Redox Reactions — Class 11 MCQs with Answers

Redox Reactions — important questions & MCQs with answers (Class 11 Chemistry)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   Oxidation involves:

   A.Loss of electrons✓ Correct

   B.Gain of electrons

   C.No change

   D.Loss of mass

   Answer: A. Loss of electrons

   Explanation: OIL RIG: Oxidation Is Loss of electrons, Reduction Is Gain. Losing electrons raises oxidation number, e.g. Zn → Zn²⁺ + 2e⁻. Confusing this with mass loss is the usual trap in the mass-loss option.

2. Q2Easy

   Oxidation number of an element in its free (elemental) state is:

   A.+1

   B.0✓ Correct

   C.−1

   D.Equal to valency

   Answer: B. 0

   Explanation: Any element in its free, uncombined state has oxidation number 0, since no electrons transfer between identical atoms — true for Na, O₂, Fe or P₄ alike. Assigning a nonzero value here, based on the element's usual valency, is the trap.

3. Q3Easy

   Reduction is:

   A.Loss of electrons

   B.Gain of electrons✓ Correct

   C.Same as combustion

   D.None

   Answer: B. Gain of electrons

   Explanation: Reduction Is Gain (the RIG in OIL RIG): a species accepts electrons and its oxidation number falls, e.g. Cu²⁺ + 2e⁻ → Cu. Mixing this up with oxidation (electron loss) is the frequent slip.

4. Q4Easy

   Oxidation number of hydrogen in most compounds is:

   A.+1✓ Correct

   B.−1

   C.0

   D.+2

   Answer: A. +1

   Explanation: Hydrogen is assigned +1 in most compounds (H₂O, HCl, NH₃) because it bonds to more electronegative atoms; it flips to −1 only in metal hydrides like NaH. Forgetting that exception is what causes errors.

5. Q5Easy

   A reducing agent:

   A.Always a metal oxide

   B.Gets reduced; accepts electrons

   C.Neither oxidised nor reduced

   D.Gets oxidised; donates electrons✓ Correct

   Answer: D. Gets oxidised; donates electrons

   Explanation: A reducing agent donates electrons to another species and is itself oxidised — its own oxidation number rises, e.g. Zn → Zn²⁺+2e⁻ reduces Cu²⁺. Saying it 'gets reduced' swaps the agent with the species it acts on.

6. Q6Easy

   Oxidation number of oxygen in most compounds is:

   A.−1

   B.+2

   C.−2✓ Correct

   D.0

   Answer: C. −2

   Explanation: Oxygen is −2 in most compounds (H₂O, CO₂), rising to −1 in peroxides (H₂O₂), −1/2ish in superoxides, and +2 only in OF₂ where fluorine is more electronegative. Defaulting to −2 everywhere misses those exceptions.

7. Q7Medium

   Oxidising agent does:

   A.Gets reduced (gains electrons)✓ Correct

   B.Gets oxidised

   C.Stays the same

   D.Disappears

   Answer: A. Gets reduced (gains electrons)

   Explanation: An oxidising agent causes another species to be oxidised by accepting its electrons, so the agent itself gains electrons and is reduced — its own oxidation number falls. Saying it 'gets oxidised' swaps the agent with the species it acts on.

8. Q8Medium

   Oxidation number of S in H₂SO₄:

   A.0

   B.−2

   C.+4

   D.+6✓ Correct

   Answer: D. +6

   Explanation: H is +1 and O is −2, so 2(+1)+S+4(−2)=0 gives S=+6, balancing the neutral formula. This is sulfur's maximum oxidation state, far above the +4 it shows in SO₂.

9. Q9Medium

   In 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O, manganate undergoes:

   A.No redox

   B.Only oxidation

   C.Only reduction

   D.Disproportionation✓ Correct

   Answer: D. Disproportionation

   Explanation: In 3MnO₄²⁻+4H⁺→2MnO₄⁻+MnO₂+2H₂O, manganate's Mn(+6) splits: two-thirds rise to +7 in MnO₄⁻ (oxidised) while one-third falls to +4 in MnO₂ (reduced) — the same element going both ways defines disproportionation.

10. Q10Medium

    Average oxidation state of S in Na₂S₂O₃ (thiosulphate):

    A.+6

    B.+4

    C.+2✓ Correct

    D.0

    Answer: C. +2

    Explanation: Na is +1 and O is −2, so 2(+1)+2S+3(−2)=0 gives 2S=+4, so the average S=+2. The real structure has one S at −1 and one at +5, but this question asks only for the mean via the same sum-to-zero rule.

11. Q11Medium

    Which species cannot act as a reducing agent?

    A.F⁻✓ Correct

    B.I⁻

    C.Br⁻

    D.Cl⁻

    Answer: A. F⁻

    Explanation: F⁻ is the conjugate base of the most electronegative element, so the F₂/F⁻ couple has the highest reduction potential of the halogens — meaning F⁻ has essentially no tendency to lose an electron back to F₂, so it can't act as a reducing agent.

12. Q12Hard

    Which couple has the highest standard reduction potential?

    A.Cl₂/Cl⁻

    B.F₂/F⁻✓ Correct

    C.Br₂/Br⁻

    D.I₂/I⁻

    Answer: B. F₂/F⁻

    Explanation: Standard reduction potentials for halogens run F₂/F⁻≈+2.87V > Cl₂/Cl⁻≈+1.36V > Br₂/Br⁻≈+1.09V > I₂/I⁻≈+0.54V, tracking electronegativity down the group. Assuming Cl₂ is strongest, since it's the familiar bleach oxidiser, misses that F₂ tops the series.

13. Q13Hard

    Disproportionation reaction:

    A.No redox

    B.Two different elements oxidised

    C.Same element is both oxidised and reduced✓ Correct

    D.Pure precipitation

    Answer: C. Same element is both oxidised and reduced

    Explanation: Disproportionation is when one species of a single element is simultaneously oxidised and reduced: e.g. in 2H₂O₂→2H₂O+O₂, oxygen goes from −1 to −2 (reduced) in H₂O and to 0 (oxidised) in O₂ — one starting state splitting two ways.

14. Q14Hard

    H₂O₂ in the reaction PbS + 4H₂O₂ → PbSO₄ + 4H₂O acts as:

    A.Oxidising agent✓ Correct

    B.Reducing agent

    C.Both

    D.Neither

    Answer: A. Oxidising agent

    Explanation: S rises from −2 in PbS to +6 in PbSO₄ (oxidised, loses 8 electrons) while O in H₂O₂ falls from −1 to −2 (reduced) — since H₂O₂ is the species being reduced here, it is the oxidising agent, unlike its more familiar dual role.

Redox Reactions — FAQs

More Class 11 Chemistry chapters

• 01Some Basic Concepts of Chemistry
• 02Structure of Atom
• 03Classification of Elements and Periodicity
• 04Chemical Bonding and Molecular Structure
• 05Thermodynamics
• 06Equilibrium

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