Redox Reactions JEE Main Questions 2025: Important Practice Questions with PYQs

Soham Mitra

Updated On: January 21, 2025 11:18 AM

The Important JEE Main Questions 2025 for Redox Reactions cover various topics of Redox Reactions Electronic Concepts of Oxidation and Reduction, Electrochemical Cells, Electrolytic and Metallic Conduction, Kohlrausch’s Law etc. Candidates can expect 1 or 2 questions from this topic. 
Redox Reactions JEE Main Questions 2025 Important Practice Questions

Redox Reactions JEE Main Questions 2025 holds a weightage of 3.3% of the total Chemistry Section. The Redox Reactions section of the JEE Main 205 Chemistry Syllabus is an important chapter and it deals with chemical reactions in which electrons are transferred between two reactants participating in it. The Important JEE Main 2025 questions for Redox Reactions cover various topics of Redox Reactions such as Electronic Concepts of Oxidation and Reduction, Electrochemical Cells, Electrolytic and Metallic Conduction, Kohlrausch’s Law etc. Candidates can find 1 question from the Redox Reactions chapter in the JEE Main 2025 exam. Therefore, candidates must accustom themselves with the important practice questions of the Redox Reactions chapter for JEE Main 2025. Students are advised to go through this article thoroughly for the JEE Main Redox Reactions questions along with their solutions.

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JEE Main 2025 Redox Reactions Important Questions

Candidates can check the practice questions for JEE Main Redox Reactions 2025 along with their solutions as provided in the table below.

Serial No.

Question

Solution

1.

Consider the following reaction:

xMnO4–+ yC2O42- +zH+ → xMn2+ + 2yCO2+(z/2)H2O

The values of x, y and z in the reaction are respectively:-

(1) 5, 2 and 16

(2) 2, 5 and 8

(3) 2, 5 and 16

(4) 5, 2 and 8

The balanced equation is given below.

2MnO4–+ 5C2O42- + 16 H+ → 2Mn2+ + 10CO2+8H2O

The values of x, y and z are 2, 5 and 16, respectively.

Hence option (3) is the answer.

2.

How many electrons are involved in the following redox reaction?

Cr2O72- + Fe2+ + C2O42- → Cr3+ + Fe3+ + CO2 (Unbalanced)

(1) 3

(2) 4

(3) 5

(4) 6

A redox reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron. Chromium and iron are involved in the reaction which is oxidised and reduced. So, a total of 6 electrons are involved in this redox reaction.

Hence option (4) is the answer.

3.

When KMn04 acts as an oxidising agent and ultimately forms [MnO42-, MnO2, Mn2O3 and Mn+2. Then the number of electrons trans­ferred in each case respectively is

(1) 4, 3, 1, 5

(2) 1, 5, 3, 7

(3) 1,3, 4, 5

(4) 3, 5, 7,1

The oxidation number of Mn in KMn04, MnO42-, MnO2, Mn2O3 and Mn+2 7, 6, 4, 3 and 2 respectively. The number of electrons transferred corresponds to the change in the oxidation number. When KMn04 acts as an oxidising agent and ultimately forms MnO42-, MnO2, Mn2O3 and Mn+2, then the number of electrons transferred in each case are 1,3,4,5 respectively.

Hence option (3) is the answer.

4.

Excess of KI reacts with CuSO4 solution and then Na2S2O3solution is added to it. Which of the statements is incorrect for this reaction?

(1) Cu2I2 is reduced

(2) Evolved I2 is reduced

(3) Na2S2O3is oxidized

(4) CuI2 is formed

2CuSo2 + 4KI → Cu2I2 +2K2SO4 + I2

I2 +2Na2S2O3 → Na2S4O6 + 2NaI

Here statement (4) is incorrect.

Hence option (4) is the answer.

5.

The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is

(1) +4

(2) +6

(3) +2

(4) +3

K2Cr2O7 + 7H2SO4 + 6KI → Cr2(SO4) + 3I2 + 7H2O + 4K2SO4

Cr gets reduced from +6 Oxidation state to +3 oxidation state.

Hence option (4) is the answer.

6.

Consider the reaction

H2SO3(aq) + Sn4+(aq) + H2O(l) → Sn2+(aq) + HSO4–(aq) + 3H+(aq)

Which of the following statements is correct?

(1) H2SO3 is the reducing agent because it undergoes oxidation

(2) H2SO3 is the reducing agent because it undergoes reduction

(3) Sn4+ is the reducing agent because it undergoes oxidation

(4) Sn4+ is the oxidizing agent because it undergoes oxidation

Oxidation is the loss of electrons during a reaction by a molecule. In the given equation, H2SO3 is the reducing agent because it undergoes oxidation.

Hence option (1) is the answer.

7.

Which of the following is a redox reaction?

(1) NaCl + KN03 → NaN03 + KC1

(2) CaC204 + 2HC1 → CaCl2 + H2C20,

(3) Mg(OH)2 + 2NH4C1 → MgCl2 + 2NH4OH

(4) Zn + 2AgCN → 2Ag + Zn(CN)2

A redox reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron. The oxidation state shows a change only in a reaction between zinc and cyanide.

Hence option (4) is the answer.

8.

What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

(1) Cr2O72- and H2O are formed

(2) Cr2O72- is reduced to +3 state of Cr

(3) Cr2O72- is oxidises to +7 state of Cr

(4) Cr3+ and Cr2O72- are formed

Dilute HNO3 is an oxidising agent.

2K2CrO4 + 2HNO3(dil) → K2Cr2O72KNO3 + H2O

CrO42- + 2HNO3 (dil) → Cr2O72- + 2NO3– + H2O

Hence option (1) is the answer.

9.

In which of the following reaction H2O2 acts as a reducing agent?

(1) H2O2 + 2H+ + 2e– → 2H2O

(2) H2O2 -2e– → O2+2H+

(3) H2O2 + 2e– → 2OH–

(4) H2O2 + 2OH– -2e– → O2 + 2H2O

(1) (1), (3)

(2) (2), (4)

(3) (1), (2)

(4) (3), (4)

Reducing agent is an element or compound that loses an electron to an electron recipient in a redox chemical reaction. In (2) and (4) , H2O2 acts as a reducing agent.

Hence option (2) is the answer.

10.

Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl because HCl

(1) gets oxidised by oxalic acid to chlorine

(2) furnishes H+ ions in addition to those from oxalic acid

(3) reduces permanganate to Mn2+

(4) Oxidises oxalic acid to carbon dioxide and water

HCl is a strong reducing agent. It reduces permanganate to Mn2+.

Hence option (3) is the answer.

11.

Which of the following reactions is an example of a redox reaction ?

(1) XeF4 + O2F2 → XeF6 + O2

(2) XeF2 + PF5 → [XeF]+ PF6–

(3) XeF6 + H2O → XeOF4 + 2HF

(4) XeF6 + 2H2O → XeO2F2 + 4HF

In equation (1) Xe undergoes oxidation and oxygen undergoes reduction.

Hence option (1) is the answer.

12.

For the redox reaction: Zn(s) + Cu2+ (0.1 M)→ Zn+ (1M) + Cu(s) taking place in a cell, E°cell is 1.10 volt. Ecell for the cell will be (2.303 RT / F = 0.0591)

(1) 2.14 V

(2) 1.80 V

(3) 1.07 V

(4) 0.82 V

Ecell = E0cell – (0.0591/n) log(1/0.1)

E0cell = 1.10 V

n = 2

Ecell = 1.10 – (0.0591/2) log(10)

= 1.10 – 0.0295

= 1.0705 V

Hence option (3) is the answer.

13.

The highest electrical conductivity of the following aqueous solutions is of

(1) 1 M acetic acid

(2) 1 M chloroacetic acid

(3) 1 M fluoroacetic acid

(4) 1 M difluoroacetic acid

More the acidity more will be the tendency to release protons. So lighter will be the electrical conductivity. Difluoroacetic acid will be the strongest acid because of the electron-withdrawing effect of two fluorine atoms so as it will show maximum electrical conductivity.

Hence option (4) is the answer.

14.

An alkali is titrated against an acid with methyl orange as an indicator, which of the following is a correct combination?

Base

Acid

Endpoint

1

strong

strong

Pinkish red to yellow

2

weak

strong

Yellow to pinkish-red

3

strong

strong

Pink to colourless

4

weak

strong

Colourless to pink

When methyl orange is added to a weak base solution, the solution becomes yellow. When the solution is titrated with a strong acid, after the endpoint, the solution is acidic. So the solution becomes pinkish-red.

Hence option (2) is the answer.

15.

Given : XNa2HAsO3 +YNaBrO3+ZHCl → NaBr + H3AsO4 + NaCl

The values of X, Y and Z in the above redox reaction are respectively :

(1) 2, 1, 3

(2) 3, 1, 6

(3) 2, 1, 2

(4) 3, 1, 4

The balanced equation is given below.

3Na2HAsO3 + NaBrO3 + 6HCl → NaBr + 3H3AsO4 + 6NaCl

The value of X, Y and Z are 3, 1 and 6 respectively.

Hence option (2) is the answer.


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We hope this article has been helpful for the candidates to be aware of the important Redox Reactions JEE Main Questions 2025. You can just fill out the Common Application Form on our website, it is a single application form for securing admission into various universities. For more articles and updates on JEE Main 2025, stay tuned with CollegeDekho.

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FAQs

Is redox reaction important for the JEE Mains exam?

Yes, JEE Main Chemistry Redox Reactions chapter is important for the JEE Main exam. Candidates should study the essential concepts, formulas, and key points, to better their preparations before the JEE Main exam.

Which is the hardest topic in the JEE Mains exam?

Mathematics is often considered to be the most challenging subject in the JEE Main exam, but its difficulty level varies from person to person. Some students may find other subjects more challenging based upon their strengths and weaknesses.

Is 80 marks in chemistry easy to score in the JEE Mains exam?

In the JEE Main Chemistry section, the questions asked are generally straightforward and the difficulty level is relatively lower. With the correct preparation approach in place, candidates can easily score 80+ in the JEE Mains Chemistry section.

Is the JEE Mains chemistry section easy?

Chemistry is considered to be the rank decider for the majority of candidates in any engineering entrance examination because it mostly deals with the basics and. In comparison to the Mathematics and Physics questions asked in the JEE Main exam, Chemistry questions are more theoretical than calculative, making it relatively easier for candidates to secure good marks.

Is NCERT enough for the JEE Mains 2025 preparation?

NCERT books are very important to prepare for the JEE Main 2025 exam. The JEE Main exam follows the CBSE syllabus for 11th and 12th classes, so NCERT books are considered to be the most reliable and necessary study materials. NCERT books explain basic fundamental concepts in a simple and clear way.

Is it easy to score 99 percentile in the JEE Mains exam?

To score a 99+ percentile in the JEE Main 2025 exam, candidates require more than just knowing the JEE Main syllabus, as it required consistent practice, strategic revision, time management, and the ability to solve complex problems under pressure. However, with the appropriate resources, one can achieve these requirements.

Can I leave chemistry for the JEE Main exam?

All sections namely Mathematics, Physics, and Chemistry are mandatory in the JEE Main exam, with questions covering the entire JEE Main syllabus. To be eligible for admission, candidates must achieve the minimum qualifying marks in each subject and the overall aggregate.

Will the JEE Mains syllabus change in 2025?

Candidates can study from the NCERT textbooks of class 11 and 12 Physics, Chemistry and Mathematics to cover the JEE Mains syllabus. No changes have been made to the JEE Main Syllabus 2025. NTA removed several topics of Physics, Chemistry and Mathematics last year from the JEE Main syllabus.

Is 75% required for the JEE Main exam?

Candidates must keep in mind that there is no requirement of 75% marks in class 12 for appearing in the JEE Mains exam. The JEE Main eligibility criteria of minimum 75% marks in class 12 is required at the time of securing admission across NITs, IIITs and GFTIs. Candidates can apply and appear in the JEE Mains 2025 exam irrespective of their class 12 marks.

Is PYQ enough for the JEE Mains preparation?

Previous year question papers are an important source for the JEE Main preparations. However, relying solely upon PYQs is not enough to fully prepare for the JEE Mains exam. Candidates must also focus on covering the entire syllabus, developing strong conceptual understanding, practising time management, and attempting mock tests.

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