Redox Reactions: JEE Main PYQ, 2020–2025

Answer: (B)

The number of electrons in a half-reaction equals the change in oxidation number of the element being reduced. Mn goes from +7 to +2, so 5 electrons are gained. Check the charge: left side −1 + 8 − 5 = +2, which equals the charge of Mn²⁺. Hence n = 5.

Answer: (B)

This is a comproportionation reaction: Br in BrO₃⁻ (+5) and Br⁻ (−1) both become Br₂ (0). BrO₃⁻ gains 5 electrons and 5 Br⁻ each lose 1 electron, so BrO₃⁻ + 5Br⁻ → 3Br₂ (6 Br atoms, 6 electrons). The oxygen of BrO₃⁻ (3 O) goes into 3H₂O, which needs 6 H⁺. The hydrogen balance gives 6 H⁺ on the reactant side. Charge check: −1 − 5 + 6 = 0 on the left, which matches the neutral Br₂ and H₂O on the right. So the coefficient of H⁺ is 6.

Answer: (A)

Count electrons lost by one FeC₂O₄: Fe²⁺ → Fe³⁺ gives 1 electron, and C₂O₄²⁻ → 2CO₂ gives 2 electrons (each C, +3 → +4). Total n-factor = 3. MnO₄⁻ → Mn²⁺ gains 5 electrons. LCM(3, 5) = 15, so 5 FeC₂O₄ react with 3 MnO₄⁻, producing 5 Fe³⁺, 10 CO₂ and 3 Mn²⁺. Charge balance: left = 0 + 3(−1) = −3 before H⁺; right = 5(+3) + 3(+2) = +21. Hence 24 H⁺ are needed on the left (−3 + 24 = +21). Oxygen check: left 5(4) + 3(4) = 32; right 20 (CO₂) + 12 (H₂O) = 32, and 24 H give 12 H₂O. Coefficient of H⁺ = 24.

Answer: (C)

2(+1) + 2x + 7(−2) = 0, so 2x = 12 and x = +6.

Answer: (B)

Principle: in a molecular redox equation, first balance electrons using the oxidation-state change, then balance the remaining atoms. Copper goes from 0 to +2 and sulphur from +6 (in H₂SO₄) to +4 (in SO₂). Electron balance gives 1 Cu for 1 SO₂. Sulphur is conserved by using 2 H₂SO₄: one goes to CuSO₄ and one to SO₂. Hydrogen from the 2 H₂SO₄ (4 atoms) forms 2 H₂O. Check oxygen: 8 on each side. The water coefficient on the product side is 2.

Answer: (A)

Oxidation: in Cu₂S, 2Cu go from +1 to +2 (2e⁻) and S goes from −2 to +6 (8e⁻), so each Cu₂S loses 10e⁻. Reduction: N goes from +5 to +2 (NO), gaining 3e⁻. The lowest common multiple is 30, so use 3 Cu₂S and 10 NO. Then Cu: 6 Cu(NO₃)₂, S: 3 H₂SO₄. Nitrogen on the right: 12 + 10 = 22, so HNO₃ = 22. Hydrogen: 22 on the left means 6 in H₂SO₄ plus 16 in water, giving 8H₂O. Oxygen check: 66 on both sides. Of the 22 HNO₃, 10 are reduced to NO (oxidising agent) and 12 appear as nitrate in 6Cu(NO₃)₂ (acid role), so the ratio is 12 : 10 = 6 : 5.

Answer: (C)

Assign oxidation numbers. In H₂S, S is −2; in SO₂, S is +4; in elemental S, S is 0. Each H₂S gives up 2 electrons (−2 to 0), and each SO₂ gains 4 electrons (+4 to 0). To balance electrons, 2 H₂S give 4 electrons, which 1 SO₂ accepts. This makes 2H₂S + SO₂ → 3S + 2H₂O, where the 3 sulphur atoms in the product come from 2 sulphur atoms of H₂S (oxidised) plus 1 from SO₂ (reduced). Check: H 4 = 4; O 2 = 2; S 3 = 3. The sulphur coefficient is 3.

Answer: (D)

With H = +1 and O = −2, the oxidation numbers of the neutral molecule add to zero: 3(+1) + x + 2(−2) = 0, so x = +1. (Two of the hydrogens are bonded directly to phosphorus, which is why H₃PO₂ is monobasic.)

Answer: (A)

The reducing agent FeC₂O₄ loses electrons in two places. Fe²⁺ → Fe³⁺ loses 1 e⁻. C₂O₄²⁻ → 2CO₂ loses 2 e⁻, since each carbon goes from +3 to +4. The total is 3 e⁻ per mole of FeC₂O₄. MnO₄⁻ → Mn²⁺ gains 5 e⁻ per mole. Equate electrons: 5 FeC₂O₄ (15 e⁻) with 3 MnO₄⁻ (15 e⁻). The full equation is 3MnO₄⁻ + 5FeC₂O₄ + 24H⁺ → 3Mn²⁺ + 5Fe³⁺ + 10CO₂ + 12H₂O, with charge +21 on each side. Moles of MnO₄⁻ per mole of FeC₂O₄ = 3/5 = 0.6.

Answer: (A)

Balance the half-reaction by atoms first, then charge. The 4 O atoms of MnO₄⁻ become 4 H₂O, which needs 8 H⁺ on the left. The left side then has charge −1 + 8 = +7, and the right side (Mn²⁺) has +2. Adding 5 electrons to the left makes it +2 on both sides, so n = 5. This matches the change in oxidation number of Mn from +7 to +2.