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Class XI · Chemistry

Redox Reactions: NEET MCQ practice

174 practice questions on Redox Reactions, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

174

Practice questions

56

Easy

110

Medium

8

Hard

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What Redox Reactions covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Electronic concepts of oxidation and reduction
  • Redox reactions, rules for assigning oxidation number
  • Balancing of redox reactions

Sample questions on Redox Reactions

Try each one, then open the answer and explanation.

Question 1 · Practice question

One gas bleaches the colour of flowers by reduction and another gas by oxidation. The gases respectively are

  1. A.SO₂ and CI₂
  2. B.CO₂ and CI₂
  3. C.NO and CI₂
  4. D.H₂S and Br₂
Show answer

Answer: A. SO₂ and CI₂

Sulfur dioxide (SO₂) bleaches by reduction, as it adds hydrogen to the coloured material making it colourless. Chlorine (Cl₂) bleaches by oxidation, breaking chromophores via oxidation. Carbon dioxide does not bleach, so other options are incorrect. This is a direct recall from redox and p‑block chemistry.

Question 2 · Practice question

Oxidation state of chromium in CrO₅ is +x. Here, value of x is

  1. A.12
  2. B.9
  3. C.11
  4. D.6
Show answer

Answer: D. 6

CrO₅ has one Cr atom, one double-bonded oxygen (O²⁻, oxidation state -2) and two peroxide groups (each O₂²⁻ with each O at -1). Setting sum of oxidation states to zero gives Cr oxidation state = +6. The tempting wrong options (12, 9, 11) would come from incorrectly counting all oxygens as -2 or misidentifying peroxide oxygens.

Question 3 · Practice question

H₂O₂ acts as both oxidising as well as reducing agent its product is H₂O, when act as oxidising agent and its product is 0 2 when act as of reducing agent. The strength of '10P means one liter of H₂O₂ solution on decomposition at S.T.P. condition liberate 10 liters of oxygen gas (H₂O → H₂O + 1/2 O₂) 15 gm Mn0 4 ) 2 sample containing inert impurity is completely reacting with 100 ml of '11.2 V' H₂O₂, then what will be the % purity of Ba (Mn0 4 ) 2 in the sample? (Atomic mass Ba = 137, Mn = 55)

  1. A.5%
  2. B.10%
  3. C.50%
  4. D.None
Show answer

Answer: C. 50%

The key is converting 'volume strength' of H₂O₂ to molarity (1 M for 11.2 V), then using the balanced redox reaction stoichiometry where MnO₄⁻ oxidizes H₂O₂: 2MnO₄⁻ + 5H₂O₂ → 2Mn²⁺ + 5O₂ + ... in acidic medium. From 0.1 mol H₂O₂, moles of Ba(MnO₄)₂ = moles H₂O₂ / 5 = 0.02 mol, mass pure = 7.5 g, giving 50% purity. The main trick is recognizing the correct stoichiometry; a wrong ratio (like 2:2) would give a different percentage.

Question 4 · Practice question

X gram of pure As 2 S₃ is completely oxidised to respective highest oxidation states by 50 ml of 0.1 M hot acidified KMnO₄ then x mass of As 2 S₃ taken is : (Molar mass of As 2 S₃= 246)

  1. A.22.4 g
  2. B.43.92 g
  3. C.64.23 g
  4. D.None
Show answer

Answer: D. None

As2S3 loses 28 electrons per mole upon oxidation to As(+5) and S(+6). The given KMnO4 in acidic medium gains 5 electrons per mole. Moles of KMnO4 = 0.005, total electrons gained = 0.025 mol. Moles of As2S3 required = 0.025/28 ≈ 0.000893 mol, giving mass ≈ 0.22 g, which does not match any of the given options (all ~20–60 g). Hence, 'None' is correct. A common mistake is miscalculating the n‑factor of As2S3 as 2 or 10, which could lead to one of the listed masses, but the correct electron‑balance yields a very small mass.

Question 5 · Practice question

The number of moles of ferrous oxalate oxidised by one mole of KMnO₄ is

  1. A.5/2
  2. B.2/5
  3. C.3/5
  4. D.5/3
Show answer

Answer: D. 5/3

One mole of FeC₂O₄ provides 3 electrons (Fe²⁺ gives 1, C₂O₄²⁻ gives 2). In acidic medium, one mole of KMnO₄ accepts 5 electrons. So electrons lost = moles of FeC₂O₄ × 3 = electrons gained = 5, giving moles of FeC₂O₄ = 5/3. The tempting wrong option 5/2 comes from incorrectly considering FeC₂O₄ as giving only 2 electrons, forgetting Fe²⁺ contribution.

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