Take timed practice tests on Redox Reactions for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus a 60-question chapter module. Each item is original and reframed for copyright safety.
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1. The oxidation state of Cr in K₂Cr₂O₇ is:
Explanation: K₂Cr₂O₇: 2(+1) + 2x + 7(−2) = 0 → 2 + 2x − 14 = 0 → x = +6.
2. The oxidation state of S in H₂SO₄ is:
Explanation: H₂SO₄: 2(+1) + x + 4(−2) = 0 → 2 + x − 8 = 0 → x = +6.
3. In the reaction Zn + CuSO₄ → ZnSO₄ + Cu, which species is oxidised?
Explanation: Zn goes from 0 → +2 (loses electrons = oxidised). Cu²⁺ goes from +2 → 0 (gains electrons = reduced).
4. In the reaction 2H₂O₂ → 2H₂O + O₂, oxygen undergoes:
Explanation: In H₂O₂, O is in −1 state. In H₂O it becomes −2 (reduced), and in O₂ it becomes 0 (oxidised). Same element both oxidised and reduced — disproportionation.
5. The number of electrons transferred in the half-reaction MnO₄⁻ → Mn²⁺ in acid is:
Explanation: Mn in MnO₄⁻ is +7. In Mn²⁺ it is +2. Change = 5. So 5 electrons are gained per MnO₄⁻: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
6. Which oxidant is used in acidic medium for oxidising Fe²⁺ to Fe³⁺ in volumetric analysis?
Explanation: KMnO₄ (permanganate) is a strong oxidising agent used in acidic medium (dilute H₂SO₄) to oxidise Fe²⁺ to Fe³⁺. It is self-indicating (pink → colourless).
7. The oxidation state of N in N₂H₄ (hydrazine) is:
Explanation: N₂H₄: 2x + 4(+1) = 0 → x = −2. Each N is in −2 state.
8. In a redox reaction, 1 mole of KMnO₄ in acidic medium reacts with Fe²⁺. The n-factor of KMnO₄ is:
Explanation: MnO₄⁻ (Mn = +7) → Mn²⁺ (Mn = +2). Change in oxidation state = 5. n-factor = 5 per mole of KMnO₄.
9. In the reaction Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ (acidic medium), ratio of Cr₂O₇²⁻ to Fe²⁺ is:
Explanation: Each Cr₂O₇²⁻ contains 2 Cr atoms changing from +6 to +3: n-factor = 2×3 = 6. Each Fe²⁺ has n-factor = 1. By equivalents: 6 Fe²⁺ per Cr₂O₇²⁻. Ratio = 1:6.
10. In the reaction I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, I₂ acts as:
Explanation: I in I₂ is 0. In I⁻ it is −1 (gained electrons = reduced). So I₂ is the oxidising agent. S in S₂O₃²⁻ is +2; in S₄O₆²⁻ it is +2.5 (lost electrons = oxidised). So S₂O₃²⁻ is the reducing agent.
11. For spontaneous redox reaction, the standard EMF (E°cell) must be:
Explanation: ΔG° = −nFE°cell. For spontaneous reaction, ΔG°
12. The oxidation state of Xe in XeO₄ is:
Explanation: XeO₄: x + 4(−2) = 0 → x = +8. Xenon in XeO₄ has the highest common oxidation state of +8.
13. In the acidic permanganate-oxalate titration (KMnO₄ + H₂C₂O₄), the volume of 0.1 M KMnO₄ needed to react with 25 mL of 0.1 M H₂C₂O₄ is:
Explanation: n-factor of KMnO₄ = 5 (Mn: +7→+2). n-factor of H₂C₂O₄ = 2 (C: +3→+4). meq of oxalate = 25×0.1×2 = 5 meq. meq of KMnO₄ = V×0.1×5 = 5 → V = 10 mL.
14. In the reaction: 3Br₂ + 6NaOH → 5NaBr + NaBrO₃ + 3H₂O, the fraction of Br₂ acting as reducing agent is:
Explanation: Br in Br₂ is 0. Br in NaBr is −1 (reduced). Br in NaBrO₃ is +5 (oxidised). 3 Br₂ = 6 Br atoms. 5 go to −1 (reduced) and 1 goes to +5 (oxidised). Br₂ that provides the Br going to +5: 1 Br atom comes from 0.5 Br₂. So 0.5 of 3 Br₂ = 1/6 acts as reducing agent. Wait: from 3Br₂, 1 Br atom (=0.5 Br₂) is oxidised (reducing agent) and 5 Br atoms (=2.5 Br₂) are reduced. Fraction acting as reducing agent = 0.5/3 = 1/6. But the disproportionation means each Br₂ simultaneously provides one Br to each. From stoichiometry: 1/6 of Br₂ acts as reducing agent.
15. The reducing agent in a redox reaction is the species that:
Explanation: A reducing agent donates electrons to the other species. It gets oxidised in the process (its oxidation state increases).
16. The oxidation state of O in OF₂ is:
Explanation: F is more electronegative than O. F takes −1, and with 2 F atoms: x + 2(−1) = 0 → x = +2. In OF₂, oxygen has +2 oxidation state — one of the rare cases where O is positive.
17. How many moles of electrons are transferred when 1 mole of Cu is oxidised to Cu²⁺?
Explanation: Cu → Cu²⁺ + 2e⁻. One mole of Cu releases 2 moles of electrons (2 Faradays).