Redox Reactions — Class 11 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Redox Reactions" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Redox Reactions" — 8 important questions with detailed answers for CBSE board exam p…
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NCERT Solutions for Class 11 Chemistry — Redox Reactions. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Redox Reactions — All 8 Questions Solved
Q1. Define oxidation and reduction. Are they always simultaneous?
Oxidation: Loss of electrons or increase in oxidation number.
Reduction: Gain of electrons or decrease in oxidation number.
Redox reaction: A reaction in which both oxidation and reduction occur simultaneously.
They are ALWAYS simultaneous because:
- Electrons lost by one species must be gained by another
- Total electrons lost = Total electrons gained (law of conservation of electrons)
- Oxidizing agent: Species that gets reduced (gains electrons)
- Reducing agent: Species that gets oxidized (loses electrons)
Example: Zn + Cu2+ -> Zn2+ + Cu
- Zn loses 2 electrons (oxidized), acts as reducing agent
- Cu2+ gains 2 electrons (reduced), acts as oxidizing agent
- Electrons lost (2) = Electrons gained (2)
Q2. Determine oxidation numbers of all atoms in K2Cr2O7 and KMnO4
Oxidation number rules:
- Alkali metals (Group 1) = +1
- Alkaline earth metals (Group 2) = +2
- Oxygen in compounds = -2 (except peroxides)
- Sum = charge on compound (for neutral: sum = 0)
In K2Cr2O7:
K: +1 (Group 1)
Cr: ?
O: -2
2(+1) + 2(Cr) + 7(-2) = 0
2 + 2Cr - 14 = 0
2Cr = 12
Cr = +6
In KMnO4:
K: +1
Mn: ?
O: -2
(+1) + Mn + 4(-2) = 0
1 + Mn - 8 = 0
Mn = +7
Summary:
K2Cr2O7: K = +1, Cr = +6, O = -2
KMnO4: K = +1, Mn = +7, O = -2
Q3. Balance the redox equation: MnO4- + Fe2+ -> Mn2+ + Fe3+ in acidic medium
Step 1: Identify oxidation states
Mn in MnO4-: +7 (reduction, gains 5e-)
Fe2+: +2 (oxidation, loses 1e-)
Step 2: Write half-reactions
Reduction: MnO4- -> Mn2+
Oxidation: Fe2+ -> Fe3+
Step 3: Balance half-reactions (acidic medium)
Reduction: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
Oxidation: Fe2+ -> Fe3+ + 1e-
Step 4: Equalize electrons
Multiply reduction by 1, oxidation by 5
Reduction: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
Oxidation: 5Fe2+ -> 5Fe3+ + 5e-
Step 5: Add and cancel electrons
MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+
Balanced equation:
MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O
Q4. Balance redox equation: Cr2O7^2- + I- -> Cr^3+ + I2 in acidic medium
Step 1: Identify oxidation states
Cr in Cr2O7^2-: +6 (reduction, gains 3e- per Cr, total 6e-)
I-: -1 (oxidation, loses 1e-)
Step 2: Write half-reactions
Reduction: Cr2O7^2- -> Cr^3+
Oxidation: I- -> I2
Step 3: Balance half-reactions (acidic medium)
Reduction: Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O
Oxidation: 2I- -> I2 + 2e-
Step 4: Equalize electrons (LCM of 6 and 2 = 6)
Multiply reduction by 1, oxidation by 3
Reduction: Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O
Oxidation: 6I- -> 3I2 + 6e-
Step 5: Add and cancel
Cr2O7^2- + 14H+ + 6I- -> 2Cr^3+ + 7H2O + 3I2
Balanced equation:
Cr2O7^2- + 6I- + 14H+ -> 2Cr^3+ + 3I2 + 7H2O
Q5. What is electrochemistry? Explain galvanic and electrolytic cells.
Electrochemistry: Study of chemical reactions involving electron transfer and their relationship with electrical energy.
Galvanic (Voltaic) Cell:
- Spontaneous redox reaction generates electrical energy
- Oxidation at anode (negative electrode), reduction at cathode (positive electrode)
- Electrons flow from anode to cathode externally through wire
- Salt bridge allows ion flow to complete circuit
- Example: Zn-Cu cell (Daniel cell)
Anode: Zn -> Zn2+ + 2e- (oxidation)
Cathode: Cu2+ + 2e- -> Cu (reduction)
Electrolytic Cell:
- Non-spontaneous redox reaction driven by external electrical energy
- Oxidation at anode (positive electrode), reduction at cathode (negative electrode)
- Electrons forced from external power source
- Example: Electrolysis of NaCl solution
Anode: 2Cl- -> Cl2 + 2e- (oxidation)
Cathode: 2H2O + 2e- -> H2 + 2OH- (reduction)
Key Difference:
- Galvanic: Spontaneous, generates electricity, anode negative
- Electrolytic: Non-spontaneous, consumes electricity, anode positive
Q6. Calculate number of electrons transferred when 0.5 mol of MnO4- is completely reduced to Mn2+
From the half-reaction for permanganate reduction:
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
This shows: 1 mol MnO4- gains 5 mol electrons
For 0.5 mol MnO4-:
Electrons transferred = 0.5 mol × 5 e-/mol = 2.5 mol electrons
In terms of Faraday's constant (F = 96500 C/mol e-):
Charge = 2.5 mol e- × 96500 C/mol = 241,250 C ≈ 241.25 kC
Or in terms of number of electrons:
2.5 mol × 6.022 × 10^23 = 1.5 × 10^24 electrons
Q7. Balance redox equation: H2S + HNO3 -> S + NO + H2O in acidic medium
Step 1: Identify oxidation states
S in H2S: -2 (oxidation to 0)
N in HNO3: +5 (reduction to +2 in NO)
Step 2: Write half-reactions
Oxidation: H2S -> S
Reduction: HNO3 -> NO
Step 3: Balance (acidic medium)
Oxidation: H2S -> S + 2H+ + 2e-
Reduction: HNO3 + 3H+ + 3e- -> NO + 2H2O
Step 4: Equalize electrons (LCM of 2 and 3 = 6)
Oxidation: 3H2S -> 3S + 6H+ + 6e-
Reduction: 2HNO3 + 6H+ + 6e- -> 2NO + 4H2O
Step 5: Add and simplify
3H2S + 2HNO3 + 6H+ -> 3S + 2NO + 4H2O + 6H+
3H2S + 2HNO3 -> 3S + 2NO + 4H2O
Balanced equation:
3H2S + 2HNO3 -> 3S + 2NO + 4H2O
Q8. What is a redox couple? Give examples and explain its significance.
Redox Couple: Oxidized and reduced forms of a substance differing by electrons.
General notation: Oxidized form / Reduced form
Examples:
1. Fe^3+ / Fe^2+ (ferric-ferrous couple)
Fe^3+ + e- ⇌ Fe^2+
2. MnO4- / Mn^2+ (permanganate couple in acidic medium)
MnO4- + 8H+ + 5e- ⇌ Mn^2+ + 4H2O
3. Cl2 / Cl- (chlorine couple)
Cl2 + 2e- ⇌ 2Cl-
4. O2 / H2O (oxygen couple in acidic medium)
O2 + 4H+ + 4e- ⇌ 2H2O
Significance:
- Shows relative strength of oxidizing and reducing agents
- Higher reduction potential = stronger oxidizing agent
- Helps predict direction of redox reaction
- Used in electrochemistry and galvanic cells
- Standard reduction potentials (E°) allow calculation of cell EMF
In electrochemistry, comparisons of E° values determine which species oxidizes and which reduces.
Redox Reactions — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What happens during an oxidation process according to the electronic concept?
- A. Gain of electrons
- B. Loss of electrons
- C. Decrease in oxidation number
- D. Gain of hydrogen
Answer: B. Loss of electrons — Oxidation is defined as the loss of electrons by a species, leading to an increase in its oxidation state.
MCQ 2. Which of the following best describes reduction?
- A. Loss of electrons
- B. Increase in oxidation number
- C. Gain of electrons
- D. Removal of oxygen
Answer: C. Gain of electrons — Reduction is the process where a chemical species gains electrons, resulting in a decrease in its oxidation number.
MCQ 3. What is the oxidation number of sulfur in H₂SO₄?
- A. +2
- B. +4
- C. +6
- D. -2
Answer: C. +6 — In H₂SO₄, hydrogen has an oxidation number of +1 (2x+1 = +2) and oxygen has -2 (4x-2 = -8). For a neutral molecule, the sum must be zero: 2 + S + (-8) = 0, so S = +6.
MCQ 4. In the reaction 2Mg(s) + O₂(g) → 2MgO(s), which species is the oxidizing agent?
- A. Mg
- B. O₂
- C. MgO
- D. Both Mg and O₂
Answer: B. O₂ — An oxidizing agent is a substance that causes oxidation and itself gets reduced. Oxygen gains electrons (changes from 0 to -2) and oxidizes Magnesium (from 0 to +2), thus O₂ is the oxidizing agent.
MCQ 5. In the reaction Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s), which species is the reducing agent?
- A. Zn
- B. CuSO₄
- C. Cu
- D. ZnSO₄
Answer: A. Zn — A reducing agent is a substance that causes reduction and itself gets oxidized. Zinc loses electrons (changes from 0 to +2) and reduces Cu²⁺ (from +2 to 0), thus Zn is the reducing agent.
MCQ 6. When balancing a redox reaction by the ion-electron method in an acidic medium, what is typically added to balance oxygen atoms?
- A. OH⁻ ions
- B. H₂O molecules
- C. H⁺ ions
- D. Electrons
Answer: B. H₂O molecules — In acidic medium, water molecules (H₂O) are added to the side deficient in oxygen atoms, and then H⁺ ions are added to the opposite side to balance hydrogen atoms.
MCQ 7. When balancing a redox reaction by the ion-electron method in a basic medium, after balancing oxygen atoms with H₂O, how are hydrogen atoms balanced?
- A. Add H⁺ ions to the side deficient in hydrogen.
- B. Add OH⁻ ions to the side deficient in hydrogen, and H₂O molecules to the opposite side.
- C. Add H₂O molecules to the side deficient in hydrogen, and OH⁻ ions to the opposite side.
- D. Add H₂O molecules to the side with excess hydrogen, and OH⁻ ions to the same side.
Answer: C. Add H₂O molecules to the side deficient in hydrogen, and OH⁻ ions to the opposite side. — In a basic medium, after balancing oxygen atoms with H₂O, hydrogen atoms are balanced by adding H₂O molecules to the side deficient in hydrogen and an equal number of OH⁻ ions to the opposite side.
MCQ 8. What is the oxidation number of Mn in KMnO₄?
- A. +2
- B. +4
- C. +6
- D. +7
Answer: D. +7 — Potassium (K) has an oxidation number of +1, and oxygen (O) has -2. For KMnO₄, the sum must be zero: +1 + Mn + (4 x -2) = 0, so +1 + Mn - 8 = 0, which means Mn = +7.
MCQ 9. What is the oxidation number of N in NO₃⁻?
- A. +3
- B. +5
- C. +6
- D. -3
Answer: B. +5 — For the nitrate ion NO₃⁻, the sum of oxidation numbers must equal the charge of the ion, -1. Oxygen has an oxidation number of -2. So, N + (3 x -2) = -1, which gives N - 6 = -1, thus N = +5.
MCQ 10. Which of the following is NOT a redox reaction?
- A. H₂ + Cl₂ → 2HCl
- B. NaCl + AgNO₃ → AgCl + NaNO₃
- C. Zn + H₂SO₄ → ZnSO₄ + H₂
- D. 2Na + S → Na₂S
Answer: B. NaCl + AgNO₃ → AgCl + NaNO₃ — In the reaction NaCl + AgNO₃ → AgCl + NaNO₃, it's a double displacement (precipitation) reaction. The oxidation states of all elements remain unchanged, so it is not a redox reaction.
MCQ 11. Which of the following elements can exhibit disproportionation reactions?
- A. Fluorine
- B. Oxygen
- C. Chlorine
- D. Sodium
Answer: C. Chlorine — Disproportionation reactions occur when an element in a particular oxidation state is simultaneously oxidized and reduced. Chlorine can exist in various oxidation states (e.g., -1, 0, +1, +3, +5, +7) and thus can undergo disproportionation.
MCQ 12. What is the oxidation number of oxygen in H₂O₂?
- A. 0
- B. -1
- C. -2
- D. +1
Answer: B. -1 — H₂O₂ is hydrogen peroxide. In peroxides, oxygen has an oxidation number of -1, while hydrogen has +1. (2 x +1) + (2 x O) = 0, so O = -1.
MCQ 13. In the reaction 2FeCl₃ + SnCl₂ → 2FeCl₂ + SnCl₄, what is the role of SnCl₂?
- A. Oxidizing agent
- B. Reducing agent
- C. Catalyst
- D. Spectator ion
Answer: B. Reducing agent — SnCl₂ (Sn is +2) is oxidized to SnCl₄ (Sn is +4), meaning it loses electrons. Therefore, SnCl₂ acts as a reducing agent, causing Fe³⁺ (in FeCl₃) to be reduced to Fe²⁺ (in FeCl₂).
MCQ 14. What is the oxidation number of chromium in K₂Cr₂O₇?
- A. +2
- B. +3
- C. +6
- D. +7
Answer: C. +6 — In K₂Cr₂O₇, potassium (K) is +1 (2x+1 = +2), and oxygen (O) is -2 (7x-2 = -14). For a neutral molecule: 2 + 2Cr + (-14) = 0, so 2Cr - 12 = 0, which means 2Cr = 12, so Cr = +6.
MCQ 15. How many electrons are involved in the reduction of one mole of MnO₄⁻ to Mn²⁺ in an acidic medium?
- A. 2
- B. 3
- C. 5
- D. 7
Answer: C. 5 — The oxidation state of Mn in MnO₄⁻ is +7 (Mn - 8 = -1, so Mn = +7). The oxidation state of Mn²⁺ is +2. The change in oxidation number is from +7 to +2, which is a decrease of 5. Therefore, 5 electrons are gained.
MCQ 16. A redox reaction is characterized by:
- A. Formation of a precipitate
- B. Exchange of ions without change in oxidation states
- C. Transfer of electrons
- D. Evolution of gas only
Answer: C. Transfer of electrons — A redox reaction involves the transfer of electrons between reactants, leading to changes in their oxidation states.
MCQ 17. What is the oxidation number of carbon in methane (CH₄)?
- A. +4
- B. +2
- C. 0
- D. -4
Answer: D. -4 — In CH₄, hydrogen has an oxidation number of +1. Since it's a neutral molecule, C + (4 x +1) = 0, so C + 4 = 0, which means C = -4.
MCQ 18. Which of the following reactions is an example of a disproportionation reaction?
- A. 2Na + Cl₂ → 2NaCl
- B. HCl + NaOH → NaCl + H₂O
- C. 2H₂O₂ → 2H₂O + O₂
- D. Ag⁺ + Cl⁻ → AgCl
Answer: C. 2H₂O₂ → 2H₂O + O₂ — In the decomposition of hydrogen peroxide (2H₂O₂ → 2H₂O + O₂), oxygen in H₂O₂ (oxidation state -1) is simultaneously reduced to O in H₂O (oxidation state -2) and oxidized to O in O₂ (oxidation state 0).
MCQ 19. Rusting of iron is an example of:
- A. Reduction
- B. Oxidation
- C. Neutralization
- D. Decomposition
Answer: B. Oxidation — Rusting is the corrosion of iron, which involves the oxidation of iron (Fe) to iron oxides in the presence of oxygen and water.
MCQ 20. According to the classical concept, reduction involves:
- A. Removal of hydrogen
- B. Addition of oxygen
- C. Addition of an electropositive element
- D. Loss of electrons
Answer: C. Addition of an electropositive element — In the classical concept, reduction is defined as the addition of hydrogen or an electropositive element, or the removal of oxygen or an electronegative element.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.