Redox Reactions

CLASSICAL IDEA OF REDOX REACTIONS – OXIDATION AND REDUCTION

**Oxidation** was originally defined as the addition of oxygen to an element or compound. As chemical knowledge expanded, this definition was broadened to include:

Addition of oxygen/electronegative elements to a substance

Removal of hydrogen/electropositive elements from a substance

**Examples of oxidation reactions:**

2Mg(s) + O₂(g) → 2MgO(s) — oxygen added to magnesium

S(s) + O₂(g) → SO₂(g) — oxygen added to sulfur

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) — hydrogen replaced by oxygen

Mg(s) + F₂(g) → MgF₂(s) — more electronegative element added

**Reduction** was initially defined as removal of oxygen from a compound. The modern definition includes:

Removal of oxygen/electronegative elements from a substance

Addition of hydrogen/electropositive elements to a substance

**Examples of reduction reactions:**

2HgO(s) → 2Hg(l) + O₂(g) — oxygen removed

2FeCl₃(aq) + H₂(g) → 2FeCl₂(aq) + 2HCl(aq) — electronegative chlorine removed

CH₂=CH₂(g) + H₂(g) → CH₃-CH₃(g) — hydrogen added

2HgCl₂(aq) + SnCl₂(aq) → Hg₂Cl₂(s) + SnCl₄(aq) — electropositive element added

**Key principle:** Oxidation and reduction always occur simultaneously in a chemical reaction. This class of reactions is called **redox reactions** (reduction-oxidation).

REDOX REACTIONS IN TERMS OF ELECTRON TRANSFER

Modern understanding defines redox reactions through electron transfer:

**Oxidation:** Loss of electron(s) by any species

**Reduction:** Gain of electron(s) by any species

**Oxidising agent (Oxidant):** Species that accepts electron(s); becomes reduced

**Reducing agent (Reductant):** Species that donates electron(s); becomes oxidised

**Half-reaction method:** Any redox reaction can be split into two half-reactions:

For the formation of sodium chloride:

Oxidation half-reaction: 2Na(s) → 2Na⁺(g) + 2e⁻

Reduction half-reaction: Cl₂(g) + 2e⁻ → 2Cl⁻(g)

Overall reaction: 2Na(s) + Cl₂(g) → 2NaCl(s)

**Important distinction:** In redox reactions, the species being oxidised is the reducing agent, and the species being reduced is the oxidising agent.

**Exam example:** In the reaction 2Na(s) + H₂(g) → 2NaH(s):

Half-reactions: 2Na(s) → 2Na⁺(g) + 2e⁻ and H₂(g) + 2e⁻ → 2H⁻(g)

Sodium is oxidised (loses electrons) and acts as reducing agent

Hydrogen is reduced (gains electrons) and acts as oxidising agent

This confirms it is a redox change despite not involving oxygen

Competitive Electron Transfer Reactions

The tendency of metals to release electrons can be compared through test reactions:

**Reaction 1:** Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) — reaction occurs readily

Zinc strip placed in copper nitrate solution becomes coated with red copper

Blue colour of Cu²⁺ solution disappears

Equilibrium greatly favours products

**Reaction 2:** Cu(s) + Ni²⁺(aq) → No appreciable reaction

Copper does not release electrons to nickel ions

**Reaction 3:** Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s) — reaction occurs readily

Blue colour develops due to Cu²⁺ formation

Silver deposits as shiny metal

**Electron-releasing order:** Zn > Cu > Ag

This competitive electron transfer demonstrates that metals have different tendencies to release electrons, forming the basis of the **electrochemical series** or **activity series of metals**. This principle is used in designing **galvanic cells** where chemical reactions produce electrical energy.

OXIDATION NUMBER

**Definition:** Oxidation number denotes the oxidation state of an element in a compound, assigned according to a set of rules based on the assumption that electron pairs in covalent bonds belong entirely to the more electronegative element.

**Purpose:** Oxidation numbers provide a systematic method for tracking electron shifts in reactions, especially those involving covalent compounds where complete electron transfer does not occur.

Rules for Assigning Oxidation Numbers

**Rule 1 – Free elements:** Each atom in an uncombined element has an oxidation number of zero.

Examples: H₂, O₂, Cl₂, O₃, P₄, S₈, Na, Mg, Al all have oxidation number = 0

**Rule 2 – Monatomic ions:** The oxidation number equals the charge on the ion.

Na⁺ has oxidation number +1

Mg²⁺ has oxidation number +2

Fe³⁺ has oxidation number +3

Cl⁻ has oxidation number –1

O²⁻ has oxidation number –2

All alkali metals (Group 1): oxidation number = +1 in compounds

All alkaline earth metals (Group 2): oxidation number = +2 in compounds

Aluminium: oxidation number = +3 in all compounds

**Rule 3 – Oxygen:** Oxidation number is –2 in most compounds, with important exceptions:

In **peroxides** (e.g., H₂O₂, Na₂O₂): each oxygen atom has oxidation number = –1

In **superoxides** (e.g., KO₂, RbO₂): each oxygen atom has oxidation number = –½

When bonded to fluorine (e.g., OF₂, O₂F₂): oxygen has positive oxidation number

In OF₂: oxidation number of O = +2

In O₂F₂: oxidation number of O = +1

**Rule 4 – Hydrogen:** Oxidation number is +1, except:

In binary compounds with metals (hydrides): oxidation number = –1

Examples: LiH, NaH, CaH₂ — hydrogen has oxidation number –1

**Rule 5 – Halogens:**

Fluorine: oxidation number = –1 in all compounds

Chlorine, bromine, iodine: oxidation number = –1 when present as halide ions

When combined with oxygen in oxoacids and oxoanions: halogens have positive oxidation numbers

Example: In HClO₄, chlorine has oxidation number +7

**Rule 6 – Algebraic sum:**

In a neutral compound: sum of all oxidation numbers = 0

In a polyatomic ion: sum of all oxidation numbers = charge on the ion

If multiple atoms of same element exist (e.g., Na₂S₂O₃), the oxidation number assigned is the average oxidation state of all atoms of that element

Calculation of Oxidation Numbers – Worked Examples

**Example 1:** Find oxidation number of carbon in CO₂

Each oxygen = –2, so total for 2 oxygen = –4

For neutral compound: x + (–4) = 0

Therefore, x = +4 (oxidation number of C = +4)

**Example 2:** Find oxidation number of nitrogen in NO₃⁻

Each oxygen = –2, so total for 3 oxygen = –6

For polyatomic ion with charge –1: x + (–6) = –1

Therefore, x = +5 (oxidation number of N = +5)

**Example 3:** Find oxidation number of sulfur in H₂S₂O₃

Each H = +1 (total +2), each O = –2 (total –6)

For neutral: 2 + 2x + (–6) = 0, where x is average oxidation number of 2 sulfur atoms

2x = 4, so x = +2 per sulfur atom on average

Oxidation Number Trends

**Metallic elements** display positive oxidation numbers only

**Non-metallic elements** display positive or negative oxidation numbers

**Transition elements** typically display multiple positive oxidation states

**Highest oxidation number of representative elements** equals:

Group number (for Groups 1-2)

Group number minus 10 (for other groups in long form of periodic table)

In Period 3: highest oxidation numbers range from +1 (Na) to +7 (Cl)

Oxidation State and Stock Notation

**Oxidation state** is used interchangeably with oxidation number and indicates the electronic state of an element in a compound.

**Stock notation** (named after German chemist Alfred Stock): Expresses oxidation number as a Roman numeral in parentheses after the element symbol.

Aurous chloride: Au(I)Cl

Auric chloride: Au(III)Cl₃

Stannous chloride: Sn(II)Cl₂

Stannic chloride: Sn(IV)Cl₄

Ferrous ion: Fe(II)

Ferric ion: Fe(III)

**Exam point:** Change in oxidation number indicates whether a species is oxidised (increase) or reduced (decrease).

IDENTIFYING REDOX REACTIONS USING OXIDATION NUMBERS

**Method:**

Assign oxidation numbers to all elements in reactants and products

If oxidation numbers change, it is a redox reaction

Element with increased oxidation number = oxidised (reducing agent)

Element with decreased oxidation number = reduced (oxidising agent)

**Worked example:** Analyze 2H₂(g) + O₂(g) → 2H₂O(l)

Reactants: H in H₂ = 0, O in O₂ = 0

Products: H in H₂O = +1, O in H₂O = –2

H: oxidation number changes 0 → +1 (oxidised, reducing agent)

O: oxidation number changes 0 → –2 (reduced, oxidising agent)

Therefore, this is a redox reaction

**Exam classification:** H₂ + Cl₂ → 2HCl is also redox because H changes from 0 to +1 and Cl changes from 0 to –1, despite being a covalent compound.

CLASSIFICATION OF REDOX REACTIONS

Redox reactions are classified into four major types:

1. Combination (Synthesis) Reactions

**Definition:** Two or more substances combine to form a single compound with simultaneous oxidation and reduction.

**General form:** A + B → AB

**Examples:**

2H₂(g) + O₂(g) → 2H₂O(l) — hydrogen oxidised, oxygen reduced

2Na(s) + Cl₂(g) → 2NaCl(s) — sodium oxidised, chlorine reduced

4P(s) + 5O₂(g) → 2P₂O₅(s) — phosphorus oxidised, oxygen reduced

**Exam point:** Not all combination reactions are redox reactions (e.g., 2Na + Cl₂ → 2NaCl is redox, but CaO + CO₂ → CaCO₃ is not).

2. Decomposition Reactions

**Definition:** A single compound breaks down into two or more substances, involving simultaneous oxidation and reduction.

**General form:** AB → A + B

**Examples:**

2HgO(s) → 2Hg(l) + O₂(g) — mercury reduced, oxygen oxidised

2H₂O(l) → 2H₂(g) + O₂(g) — hydrogen oxidised and reduced simultaneously (disproportionation variant)

2KMnO₄(s) → K₂MnO₄(s) + MnO₂(s) + O₂(g)

**Exam point:** Thermal decomposition of certain salts produces redox reactions.

3. Displacement Reactions

**Definition:** An element replaces another element from its compound, involving electron transfer between elements with different reactivities.

**General forms:**

**Single displacement:** A + BC → AC + B

**Metal displacement:** More reactive metal displaces less reactive metal

**Non-metal displacement:** More electronegative non-metal displaces less electronegative non-metal

**Metal displacement examples:**

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) — Zn more reactive, oxidised

Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s) — Fe more reactive than Cu

Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s) — Cu more reactive than Ag

Cl₂(g) + 2NaBr(aq) → 2NaCl(aq) + Br₂(l) — Cl more electronegative, oxidised

**Non-metal displacement:**

H₂(g) + 2AgNO₃(aq) → 2HNO₃(aq) + 2Ag(s) — not displacement but redox

Cl₂(g) + 2KI(aq) → 2KCl(aq) + I₂(l) — Cl more electronegative than I

**Exam point:** Displacement reactions follow the activity series; a more reactive element always displaces a less reactive element.

4. Disproportionation Reactions

**Definition:** A single element simultaneously undergoes both oxidation and reduction in the same reaction, producing oxidised and reduced forms of that element.

**Key characteristic:** One element exhibits both positive and negative changes in oxidation number, existing in two different oxidation states in products.

**Examples:**

**Cl₂ + 2NaOH → NaCl + NaClO + H₂O** (cold, dilute solution)

Cl in Cl₂: 0

Cl in NaCl: –1 (reduced)

Cl in NaClO: +1 (oxidised)

One Cl atom reduced, other oxidised

**3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O** (hot, concentrated solution)

Cl in Cl₂: 0

Cl in NaCl: –1 (reduced)

Cl in NaClO₃: +5 (oxidised)

**2H₂O₂ → 2H₂O + O₂** (catalytic decomposition)

O in H₂O₂: –1

O in H₂O: –2 (reduced)

O in O₂: 0 (oxidised)

**P₄ + 3NaOH + 3H₂O → 3NaH₂PO₂ + PH₃**

P in P₄: 0

P in NaH₂PO₂: +1 (oxidised)

P in PH₃: –3 (reduced)

**2KMnO₄ → K₂MnO₄ + MnO₂ + O₂** (thermal decomposition)

Mn in KMnO₄: +7

Mn in K₂MnO₄: +6 (reduced)

Mn in MnO₂: +4 (reduced)

O undergoes oxidation

**Exam distinction:** In disproportionation, the **same element** must appear in at least three different oxidation states OR in exactly two different oxidation states in the products (created from the same initial state).

**Counter-example (not disproportionation):** 2FeCl₃ + SnCl₂ → 2FeCl₂ + SnCl₄ — different elements undergoing redox, so it is a displacement reaction

BALANCING REDOX EQUATIONS

Redox equations can be balanced using two methods:

Method 1: Oxidation Number Method

**Steps:**

1. Assign oxidation numbers to all elements in reactants and products

2. Identify which elements undergo change in oxidation number

3. Calculate the total increase and decrease in oxidation number

4. Balance increase and decrease by adjusting coefficients (lowest common multiple)

5. Balance remaining atoms (typically O and H)

**Example:** Balance Zn + HNO₃ → Zn(NO₃)₂ + NO + H₂O

Step 1 – Oxidation numbers:

Zn: 0 → +2 (oxidised, increase of 2 per Zn)

N in HNO₃: +5 → +2 in NO (reduced, decrease of 3 per N)

Step 2 – Find LCM of 2 and 3 = 6

Need 3 Zn atoms (3 × 2 = 6 increase)

Need 2 N atoms reduced (2 × 3 = 6 decrease)

Step 3 – Tentative equation: 3Zn + HNO₃ → 3Zn(NO₃)₂ + 2NO + H₂O

Step 4 – Balance nitrogen: 3Zn(NO₃)₂ requires 6 NO₃⁻, but only 2N reduced to NO, so 4N remain as NO₃⁻

Requires 8 HNO₃ total (2 reduced to NO + 6 in zinc nitrate)

Step 5 – Balanced equation:

**3Zn + 8HNO₃(dil) → 3Zn(NO₃)₂ + 2NO + 4H₂O**

**Verification:**

Zn: 3 on both sides ✓

N: 8 on left (8HNO₃), 6+2 = 8 on right ✓

O: 24 on left, 18+2+4 = 24 on right ✓

H: 8 on left, 8 on right ✓

**Exam example 2:** Balance K₂Cr₂O₇ + HCl → KCl + CrCl₃ + Cl₂ + H₂O

Oxidation numbers:

Cr in K₂Cr₂O₇: +6 → +3 in CrCl₃ (decrease of 3, × 2 for two Cr = 6 total)

Cl in HCl: –1 in some → 0 in Cl₂ (increase of 1, need 6 Cl atoms = 6 HCl)

Balanced equation:

**K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 3Cl₂ + 7H₂O**

Method 2: Half-Reaction Method (Ion-Electron Method)

**Steps:**

1. Write skeletal ionic/molecular equation

2. Separate into oxidation and reduction half-reactions

3. Balance atoms other than O and H in each half-reaction

4. Balance O by adding H₂O; balance H by adding H⁺ (or OH⁻ in basic solution)

5. Balance charge by adding electrons

6. Multiply half-reactions by appropriate factors to equalize electrons

7. Add half-reactions and cancel spectator species

**Acidic medium example:** Permanganate oxidizes ferrous ion

Reduction half-reaction:

MnO₄⁻ → Mn²⁺

Balance O: MnO₄⁻ → Mn²⁺ + 4H₂O

Balance H: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Balance charge: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Oxidation half-reaction:

Fe²⁺ → Fe³⁺

Balance charge: Fe²⁺ → Fe³⁺ + e⁻

Equalize electrons (multiply first by 1, second by 5):

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

5Fe²⁺ → 5Fe³⁺ + 5e⁻

Add and simplify:

**MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O**

**Basic medium example:** Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺

Reduction half-reaction:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

In basic medium, add OH⁻ to neutralize H⁺:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Add 14OH⁻ to both sides: Cr₂O₇²⁻ + 14H₂O + 6e⁻ → 2Cr³⁺ + 7H₂O + 14OH⁻

Simplify: **Cr₂O₇²⁻ + 7H₂O + 6e⁻ → 2Cr³⁺ + 14OH⁻**

Oxidation half-reaction: **Fe²⁺ → Fe³⁺ + e⁻** (multiply by 6)

Final balanced equation:

**Cr₂O₇²⁻ + 6Fe²⁺ + 7H₂O → 2Cr³⁺ + 6Fe³⁺ + 14OH⁻**

**Exam tip:** Half-reaction method is more systematic for complex reactions and works seamlessly in both acidic and basic solutions.

OXIDISING AND REDUCING AGENTS

**Oxidising agent (Oxidant):** Species that causes oxidation by accepting electrons; itself gets reduced

Examples: O₂, Cl₂, KMnO₄, K₂Cr₂O₇, HNO₃ (dilute)

**Reducing agent (Reductant):** Species that causes reduction by donating electrons; itself gets oxidised

Examples: H₂, C, CO, metals (Zn, Fe), H₂S, FeSO₄

**Important principle:** In any redox reaction:

The element showing increase in oxidation number is oxidised (reducing agent)

The element showing decrease in oxidation number is reduced (oxidising agent)

**Problem 7.1 analysis:**

**Reaction (i):** H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)

H in H₂S: –1 → +1 in HCl (oxidised) — H₂S is oxidising agent

Cl in Cl₂: 0 → –1 in HCl (reduced) — Cl₂ is reducing agent

Actually corrected:

S in H₂S: –2 → 0 in S (oxidised) — H₂S is reducing agent

Cl in Cl₂: 0 → –1 in HCl (reduced) — Cl₂ is oxidising agent

**Reaction (ii):** 3Fe₃O₄(s) + 8Al(s) → 9Fe(s) + 4Al₂O₃(s)

Al: 0 → +3 (oxidised) — Al is reducing agent

Fe in Fe₃O₄: average +8/3 → 0 (reduced) — Fe₃O₄ is oxidising agent

**Exam point:** In displacement reactions, the metal lower in the activity series (or having lower tendency to lose electrons) acts as the oxidising agent.

ELECTROCHEMICAL SERIES AND ELECTRODE PROCESSES

**Electrochemical series:** Arrangement of metals (and other species) in decreasing order of their tendency to lose electrons or increasing order of their tendency to gain electrons.

**Standard reduction potentials:** Quantitatively express the tendency of a species to be reduced.

**Order of electron-releasing tendency (from competitive reactions observed):**

**Zn > Cu > Ag**

**General series (partial):**

**K > Ca > Na > Mg > Al > Zn > Fe > Ni > Cu > Hg > Ag > Au**

**Applications:**

Predicts which metal will displace another from its salt solution

More reactive metal (higher in series) always displaces less reactive metal

Example: Zn displaces Cu, but Cu cannot displace Zn

**Electrode processes:**

**Oxidation at anode:** Species loses electrons

**Reduction at cathode:** Species gains electrons

Forms basis of galvanic cells and electrolytic cells

**Exam relevance:** Understanding electrochemical series helps predict spontaneity of redox reactions and electrode reactions in electrochemistry (Class XII topic).

IMPORTANT EXAM POINTS AND TIPS

1. **Distinction between oxidation number and formal charge:** Oxidation number assumes complete electron transfer; formal charge from Lewis structures gives partial electron shift

2. **Redox vs non-redox reactions:**

Redox: Change in oxidation numbers occurs

Non-redox: No change in oxidation numbers (acid-base, displacement without electron transfer)

3. **Peroxides and superoxides:** Special cases where oxygen has non-standard oxidation numbers; frequently tested in boards

4. **Disproportionation identification:** Same element in initial state must produce at least two different oxidation states in products

5. **Balancing tips:**

Oxidation number method: faster for simple reactions

Half-reaction method: more reliable for complex ions, especially in basic/acidic solutions

6. **Stock notation:** Modern IUPAC method for expressing oxidation states; must write correctly (e.g., Cu(II)O not CuO(II))

7. **Calculation practice:** Oxidation number assignments frequently tested; students must practice assigning numbers to polyatomic ions

8. **Real-world applications:** Combustion reactions, corrosion, batteries, chlor-alkali process, steel production all involve redox chemistry

9. **Spontaneity connection:** Reactions that increase total oxidation number changes (larger electron transfer) are typically more spontaneous; related to ΔG = ΔH – TΔS from thermodynamics

10. **Common reducing agents:** H₂, C, CO, metals, H₂S, SO₂, I⁻, Fe²⁺

**Common oxidising agents:** O₂, Cl₂, F₂, KMnO₄, K₂Cr₂O₇, HNO₃, H₂O₂, H₂SO₄ (conc.)