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Redox Reaction: JEE Main Previous Year Questions (2020-2026)
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Page 1 JEE Main Previous Year Questions (2021-2026): Redox Reactions (January 2026) Q1: One mole of Cl 2(g) was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of Cl ?, ClO ? and OH ? are respectively (assume volume remains constant): A: 0.5 M, 0.5 M, 0.5 M B: 1 M, 1 M, 1 M C: 0.5 M, 0.5 M, 1 M D: 0.75 M, 0.75 M, 1 M Answer: C Explanation: In cold KOH, chlorine disproportionates: Cl 2 +2OH - ? Cl - + ClO - + H 2 O Initial moles of OH ? in 2 L of 2 M KOH: n(OH-) = 2 × 2 = 4 mol Given 1 mol Cl 2: OH - consumed = 2 × 1 = 2 mol Cl - formed = 1 mol ClO - formed = 1 mol OH - remaining = 4 - 2 = 2 mol Volume = 2 L Answer is: 0.5 M, 0.5 M, 1 M. Q2: The oxidation state of chromium in the final product formed in the reaction between KI and acidified K 2Cr 2O 7 solution is: A: +2 B: +3 C: +4 D: +6 Answer: B Explanation: In acidified K 2Cr 2O 7, chromium is in the +6 oxidation state (in Cr 2O 7² ?). When KI is added in acidic medium, I ? is oxidised to I 2 and dichromate is reduced to Cr³ ?. Reduction half-reaction: Page 2 JEE Main Previous Year Questions (2021-2026): Redox Reactions (January 2026) Q1: One mole of Cl 2(g) was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of Cl ?, ClO ? and OH ? are respectively (assume volume remains constant): A: 0.5 M, 0.5 M, 0.5 M B: 1 M, 1 M, 1 M C: 0.5 M, 0.5 M, 1 M D: 0.75 M, 0.75 M, 1 M Answer: C Explanation: In cold KOH, chlorine disproportionates: Cl 2 +2OH - ? Cl - + ClO - + H 2 O Initial moles of OH ? in 2 L of 2 M KOH: n(OH-) = 2 × 2 = 4 mol Given 1 mol Cl 2: OH - consumed = 2 × 1 = 2 mol Cl - formed = 1 mol ClO - formed = 1 mol OH - remaining = 4 - 2 = 2 mol Volume = 2 L Answer is: 0.5 M, 0.5 M, 1 M. Q2: The oxidation state of chromium in the final product formed in the reaction between KI and acidified K 2Cr 2O 7 solution is: A: +2 B: +3 C: +4 D: +6 Answer: B Explanation: In acidified K 2Cr 2O 7, chromium is in the +6 oxidation state (in Cr 2O 7² ?). When KI is added in acidic medium, I ? is oxidised to I 2 and dichromate is reduced to Cr³ ?. Reduction half-reaction: Thus, chromium in the final product is Cr³ ?, so the oxidation state is +3. Correct option: B (+3) Q3: 500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO4 solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na2S2O3 solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of Na2S2O3 consumed is ______. (Nearest integer) Answer: 3 Explanation: In basic medium, KMnO4 is reduced to MnO2 and I? is oxidised to I2. Step 1: Balance the redox reaction in basic medium Half-reactions: Reduction: Step 2: Calculate moles of reactants Moles of KI: n(I?) = 0.5 × 1.2 = 0.6 mol Moles of KMnO4: Page 3 JEE Main Previous Year Questions (2021-2026): Redox Reactions (January 2026) Q1: One mole of Cl 2(g) was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of Cl ?, ClO ? and OH ? are respectively (assume volume remains constant): A: 0.5 M, 0.5 M, 0.5 M B: 1 M, 1 M, 1 M C: 0.5 M, 0.5 M, 1 M D: 0.75 M, 0.75 M, 1 M Answer: C Explanation: In cold KOH, chlorine disproportionates: Cl 2 +2OH - ? Cl - + ClO - + H 2 O Initial moles of OH ? in 2 L of 2 M KOH: n(OH-) = 2 × 2 = 4 mol Given 1 mol Cl 2: OH - consumed = 2 × 1 = 2 mol Cl - formed = 1 mol ClO - formed = 1 mol OH - remaining = 4 - 2 = 2 mol Volume = 2 L Answer is: 0.5 M, 0.5 M, 1 M. Q2: The oxidation state of chromium in the final product formed in the reaction between KI and acidified K 2Cr 2O 7 solution is: A: +2 B: +3 C: +4 D: +6 Answer: B Explanation: In acidified K 2Cr 2O 7, chromium is in the +6 oxidation state (in Cr 2O 7² ?). When KI is added in acidic medium, I ? is oxidised to I 2 and dichromate is reduced to Cr³ ?. Reduction half-reaction: Thus, chromium in the final product is Cr³ ?, so the oxidation state is +3. Correct option: B (+3) Q3: 500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO4 solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na2S2O3 solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of Na2S2O3 consumed is ______. (Nearest integer) Answer: 3 Explanation: In basic medium, KMnO4 is reduced to MnO2 and I? is oxidised to I2. Step 1: Balance the redox reaction in basic medium Half-reactions: Reduction: Step 2: Calculate moles of reactants Moles of KI: n(I?) = 0.5 × 1.2 = 0.6 mol Moles of KMnO4: Hence, KMnO4 is the limiting reagent. Step 3: Moles of I 2 liberated Step 4: Titration with thiosulfate Reaction: Answer (nearest integer) Q4: X and Y are the number of electrons involved, respectively during the oxidation of I? to I2 and S²? to S by acidi?ed K2Cr2O7. The value of X + Y is ______. Answer: 12 Explanation: In this reaction, 6I - ions get oxidised to 3I 2 Each I - loses 1 electron during oxidation, so total electrons lost (in moles) = X = 6. Here, 3S 2- ions get oxidised to 3S. Each S2 - loses 2 electrons to become S, so total electrons lost (in moles) = Y = 3 x 2 = 6. Therefore, x + y = 6 + 6 = 12. Q5: 200 cc of x × 10?³M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's Page 4 JEE Main Previous Year Questions (2021-2026): Redox Reactions (January 2026) Q1: One mole of Cl 2(g) was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of Cl ?, ClO ? and OH ? are respectively (assume volume remains constant): A: 0.5 M, 0.5 M, 0.5 M B: 1 M, 1 M, 1 M C: 0.5 M, 0.5 M, 1 M D: 0.75 M, 0.75 M, 1 M Answer: C Explanation: In cold KOH, chlorine disproportionates: Cl 2 +2OH - ? Cl - + ClO - + H 2 O Initial moles of OH ? in 2 L of 2 M KOH: n(OH-) = 2 × 2 = 4 mol Given 1 mol Cl 2: OH - consumed = 2 × 1 = 2 mol Cl - formed = 1 mol ClO - formed = 1 mol OH - remaining = 4 - 2 = 2 mol Volume = 2 L Answer is: 0.5 M, 0.5 M, 1 M. Q2: The oxidation state of chromium in the final product formed in the reaction between KI and acidified K 2Cr 2O 7 solution is: A: +2 B: +3 C: +4 D: +6 Answer: B Explanation: In acidified K 2Cr 2O 7, chromium is in the +6 oxidation state (in Cr 2O 7² ?). When KI is added in acidic medium, I ? is oxidised to I 2 and dichromate is reduced to Cr³ ?. Reduction half-reaction: Thus, chromium in the final product is Cr³ ?, so the oxidation state is +3. Correct option: B (+3) Q3: 500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO4 solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na2S2O3 solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of Na2S2O3 consumed is ______. (Nearest integer) Answer: 3 Explanation: In basic medium, KMnO4 is reduced to MnO2 and I? is oxidised to I2. Step 1: Balance the redox reaction in basic medium Half-reactions: Reduction: Step 2: Calculate moles of reactants Moles of KI: n(I?) = 0.5 × 1.2 = 0.6 mol Moles of KMnO4: Hence, KMnO4 is the limiting reagent. Step 3: Moles of I 2 liberated Step 4: Titration with thiosulfate Reaction: Answer (nearest integer) Q4: X and Y are the number of electrons involved, respectively during the oxidation of I? to I2 and S²? to S by acidi?ed K2Cr2O7. The value of X + Y is ______. Answer: 12 Explanation: In this reaction, 6I - ions get oxidised to 3I 2 Each I - loses 1 electron during oxidation, so total electrons lost (in moles) = X = 6. Here, 3S 2- ions get oxidised to 3S. Each S2 - loses 2 electrons to become S, so total electrons lost (in moles) = Y = 3 x 2 = 6. Therefore, x + y = 6 + 6 = 12. Q5: 200 cc of x × 10?³M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here x = ______ . Answer: 375 Explanation: In acidic medium, dichromate oxidises Fe 2+ to Fe 3+ Half-reactions (NCERT method): Page 5 JEE Main Previous Year Questions (2021-2026): Redox Reactions (January 2026) Q1: One mole of Cl 2(g) was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of Cl ?, ClO ? and OH ? are respectively (assume volume remains constant): A: 0.5 M, 0.5 M, 0.5 M B: 1 M, 1 M, 1 M C: 0.5 M, 0.5 M, 1 M D: 0.75 M, 0.75 M, 1 M Answer: C Explanation: In cold KOH, chlorine disproportionates: Cl 2 +2OH - ? Cl - + ClO - + H 2 O Initial moles of OH ? in 2 L of 2 M KOH: n(OH-) = 2 × 2 = 4 mol Given 1 mol Cl 2: OH - consumed = 2 × 1 = 2 mol Cl - formed = 1 mol ClO - formed = 1 mol OH - remaining = 4 - 2 = 2 mol Volume = 2 L Answer is: 0.5 M, 0.5 M, 1 M. Q2: The oxidation state of chromium in the final product formed in the reaction between KI and acidified K 2Cr 2O 7 solution is: A: +2 B: +3 C: +4 D: +6 Answer: B Explanation: In acidified K 2Cr 2O 7, chromium is in the +6 oxidation state (in Cr 2O 7² ?). When KI is added in acidic medium, I ? is oxidised to I 2 and dichromate is reduced to Cr³ ?. Reduction half-reaction: Thus, chromium in the final product is Cr³ ?, so the oxidation state is +3. Correct option: B (+3) Q3: 500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO4 solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na2S2O3 solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of Na2S2O3 consumed is ______. (Nearest integer) Answer: 3 Explanation: In basic medium, KMnO4 is reduced to MnO2 and I? is oxidised to I2. Step 1: Balance the redox reaction in basic medium Half-reactions: Reduction: Step 2: Calculate moles of reactants Moles of KI: n(I?) = 0.5 × 1.2 = 0.6 mol Moles of KMnO4: Hence, KMnO4 is the limiting reagent. Step 3: Moles of I 2 liberated Step 4: Titration with thiosulfate Reaction: Answer (nearest integer) Q4: X and Y are the number of electrons involved, respectively during the oxidation of I? to I2 and S²? to S by acidi?ed K2Cr2O7. The value of X + Y is ______. Answer: 12 Explanation: In this reaction, 6I - ions get oxidised to 3I 2 Each I - loses 1 electron during oxidation, so total electrons lost (in moles) = X = 6. Here, 3S 2- ions get oxidised to 3S. Each S2 - loses 2 electrons to become S, so total electrons lost (in moles) = Y = 3 x 2 = 6. Therefore, x + y = 6 + 6 = 12. Q5: 200 cc of x × 10?³M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here x = ______ . Answer: 375 Explanation: In acidic medium, dichromate oxidises Fe 2+ to Fe 3+ Half-reactions (NCERT method): Q1: The species which does not undergo disproportionation reaction is: JEE Main 2025 (Online) 22nd January Evening Shift Options: A. ClO 4 - B. ClO 3 - C. ClO - D. ClO 2 - Ans: A Solution: Oxidation States: In ClO 4 - (perchlorate), chlorine is in the +7 oxidation state. In ClO 3 - (chlorate), chlorine is in the +5 oxidation state. In ClO 2 - (chlorite), chlorine is in the +3 oxidation state. In ClO - (hypochlorite), chlorine is in the +1 oxidation state. Disproportionation Reaction: A disproportionation reaction is one in which a species simultaneously undergoes oxidation and reduction. For this to occur, the element must be in an intermediate oxidation state such that it can be oxidized to a higher state and reduced to a lower one. In ClO - and ClO 2 - , chlorine is in lower oxidation states ( +1 and +3 , respectively), making them susceptible to disproportionation. For example, hypochlorite can disproportionate in basic solution as follows: 3ClO - ? 2Cl - + ClO 3 - In ClO 3 - , chlorine is in an intermediate oxidation state ( +5 ) that, under certain conditions, can undergo disproportionation. However, in ClO 4 - , chlorine is in its highest possible oxidation state ( +7 ) and cannot be oxidized further. Since disproportionation requires one part of the species to be oxidized and the other reduced, ClO 4 - is thermodynamically stable and does not undergo disproportionation. ClO 4 - (perchlorate ion) does not undergo disproportionation. Q2: Which of the following oxidation reactions are carried out by both ?? ?? ???? ?? ?? ?? and ???????? ?? in acidic medium? A. ?? - ? ?? ?? B. ?? ?? - ? ?? C. ???? ?? + ? ???? ?? + D. ?? - ? ???? ?? - E. ?? ?? ?? ?? ?? - ? ???? ?? ?? - Choose the correct answer from the options given below :Read More
FAQs on Redox Reaction: JEE Main Previous Year Questions (2020-2026)
| 1. What is a redox reaction and why is it important in chemistry? |  |
Ans. A redox reaction, short for reduction-oxidation reaction, involves the transfer of electrons between two species. In such reactions, one species gets oxidized (loses electrons) while the other gets reduced (gains electrons). Redox reactions are important in chemistry as they are fundamental to processes such as combustion, respiration, and corrosion, and are also utilized in batteries and electroplating.
| 2. How can I identify the oxidizing and reducing agents in a redox reaction? | |
Ans. To identify the oxidizing and reducing agents, one must first determine the oxidation states of the reactants and products. The species that undergoes an increase in oxidation state is the reducing agent (since it donates electrons), while the species that undergoes a decrease in oxidation state is the oxidizing agent (since it accepts electrons).
| 3. What are some common examples of redox reactions that appear in JEE Mains exams? | |
Ans. Common examples of redox reactions that might appear in JEE Mains include the reaction between zinc and copper sulfate (Zn + CuSO4 → ZnSO4 + Cu), combustion reactions (e.g., burning of hydrocarbons), and the reactions involving halogens (e.g., Cl2 + 2NaBr → 2NaCl + Br2). These reactions illustrate the transfer of electrons and changes in oxidation states.
| 4. How do you balance redox reactions, especially in acidic and basic solutions? | |
Ans. To balance redox reactions, one can use the half-reaction method. First, separate the oxidation and reduction half-reactions. Then, balance the atoms and charges by adding electrons, protons (H⁺), and water (H₂O) as needed. In acidic solutions, protons are added, while in basic solutions, hydroxide ions (OH⁻) are used to balance the reaction. Finally, combine the half-reactions to get the balanced equation.
| 5. What role do redox reactions play in biological systems, and can you give an example relevant to JEE Mains? | |
Ans. Redox reactions are crucial in biological systems, particularly in cellular respiration and photosynthesis. For instance, during cellular respiration, glucose (C6H12O6) is oxidized to carbon dioxide (CO2) while oxygen (O2) is reduced to water (H2O), releasing energy. This process is often represented as: C6H12O6 + 6O2 → 6CO2 + 6H2O. Understanding these reactions can help in solving related problems in JEE Mains.
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