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Chapter 7: Redox Reactions — Online MCQ Test

CHEMISTRY · CLASS 11 FIRST PUC · Karnataka State Board
Practice Chapter 7: Redox Reactions with a free chapter-wise online MCQ test. This chapter covers: Oxidation Reduction Oxidation number Reducing agent Oxidising agent Electron transfer Redox equation Balancing reactions Disproportionation. AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter 7: Redox Reactions — Important Questions & Answers

Which of the following best defines oxidation in terms of electron transfer?
  • A. Gain of electrons
  • B. Loss of electrons
  • C. Gain of protons
  • D. Loss of neutrons
Answer: B. Loss of electrons
Oxidation is defined as the loss of electrons. In redox reactions, the species that loses electrons is oxidized.
Which species acts as a reducing agent in a redox reaction?
  • A. Species that gains electrons
  • B. Species that loses electrons
  • C. Species that is not changed
  • D. Species that always has oxidation number zero
Answer: B. Species that loses electrons
A reducing agent donates electrons to another species and itself gets oxidized. Hence, it loses electrons.
In the reaction CuO + H2 → Cu + H2O, which species is oxidised?
  • A. CuO
  • B. Cu
  • C. H2
  • D. H2O
Answer: C. H2
Hydrogen changes from oxidation number 0 in H2 to +1 in H2O, so it is oxidized. CuO is reduced to Cu.
Which of the following represents the correct oxidation number change in the reaction 2FeCl2 + Cl2 → 2FeCl3?
  • A. Fe: +2 to +3; Cl: 0 to -1
  • B. Fe: +3 to +2; Cl: -1 to 0
  • C. Fe: 0 to +2; Cl: 0 to +1
  • D. Fe: +2 to 0; Cl: -1 to 0
Answer: A. Fe: +2 to +3; Cl: 0 to -1
Iron is oxidized from +2 to +3, while chlorine is reduced from 0 in Cl2 to -1 in FeCl3. This is a redox reaction.
Which one of the following is the correct balanced equation for the reaction of iron(II) with dichromate ion in acidic medium?
  • A. Cr2O7^2- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
  • B. Cr2O7^2- + 3Fe2+ + 14H+ → 2Cr3+ + 3Fe3+ + 7H2O
  • C. Cr2O7^2- + 6Fe3+ + 14H+ → 2Cr3+ + 6Fe2+ + 7H2O
  • D. 2Cr2O7^2- + 6Fe2+ + 14H+ → 4Cr3+ + 6Fe3+ + 7H2O
Answer: A. Cr2O7^2- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
Dichromate ion is reduced to Cr3+, and each Fe2+ is oxidized to Fe3+. The correct stoichiometric balance in acidic medium is Cr2O7^2- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O.