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Vidaara.orgClass 11 · Chemistry
CodeVID-C11-08-CH-01
Chapter Assignment — Redox Reactions
Chapter: Redox Reactions
Topic: All Topics
Maximum Marks: 40
Time: 90 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • All questions are compulsory.
  • Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
  • Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions 6 × 1 = 6 marks
1.
Reduction is the:
  • A.loss of electrons
  • B.gain of electrons
  • C.gain of oxygen
  • D.loss of hydrogen
2.
The oxidation number of Cr in K2CrO4 is:
  • A.+3
  • B.+4
  • C.+6
  • D.+7
3.
An oxidising agent always:
  • A.loses electrons
  • B.gains electrons
  • C.is oxidised
  • D.increases its oxidation number
4.
Decomposition of KClO3 to KCl and O2 is a redox reaction of type:
  • A.combination
  • B.decomposition
  • C.displacement
  • D.comproportionation
5.
The n-factor of KMnO4 in acidic medium is:
  • A.1
  • B.3
  • C.5
  • D.7
6.
The indicator used in a KMnO4 titration is:
  • A.starch
  • B.phenolphthalein
  • C.none — it is self-indicating
  • D.methyl orange
Section B — Short Answer (2 marks) 4 × 2 = 8 marks
7.
Define oxidation and reduction in terms of oxidation number.
8.
Find the oxidation number of P in H3PO4.
9.
Give one example each of a displacement and a disproportionation reaction.
10.
Why must dilute H2SO4 (not HCl) be used in a permanganate titration?
Section C — Short Answer (3 marks) 2 × 3 = 6 marks
11.
Balance MnO4- + C2O42- → Mn2+ + CO2 in acidic medium.
12.
20 mL of 0.1 M oxalic acid solution required V mL of 0.02 M KMnO4 in acid. Find V.
Section D — Long Answer (5 marks) 2 × 5 = 10 marks
13.
Balance the reaction Cr2O72- + Fe2+ → Cr3+ + Fe3+ in acidic medium by the half-reaction method, showing both halves.
14.
Explain the electrochemical series and use Zn²⁺/Zn (-0.76 V), Cu²⁺/Cu (+0.34 V) and Ag⁺/Ag (+0.80 V) to rank the three metals as reducing agents and predict whether Cu displaces Ag from AgNO3.

Answer Key

Section A — Multiple Choice Questions
  1. (B) gain of electrons
  2. (C) +6
  3. (B) gains electrons
  4. (B) decomposition
  5. (C) 5
  6. (C) none — it is self-indicating
Section B — Short Answer (2 marks)
  1. Oxidation is an increase in oxidation number; reduction is a decrease in oxidation number.
  2. 3(+1) + x + 4(-2) = 0 gives x = +5.
  3. Displacement: Zn + CuSO₄ → ZnSO₄ + Cu. Disproportionation: 2H₂O₂ → 2H₂O + O₂.
  4. MnO₄⁻ would oxidise Cl⁻ from HCl to Cl₂, consuming extra permanganate and giving wrong results; H₂SO₄ is not oxidised.
Section C — Short Answer (3 marks)
  1. 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.
  2. Moles oxalate = 0.1 × 20/1000 = 2×10⁻³; ratio MnO₄⁻ : C₂O₄²⁻ = 2 : 5, so moles KMnO₄ = (2/5)×2×10⁻³ = 8×10⁻⁴; V = 8×10⁻⁴ / 0.02 = 0.04 L = 40 mL.
Section D — Long Answer (5 marks)
  1. Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (×6 gives 6e⁻). Adding and cancelling electrons: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.
  2. The electrochemical series ranks redox couples by standard reduction potential (vs SHE = 0 V); more negative means a stronger reducing agent. Order of reducing strength: Zn > Cu > Ag (Zn most negative). Since Cu²⁺/Cu (+0.34 V) is below (less positive than) Ag⁺/Ag (+0.80 V), Cu is the stronger reductant and displaces Ag: Cu + 2Ag⁺ → Cu²⁺ + 2Ag, with EMF = +0.80 - (+0.34) = +0.46 V, confirming spontaneity.
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