Balancing of Redox Equations balancing of redox reactions.: NEET MCQs with Answers

8 free questions on Balancing of Redox Equations balancing of redox reactions. for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
When the redox reaction given below is balanced, the values of the coefficients a, b and c are, respectively: [NEET 2023]
aCr₂O₇²⁻ + bSO₃²⁻(aq) + cH⁺(aq) → 2aCr³⁺(aq) + bSO₄²⁻(aq) + (c/2)H₂O(l)
  1. 3, 8, 1
  2. 1, 8, 3
  3. 8, 1, 3
  4. 1, 3, 8
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Answer: (D) 1, 3, 8
Explanation: Cr changes +6 → +3 (2 Cr gain 6e⁻ per Cr₂O₇²⁻) and S changes +4 → +6 (each SO₃²⁻ loses 2e⁻), so 1 Cr₂O₇²⁻ reacts with 3 SO₃²⁻. Balanced: Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O. Hence a = 1, b = 3 and c = 8 (c/2 = 4 H₂O).

Question 2

Previous-year question
For the redox reaction MnO₄⁻ + C₂O₄²⁻ + H⁺ → Mn²⁺ + CO₂ + H₂O, the correct coefficients of the reactants for the balanced equation are [NEET 2018]
  1. MnO₄⁻ = 16, C₂O₄²⁻ = 5, H⁺ = 2
  2. MnO₄⁻ = 2, C₂O₄²⁻ = 5, H⁺ = 16
  3. MnO₄⁻ = 5, C₂O₄²⁻ = 16, H⁺ = 2
  4. MnO₄⁻ = 2, C₂O₄²⁻ = 16, H⁺ = 5
Show answer
Answer: (B) MnO₄⁻ = 2, C₂O₄²⁻ = 5, H⁺ = 16
Explanation: Each MnO₄⁻ gains 5 electrons (+7 → +2) and each C₂O₄²⁻ loses 2 electrons, so the ratio must be 2 : 5. Balancing oxygen with water then fixes H⁺ at 16: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. This is the basis of the standard oxalate–permanganate titration.

Question 3

In basic solution, permanganate oxidises sulphite: MnO₄⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻. When this is balanced with 2MnO₄⁻ and the smallest whole-number coefficients, which option is correct?
  1. 3SO₃²⁻ and 2OH⁻ on the reactant side, with 1H₂O on the product side
  2. 3SO₃²⁻ and 4H₂O on the reactant side, with 8OH⁻ on the product side
  3. 5SO₃²⁻ and 6OH⁻ on the reactant side, with 3H₂O on the product side
  4. 3SO₃²⁻ and 1H₂O on the reactant side, with 2OH⁻ on the product side
Show answer
Answer: (D) 3SO₃²⁻ and 1H₂O on the reactant side, with 2OH⁻ on the product side
Explanation: Manganese goes from +7 to +4, gaining 3 electrons, and sulphur goes from +4 to +6, losing 2 electrons, so 2MnO₄⁻ pair with 3SO₃²⁻. The charge is then −8 on the left and −6 on the right, so 2OH⁻ are added on the right. One H₂O on the left balances hydrogen and oxygen, giving 2MnO₄⁻ + 3SO₃²⁻ + H₂O → 2MnO₂ + 3SO₄²⁻ + 2OH⁻.

Question 4

White phosphorus reacts with hot aqueous alkali to give PH₃ and H₂PO₂⁻. In the balanced equation, what is the ratio of P atoms oxidised to P atoms reduced, and what mass of PH₃ forms from 12.4 g of P₄? (P = 31, H = 1)
  1. 1 : 3 and 3.4 g
  2. 3 : 1 and 13.6 g
  3. 3 : 1 and 3.4 g
  4. 1 : 3 and 10.2 g
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Answer: (C) 3 : 1 and 3.4 g
Explanation: Phosphorus goes from 0 to −3 in PH₃ (gain of 3 electrons) and from 0 to +1 in H₂PO₂⁻ (loss of 1 electron). Electron balance needs three atoms oxidised for each atom reduced: P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻. Here 12.4 g P₄ is 0.10 mol, which gives 0.10 mol PH₃ or 3.4 g.

Question 5

In faintly alkaline solution, permanganate quantitatively oxidises thiosulphate to sulphate and is itself reduced to MnO₂. How many moles of MnO₄⁻ are needed to oxidise 0.30 mol of S₂O₃²⁻ completely?
  1. 1.60 mol
  2. 0.80 mol
  3. 0.30 mol
  4. 0.48 mol
Show answer
Answer: (B) 0.80 mol
Explanation: Sulphur rises from an average of +2 to +6, so each S₂O₃²⁻ loses 8 electrons, and manganese falls from +7 to +4, gaining 3 electrons. The balanced equation is 8MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 6SO₄²⁻ + 2OH⁻. So 0.30 mol of thiosulphate needs 0.30 × 8/3 = 0.80 mol of permanganate.

Question 6

Permanganate ion oxidises sulphur dioxide gas in acidic medium to give Mn²⁺ and hydrogen sulphate ion: MnO₄⁻ + SO₂ + H₂O + H⁺ → Mn²⁺ + HSO₄⁻. In the balanced equation with smallest whole numbers, the coefficients of SO₂, H₂O and H⁺ are respectively:
  1. 2, 5 and 1
  2. 5, 6 and 2
  3. 5, 2 and 1
  4. 5, 2 and 4
Show answer
Answer: (C) 5, 2 and 1
Explanation: Mn falls by 5 units and S rises by 2 units, so 2MnO₄⁻ react with 5SO₂. Oxygen balance needs 2H₂O on the left, and the hydrogen and charge then need only 1H⁺. The equation 2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻ has charge −1 on both sides.

Question 7

Excess chlorine left in treated drinking water is removed with sulphur dioxide, which forms chloride and sulphate ions in water. What mass of SO₂ is needed to remove 7.1 g of dissolved Cl₂? (Cl = 35.5, S = 32, O = 16)
  1. 1.6 g
  2. 6.4 g
  3. 12.8 g
  4. 3.2 g
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Answer: (B) 6.4 g
Explanation: Cl₂ gains 2 electrons to form 2Cl⁻, and S goes from +4 in SO₂ to +6 in SO₄²⁻, losing 2 electrons. The balanced equation is Cl₂ + SO₂ + 2H₂O → 2Cl⁻ + SO₄²⁻ + 4H⁺, a 1 : 1 mole ratio. Here 7.1 g Cl₂ is 0.10 mol, so 0.10 mol SO₂ or 6.4 g is required.
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Question 8

In solution, Mn³⁺ is unstable and disproportionates into Mn²⁺ and MnO₂ with liberation of H⁺. In the balanced ionic equation, how many moles of H⁺ are formed per mole of Mn³⁺ consumed?
  1. 2
  2. 4
  3. 1
  4. 3
Show answer
Answer: (A) 2
Explanation: One Mn³⁺ is reduced to Mn²⁺ (gain of 1 electron) and one is oxidised to MnO₂ (loss of 1 electron), so two Mn³⁺ are needed. Balancing oxygen and hydrogen gives 2Mn³⁺ + 2H₂O → Mn²⁺ + MnO₂ + 4H⁺. Thus 4 mol H⁺ form per 2 mol Mn³⁺, which is 2 mol H⁺ per mole of Mn³⁺.

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