 8 total N; H: 8 H⁺ balance to 4 H₂O.)

Step 4 — Verify: Cu 3=3 ✓. N: 8 = 6+2 ✓. H: 8 = 8 ✓. O: 24 = 18(Cu(NO3)2)+2(NO)+4(H2O)=24 ✓.

Balanced equation: 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO↑ + 4H₂O

Step 1 — Half reactions. Reduction: Cr₂O₄₄²⁻ → Cr³⁺. Oxidation: SO₂ → SO₄²⁻.

Step 2 — Balance atoms other than O, H. Cr₂O₄₄²⁻ → 2Cr³⁺ (Cr balanced). S is already 1 = 1.

Step 3 — Balance O with H₂O, then H with H⁺.
Reduction: Cr₂O₄₄²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O.
Oxidation: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺.

Step 4 — Balance charge with electrons.
Reduction: left charge = −2+14 = +12; right = +6. Add 6e⁻ to left: Cr₂O₄₄²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
Oxidation: left charge = 0; right = −2+4 = +2. Add 2e⁻ to right: SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻.

Step 5 — Equalise electrons (LCM 6) and add. Multiply the oxidation half by 3, then add to the reduction half; 12H⁺ and 6H₂O cancel partially.

Balanced equation: Cr₂O₄₄²⁻ + 3SO₂ + 2H⁺ → 2Cr³⁺ + 3SO₄²⁻ + H₂O

Charge check: left: −2+0+2 = 0. right: +6−6+0 = 0. ✓

Step 1 — Recognise disproportionation and split. Reduction: Br₂ → Br⁻. Oxidation: Br₂ → BrO₃⁻.

Step 2 — Balance Br atoms. Reduction: Br₂ → 2Br⁻. Oxidation: Br₂ → 2BrO₃⁻.

Step 3 — Balance O with H₂O, H with H⁺ (treat as acidic first). Oxidation: Br₂ + 6H₂O → 2BrO₃⁻ + 12H⁺.

Step 4 — Balance charge with electrons. Reduction: Br₂ + 2e⁻ → 2Br⁻ (charge: 0 → −2, balanced by 2e⁻). Oxidation: Br₂ + 6H₂O → 2BrO₃⁻ + 12H⁺ + 10e⁻ (charge left 0, right −2+12−10=0 ✓).

Step 5 — Equalise electrons (LCM 10). Multiply reduction half by 5: 5Br₂ + 10e⁻ → 10Br⁻. Add to oxidation half: 6Br₂ + 6H₂O → 10Br⁻ + 2BrO₃⁻ + 12H⁺, i.e. 3Br₂ + 3H₂O → 5Br⁻ + BrO₃⁻ + 6H⁺ (divide by 2).

Step 6 — Basic medium: neutralise H⁺ with OH⁻. Add 6OH⁻ to both sides; 6H⁺+6OH⁻ → 6H₂O on the right, cancel 3H₂O common to both sides.

Balanced equation: 3Br₂ + 6OH⁻ → 5Br⁻ + BrO₃⁻ + 3H₂O

C5 · Redox Titrations & Applications

Step 1 — n-factors. KMnO₄ in acidic medium: n = 5. FeSO₄ (Fe²⁺ → Fe³⁺): n = 1.

Step 2 — Normality of KMnO₄. N = M × n = 0.02 × 5 = 0.1 N.

Step 3 — Apply N₁V₁ = N₂V₂. N(FeSO₄) × 25.0 = 0.1 × 20.0 ⇒ N(FeSO₄) = 2.0/25.0 = 0.08 N.

Step 4 — Convert to molarity. M = N / n = 0.08 / 1 = 0.08 M.

Step 1 — Colours of the two states. MnO₄⁻ (Mn, +7) is intensely purple/violet in solution. Mn²⁺ (+2), the reduced product in acidic medium, is almost colourless (very pale pink at the concentrations used).

Step 2 — During titration. As KMnO₄ is added drop by drop to the reducing agent, each drop is reduced and decolourised instantly, as long as reducing agent remains in excess.

Step 3 — End point. Once all reducing agent is consumed, the very next drop of KMnO₄ has nothing left to reduce it, so the solution retains a faint permanent pink/purple tinge — this is the end point.

Conclusion: KMnO₄ is called a self-indicating titrant because its own colour change marks the equivalence point, with no need for starch, methyl orange, or any other external indicator.

Step 1 — Moles of thiosulphate used. moles S₂O₃²⁻ = 0.1 M × 0.020 L = 2.0 × 10⁻³ mol.

Step 2 — Link thiosulphate to I₂. From I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, moles I₂ = ½ × moles S₂O₃²⁻ = 1.0 × 10⁻³ mol.

Step 3 — Link I₂ to Cu²⁺. From 2Cu²⁺ + 4I⁻ → 2CuI + I₂, moles Cu²⁺ = 2 × moles I₂ = 2.0 × 10⁻³ mol.

Conclusion: the sample contains 2.0 × 10⁻³ mol of Cu²⁺ — this same "chain the equivalences" strategy is exactly how the n-factor/equivalence approach short-cuts multi-step iodometric problems without writing every intermediate equation.

Interactive Lab

Five hands-on modules to build intuition — not just formulas.

Oxidation Number Trainer

A highlighted atom, a compound, four options. Score builds as you go — no repeats until the set is exhausted.

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Reaction Type Sorter

Read the reaction, then click the correct category. Instant feedback with reasoning.

Score: 0
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Electron Transfer Visualiser

Watch electrons physically animate from the reducing agent to the oxidising agent. Pick a reaction and press play.

Half-Reaction Balancing Simulator

Step through the ion-electron method one click at a time. Try to predict the next step before revealing it.

Disproportionation Spotter

Yes or no: is the highlighted element disproportionating in this reaction?

Score: 0
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Academia Aeternum · NCERT Class XI Chemistry · Chapter 7 — Redox Reactions

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    Redox Reactions are among the most fundamental processes in chemistry, explaining how electrons are transferred between substances during chemical reactions. From the burning of fuels and corrosion of metals to cellular respiration, photosynthesis, batteries and industrial metallurgy, redox reactions play a vital role in both nature and technology. In this chapter, you will develop a clear understanding of oxidation, reduction, oxidation number, oxidising and reducing agents, and the electronic…
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    redox reactions — Learning Resources

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    redox reactions-exercises

    Frequently Asked Questions

    A redox reaction is a chemical reaction in which oxidation and reduction occur simultaneously through the transfer of electrons or a change in the oxidation numbers of the reacting species. One substance is oxidised while another is reduced.

    Oxidation is the increase in oxidation number or loss of electrons, whereas reduction is the decrease in oxidation number or gain of electrons. These two processes always occur together in a redox reaction.

    Oxidation number is the hypothetical charge assigned to an atom assuming complete transfer of bonding electrons to the more electronegative atom. It helps identify oxidation, reduction, oxidising agents, reducing agents and is essential for balancing redox equations.

    An oxidising agent accepts electrons and undergoes reduction while causing another substance to oxidise. A reducing agent donates electrons and undergoes oxidation while causing another substance to reduce.

    Redox reactions are classified into combination reactions, decomposition reactions, metal displacement reactions, non-metal displacement reactions and disproportionation reactions. Each type involves a change in oxidation numbers of the participating elements.

    Redox reactions are balanced using either the Oxidation Number Method or the Half-Reaction (Ion-Electron) Method. Both methods ensure that the number of atoms and the total charge are equal on both sides of the equation.

    Stock notation is the IUPAC system of representing the oxidation state of a metal using Roman numerals in parentheses. For example, FeCl3 is named Iron(III) chloride, indicating that iron has an oxidation number of +3.

    A disproportionation reaction is a special type of redox reaction in which the same element undergoes both oxidation and reduction simultaneously, producing compounds with higher and lower oxidation states. The decomposition of hydrogen peroxide is a common example.

    A Daniell cell is a galvanic cell consisting of zinc and copper electrodes connected through a salt bridge. It converts chemical energy into electrical energy and demonstrates the practical application of redox reactions and electrode potentials.

    Redox Reactions is a foundational chapter for understanding electrochemistry, metallurgy and inorganic chemistry. Questions on oxidation number, balancing redox equations, oxidising and reducing agents, electrode potential and redox titrations are frequently asked in CBSE Board examinations, JEE Main, NEET and CUET.

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