NEETAceToday's NEET Daily

Balancing redox reactions: NEET MCQs with answers

90 questions on this topic of Redox Reactions and Electrochemistry (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Redox Reactions and Electrochemistry freeAll Redox Reactions and Electrochemistry questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. Given the following two statements regarding redox reaction balancing: Statement I: While balancing redox half-reactions in basic medium by the ion-electron method, for each oxygen atom excess on one side, one H₂O molecule is added to that side and two OH⁻ ions to the opposite side. Statement II: The oxidation state of chlorine in bleaching powder, CaOCl₂, is zero for all chlorine atoms present. In the light of the above statements, choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct: in basic medium, an excess of O on one side is routinely balanced by adding 1 H₂O to that side and 2 OH⁻ to the other. Statement II is incorrect: bleaching powder contains Ca²⁺, Cl⁻ (oxidation state -1), and OCl⁻ (oxidation state +1), so the two chlorine atoms have oxidation states of -1 and +1, not zero.

medium

2. Consider the disproportionation reaction of elemental sulphur in hot alkaline medium: S₈ + OH⁻ → S²⁻ + S₂O₃²⁻ + H₂O (unbalanced) When balanced with the smallest whole-number coefficients, what is the coefficient of OH⁻?

  1. 8
  2. 24
  3. 16
  4. 12
Show answer and explanation

Answer: (D) 12

The reduction half-reaction is S + 2e⁻ → S²⁻ and the oxidation half-reaction is 2S + 3H₂O → S₂O₃²⁻ + 6H⁺ + 4e⁻. Balancing in basic medium per mole of S gives 4S + 6OH⁻ → 2S²⁻ + S₂O₃²⁻ + 3H₂O. Scaling to S₈ gives: S₈ + 12OH⁻ → 4S²⁻ + 2S₂O₃²⁻ + 6H₂O. Therefore, the stoichiometric coefficient of OH⁻ is 12.

hard

3. In the balanced redox reaction: a SCN⁻ + b IO₄⁻ + c H₂O → d SO₄²⁻ + e HCN + f IO₃⁻ + g H⁺ taking place in acidic solution, what is the value of the coefficient b when balanced with the smallest whole-number coefficients?

  1. 4
  2. 5
  3. 7
  4. 3
Show answer and explanation

Answer: (D) 3

The oxidation half-reaction: SCN⁻ + 4H₂O → SO₄²⁻ + HCN + 7H⁺ + 6e⁻ (S goes from -2 to +6, losing 6e⁻). The reduction half-reaction: IO₄⁻ + 2H⁺ + 2e⁻ → IO₃⁻ + H₂O. Multiplying the reduction half-reaction by 3: 3IO₄⁻ + 6H⁺ + 6e⁻ → 3IO₃⁻ + 3H₂O. Adding both gives: SCN⁻ + 3IO₄⁻ + H₂O → SO₄²⁻ + HCN + 3IO₃⁻ + H⁺. Hence, b = 3.

medium

4. In acidic medium, iodate ion oxidises arsenite ion according to: a IO₃⁻ + b H₃AsO₃ → c I⁻ + d H₃AsO₄. When balanced with the smallest whole-number coefficients, the coefficient of H₃AsO₃ (b) is:

  1. 1
  2. 2
  3. 3
  4. 5
Show answer and explanation

Answer: (C) 3

Reduction: IO₃⁻ + 6H⁺ + 6e⁻ → I⁻ + 3H₂O (6 electrons per IO₃⁻). Oxidation: H₃AsO₃ + H₂O → H₃AsO₄ + 2H⁺ + 2e⁻ (2 electrons per H₃AsO₃). Multiplying the oxidation half-reaction by 3 gives 6 electrons, yielding: IO₃⁻ + 3H₃AsO₃ → I⁻ + 3H₃AsO₄. Thus, the coefficient b is 3.

easy

5. Redox reactions are commonly balanced by the oxidation-number method or by the:

  1. Hess's law method
  2. Le Chatelier method
  3. Half-reaction (ion-electron) method
  4. Trial and error method only
Show answer and explanation

Answer: (C) Half-reaction (ion-electron) method

Both the oxidation-number method and the half-reaction method ensure that electrons lost in oxidation equal electrons gained in reduction.

NCERT Class 11

hard

6. In alkaline solution, hypochlorite ion (ClO⁻) oxidises chromium(III) hydroxide [Cr(OH)₃] to chromate ion (CrO₄²⁻), while being reduced to chloride (Cl⁻). What is the coefficient of OH⁻ in the balanced ionic equation using smallest integers?

  1. 2
  2. 5
  3. 4
  4. 8
Show answer and explanation

Answer: (C) 4

Oxidation: Cr(OH)₃ + 5OH⁻ → CrO₄²⁻ + 4H₂O + 3e⁻. Reduction: ClO⁻ + H₂O + 2e⁻ → Cl⁻ + 2OH⁻. Multiply oxidation by 2 and reduction by 3: 2Cr(OH)₃ + 10OH⁻ + 3ClO⁻ + 3H₂O → 2CrO₄²⁻ + 8H₂O + 3Cl⁻ + 6OH⁻. Net OH⁻ on the reactant side is 10 - 6 = 4 OH⁻.

medium

7. Consider the following two statements regarding balancing of redox reactions: Statement I: In balancing a redox reaction in basic medium using the ion-electron method, for each excess oxygen atom on one side, one H₂O molecule is added to that side and two OH⁻ ions to the opposite side. Statement II: Oxidation number method cannot be applied to reactions where disproportionation occurs. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I describes the standard shortcut rule for balancing oxygen in basic medium: add H₂O to the side with excess O and 2OH⁻ to the other side (net addition of O). Statement II is incorrect because disproportionation reactions can readily be balanced using the oxidation number method by writing the reactant twice (once as oxidising agent, once as reducing agent).

medium

8. When copper reacts with concentrated nitric acid, it produces nitrogen dioxide gas and copper(II) nitrate: Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O. What fraction of the total nitric acid acts strictly as an oxidising agent?

  1. 1/4
  2. 1/2
  3. 3/4
  4. 1
Show answer and explanation

Answer: (B) 1/2

Out of 4 moles of HNO₃, 2 moles are reduced to NO₂ (acting as an oxidising agent), while the remaining 2 moles provide the nitrate ions for Cu(NO₃)₂ without undergoing change in oxidation number (acting as an acid). Thus, the fraction is 2/4 = 1/2.

medium

9. Consider the oxidation of oxalate ion by permanganate in acidic solution: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. If 0.1 mol of C₂O₄²⁻ is completely oxidised, what volume of CO₂ gas is produced at STP (22.7 L/mol)?

  1. 2.27 L
  2. 4.54 L
  3. 1.14 L
  4. 22.7 L
Show answer and explanation

Answer: (B) 4.54 L

From the stoichiometry, 5 moles of C₂O₄²⁻ yield 10 moles of CO₂ (a 1:2 ratio). Hence, 0.1 mol of C₂O₄²⁻ produces 0.2 mol of CO₂. Volume at STP = 0.2 × 22.7 L = 4.54 L.

medium

10. Consider the oxidation of copper(II) sulphide by nitric acid in acidic medium: a CuS + b NO₃⁻ + c H⁺ → d Cu²⁺ + e S + f NO + g H₂O. Which set of coefficients (a, b, f) is correct for the balanced equation?

  1. a = 2, b = 3, f = 3
  2. a = 1, b = 2, f = 1
  3. a = 3, b = 4, f = 4
  4. a = 3, b = 2, f = 2
Show answer and explanation

Answer: (D) a = 3, b = 2, f = 2

CuS oxidises to Cu²⁺ + S + 2e⁻ (each CuS loses 2e⁻). NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O (gains 3e⁻). Multiplying the CuS half-reaction by 3 and the nitrate half-reaction by 2 yields: 3CuS + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 3S + 2NO + 4H₂O. Thus, a = 3, b = 2, and f = 2.

easy

11. Consider the following statements regarding the balancing of redox reactions: Statement I: In balancing redox half-reactions in basic medium using the ion-electron method, for each excess oxygen atom on one side, one H₂O molecule is added to that side and two OH⁻ ions are added to the opposite side. Statement II: The oxidation number method relies on balancing the total increase in oxidation number with the total decrease in oxidation number. Which of the following is correct?

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Both statements are standard rules outlined in the NCERT textbook for balancing redox reactions by ion-electron and oxidation number methods.

medium

12. In acidic medium, hydrazine (N₂H₄) is oxidised to N₂ by bromate ions (BrO₃⁻), which are reduced to bromide ions (Br⁻). How many moles of N₂H₄ are required to react completely with 2 moles of BrO₃⁻?

  1. 3 moles
  2. 2 moles
  3. 1 mole
  4. 5 moles
Show answer and explanation

Answer: (A) 3 moles

Reduction half-reaction: BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O (6 electrons gained per BrO₃⁻). Oxidation half-reaction: N₂H₄ → N₂ + 4H⁺ + 4e⁻ (4 electrons lost per N₂H₄). For 2 moles of BrO₃⁻, total electrons required = 2 × 6 = 12 electrons. Since each mole of N₂H₄ provides 4 electrons, moles of N₂H₄ needed = 12 / 4 = 3 moles.

Keep going with Redox Reactions and Electrochemistry. All 596 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Redox Reactions and Electrochemistry