Redox Reactions JEE Main Questions 2025: Important Practice Questions with PYQs
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
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.
Quick Links:
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?
| 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. |
JEE Main Preparatory Articles: