Redox Reactions: JEE Advanced PYQ, 2007–2026

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: (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)

Principle: balance electrons first, then atoms other than O and H, then O with H₂O, then H with H⁺, and finally check charge. Step 1: the electron balance is 4 Zn for 1 NO₃⁻, since Zn loses 2 e⁻ and N gains 8 e⁻. Step 2: 4Zn + NO₃⁻ → 4Zn²⁺ + NH₄⁺. Step 3: 3 O on the reactant side require 3 H₂O on the product side. Step 4: products carry 6 H from H₂O and 4 H from NH₄⁺, so 10 H⁺ go on the reactant side. Final balanced equation: 4Zn + NO₃⁻ + 10H⁺ → 4Zn²⁺ + NH₄⁺ + 3H₂O. Charge check: +9 on each side.

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: (C)

Principle: in a disproportionation, the same element is both oxidised and reduced. Chlorine goes from 0 in Cl₂ to −1 in Cl⁻ (gain of 1 per atom) and to +5 in ClO₃⁻ (loss of 5 per atom). For electron balance, one ClO₃⁻ requires five Cl⁻ to be formed, so the ratio is 5 Cl⁻ : 1 ClO₃⁻. Then the total chlorine atoms (6) fix 3 Cl₂, the charge balance fixes 6 OH⁻, and the hydrogen balance fixes 3 H₂O. The coefficient of Cl⁻ is therefore 5.

Answer: (A)

Assign oxidation numbers in As₂S₃: As = +3, S = −2. On oxidation, As(+3) → As(+5) in H₃AsO₄: 2e⁻ per As, so 4e⁻ for two As; S(−2) → S(+6) in H₂SO₄: 8e⁻ per S, so 24e⁻ for three S. Total loss per formula unit = 28e⁻. Each N(+5) in HNO₃ → N(+4) in NO₂ gains 1e⁻, hence 28 HNO₃ (and 28 NO₂). Equation: As₂S₃ + 28HNO₃ → 2H₃AsO₄ + 3H₂SO₄ + 28NO₂ + 8H₂O. Check H: 28 = 6 + 6 + 16 ✓; O: 84 = 8 + 12 + 56 + 8 ✓. Coefficient of HNO₃ = 28.

Answer: (B)

Principle: in basic medium, balance the half-reactions in acid form using H₂O and H⁺, then convert each H⁺ into OH⁻ by adding OH⁻ to both sides to neutralise it. Step 1: the electron balance is 2MnO₄⁻ for 3SO₃²⁻ (Mn gains 3 e⁻, S loses 2 e⁻ each). Step 2: the oxygen and hydrogen balance requires 1 H₂O on the reactant side, and the leftover H ends up as 2 OH⁻ on the product side. Step 3: check O and charge as above. The final coefficient of OH⁻ on the product side is 2.

Answer: (B)

Principle: balance electrons using the change in oxidation state per atom, then balance the remaining atoms and oxygen, then hydrogen, and finally check charge. Step 1: oxidation states. Ethanol CH₃CH₂OH has carbons at −3 and −1. Acetic acid CH₃COOH has carbons at −3 and +3. Overall the two carbons change by 4 units, so each ethanol loses 4 e⁻. Each Cr₂O₇²⁻ gains 6 e⁻ (2 Cr from +6 to +3). Step 2: 3 ethanol per 2 dichromate, so 3CH₃COOH and 4Cr³⁺ appear on the product side. Step 3: oxygen (17 on each side once water is added) gives 11 H₂O. Step 4: hydrogen gives 16 H⁺. Careful counting is the point of the question: a student who uses 1 ethanol per dichromate gets fractional or inconsistent results.