JEE Chemistry Redox Reactions MCQs Set 11

Practice 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.

Access Redox Reactions Questions and Solutions

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 acid solution, the change \( \ce{MnO4^- -> Mn^2+} \) involves
(a) oxidation by 3 electrons
(b) reduction by 3 electrons
(c) oxidation by 5 electrons
(d) reduction by 5 electrons
Show Answer & Explanation

Answer: (d) reduction by 5 electrons

Explanation:
1. Mn goes from +7 to +2, a decrease of 5 units.
2. So it is reduction, gaining 5 electrons.

Teacher's Note:
1. Oxidation is loss of electrons (oxidation number increases); reduction is gain of electrons (oxidation number decreases).
2. The oxidising agent is itself reduced and the reducing agent is itself oxidised.

Question 2: In the reaction \( \ce{4Fe + 3O2 -> 4Fe^3+ + 6O^2-} \), which statement is incorrect?
(a) It is a redox reaction
(b) Metallic iron is a reducing agent
(c) \( \ce{Fe^3+} \) is an oxidising agent
(d) Metallic iron is reduced to \( \ce{Fe^3+} \)
Show Answer & Explanation

Answer: (d) Metallic iron is reduced to \( \ce{Fe^3+} \)

Explanation:
1. Iron goes from 0 to +3, so it is oxidised (not reduced) and acts as a reducing agent.
2. Statement (d) is therefore wrong.

Teacher's Note:
1. Oxidation is loss of electrons (oxidation number increases); reduction is gain of electrons (oxidation number decreases).
2. The oxidising agent is itself reduced and the reducing agent is itself oxidised.

Question 3: When iron or zinc is added to \( \ce{CuSO4} \) solution, copper is precipitated. It is due to
(a) oxidation of \( \ce{Cu^2+} \)
(b) reduction of \( \ce{Cu^2+} \)
(c) hydrolysis of \( \ce{CuSO4} \)
(d) ionisation of \( \ce{CuSO4} \)
Show Answer & Explanation

Answer: (b) reduction of \( \ce{Cu^2+} \)

Explanation:
1. \( \ce{Cu^2+ + 2e^- -> Cu} \): the copper ions gain electrons and are reduced to metallic copper.

Teacher's Note:
1. Oxidation is loss of electrons (oxidation number increases); reduction is gain of electrons (oxidation number decreases).
2. The oxidising agent is itself reduced and the reducing agent is itself oxidised.

Question 4: In the reaction \( \ce{C2O4^2- + MnO4^- + H^+ -> Mn^2+ + CO2 + H2O} \) the reductant is
(a) \( \ce{C2O4^2-} \)
(b) \( \ce{MnO4^-} \)
(c) \( \ce{Mn^2+} \)
(d) \( \ce{H^+} \)
Show Answer & Explanation

Answer: (a) \( \ce{C2O4^2-} \)

Explanation:
1. Carbon goes from +3 (oxalate) to +4 (CO2): it is oxidised.
2. The species oxidised is the reductant: oxalate ion.

Teacher's Note:
1. Oxidation is loss of electrons (oxidation number increases); reduction is gain of electrons (oxidation number decreases).
2. The oxidising agent is itself reduced and the reducing agent is itself oxidised.

Question 5: Which of the following is the most powerful oxidising agent?
(a) \( \ce{F2} \)
(b) \( \ce{Cl2} \)
(c) \( \ce{I2} \)
(d) \( \ce{Br2} \)
Show Answer & Explanation

Answer: (a) \( \ce{F2} \)

Explanation:
1. Oxidising power decreases down the halogen group.
2. Fluorine has the highest reduction potential (+2.87 V).

Teacher's Note:
1. Oxidation is loss of electrons (oxidation number increases); reduction is gain of electrons (oxidation number decreases).
2. The oxidising agent is itself reduced and the reducing agent is itself oxidised.

Question 6: Of the four oxyacids of chlorine, the strongest oxidising agent in dilute aqueous solution is
(a) \( \ce{HClO4} \)
(b) \( \ce{HClO3} \)
(c) \( \ce{HClO2} \)
(d) \( \ce{HOCl} \)
Show Answer & Explanation

Answer: (d) \( \ce{HOCl} \)

Explanation:
1. In dilute aqueous solution the oxidising power of the oxyacids of chlorine decreases with increasing oxidation state of Cl.
2. HOCl (Cl = +1) is the strongest oxidising agent.

Teacher's Note:
1. A species whose element is in an intermediate oxidation state can act both as an oxidising and as a reducing agent.
2. In a balanced equation both atoms and charge must balance on the two sides.

Question 7: Identify the correct statement about \( \ce{H2O2} \).
(a) It acts as a reducing agent only
(b) It acts as both oxidising and reducing agent
(c) It is neither an oxidiser nor a reducer
(d) It acts as an oxidising agent only
Show Answer & Explanation

Answer: (b) It acts as both oxidising and reducing agent

Explanation:
1. Oxygen in \( \ce{H2O2} \) is in the intermediate oxidation state -1 (between -2 and 0).
2. It can therefore be reduced to -2 or oxidised to 0.

Teacher's Note:
1. A species whose element is in an intermediate oxidation state can act both as an oxidising and as a reducing agent.
2. In a balanced equation both atoms and charge must balance on the two sides.

Question 8: Which of the following behaves as both an oxidising and a reducing agent?
(a) \( \ce{H2SO4} \)
(b) \( \ce{SO2} \)
(c) \( \ce{H2S} \)
(d) \( \ce{HNO3} \)
Show Answer & Explanation

Answer: (b) \( \ce{SO2} \)

Explanation:
1. Sulphur in \( \ce{SO2} \) is +4, intermediate between -2 and +6.
2. In \( \ce{H2SO4} \) (+6) and \( \ce{HNO3} \) (+5) the element is at its maximum, and in \( \ce{H2S} \) (-2) at its minimum.

Teacher's Note:
1. A species whose element is in an intermediate oxidation state can act both as an oxidising and as a reducing agent.
2. In a balanced equation both atoms and charge must balance on the two sides.

Question 9: Which of the following is NOT a reducing agent?
(a) \( \ce{NaNO2} \)
(b) \( \ce{NaNO3} \)
(c) \( \ce{HI} \)
(d) \( \ce{SnCl2} \)
Show Answer & Explanation

Answer: (b) \( \ce{NaNO3} \)

Explanation:
1. Nitrogen in \( \ce{NaNO3} \) is already at its highest oxidation state (+5), so it can only be reduced.
2. The other three can be oxidised.

Teacher's Note:
1. A species whose element is in an intermediate oxidation state can act both as an oxidising and as a reducing agent.
2. In a balanced equation both atoms and charge must balance on the two sides.

Question 10: The strongest reducing agent among the halide ions is
(a) \( \ce{F^-} \)
(b) \( \ce{Cl^-} \)
(c) \( \ce{Br^-} \)
(d) \( \ce{I^-} \)
Show Answer & Explanation

Answer: (d) \( \ce{I^-} \)

Explanation:
1. The ease of oxidation of halide ions increases down the group.
2. Iodide is the strongest reducing agent.

Teacher's Note:
1. Oxidation is loss of electrons (oxidation number increases); reduction is gain of electrons (oxidation number decreases).
2. The oxidising agent is itself reduced and the reducing agent is itself oxidised.

Question 11: In which of the following reactions is \( \ce{H2O2} \) a reducing agent?
(a) \( \ce{2FeCl2 + 2HCl + H2O2 -> 2FeCl3 + 2H2O} \)
(b) \( \ce{Cl2 + H2O2 -> 2HCl + O2} \)
(c) \( \ce{2HI + H2O2 -> 2H2O + I2} \)
(d) \( \ce{H2SO3 + H2O2 -> H2SO4 + H2O} \)
Show Answer & Explanation

Answer: (b) \( \ce{Cl2 + H2O2 -> 2HCl + O2} \)

Explanation:
1. A reducing agent is oxidised. In (b) oxygen of H2O2 goes from -1 to 0 in O2 while Cl2 is reduced.
2. In (a), (c) and (d) H2O2 is reduced to water.

Teacher's Note:
1. Oxidation is loss of electrons (oxidation number increases); reduction is gain of electrons (oxidation number decreases).
2. The oxidising agent is itself reduced and the reducing agent is itself oxidised.

Question 12: In the reaction \( \ce{H2O + Br2 -> HOBr + HBr} \), the best description of the behaviour of bromine is
(a) oxidised only
(b) reduced only
(c) proton acceptor only
(d) both oxidised and reduced
Show Answer & Explanation

Answer: (d) both oxidised and reduced

Explanation:
1. Bromine goes from 0 to +1 (HOBr) and from 0 to -1 (HBr).
2. It is oxidised and reduced at the same time: a disproportionation.

Teacher's Note:
1. A species whose element is in an intermediate oxidation state can act both as an oxidising and as a reducing agent.
2. In a balanced equation both atoms and charge must balance on the two sides.

Question 13: In the reaction \( \ce{Ag2O + H2O2 -> 2Ag + H2O + O2} \), \( \ce{H2O2} \) acts as
(a) reducing agent
(b) oxidising agent
(c) bleaching agent
(d) none of the above
Show Answer & Explanation

Answer: (a) reducing agent

Explanation:
1. Ag(+1) is reduced to Ag(0) and oxygen of H2O2 is oxidised from -1 to 0.
2. So H2O2 is the reducing agent.

Teacher's Note:
1. Oxidation is loss of electrons (oxidation number increases); reduction is gain of electrons (oxidation number decreases).
2. The oxidising agent is itself reduced and the reducing agent is itself oxidised.

Question 14: In the reaction \( \ce{P + NaOH -> PH3 + NaH2PO2} \) (balanced form: \( \ce{4P + 3NaOH + 3H2O -> PH3 + 3NaH2PO2} \))
(a) P is oxidised only
(b) P is reduced only
(c) P is oxidised as well as reduced
(d) Na is reduced
Show Answer & Explanation

Answer: (c) P is oxidised as well as reduced

Explanation:
1. P goes from 0 to -3 in \( \ce{PH3} \) (reduction) and from 0 to +1 in \( \ce{NaH2PO2} \) (oxidation).

Teacher's Note:
1. A species whose element is in an intermediate oxidation state can act both as an oxidising and as a reducing agent.
2. In a balanced equation both atoms and charge must balance on the two sides.

Question 15: The oxidation number of Mn in \( \ce{MnO4^-} \) is
(a) +7
(b) -5
(c) +6
(d) +5
Show Answer & Explanation

Answer: (a) +7

Explanation:
1. x + 4(-2) = -1, so x = +7.

Teacher's Note:
1. Assign oxidation numbers with the rules: sum = charge of the species, O is usually -2, H is +1, halogens in halides -1.
2. Write x for the unknown element, then solve the sum of all oxidation numbers.

Question 16: The oxidation number of Ba in barium peroxide is
(a) +6
(b) +2
(c) 1
(d) +4
Show Answer & Explanation

Answer: (b) +2

Explanation:
1. Barium is an alkaline earth metal, always +2; the peroxide ion is \( \ce{O2^2-} \).

Teacher's Note:
1. Assign oxidation numbers with the rules: sum = charge of the species, O is usually -2, H is +1, halogens in halides -1.
2. Write x for the unknown element, then solve the sum of all oxidation numbers.

Question 17: Chlorine is in the +1 oxidation state in
(a) \( \ce{HCl} \)
(b) \( \ce{HClO4} \)
(c) \( \ce{ICl} \)
(d) \( \ce{Cl2O} \)
Show Answer & Explanation

Answer: (d) \( \ce{Cl2O} \)

Explanation:
1. In HCl and ICl chlorine is the more electronegative atom, so it is -1.
2. In \( \ce{HClO4} \) it is +7. In \( \ce{Cl2O} \): 2x + (-2) = 0, so x = +1.

Teacher's Note:
1. Assign oxidation numbers with the rules: sum = charge of the species, O is usually -2, H is +1, halogens in halides -1.
2. Write x for the unknown element, then solve the sum of all oxidation numbers.

Question 18: In the reaction \( \ce{Cl2 + H2S -> 2HCl + S} \), the oxidation number of sulphur changes from
(a) 0 to 2
(b) 2 to 0
(c) -2 to 0
(d) -2 to -1
Show Answer & Explanation

Answer: (c) -2 to 0

Explanation:
1. Sulphur in H2S is -2 and in elemental S it is 0: it is oxidised.

Teacher's Note:
1. Assign oxidation numbers with the rules: sum = charge of the species, O is usually -2, H is +1, halogens in halides -1.
2. Write x for the unknown element, then solve the sum of all oxidation numbers.

Question 19: The oxidation number of cobalt in \( \ce{K[Co(CO)4]} \) is
(a) +1
(b) +3
(c) -1
(d) -3
Show Answer & Explanation

Answer: (c) -1

Explanation:
1. K is +1, CO is neutral, and the complex ion carries a charge of -1.
2. So Co = -1.

Teacher's Note:
1. Assign oxidation numbers with the rules: sum = charge of the species, O is usually -2, H is +1, halogens in halides -1.
2. Write x for the unknown element, then solve the sum of all oxidation numbers.

Question 20: When \( \ce{K2Cr2O7} \) is converted to \( \ce{K2CrO4} \), the change in the oxidation state of chromium is
(a) 0
(b) 6
(c) 4
(d) 3
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. Chromium is +6 in both dichromate and chromate: the change is zero (no redox).

Teacher's Note:
1. Assign oxidation numbers with the rules: sum = charge of the species, O is usually -2, H is +1, halogens in halides -1.
2. Write x for the unknown element, then solve the sum of all oxidation numbers.

Question 21: In the conversion \( \ce{Br2 -> BrO3^-} \), the oxidation state of bromine changes from
(a) -1 to -1
(b) 0 to -1
(c) 0 to +5
(d) 0 to -5
Show Answer & Explanation

Answer: (c) 0 to +5

Explanation:
1. Br2 is 0; in \( \ce{BrO3^-} \): x + 3(-2) = -1, so x = +5.

Teacher's Note:
1. Assign oxidation numbers with the rules: sum = charge of the species, O is usually -2, H is +1, halogens in halides -1.
2. Write x for the unknown element, then solve the sum of all oxidation numbers.

Question 22: The oxidation state of sulphur in \( \ce{H2S2O8} \) (peroxodisulphuric acid) is
(a) +2
(b) +4
(c) +6
(d) +7
Show Answer & Explanation

Answer: (c) +6

Explanation:
1. The molecule has one peroxide link, so two oxygen atoms are -1 and six are -2.
2. 2(+1) + 2x + 2(-1) + 6(-2) = 0 gives x = +6.

Teacher's Note:
1. Assign oxidation numbers with the rules: sum = charge of the species, O is usually -2, H is +1, halogens in halides -1.
2. Write x for the unknown element, then solve the sum of all oxidation numbers.

Question 23: The oxidation state of nitrogen in \( \ce{N3H} \) (hydrazoic acid) is
(a) +1/3
(b) +3
(c) -1
(d) -1/3
Show Answer & Explanation

Answer: (d) -1/3

Explanation:
1. +1 + 3x = 0, so x = -1/3 (average oxidation state).

Teacher's Note:
1. Assign oxidation numbers with the rules: sum = charge of the species, O is usually -2, H is +1, halogens in halides -1.
2. Write x for the unknown element, then solve the sum of all oxidation numbers.

Question 24: The oxidation state of Pt in \( \ce{[Pt(C2H4)Cl3]^-} \) is
(a) +1
(b) +2
(c) +3
(d) +4
Show Answer & Explanation

Answer: (b) +2

Explanation:
1. Ethene is neutral and each Cl is -1: x + 0 + 3(-1) = -1, so x = +2.

Teacher's Note:
1. Assign oxidation numbers with the rules: sum = charge of the species, O is usually -2, H is +1, halogens in halides -1.
2. Write x for the unknown element, then solve the sum of all oxidation numbers.

Question 25: Sulphur has the highest oxidation state in
(a) \( \ce{SO2} \)
(b) \( \ce{H2SO4} \)
(c) \( \ce{Na2S2O3} \)
(d) \( \ce{Na2S4O6} \)
Show Answer & Explanation

Answer: (b) \( \ce{H2SO4} \)

Explanation:
1. SO2: +4; H2SO4: +6; Na2S2O3: +2; Na2S4O6: +2.5 (average).
2. The highest is +6 in H2SO4.

Teacher's Note:
1. Assign oxidation numbers with the rules: sum = charge of the species, O is usually -2, H is +1, halogens in halides -1.
2. Write x for the unknown element, then solve the sum of all oxidation numbers.

Free study material for Chemistry

Download Chapter MCQs: JEE Chemistry Redox Reactions

About Redox Reactions MCQs for 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.

How to Verify Your MCQ Answers

Each question includes structured solution keys mapped directly to standard JEE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

Enhance Speed with Online MCQ Tests

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 11?

You can get most exhaustive JEE Chemistry Redox Reactions MCQs Set 11 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 11 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 11, 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 11?

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