Redox Reactions

7 NEET-UG practice questions on Redox Reactions, part of Chemistry. The first 7 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Medium

   Given below are two statements: Statement I: The oxidation number of oxygen in OF₂ is −2. Statement II: The oxidation number of hydrogen in NaH is +1. Choose the correct answer.

   1. ABoth Statement I and Statement II are correct
   2. BBoth Statement I and Statement II are incorrect
   3. CStatement I is correct but Statement II is incorrect
   4. DStatement I is incorrect but Statement II is correct

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   Answer: B · Both Statement I and Statement II are incorrect

   Fluorine is always −1, so oxygen in OF₂ is +2. In metal hydrides, hydrogen is −1, as sodium is +1. Both statements are therefore false.

   • A: Both statements are false, as both use the usual values wrongly.
   • C: Statement I is false, since F forces oxygen to +2.
   • D: Statement II is false, since hydrogen is −1 with a metal.

   Remember this

   Oxygen is positive in fluorides, and hydrogen is −1 in metal hydrides.

2. Question 2 · Hard

   Exactly 1.26 g of H₂C₂O₄·2H₂O is dissolved to make 250 mL of solution. A 25 mL portion is acidified and titrated with 0.02 M KMnO₄: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O What volume of KMnO₄ is needed? (H₂C₂O₄·2H₂O = 126, H₂C₂O₄ = 90)

   1. A28 mL
   2. B125 mL
   3. C20 mL
   4. D50 mL

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   Answer: C · 20 mL

   Oxalic acid = 1.26/126 = 0.01 mol in 250 mL, so 25 mL holds 1 mmol. KMnO₄ needed = 1 × 2/5 = 0.4 mmol, and the volume = 0.4/0.02 = 20 mL. Check by equivalents: 1 mmol × 2 = 2 meq, and V × 0.02 × 5 = 2 gives V = 20 mL.

   • A: This uses the anhydrous molar mass for hydrated crystals.
   • B: This inverts the ratio, using 5/2 instead of 2/5.
   • D: This takes a 1 : 1 mole ratio.

   Remember this

   In acid, 2 mol KMnO₄ react with 5 mol oxalate; use the hydrate's molar mass for the weighed crystals.

3. Question 3 · Easy

   In the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), which species acts as the oxidising agent?

   1. ACu²⁺
   2. BZn
   3. CCu
   4. DZn²⁺

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   Answer: A · Cu²⁺

   Cu²⁺ gains two electrons and is reduced to Cu, while it oxidises Zn to Zn²⁺. The species that is reduced is the oxidising agent.

   • B: Zn loses electrons, so it is the reducing agent.
   • C: Cu is a product, formed by reduction of Cu²⁺.
   • D: Zn²⁺ is a product, formed by oxidation of Zn.

   Remember this

   The oxidising agent is reduced; the reducing agent is oxidised.

4. Question 4 · Medium

   An acidified permanganate solution oxidises iron(II): a MnO₄⁻ + b Fe²⁺ + c H⁺ → Mn²⁺ + Fe³⁺ + H₂O (not balanced) For the balanced ionic equation in lowest whole numbers, what are a, b and c?

   1. A1, 3, 8
   2. B1, 5, 4
   3. C1, 5, 8
   4. D2, 5, 16

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   Answer: C · 1, 5, 8

   Mn goes from +7 to +2, gaining 5 electrons, and each Fe²⁺ loses 1, so b = 5 per MnO₄⁻. Four O need 8 H⁺ to form 4 H₂O. The charges match: −1 + 10 + 8 = +17 and +2 + 15 = +17.

   • A: This takes Mn as changing by three units, as it does in going to MnO₂.
   • B: This gives H⁺ only half of what the four oxygen atoms need.
   • D: This doubles MnO₄⁻ and H⁺ but not Fe²⁺.

   Remember this

   Balance electrons first, then O with H₂O, then H with H⁺, and confirm by charge.

5. Question 5 · Easy

   What is the oxidation number of phosphorus in H₃PO₂?

   1. A−1
   2. B+5
   3. C+3
   4. D+1

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   Answer: D · +1

   With H = +1 and O = −2, the sum is 3(+1) + x + 2(−2) = 0, so x = +1. Only the sum over all atoms matters, so count each atom in the formula.

   • A: This has the sign reversed.
   • B: This is the value in H₃PO₄.
   • C: This is the value in H₃PO₃.

   Remember this

   The oxidation numbers in a neutral molecule add up to zero.

6. Question 6 · Medium

   Disproportionation occurs in which of the following reactions?

   1. ACaCO₃(s) → CaO(s) + CO₂(g)
   2. B2H₂O₂(aq) → 2H₂O(l) + O₂(g)
   3. C2Mg(s) + O₂(g) → 2MgO(s)
   4. D2KClO₃(s) → 2KCl(s) + 3O₂(g)

   Answer and explanation in LearnFree account

7. Question 7 · Medium

   Arrange the following in increasing order of the oxidation number of nitrogen. A. NO₂ B. HNO₃ C. N₂O D. NH₃

   1. AD, A, C, B
   2. BD, C, A, B
   3. CB, A, C, D
   4. DC, D, A, B

   Answer and explanation in LearnFree account