JEE Chemistry Redox Reactions MCQs Set 21

Download JEE MCQs for JEE Chemistry: Redox Reactions

Review structured MCQ sets for JEE Chemistry Redox Reactions. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Redox Reactions

View or download the dedicated Redox Reactions MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: Given E° values: O₂/H₂O = +1.23 V, S₂O₈²⁻/SO₄²⁻ = +2.05 V, Br₂/Br⁻ = +1.09 V, Au³⁺/Au = +1.40 V. The strongest oxidising agent is:
(a) Au³⁺
(b) O₂
(c) S₂O₈²⁻
(d) Br₂
Show Answer & Explanation

Answer: (c) S₂O₈²⁻

Explanation:
1. A higher standard reduction potential means a stronger oxidising agent.
2. S₂O₈²⁻ has the highest value, +2.05 V.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 2: Which of the following is an example of a disproportionation reaction?
(a) 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O
(b) 2NaBr + Cl₂ → 2NaCl + Br₂
(c) 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂
(d) 2CuBr → CuBr₂ + Cu
Show Answer & Explanation

Answer: (d) 2CuBr → CuBr₂ + Cu

Explanation:
1. In disproportionation the same element is both oxidised and reduced.
2. In 2CuBr → CuBr₂ + Cu, Cu(+1) goes to Cu(+2) and to Cu(0).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 3: Given Co³⁺/Co²⁺ = +1.81 V, Pb⁴⁺/Pb²⁺ = +1.67 V, Ce⁴⁺/Ce³⁺ = +1.61 V and Bi³⁺/Bi = +0.20 V, the oxidising power of the species increases in the order:
(a) Ce⁴⁺ < Pb⁴⁺ < Bi³⁺ < Co³⁺
(b) Bi³⁺ < Ce⁴⁺ < Pb⁴⁺ < Co³⁺
(c) Co³⁺ < Ce⁴⁺ < Bi³⁺ < Pb⁴⁺
(d) Co³⁺ < Pb⁴⁺ < Ce⁴⁺ < Bi³⁺
Show Answer & Explanation

Answer: (b) Bi³⁺ < Ce⁴⁺ < Pb⁴⁺ < Co³⁺

Explanation:
1. Oxidising power rises with the reduction potential.
2. Order of E°: 0.20 < 1.61 < 1.67 < 1.81.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 4: Which of the following reactions is an example of a redox reaction?
(a) XeF₄ + O₂F₂ → XeF₆ + O₂
(b) XeF₂ + PF₅ → [XeF]⁺PF₆⁻
(c) XeF₆ + H₂O → XeOF₄ + 2HF
(d) XeF₆ + 2H₂O → XeO₂F₂ + 4HF
Show Answer & Explanation

Answer: (a) XeF₄ + O₂F₂ → XeF₆ + O₂

Explanation:
1. In XeF₄ + O₂F₂ → XeF₆ + O₂, Xe goes from +4 to +6 and O from +1 to 0.
2. The other reactions involve no change in oxidation states.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 5: Given E°(Cl₂/Cl⁻) = 1.36 V, E°(Cr³⁺/Cr) = −0.74 V, E°(Cr₂O₇²⁻/Cr³⁺) = 1.33 V and E°(MnO₄⁻/Mn²⁺) = 1.51 V, the strongest reducing agent among Cr, Mn²⁺, Cr³⁺ and Cl⁻ is:
(a) Cr
(b) Mn²⁺
(c) Cr³⁺
(d) Cl⁻
Show Answer & Explanation

Answer: (a) Cr

Explanation:
1. The lowest reduction potential belongs to the strongest reducing agent.
2. Cr³⁺/Cr = −0.74 V is the lowest, so Cr is the strongest reducing agent.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 6: In which of the following reactions does H₂O₂ act as a reducing agent? (1) H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O (2) H₂O₂ − 2e⁻ → O₂ + 2H⁺ (3) H₂O₂ + 2e⁻ → 2OH⁻ (4) H₂O₂ + 2OH⁻ − 2e⁻ → O₂ + 2H₂O
(a) 1, 2
(b) 3, 4
(c) 1, 3
(d) 2, 4
Show Answer & Explanation

Answer: (d) 2, 4

Explanation:
1. A reducing agent loses electrons.
2. In (2) and (4) H₂O₂ loses electrons and goes to O₂.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 7: In the reaction x MnO₄⁻ + y C₂O₄²⁻ + z H⁺ → x Mn²⁺ + 2y CO₂ + (z/2) H₂O, the values of x, y and z are respectively:
(a) 5, 2 and 16
(b) 2, 5 and 8
(c) 2, 5 and 16
(d) 5, 2 and 8
Show Answer & Explanation

Answer: (c) 2, 5 and 16

Explanation:
1. MnO₄⁻ gains 5 e⁻ and each C₂O₄²⁻ loses 2 e⁻, so x : y = 2 : 5.
2. Balancing charge and hydrogen gives z = 16.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 8: Given E°(Cr³⁺/Cr) = −0.74 V, E°(MnO₄⁻/Mn²⁺) = 1.51 V, E°(Cr₂O₇²⁻/Cr³⁺) = 1.33 V and E°(Cl₂/Cl⁻) = 1.36 V, the strongest oxidising agent among Cl₂, Cr³⁺, Mn²⁺ and MnO₄⁻ is:
(a) Cl₂
(b) Cr³⁺
(c) Mn²⁺
(d) MnO₄⁻
Show Answer & Explanation

Answer: (d) MnO₄⁻

Explanation:
1. Choose the species with the highest reduction potential.
2. MnO₄⁻/Mn²⁺ = 1.51 V is the highest.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 9: The products expected from the disproportionation of hypochlorous acid are:
(a) HClO₃ and Cl₂O
(b) HClO₂ and HClO₄
(c) HCl and Cl₂O
(d) HCl and HClO₃
Show Answer & Explanation

Answer: (d) HCl and HClO₃

Explanation:
1. 3HOCl → 2HCl + HClO₃.
2. Cl(+1) goes to Cl(−1) in HCl and to Cl(+5) in HClO₃.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 10: Which of the following reactions depicts the oxidising behaviour of H₂SO₄?
(a) 2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O
(b) Ca(OH)₂ + H₂SO₄ → CaSO₄ + 2H₂O
(c) NaCl + H₂SO₄ → NaHSO₄ + HCl
(d) 2PCl₅ + H₂SO₄ → 2POCl₃ + 2HCl + SO₂Cl₂
Show Answer & Explanation

Answer: (a) 2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O

Explanation:
1. In the first reaction S goes from +6 in H₂SO₄ to +4 in SO₂, so H₂SO₄ is reduced and acts as the oxidising agent.
2. The others involve no change in oxidation state.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 11: The oxidation state of Cr in [Cr(NH₃)₄Cl₂]⁺ is:
(a) 0
(b) +1
(c) +2
(d) +3
Show Answer & Explanation

Answer: (d) +3

Explanation:
1. NH₃ is neutral and Cl is −1.
2. x + 0 + 2(−1) = +1, so x = +3.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 12: The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is:
(a) +3
(b) +2
(c) +6
(d) +4
Show Answer & Explanation

Answer: (a) +3

Explanation:
1. Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂.
2. Dichromate is reduced to Cr³⁺.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 13: Standard reduction potentials of three metals A, B and C are +0.5 V, −3.0 V and −1.2 V respectively. The reducing power of these metals is:
(a) B > C > A
(b) A > B > C
(c) C > B > A
(d) A > C > B
Show Answer & Explanation

Answer: (a) B > C > A

Explanation:
1. The more negative the reduction potential, the stronger the reducing agent.
2. −3.0 < −1.2 < +0.5, so B > C > A.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 14: The oxidation number of Cl in CaOCl₂ (bleaching powder) is:
(a) zero, since it contains Cl₂
(b) −1, since it contains Cl⁻
(c) +1, since it contains ClO⁻
(d) +1 and −1, since it contains ClO⁻ and Cl⁻
Show Answer & Explanation

Answer: (d) +1 and −1, since it contains ClO⁻ and Cl⁻

Explanation:
1. CaOCl₂ is Ca(OCl)Cl.
2. Cl in OCl⁻ is +1 and Cl in Cl⁻ is −1.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 15: MnO₄⁻ is a good oxidising agent in different media, changing to Mn²⁺, MnO₄²⁻, MnO₂ and Mn₂O₃ respectively. The changes in oxidation number are respectively:
(a) 1, 3, 4, 5
(b) 5, 4, 3, 2
(c) 5, 1, 3, 4
(d) 2, 6, 4, 3
Show Answer & Explanation

Answer: (c) 5, 1, 3, 4

Explanation:
1. Mn is +7 in MnO₄⁻.
2. Mn²⁺ (+2) change 5; MnO₄²⁻ (+6) change 1; MnO₂ (+4) change 3; Mn₂O₃ (+3) change 4.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 16: Which of the following is a redox reaction?
(a) NaCl + KNO₃ → NaNO₃ + KCl
(b) CaC₂O₄ + 2HCl → CaCl₂ + H₂C₂O₄
(c) Ca(OH)₂ + 2NH₄Cl → CaCl₂ + 2NH₃ + 2H₂O
(d) 2K[Ag(CN)₂] + Zn → 2Ag + K₂[Zn(CN)₄]
Show Answer & Explanation

Answer: (d) 2K[Ag(CN)₂] + Zn → 2Ag + K₂[Zn(CN)₄]

Explanation:
1. Only in the last reaction does an oxidation state change: Ag(+1) → Ag(0) and Zn(0) → Zn(+2).
2. The others are ionic or acid-base reactions.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 17: Which of the following reactions is possible at the anode?
(a) F₂ + 2e⁻ → 2F⁻
(b) 2H⁺ + ½O₂ + 2e⁻ → H₂O
(c) 2Cr³⁺ + 7H₂O → Cr₂O₇²⁻ + 14H⁺ + 6e⁻
(d) None of these
Show Answer & Explanation

Answer: (c) 2Cr³⁺ + 7H₂O → Cr₂O₇²⁻ + 14H⁺ + 6e⁻

Explanation:
1. Oxidation takes place at the anode.
2. Only 2Cr³⁺ + 7H₂O → Cr₂O₇²⁻ + 14H⁺ + 6e⁻ loses electrons.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 18: In which of the following does the oxidation state of Xe change?
(a) XeF₂ + PF₅ → [XeF]⁺[PF₆]⁻
(b) XeF₆ + H₂O → XeOF₄ + 2HF
(c) XeF₄ + O₂F₂ → XeF₆ + O₂
(d) XeF₆ + 2H₂O → XeO₂F₂ + 4HF
Show Answer & Explanation

Answer: (c) XeF₄ + O₂F₂ → XeF₆ + O₂

Explanation:
1. In XeF₄ (+4) the Xe goes to XeF₆ (+6) when O₂F₂ is reduced to O₂.
2. In the other three Xe keeps its oxidation state.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Free study material for Chemistry

Redox Reactions Objective Questions & Solutions for JEE Chemistry

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

Follow up your worksheet practice by attempting the interactive online Chemistry MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Chemistry Redox Reactions MCQs Set 21?

You can get most exhaustive JEE Chemistry Redox Reactions MCQs Set 21 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 21 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 21, 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 21?

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 21 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.