JEE Chemistry Redox Reactions MCQs Set 20

Multiple Choice Questions (MCQs) for JEE Chemistry: Redox Reactions

Access targeted multiple-choice questions for Redox Reactions designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Redox Reactions MCQs for JEE Chemistry

Access the complete set of multiple-choice questions for Redox Reactions below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: In the reaction SO2 + 2H2S → 3S + 2H2O, the substance oxidised is:
(a) S
(b) SO2
(c) H2S
(d) H2O
Show Answer & Explanation

Answer: (c) H2S

Explanation:
1. In H2S sulphur is −2 and in the product S it is 0, so H2S loses electrons.
2. In SO2 sulphur goes from +4 to 0, so SO2 is reduced.
3. Answer: (c)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 2: The oxidation number of phosphorus in Ba(H2PO2)2 is:
(a) +3
(b) +2
(c) +1
(d) −1
Show Answer & Explanation

Answer: (c) +1

Explanation:
1. Ba = +2, H = +1 (4 atoms), O = −2 (4 atoms).
2. 2 + 4 + 2x − 8 = 0, so x = +1.
3. Answer: (c)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 3: Which of the following reactions do NOT involve oxidation-reduction? (I) 2Cs + 2H2O → 2CsOH + H2; (II) 2CaI2 → 2Ca + 2I2; (III) NH4Br + KOH → KBr + NH3 + H2O; (IV) 4KCN + Fe(CN)2 → K4[Fe(CN)6]
(a) I, II
(b) I, III
(c) I, III, IV
(d) III, IV
Show Answer & Explanation

Answer: (d) III, IV

Explanation:
1. (I) and (II) change oxidation numbers (Cs 0 → +1, Ca +2 → 0).
2. (III) is an acid-base reaction and (IV) is complex formation; no oxidation number changes in either.
3. Answer: (d)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 4: In the reaction 3Br2 + 6CO32− + 3H2O → 5Br− + BrO3− + 6HCO3−:
(a) bromine is oxidised and carbonate is reduced
(b) bromine is reduced and water is oxidised
(c) bromine is neither reduced nor oxidised
(d) bromine is both reduced and oxidised
Show Answer & Explanation

Answer: (d) bromine is both reduced and oxidised

Explanation:
1. Br2 (0) goes to Br− (−1), which is reduction.
2. Br2 (0) also goes to BrO3− (+5), which is oxidation, so it disproportionates.
3. Answer: (d)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 5: Which of the following is NOT a disproportionation reaction? (I) NH4NO3 → N2O + H2O; (II) P4 → PH3 + HPO2; (III) PCl5 → PCl3 + Cl2; (IV) IO3− + I− → I2
(a) I, II
(b) I, III, IV
(c) II, IV
(d) I, III
Show Answer & Explanation

Answer: (b) I, III, IV

Explanation:
1. In a disproportionation one element is oxidised and reduced from a single oxidation state.
2. (II) fits (P from 0 to −3 and +1).
3. (I) and (IV) are the reverse type (comproportionation), and (III) is a decomposition in which only Cl is oxidised.
4. Answer: (b)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 6: The number of moles of K2Cr2O7 reduced by 1 mol of Sn2+ ions will be:
(a) 1/3
(b) 3
(c) 1/6
(d) 6
Show Answer & Explanation

Answer: (a) 1/3

Explanation:
1. Sn2+ → Sn4+ gives 2 electrons per mole.
2. Cr2O72− → 2Cr3+ takes 6 electrons, so moles = 2/6 = 1/3.
3. Answer: (a)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 7: The number of moles of KMnO4 required to oxidise 2 mol of FeC2O4 in acidic medium is:
(a) 1.2
(b) 3.33
(c) 0.4
(d) 0.8
Show Answer & Explanation

Answer: (a) 1.2

Explanation:
1. Per FeC2O4: Fe2+ → Fe3+ (1 e−) and C2O42− → 2CO2 (2 e−) = 3 e−.
2. 2 mol lose 6 mol e−; MnO4− takes 5, so moles = 6/5 = 1.2.
3. Answer: (a)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 8: The number of electrons lost in the change 3Fe + 4H2O → Fe3O4 + 8H+ (balanced) is:
(a) 2
(b) 4
(c) 6
(d) 8
Show Answer & Explanation

Answer: (d) 8

Explanation:
1. In Fe3O4 the three Fe atoms total +8 (2 + 3 + 3).
2. So 3 Fe atoms lose 8 electrons together.
3. Answer: (d)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 9: Which gas is evolved when PbO2 is treated with concentrated HNO3?
(a) NO2
(b) O2
(c) N2
(d) N2O
Show Answer & Explanation

Answer: (b) O2

Explanation:
1. Pb is already at its high +4 state and HNO3 is also strongly oxidising, so neither oxidises the other; the oxide ion of PbO2 is oxidised instead.
2. 2PbO2 + 4HNO3 → 2Pb(NO3)2 + 2H2O + O2.
3. Answer: (b)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 10: For decolourisation of 1.5 moles of KMnO4 (acidic medium), the moles of H2O2 required are:
(a) 3/2
(b) 9/4
(c) 15/4
(d) 21/4
Show Answer & Explanation

Answer: (c) 15/4

Explanation:
1. 1.5 mol MnO4− accepts 1.5 × 5 = 7.5 mol e−.
2. H2O2 gives 2 e− per mole, so moles = 7.5/2 = 15/4.
3. Answer: (c)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 11: In which of the following does a transfer of five electrons take place?
(a) CrO42− → Cr3+
(b) MnO4− → MnO2
(c) Cr2O72− → 2Cr3+
(d) MnO4− → Mn2+
Show Answer & Explanation

Answer: (d) MnO4− → Mn2+

Explanation:
1. Mn changes from +7 to +2 in the last case, a gain of 5 electrons.
2. The others involve 3, 3 and 6 electrons.
3. Answer: (d)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 12: In which reaction does H2O2 act as a reducing agent?
(a) Ag2O + H2O2 → 2Ag + H2O + O2
(b) 2KI + H2O2 → 2KOH + I2
(c) PbS + 4H2O2 → PbSO4 + 4H2O
(d) H2O2 + SO2 → H2SO4
Show Answer & Explanation

Answer: (a) Ag2O + H2O2 → 2Ag + H2O + O2

Explanation:
1. H2O2 acts as a reductant when O (−1) goes to O2 (0).
2. That happens in the Ag2O reaction; in the other three, O (−1) goes to −2, so H2O2 is an oxidant.
3. Answer: (a)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 13: In the equation NO2− + H2O → NO3− + 2H+ + ne−, the value of n is:
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. N goes from +3 in NO2− to +5 in NO3−.
2. It loses 2 electrons, so n = 2.
3. Answer: (b)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 14: What mass of HNO3 is needed to convert 5 g of iodine into iodic acid according to I2 + HNO3 → HIO3 + NO2 + H2O? (I = 127, HNO3 = 63)
(a) 12.4 g
(b) 24.8 g
(c) 0.248 g
(d) 49.6 g
Show Answer & Explanation

Answer: (a) 12.4 g

Explanation:
1. Balanced: I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O.
2. Moles of I2 = 5/254 = 0.0197, so HNO3 = 0.197 mol.
3. Mass = 0.197 × 63 = 12.4 g.
4. Answer: (a)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 15: In the reduction of dichromate by Fe(II), the number of electrons involved per chromium atom is:
(a) 2
(b) 1
(c) 3
(d) 4
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. Chromium changes from +6 to +3.
2. That is 3 electrons gained per Cr atom.
3. Answer: (c)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 16: How many grams of I2 are present in a solution which requires 40 mL of 0.11 N Na2S2O3 to react with it? (I2 + 2S2O32− → S4O62− + 2I−)
(a) 12.7 g
(b) 0.558 g
(c) 25.4 g
(d) 11.4 g
Show Answer & Explanation

Answer: (b) 0.558 g

Explanation:
1. Milli-equivalents of thiosulphate = 40 × 0.11 = 4.4.
2. Equivalent weight of I2 is 127, so mass = 4.4 × 127/1000 = 0.558 g.
3. Answer: (b)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Question 17: In alkaline medium KMnO4 reacts as 2KMnO4 + 2KOH → 2K2MnO4 + H2O + O. Therefore its equivalent mass will be (molar mass 158):
(a) 31.6
(b) 52.7
(c) 79
(d) 158
Show Answer & Explanation

Answer: (d) 158

Explanation:
1. Mn goes from +7 to +6, a change of 1 electron per formula unit.
2. Equivalent mass = 158/1 = 158.
3. Answer: (d)

Teacher's Note:
1. Assign oxidation numbers first and track the electrons gained and lost. Check that electrons balance.

Free study material for Chemistry

Multiple Choice Questions (MCQs) for JEE Chemistry Redox Reactions

Download Multiple Choice Questions: Redox Reactions (JEE Chemistry)

Explore reliable practice questions for Redox Reactions tailored for JEE Chemistry learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

Concept Clarification for Redox Reactions

Cross-reference your completed choices with comprehensive NCERT solutions for JEE Chemistry to ensure absolute clarity across all sub-topics in this chapter.

Additional Study Resources for JEE Chemistry

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in JEE assessments.

FAQs

Where can I access latest JEE Chemistry Redox Reactions MCQs Set 20?

You can get most exhaustive JEE Chemistry Redox Reactions MCQs Set 20 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry Redox Reactions MCQs Set 20 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry Redox Reactions MCQs Set 20, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry Redox Reactions MCQs Set 20?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry Redox Reactions MCQs Set 20 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.