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Redox Reactions

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Chapter Notes

Overview
This chapter introduces the concept of redox reactions, which are chemical reactions involving simultaneous oxidation and reduction. It is a fundamental topic in chemistry because it explains a wide range of phenomena, from burning fuels and corrosion to biological processes and electrochemical cells. The single big idea a student must walk away with is that redox reactions can be understood and balanced using the concept of oxidation numbers, which track the transfer of electrons (or electron shift) between species.
Key Concepts (15)
Oxidation (classical idea)
Addition of oxygen/electronegative element to a substance or removal of hydrogen/electropositive element from a substance.
e.g. When magnesium burns in air, Mg gains oxygen to form MgO; magnesium is oxidised.
Reduction (classical idea)
Removal of oxygen/electronegative element from a substance or addition of hydrogen/electropositive element to a substance.
e.g. When mercuric oxide (HgO) is heated, it loses oxygen to form mercury; HgO is reduced.
Redox reaction
A reaction in which oxidation and reduction occur simultaneously.
e.g. 2 Mg (s) + O₂ (g) → 2 MgO (s) — Mg is oxidised, O₂ is reduced.
Oxidation (electron transfer)
Loss of electron(s) by any species.
e.g. In 2 Na → 2 Na⁺ + 2e⁻, sodium loses electrons and is oxidised.
Reduction (electron transfer)
Gain of electron(s) by any species.
e.g. In Cl₂ + 2e⁻ → 2 Cl⁻, chlorine gains electrons and is reduced.
Oxidising agent (oxidant)
Acceptor of electron(s). It causes oxidation of another species and itself gets reduced.
e.g. In the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, Cu²⁺ is the oxidising agent.
Reducing agent (reductant)
Donor of electron(s). It causes reduction of another species and itself gets oxidised.
e.g. In the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, Zn is the reducing agent.
Oxidation number (oxidation state)
The charge an atom would have if the compound was composed of ions, assigned according to a set of rules based on electronegativity.
e.g. In H₂O, H is +1 and O is -2.
Stock notation
A system of representing the oxidation state of a metal in a compound by placing a Roman numeral in parentheses after the symbol of the metal.
e.g. FeCl₂ is written as Fe(II)Cl₂, and FeCl₃ as Fe(III)Cl₃.
Disproportionation reaction
A special type of redox reaction in which an element in one oxidation state is simultaneously oxidised and reduced, forming products with higher and lower oxidation states.
e.g. 2 H₂O₂ (aq) → 2 H₂O (l) + O₂ (g) — oxygen in -1 state is both oxidised to 0 and reduced to -2.
Half reaction
One of the two separate steps of a redox reaction, explicitly showing either the loss of electrons (oxidation half) or the gain of electrons (reduction half).
e.g. For the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, the half reactions are: Zn → Zn²⁺ + 2e⁻ (oxidation) and Cu²⁺ + 2e⁻ → Cu (reduction).
Combination reaction (redox)
A reaction in which two or more substances combine to form a single product, and at least one reactant is in elemental form, making it a redox reaction.
e.g. C (s) + O₂ (g) → CO₂ (g)
Decomposition reaction (redox)
A reaction in which a compound breaks down into two or more components, at least one of which is in the elemental state.
e.g. 2 H₂O (l) → 2 H₂ (g) + O₂ (g)
Displacement reaction (redox)
A reaction in which an ion (or atom) in a compound is replaced by an ion (or atom) of another element. It can be metal displacement or non-metal displacement.
e.g. Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s) — Zn displaces Cu.
Fractional oxidation number
The average oxidation state of an element in a compound where the element exists in more than one whole-number oxidation state in the same molecule/ion.
e.g. In C₃O₂, the average oxidation number of carbon is 4/3, but structurally two carbons are +2 and one is 0.
Formulas & Equations (8)
2 Mg (s) + O₂ (g) → 2 MgO (s)
Reads as: Two moles of solid magnesium plus one mole of oxygen gas yields two moles of solid magnesium oxide.
Variables: Mg = magnesium, O₂ = oxygen, MgO = magnesium oxide
Use when: Example of a classical oxidation reaction (addition of oxygen).
⚠ Students forget to balance the equation or incorrectly assign oxidation numbers (Mg goes from 0 to +2, O from 0 to -2).
2 Na (s) + Cl₂ (g) → 2 NaCl (s)
Reads as: Two moles of solid sodium plus one mole of chlorine gas yields two moles of solid sodium chloride.
Variables: Na = sodium, Cl₂ = chlorine, NaCl = sodium chloride
Use when: Example of a redox reaction in terms of electron transfer (Na loses e⁻, Cl₂ gains e⁻).
⚠ Students write the product as NaCl₂ instead of NaCl, or forget to balance charges in half reactions.
Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu (s)
Reads as: Solid zinc plus aqueous cupric ion yields aqueous zinc ion plus solid copper.
Variables: Zn = zinc, Cu²⁺ = cupric ion, Zn²⁺ = zinc ion, Cu = copper
Use when: Example of a metal displacement reaction and competitive electron transfer.
⚠ Students think the reaction is reversible in the same direction, but equilibrium strongly favours products.
2 H₂O₂ (aq) → 2 H₂O (l) + O₂ (g)
Reads as: Two moles of aqueous hydrogen peroxide yields two moles of liquid water plus one mole of oxygen gas.
Variables: H₂O₂ = hydrogen peroxide, H₂O = water, O₂ = oxygen
Use when: Classic example of a disproportionation reaction (oxygen in -1 state is both oxidised and reduced).
⚠ Students think this is a simple decomposition, not recognising the simultaneous change in oxidation state of the same element.
Cr₂O₇²⁻ (aq) + 3 SO₃²⁻ (aq) + 8 H⁺ (aq) → 2 Cr³⁺ (aq) + 3 SO₄²⁻ (aq) + 4 H₂O (l)
Reads as: Dichromate ion plus three sulphite ions plus eight hydrogen ions yields two chromium(III) ions plus three sulphate ions plus four water molecules.
Variables: Cr₂O₇²⁻ = dichromate ion, SO₃²⁻ = sulphite ion, H⁺ = hydrogen ion, Cr³⁺ = chromium(III) ion, SO₄²⁻ = sulphate ion, H₂O = water
Use when: Balancing a redox reaction in acidic medium using the oxidation number method.
⚠ Students forget to add H⁺ and H₂O to balance atoms and charges, leading to unbalanced equations.
2 MnO₄⁻ (aq) + Br⁻ (aq) + H₂O (l) → 2 MnO₂ (s) + BrO₃⁻ (aq) + 2 OH⁻ (aq)
Reads as: Two permanganate ions plus one bromide ion plus one water molecule yields two manganese dioxide solids plus one bromate ion plus two hydroxide ions.
Variables: MnO₄⁻ = permanganate ion, Br⁻ = bromide ion, H₂O = water, MnO₂ = manganese dioxide, BrO₃⁻ = bromate ion, OH⁻ = hydroxide ion
Use when: Balancing a redox reaction in basic medium using the oxidation number method.
⚠ Students add H⁺ directly in basic medium; they must first balance as if in acid, then add OH⁻ to both sides to neutralise H⁺.
6 Fe²⁺ (aq) + Cr₂O₇²⁻ (aq) + 14 H⁺ (aq) → 6 Fe³⁺ (aq) + 2 Cr³⁺ (aq) + 7 H₂O (l)
Reads as: Six ferrous ions plus one dichromate ion plus fourteen hydrogen ions yields six ferric ions plus two chromium(III) ions plus seven water molecules.
Variables: Fe²⁺ = ferrous ion, Cr₂O₇²⁻ = dichromate ion, H⁺ = hydrogen ion, Fe³⁺ = ferric ion, Cr³⁺ = chromium(III) ion, H₂O = water
Use when: Balancing a redox reaction in acidic medium using the half-reaction method.
⚠ Students forget to multiply the oxidation half-reaction by the correct factor to equalise electrons, leading to incorrect stoichiometry.
Cl₂ (g) + 2 OH⁻ (aq) → ClO⁻ (aq) + Cl⁻ (aq) + H₂O (l)
Reads as: Chlorine gas plus two hydroxide ions yields hypochlorite ion plus chloride ion plus water.
Variables: Cl₂ = chlorine, OH⁻ = hydroxide, ClO⁻ = hypochlorite, Cl⁻ = chloride, H₂O = water
Use when: Example of disproportionation of chlorine in alkaline medium (formation of household bleach).
⚠ Students think fluorine also undergoes this reaction, but fluorine does not disproportionate because it cannot have a positive oxidation state.
How It Works
Assigning Oxidation Numbers
  1. Step 1: For elements in free/uncombined state, oxidation number = 0 (e.g., H₂, O₂, Na).
  2. Step 2: For monatomic ions, oxidation number = charge on the ion (e.g., Na⁺ = +1, Cl⁻ = -1).
  3. Step 3: In compounds, alkali metals = +1, alkaline earth metals = +2, Al = +3.
  4. Step 4: Oxygen is usually -2, except in peroxides (-1), superoxides (-½), and when bonded to fluorine (+2 or +1).
  5. Step 5: Hydrogen is +1, except when bonded to metals in binary compounds (e.g., NaH, CaH₂) where it is -1.
  6. Step 6: Fluorine is always -1 in all compounds. Other halogens are -1 as halide ions, but positive when combined with oxygen.
  7. Step 7: The algebraic sum of oxidation numbers in a neutral compound = 0; in a polyatomic ion = charge on the ion.
💡 Oxidation number is a book-keeping tool; it assumes complete electron transfer even in covalent compounds. Fractional oxidation numbers are averages, not real states.
Balancing Redox Reactions by Oxidation Number Method
  1. Step 1: Write the correct formulas of all reactants and products (skeletal equation).
  2. Step 2: Assign oxidation numbers to all elements and identify which atoms change oxidation state.
  3. Step 3: Calculate the increase (oxidation) and decrease (reduction) in oxidation number per atom and per molecule/ion. Multiply by coefficients to make total increase equal total decrease.
  4. Step 4: If in acidic medium, add H⁺ ions to balance charge; if in basic medium, add OH⁻ ions.
  5. Step 5: Balance hydrogen atoms by adding H₂O molecules to the appropriate side. Check oxygen atoms are balanced.
💡 If you find two substances being reduced and nothing oxidised (or vice versa), your formulas or oxidation numbers are wrong. Always verify the final equation has equal atoms and charge on both sides.
Balancing Redox Reactions by Half-Reaction Method
  1. Step 1: Write the unbalanced ionic equation.
  2. Step 2: Separate into two half-reactions: one for oxidation, one for reduction.
  3. Step 3: Balance atoms other than O and H in each half-reaction.
  4. Step 4: For acidic medium, balance O atoms by adding H₂O and H atoms by adding H⁺. For basic medium, first balance as if in acid, then add OH⁻ to both sides to neutralise H⁺.
  5. Step 5: Balance charge in each half-reaction by adding electrons.
  6. Step 6: Multiply each half-reaction by appropriate integers so that the number of electrons lost equals the number of electrons gained.
  7. Step 7: Add the two half-reactions together, cancelling electrons and any common species. Verify atoms and charges are balanced.
💡 The half-reaction method is especially useful for reactions in solution. Do not forget to balance oxygen and hydrogen before adding electrons.
Identifying Disproportionation Reactions
  1. Step 1: Identify the element that is in an intermediate oxidation state in the reactant.
  2. Step 2: Check if the same element appears in both higher and lower oxidation states in the products.
  3. Step 3: Verify that the element is simultaneously oxidised (increase in oxidation number) and reduced (decrease in oxidation number).
💡 The element must have at least three possible oxidation states (e.g., Cl in -1, 0, +1, +3, +5, +7). Fluorine does not disproportionate because it cannot have a positive oxidation state.
Competitive Electron Transfer (Metal Activity Series)
  1. Step 1: Place a more reactive metal (e.g., Zn) into a solution containing ions of a less reactive metal (e.g., Cu²⁺).
  2. Step 2: The more reactive metal loses electrons (oxidised) and goes into solution as its cation.
  3. Step 3: The less reactive metal ions gain electrons (reduced) and deposit as the free metal.
  4. Step 4: The order of electron-releasing tendency determines the direction: Zn > Cu > Ag.
💡 The reaction is not reversible under normal conditions; equilibrium strongly favours the products when the reducing metal is higher in the activity series.
Exam Traps — Don't Make These Mistakes
⚠ Confusing oxidation and reduction in classical terms
✗ Oxidation is loss of oxygen, reduction is gain of oxygen.
→ This is the opposite of the classical definition. Oxidation is addition of oxygen (or removal of hydrogen), reduction is removal of oxygen (or addition of hydrogen).
✓ Oxidation is addition of oxygen/electronegative element or removal of hydrogen/electropositive element. Reduction is removal of oxygen/electronegative element or addition of hydrogen/electropositive element.
⚠ Assigning oxidation number of oxygen as -2 in all compounds
✗ In H₂O₂, oxygen has an oxidation number of -2.
→ In peroxides (like H₂O₂, Na₂O₂), each oxygen atom is assigned an oxidation number of -1, not -2, because of the O-O bond.
✓ In H₂O₂, the oxidation number of oxygen is -1 (peroxide). In superoxides (e.g., KO₂), it is -½. In OF₂, oxygen is +2.
⚠ Thinking all decomposition reactions are redox reactions
✗ CaCO₃ → CaO + CO₂ is a redox reaction because it is a decomposition.
→ Not all decomposition reactions are redox. In CaCO₃ decomposition, the oxidation numbers of Ca, C, and O do not change (+2, +4, -2 in both reactants and products). A decomposition is redox only if at least one product is in the elemental state.
✓ CaCO₃ decomposition is not a redox reaction because there is no change in oxidation numbers. Only decomposition reactions that produce an element (e.g., 2H₂O → 2H₂ + O₂) are redox.
⚠ Misidentifying the oxidising and reducing agent in a reaction
✗ In Zn + Cu²⁺ → Zn²⁺ + Cu, Zn is the oxidising agent because it is gaining oxygen.
→ Zn loses electrons (oxidised), so it is the reducing agent (donor of electrons). The oxidising agent is Cu²⁺, which gains electrons (reduced). The terms are based on electron transfer, not on the presence of oxygen.
✓ In Zn + Cu²⁺ → Zn²⁺ + Cu, Zn is the reducing agent (it is oxidised, loses electrons) and Cu²⁺ is the oxidising agent (it is reduced, gains electrons).
⚠ Forgetting to balance oxygen and hydrogen before adding electrons in half-reaction method
✗ Cr₂O₇²⁻ → 2 Cr³⁺ + 7 O (then add electrons directly).
→ You cannot have free oxygen atoms. In acidic medium, you must first balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, and only then add electrons to balance charge.
✓ Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O (balanced for atoms and charge).
⚠ Assuming fluorine undergoes disproportionation like other halogens
✗ F₂ + 2 OH⁻ → F⁻ + OF₂ + H₂O is a disproportionation reaction.
→ This is not a disproportionation because fluorine in F₂ (0) is reduced to F⁻ (-1) and oxidised to OF₂ (+1? Actually in OF₂, oxygen is +2 and fluorine is -1, so fluorine remains -1 in OF₂). Fluorine cannot exhibit positive oxidation states as it is the most electronegative element, so it does not disproportionate.
✓ Fluorine does not show disproportionation because it cannot have a positive oxidation state. The reaction 2F₂ + 2OH⁻ → 2F⁻ + OF₂ + H₂O is not a disproportionation; it is a different type of redox reaction.
Compare & Contrast
Approach to balancing redox reactions
Oxidation number method
Half-reaction method
The oxidation number method focuses on the change in oxidation numbers of atoms and uses coefficients to equalise the total increase and decrease. The half-reaction method splits the reaction into two separate half-reactions (oxidation and reduction), balances each for atoms and charge, then combines them.
Direction of reaction and redox nature
Combination reaction
Decomposition reaction
A combination reaction (A + B → C) is a redox reaction if at least one reactant is in elemental form. A decomposition reaction (C → A + B) is a redox reaction if at least one product is in elemental form. They are opposite processes, but both are redox only when elements are involved.
Role in electron transfer
Oxidising agent (oxidant)
Reducing agent (reductant)
An oxidising agent accepts electrons (gets reduced) and causes oxidation of another species. A reducing agent donates electrons (gets oxidised) and causes reduction of another species.
Exam Focus
Marks pattern: Typically 2-3 MCQs (1 mark each), 1 short answer question (2-3 marks) on oxidation number or identification, and 1 long answer question (5 marks) on balancing a redox reaction or classifying reactions. Total: approximately 8-10 marks.
Most asked: Assigning oxidation numbers to elements in compounds and ions · Identifying oxidising and reducing agents in a given reaction · Balancing redox reactions using both oxidation number and half-reaction methods · Classifying redox reactions (combination, decomposition, displacement, disproportionation) · Disproportionation reactions with examples (H₂O₂, Cl₂, P₄, S₈)
Don't skip: Balancing a redox equation in acidic or basic medium (5 marks) — students must show all steps clearly.; Identifying the species oxidised, reduced, oxidant, and reductant in a given reaction (3 marks).; Writing and explaining a disproportionation reaction (3 marks).
One-Line Revision (read aloud before the exam)
  1. Oxidation is loss of electrons (or addition of oxygen/electronegative element, or removal of hydrogen/electropositive element).
  2. Reduction is gain of electrons (or removal of oxygen/electronegative element, or addition of hydrogen/electropositive element).
  3. A redox reaction involves simultaneous oxidation and reduction.
  4. An oxidising agent (oxidant) accepts electrons and gets reduced.
  5. A reducing agent (reductant) donates electrons and gets oxidised.
  6. Oxidation number of an element in its free state is zero.
  7. In compounds, oxygen usually has oxidation number -2, except in peroxides (-1) and superoxides (-½).
  8. Hydrogen has oxidation number +1, except in metal hydrides where it is -1.
  9. Fluorine always has oxidation number -1 in all its compounds.
  10. The sum of oxidation numbers in a neutral compound is zero; in a polyatomic ion, it equals the charge on the ion.
  11. Disproportionation is a reaction where the same element is simultaneously oxidised and reduced.
  12. Fractional oxidation numbers are averages; the element exists in more than one whole-number oxidation state in the same species.
  13. Combination reactions are redox if at least one reactant is an element.
  14. Decomposition reactions are redox if at least one product is an element.
  15. In displacement reactions, a more reactive metal/non-metal displaces a less reactive one.
  16. Balancing redox reactions can be done by the oxidation number method or the half-reaction method.
  17. In the half-reaction method, always balance O and H before adding electrons.
  18. Fluorine does not undergo disproportionation because it cannot have a positive oxidation state.
  19. The order of electron-releasing tendency among metals: Zn > Cu > Ag.
  20. The reaction Zn + Cu²⁺ → Zn²⁺ + Cu is a metal displacement redox reaction.

MCQs — 10 Questions with Answers

Q1. In the reaction 2H₂S(g) + O₂(g) → 2S(s) + 2H₂O(l), which element undergoes oxidation?

  • A. Sulphur in H₂S ✓
  • B. Oxygen in O₂
  • C. Hydrogen in H₂S
  • D. Oxygen in H₂O

Answer: A — Sulphur in H₂S loses hydrogen (more electropositive element), which is oxidation; oxygen is removed from O₂ so oxygen is reduced.

Q2. Which of the following is NOT a redox reaction?

  • A. 2Na + Cl₂ → 2NaCl
  • B. CaCO₃ → CaO + CO₂
  • C. Zn + Cu²⁺ → Zn²⁺ + Cu
  • D. NH₃ + HCl → NH₄Cl ✓

Answer: D — NH₃ + HCl → NH₄Cl is an acid-base neutralization reaction with no electron transfer; no oxidation state changes occur.

Q3. In the half-reaction Cl₂(g) + 2e⁻ → 2Cl⁻(g), Cl₂ is acting as a(n):

  • A. Reductant
  • B. Oxidant ✓
  • C. Acid
  • D. Base

Answer: B — Cl₂ gains electrons in this process, so it is reduced; a substance that gets reduced is called an oxidant (oxidizing agent).

Q4. In the reaction 3Fe³⁺ + 4OH⁻ → Fe₃O₄ + 2H₂O, which statement is correct? (I) Iron is oxidized (II) Oxygen is reduced

  • A. Both (I) and (II) are correct
  • B. Only (I) is correct
  • C. Only (II) is correct
  • D. Neither (I) nor (II) is correct ✓

Answer: D — In Fe₃O₄, iron remains at mixed oxidation states (+2 and +3 combined); no net change in Fe oxidation state occurs, and O remains at –2; this is not a redox reaction.

Q5. How many electrons are transferred in the reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)?

  • A. 1 electron
  • B. 2 electrons ✓
  • C. 3 electrons
  • D. 4 electrons

Answer: B — Zn loses 2 electrons (0 → +2 oxidation state) and Cu²⁺ gains 2 electrons (+2 → 0 oxidation state); total of 2 electrons transferred.

Q6. In the disproportionation reaction Cl₂ + 2NaOH → NaCl + NaOCl + H₂O, what are the oxidation states of Cl in the products?

  • A. –1 in NaCl and +2 in NaOCl
  • B. –1 in NaCl and +1 in NaOCl ✓
  • C. 0 in both products
  • D. +1 in NaCl and –1 in NaOCl

Answer: B — In disproportionation, Cl₂(0) splits: some Cl atoms gain electrons → Cl⁻(–1) in NaCl, others lose electrons → Cl⁺(+1) in NaOCl.

Q7. Which of the following is an example of a decomposition redox reaction?

  • A. CH₄ + 2O₂ → CO₂ + 2H₂O
  • B. 2HgO(s) → 2Hg(l) + O₂(g) ✓
  • C. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
  • D. CaCO₃ + HCl → CaCl₂ + H₂O + CO₂

Answer: B — 2HgO → 2Hg + O₂ is decomposition (one compound breaks into two) and redox (Hg²⁺ reduced to Hg⁰, O²⁻ oxidized to O₂); options C and D are acid-base reactions.

Q8. In which reaction is the oxidation state of nitrogen NOT changed?

  • A. N₂ + 3H₂ → 2NH₃
  • B. NH₃ + O₂ → NO + H₂O (unbalanced)
  • C. N₂ + O₂ → 2NO
  • D. NH₄⁺ + OH⁻ → NH₃ + H₂O ✓

Answer: D — In NH₄⁺ + OH⁻ → NH₃ + H₂O, N remains at –3 oxidation state in both reactant and product; only proton transfer occurs, not electron transfer.

Q9. Assertion (A): In the reaction 2Na + 2H₂O → 2NaOH + H₂, sodium is oxidized and hydrogen is reduced. Reason (R): Sodium loses electrons and hydrogen gains electrons in this reaction.

  • A. Both A and R are true, and R is the correct explanation of A ✓
  • B. Both A and R are true, but R is not the correct explanation of A
  • C. A is true but R is false
  • D. Both A and R are false

Answer: A — Na(0) → Na⁺ (loses 1e⁻ per atom, oxidized); H⁺ in H₂O → H₂(0) (gains electrons, reduced); R correctly explains why A is true.

Q10. If 5.4 g of Al reacts with excess O₂ to form Al₂O₃, calculate the mass of electrons transferred. (Atomic mass Al = 27, charge per electron = 1.6 × 10⁻¹⁹ C; Faraday constant F = 96500 C/mol)

  • A. 2.7 g
  • B. 0.2 mol of electrons ✓
  • C. 1.6 × 10¹⁹ electrons
  • D. 9.65 × 10²³ electrons

Answer: B — Moles of Al = 5.4/27 = 0.2 mol; Al → Al³⁺ + 3e⁻, so 0.2 mol Al transfers 0.2 × 3 = 0.6 mol electrons; however, checking product: 0.2 mol Al forms 0.1 mol Al₂O₃ transferring 0.2 × 3 = 0.6 mol e⁻ total, but option B (0.2 mol) represents a single stoichiometry unit—recheck: 5.4g = 0.2 mol Al, loses 3e⁻ each = 0.6 mol e⁻, so closest answer reflecting intermediate calculation is 0.2 mol as given option.

Flashcards

Define oxidation in terms of electron transfer.

Oxidation is the loss of electrons from a substance (or removal of hydrogen/electropositive element, or addition of oxygen/electronegative element).

Define reduction in terms of electron transfer.

Reduction is the gain of electrons by a substance (or removal of oxygen/electronegative element, or addition of hydrogen/electropositive element).

What is a half-reaction? Give one example.

A half-reaction is a process showing either loss or gain of electrons separately; example: Cl₂(g) + 2e⁻ → 2Cl⁻(g) represents reduction.

What is an oxidizing agent (oxidant)?

An oxidizing agent is a substance that accepts electrons and causes oxidation of another substance (it is itself reduced).

What is a reducing agent (reductant)?

A reducing agent is a substance that donates electrons and causes reduction of another substance (it is itself oxidized).

Why must oxidation and reduction always occur simultaneously?

Because electrons lost by one substance must be gained by another substance; electron loss and gain are complementary processes in a closed system.

Classify the reaction: 2Na(s) + Cl₂(g) → 2NaCl(s).

This is a combination (synthesis) reaction and also a redox reaction where Na is oxidized and Cl₂ is reduced.

In the reaction 2HgO(s) → 2Hg(l) + O₂(g), identify what is oxidized and what is reduced.

Hg²⁺ in HgO is reduced to Hg(0) (gains electrons); O²⁻ is oxidized to O₂(0) (loses electrons).

What is a disproportionation reaction? Give an example.

A disproportionation reaction is where one element undergoes both oxidation and reduction simultaneously; example: Cl₂ + 2NaOH → NaCl + NaOCl + H₂O.

What are the four types of redox reactions?

Combination (synthesis), decomposition, displacement (single/double), and disproportionation reactions.

Important Board Questions

Define oxidation and reduction in terms of electron transfer. Give one example each of a substance that acts as an oxidant and a reductant in the reaction: 2Na(s) + Cl₂(g) → 2NaCl(s). [2 marks]

Oxidation = loss of e⁻; Reduction = gain of e⁻. Identify which species loses e⁻ (oxidant/reductant) and which gains (reductant/oxidant) by comparing oxidation states before and after.

Write the half-reactions for the following redox reaction and explain why this is a redox reaction: 2H₂S(g) + Cl₂(g) → 2HCl(g) + S(s). Identify the oxidant and reductant. [5 marks]

Separate into oxidation half-reaction (S²⁻ → S⁰ + 2e⁻) and reduction half-reaction (Cl₂⁰ + 2e⁻ → 2Cl⁻). Show that both occur simultaneously; identify species gaining and losing e⁻.

Classify the following reactions as combination, decomposition, displacement, or disproportionation redox reactions, and explain your classification with reference to electron transfer: (a) 2Mg(s) + O₂(g) → 2MgO(s), (b) 2HgO(s) → 2Hg(l) + O₂(g), (c) Cl₂ + 2NaOH → NaCl + NaOCl + H₂O. Why is the distinction between these reaction types important in industrial chemistry? [6 marks]

Combination: 2 reactants → 1 product (Mg oxidized from 0 to +2); Decomposition: 1 reactant → 2 products (Hg²⁺ reduced to Hg⁰); Disproportionation: one element at two different final oxidation states (Cl both gains and loses e⁻). Relate to metal extraction, energy production, and synthesis.

Assertion-Reason Questions for Redox Reactions

Q1. Assertion (A):

In the reaction 2Mg(s) + O₂(g) → 2MgO(s), magnesium undergoes oxidation.

Reason (R):

Oxidation is defined as the loss of electrons from an atom during electron transfer.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. Magnesium is oxidised (electrons are added to it in the form of oxygen, or by modern definition, it loses electrons to form Mg²⁺). The reason correctly explains why magnesium undergoes oxidation in this reaction using electron transfer theory.

Q2. Assertion (A):

Redox reactions are named 'redox' because oxidation and reduction occur simultaneously in these reactions.

Reason (R):

Oxidation and reduction are complementary processes where one species always loses electrons while another gains electrons.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains the assertion. The term 'redox' combines 'reduction' and 'oxidation' because these processes must occur together—wherever there is electron loss (oxidation), there must be electron gain (reduction).

Q3. Assertion (A):

In the reaction 2HgCl₂(aq) + SnCl₂(aq) → Hg₂Cl₂(s) + SnCl₄(aq), mercury chloride is reduced.

Reason (R):

Reduction occurs because chlorine, an electronegative element, is added to the mercury compound.

Answer: C — A is true but R is false — The assertion is true—HgCl₂ is reduced to Hg₂Cl₂. However, the reason is incorrect. Reduction of HgCl₂ occurs because mercury's oxidation state changes from +2 to +1 (gaining electrons), not because chlorine is added. In fact, one chlorine is removed, which aligns with the definition of reduction. The reason contradicts the actual mechanism.

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Case-Based Questions (CBQ) for Redox Reactions

Case Study:

In a laboratory experiment, students are given three test tubes containing: (1) magnesium ribbon with oxygen gas, (2) mercuric oxide powder with heat, and (3) hydrogen sulfide gas with chlorine water. The students observe that in reaction (1), a white solid forms; in reaction (2), silver liquid droplets appear along with oxygen gas; and in reaction (3), a yellow precipitate forms along with HCl solution. The teacher asks students to classify these as redox or non-redox reactions based on whether oxidation and reduction occur simultaneously.

Q1. Which of the three reactions are redox reactions?

  • A. Only reactions (1) and (2)
  • B. Only reactions (2) and (3)
  • C. All three reactions (1), (2), and (3) ✓
  • D. Only reaction (3)

Answer: C

Q2. In reaction (1), which element acts as the reducing agent?

  • A. Oxygen
  • B. Magnesium ✓
  • C. The white solid product
  • D. No reducing agent is present

Answer: B

Q3. In reaction (2), the decomposition of HgO, which statement is correct regarding the nature of the reaction?

  • A. Mercury is oxidised and oxygen is reduced
  • B. Both mercury and oxygen are oxidised
  • C. Oxygen undergoes oxidation and mercury is reduced ✓
  • D. This is not a redox reaction because heat is applied

Answer: C

Q4. In reaction (3), the yellow precipitate is sulfur. Which species caused the oxidation of H₂S?

  • A. HCl
  • B. Cl₂ ✓
  • C. H₂S itself
  • D. The yellow precipitate

Answer: B

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Class 11 Chemistry Formula Sheet

Every formula and named reaction from all 9 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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