Redox Reactions JEE Main previous year questions with solutions

5 solved JEE Main questions on Redox Reactions, free to read — no sign-in needed. The full chapter has 71 questions; sign in to attempt the remaining 66 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (24 Jan, Shift 1)Balancing of redox reactions
   X and Y are the number of electrons involved, respectively during the oxidation of $\mathrm{I}^{-}$to $\mathrm{I}_{2}$ and $\mathrm{S}^{2-}$ to S by acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$. The value of $\mathrm{X}+\mathrm{Y}$ is $\_\_\_\_$.
   Show answer & solution

   Answer: 12

   In acidic medium, $K_2Cr_2O_7$ acts as a strong oxidizing agent where $Cr_2O_7^{2-}$ is reduced to $Cr^{3+}$. The reduction half-reaction is: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$. For the oxidation of $I^-$ to $I_2$: The balanced oxidation half-reaction is $2I^- \rightarrow I_2 + 2e^-$. To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3: $6I^- \rightarrow 3I_2 + 6e^-$. Thus, the number of electrons involved in the balanced redox reaction for $I^-$ oxidation is $X = 6$. For the oxidation of $S^{2-}$ to $S$: The balanced oxidation half-reaction is $S^{2-} \rightarrow S + 2e^-$. To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3: $3S^{2-} \rightarrow 3S + 6e^-$. Thus, the number of electrons involved in the balanced redox reaction for $S^{2-}$ oxidation is $Y = 6$. The value of $X + Y = 6 + 6 = 12$.
2. Q2JEE Main 2025 (29 Jan, Shift 2)Equivalence concept
   0.1 M solution of KI reacts with excess of $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{KIO}_3$ solutions. According to equation $5 \mathrm{I}^{-}+\mathrm{IO}_3^{-}+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}$ Identify the correct statements : (A) 200 mL of KI solution reacts with 0.004 mol of $\mathrm{KIO}_3$ (B) 200 mL of KI solution reacts with 0.006 mol of $\mathrm{H}_2 \mathrm{SO}_4$ (C) 0.5 L of KI solution produced 0.005 mol of $\mathrm{I}_2$ (D) Equivalent weight of $\mathrm{KIO}_3$ is equal to ( $\frac{\text { Molecular weight }}{5}$ ) Choose the correct answer from the options given below :
   1. A.(A) and (D) only
   2. B.(C) and (D) only
   3. C.(B) and (C) only
   4. D.(A) and (B) only
   Show answer & solution

   Answer: (A)

   $\begin{aligned} & \mathrm{E}_{\mathrm{KIO}_3}=\frac{\text { Molecular welght }}{\mathrm{n}_{\mathrm{f}}} \\ & \mathrm{n}_{\mathrm{f}}=5 \\ & \mathrm{E}_{\mathrm{KIO}_3}=\frac{\text { Molecular welght }}{5}\end{aligned}$ (D) is correct meq of $\mathrm{KI}=0.1 \times 200=20$ meq of $\mathrm{KIO}_3=4 \times 5=20$ $(\mathrm{A})$ is correct
3. Q3JEE Main 2024 (27 Jan, Shift 2)Oxidation and Reduction Reactions
   Which of the following cannot function as an oxidising agent?
   1. A.${N}^{3-}$
   2. B.${SO}_{4}^{2-}$
   3. C.${BrO}_{3}^{-}$
   4. D.${MnO}_{4}^{-}$
   Show answer & solution

   Answer: (A)

   An oxidising agent is a substance that oxidises other substances involved in the reaction by gaining or accepting electrons from them. So the lowest oxidation state substance will go to the higher oxidation state, which means it reduces others. Hence, it acts as a reducing agent. $-3$ is the lowest oxidation state for nitrogen. Hence, it only acts as a reducing agent. Hence option A is the answer.
4. Q4JEE Main 2023 (15 Apr, Shift 1)Redox titration
   The total change in the oxidation state of manganese involved in the reaction of ${KMnO}_{4}$ and potassium iodide in the acidic medium is ________.
   Show answer & solution

   Answer: 5

   In the reaction of ${KMnO}_{4}$and potassium iodide ($KI$) in an acidic medium, the ${KMnO}_{4}$ acts as an oxidizing agent, and the iodide ions (${I}^{-}$) in $KI$ act as reducing agents. ${KMnO}_{4}+KI+{H}^{+}⟶{Mn}^{+2}+{I}_{2}+{H}_{2}O$ In ${KMnO}_{4}$, the oxidation state of manganese is $+7$, which is the highest possible oxidation state for manganese in this compound. In the final product, ${MnSO}_{4}$, the oxidation state of manganese is $+2$. Therefore, the total change in the oxidation state of manganese is: $+7$ (initial)$-+2$ (final) $=5$ Hence the change in $O.S$. of $Mn$ is $\left(5\right)$.
5. Q5JEE Main 2021 (25 Jul, Shift 2)Oxidation number
   Identify the process in which change in the oxidation state is five :
   1. A.${Cr}_{2}{O}_{7}^{2-}\rightarrow 2{Cr}^{3+}$
   2. B.${MnO}_{4}^{-}\rightarrow {Mn}^{2+}$
   3. C.${CrO}_{4}^{2-}\rightarrow {Cr}^{3+}$
   4. D.${C}_{2}{O}_{4}^{2-}\rightarrow 2{CO}_{2}$
   Show answer & solution

   Answer: (B)

   ${MnO}_{4}^{-}+5e\rightarrow {Mn}^{+2}$