Vidaara.orgClass 11 · Chemistry
CodeVID-C11-08-T2-01
Assignment — Balancing Redox Reactions
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
5 × 1 = 5 marks
1.
In the half-reaction method, oxygen is balanced by adding:
- A.O₂
- B.H₂O
- C.OH⁻
- D.H⁺
2.
Hydrogen in an acidic half-reaction is balanced by adding:
- A.H₂O
- B.OH⁻
- C.H⁺
- D.H₂
3.
Cr2O72- → 2Cr3+ involves a gain of how many electrons?
- A.3
- B.4
- C.6
- D.7
4.
A correctly balanced equation must conserve:
- A.mass only
- B.charge only
- C.mass and charge
- D.neither
5.
In 2MnO4- + 5C2O42- + 16H+ → products, the number of CO2 molecules formed is:
- A.5
- B.8
- C.10
- D.16
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
List the steps of the ion–electron method in order.
7.
Write the balanced reduction half-reaction of Cr2O72- in acidic medium.
8.
How is an acidic balanced equation converted to basic medium?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Balance MnO4- + Fe2+ → Mn2+ + Fe3+ in acidic medium by the half-reaction method.
10.
Balance Cl2 → Cl- + ClO3- in basic medium.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Balance Cr2O72- + C2O42- → Cr3+ + CO2 in acidic medium, showing both half-reactions.
Answer Key
Section A — Multiple Choice Questions
- (B) H₂O
- (C) H⁺
- (C) 6
- (C) mass and charge
- (C) 10
Section B — Short Answer (2 marks)
- Split into halves; balance atoms other than O and H; balance O with H₂O; balance H with H⁺; balance charge with electrons; equalise electrons and add (for base, add OH⁻ per H⁺).
- Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
- Add one OH⁻ to both sides for every H⁺; combine H⁺ + OH⁻ into H₂O and cancel any duplicated water.
Section C — Short Answer (3 marks)
- MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
- 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O.
Section D — Long Answer (5 marks)
- Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Oxidation: C₂O₄²⁻ → 2CO₂ + 2e⁻ (×3 gives 6e⁻). Adding: Cr₂O₇²⁻ + 14H⁺ + 3C₂O₄²⁻ → 2Cr³⁺ + 6CO₂ + 7H₂O.
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