JEE Main Chemistry: Redox Reactions High-Yield Practice Questions, Formulas & PYQ Breakdown (2026-27)

Correct Answer: c

Explanation: HNO$_{2}$ ​$\overset{+ 3}{}$ Here N- atom is neither in its max oxidation number nor in its minimum oxidation number. Therefore, it can undergo oxidation as well as reduction.

Correct Answer: (b)

Explanation: Let’s balance each equation by the half-reaction method. 1) In acidic medium  Reduction:  $\mathrm{MnO_4^-} + 8\,H^+ + 5e^- \;\longrightarrow\; \mathrm{Mn^{2+}} + 4\,H_2O$  Oxidation:  $\mathrm{SO_3^{2-}} + H_2O \;\longrightarrow\; \mathrm{SO_4^{2-}} + 2\,H^+ + 2e^-$  – LCM of electrons = 10 → multiply first by 2, second by 5 and add:  $2\,\mathrm{MnO_4^-} + 5\,\mathrm{SO_3^{2-}} + 6\,H^+ \;\longrightarrow\; 2\,\mathrm{Mn^{2+}} + 5\,\mathrm{SO_4^{2-}} + 3\,H_2O$  → x=2, a=5, b=6, c=3. 2) In basic medium  Oxidation (thiosulfate → sulfate):  $\mathrm{S_2O_3^{2-}} + 5\,H_2O \;\longrightarrow\; 2\,\mathrm{SO_4^{2-}} + 10\,H^+ + 8e^-$  Reduction (permanganate → MnO₂):  $\mathrm{MnO_4^-} + 2\,H_2O + 3e^- \;\longrightarrow\; \mathrm{MnO_2} + 4\,OH^-$  – LCM of electrons = 24 → multiply oxidation by 3, reduction by 8, add, then neutralize H⁺ with OH⁻:  $3\,\mathrm{S_2O_3^{2-}} + 8\,\mathrm{MnO_4^-} + H_2O \;\longrightarrow\; 6\,\mathrm{SO_4^{2-}} + 8\,\mathrm{MnO_2} + 2\,OH^-$  → c=3, d=8 (and indeed b=6, x=2 again). Answer: a = 5, b = 6, c = 3, d = 8 (Option B)

Correct Answer: (c)

Explanation: Chlorous acid is $HClO_2$. Let the oxidation state of chlorine be $x$. In $HClO_2$: Oxidation state of hydrogen $= +1$ Oxidation state of oxygen $= -2$ Since the molecule is neutral, the sum of oxidation states is $0$. $ (+1) + x + 2(-2) = 0 $ $ 1 + x - 4 = 0 $ $ x - 3 = 0 $ $ x = +3 $ So, the oxidation state of chlorine in chlorous acid is $\boxed{+3}$. Correct option: Option C.

Correct Answer: 10

Explanation: $\overset{(0)}{\mathrm{Fe}}(\mathrm{CO})_5 \\;\\; \overset{(+4)}{\mathrm{V}} \mathrm{O}^{2+} \\;\\; \overset{(+6)}{\mathrm{W}}\mathrm{O}_3$ So, Sum of oxidation state $=0+4+6=10$

Correct Answer: (a)

Explanation: $2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \longrightarrow 2 \mathrm{CuI}_2$ This $\mathrm{Cu}$ (II) iodide immediately decomposes to liberate $\mathrm{I}_2$ and insoluble copper (I) iodide. $2 \mathrm{CuI}_2 \longrightarrow 2 \mathrm{CuI}+\mathrm{I}_2$