Redox Reactions JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Redox Reactions, free to read — no sign-in needed. The full chapter has 28 questions; sign in to attempt the remaining 23 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 1)
   One of the products formed from the reaction of permanganate ion with iodide ion in neutral aqueous medium is
   1. A.$\mathrm{I}_2$
   2. B.$\mathrm{IO}_3^{-}$
   3. C.$\mathrm{IO}_4^{-}$
   4. D.$\mathrm{IO}_2^{-}$
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   Answer: (B)

   $\mathrm{I}^{-}+2 \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Neutral solution }} 2 \mathrm{MnO}_2+\mathrm{IO}_3^{-}+2 \mathrm{OH}^{-}$
2. Q2JEE Advanced Adv 2024 (Paper 2)
   In a metal deficient oxide sample, $\mathbf{M}_{\mathbf{x}} \mathbf{Y}_2 \mathbf{O}_4$ ( $\mathbf{M}$ and $\mathbf{Y}$ are metals), $\mathbf{M}$ is present in both +2 and +3 oxidation states and $\mathbf{Y}$ is in +3 oxidation state. If the fraction of $\mathbf{M}^{2+}$ ions present in $\mathbf{M}$ is $\frac{1}{3}$, the value of $\mathbf{X}$ is ________.
   1. A.0.25
   2. B.0.33
   3. C.0.67
   4. D.0.75
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   Answer: (D)

   Average oxidation state of $\mathrm{M}=\frac{1}{3} \times 2+\frac{2}{3} \times 3=+\frac{8}{3}$ $\begin{aligned} & \therefore \text { For } \mathrm{M}_{\mathrm{X}} \mathrm{M}_{2} \mathrm{Y}_4 \\ & \frac{8}{3} \times \mathrm{x}+3 \times 2+4(-2)=0 \\ & \frac{8}{3} \times \mathrm{x}=2 \\ & \mathrm{x}=\frac{3}{4}=0.75\end{aligned}$
3. Q3JEE Advanced Adv 2024 (Paper 1)
   At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid $\left(\mathrm{HNO}_2\right)$ gives the species
   1. A.$\mathrm{H}_3 \mathrm{O}^{+}, \mathrm{NO}_3^{-}$and $\mathrm{NO}$
   2. B.$\mathrm{H}_3 \mathrm{O}^{+}, \mathrm{NO}_3{ }^{-}$and $\mathrm{NO}_2$
   3. C.$\mathrm{H}_3 \mathrm{O}^{+}, \mathrm{NO}^{-}$and $\mathrm{NO}_2$
   4. D.$\mathrm{H}_3 \mathrm{O}^{+}, \mathrm{NO}_3{ }^{-}$and $\mathrm{N}_2 \mathrm{O}$
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   Answer: (A)

   $3 \mathrm{HNO}_2(\mathrm{aq}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{NO}_3^{-}+2 \mathrm{NO}$
4. Q4JEE Advanced Adv 2023 (Paper 2)
   ${H}_{2}S$ ($5$ moles) reacts completely with acidified aqueous potassium permanganate solution. In this reaction, the number of moles of water produced is $x$, and the number of moles of electrons involved is $y$. The value of $(x+y)$ is
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   Answer: 18

   ${KMnO}_{4}⟶^{{H}^{+}}{Mn}^{2+}$ ${S}^{-2}⟶S^{0}$ The number of electrons involved per molecule is called n-factor. n-factor of ${KMnO}_{4}=5$ n-factor of ${S}^{-2}\left({H}_{2}S\right)=2$ $Moles\times n-factor=numberofequivalents$ $\left({n}_{{KMnO}_{4}}\times 5\right)=(5\times 2{)}_{{H}_{2}S}$ $∴{n}_{{KMnO}_{4}}=2$ $∴2{KMnO}_{4}+3{H}_{2}{SO}_{4}+5{H}_{2}S\rightarrow {K}_{2}{SO}_{4}+2{MnSO}_{4}+5S+8{H}_{2}O$ Number of moles of water produced $=8$ Number of moles of electrons involved $=10$ $∴x=8,y=10\Rightarrow (x+y)=18$
5. Q5JEE Advanced Adv 2020 (Paper 2)
   Choose the correct statement(s) among the following.
   1. A.${SnCl}_{2}\cdot 2{H}_{2}O$ is a reducing agent.
   2. B.${SnO}_{2}$ reacts with $KOH$ to form ${K}_{2}\left[Sn(OH{)}_{6}\right]$
   3. C.A solution of ${PbCl}_{2}$ in $HCl$ contains ${Pb}^{2+}$and ${Cl}^{-}$ ions.
   4. D.The reaction of ${Pb}_{3}{O}_{4}$ with hot dilute nitric acid to give ${PbO}_{2}$ is a redox reaction.
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   Answer: A,B

   (A) ${Sn}^{2+}$ is a good reducing agent which gets oxidise into ${Sn}^{4+}$ (B) Amphoteric nature. (C) ${PbCl}_{2}+2{Cl}^{-}⟶{\left[{PbCl}_{4}\right]}^{2-}$ (D) ${Pb}_{3}{O}_{4}{}_{(2PbO + {PbO}_{2})}+4{HNO}_{3}{}_{(Conc.)}\rightarrow {PbO}_{2}↓+2Pb{\left({NO}_{3}\right)}_{2}+6{H}_{2}O$