Redox Reactions NEET PYQs (2005–2026) with Solutions

10 solved Redox Reactions PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumRedox titration of oxalic acid with KMnO4

   In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with $\mathrm{KMnO_4}$ solution. If the volume of $\mathrm{KMnO_4}$ solution required to reach end point is 10 mL, the strength of the $\mathrm{KMnO_4}$ solution is

   1. AOption A: $0.10\ \mathrm{M}$
   2. BOption B: $0.20\ \mathrm{M}$
   3. COption C: $0.25\ \mathrm{M}$
   4. DOption D: $0.15\ \mathrm{M}$

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   Correct answer: (A) $0.10\ \mathrm{M}$

   Explanation

   Balanced ionic equation in acidic medium: $$\mathrm{2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O}$$ Moles of oxalic acid $= 10\ \mathrm{mL} \times 0.25\ \mathrm{M} = 2.5\ \mathrm{mmol}$. Moles of$\mathrm{MnO_4^-}$needed$= \frac{2}{5} \times 2.5 = 1.0\ \mathrm{mmol}$. Molarity of$\mathrm{KMnO_4} = \frac{1.0\ \mathrm{mmol}}{10\ \mathrm{mL}} = 0.10\ \mathrm{M}$. Check with electrons:$5 \times M \times 10 = 2 \times 0.25 \times 10$gives$M = 0.10\ \mathrm{M}$. Option (A).

2. Question 2 (NEET 2026)

   NEET 2026MediumMnO2 oxidation of halides (halogen preparation)

   Given below are two statements : Statement-I : Heating NaCl with concentrated $\mathrm{H_2SO_4}$ and $\mathrm{MnO_2}$ results in oxidation of Mn. Statement-II : Heating NaI with concentrated $\mathrm{H_2SO_4}$ and $\mathrm{MnO_2}$ results in reduction of Mn. In light of the above statements, choose the most appropriate answer from the options given below.

   1. AOption A: Both Statement-I and Statement-II are correct
   2. BOption B: Both Statement-I and Statement-II are incorrect
   3. COption C: Statement-I is correct but Statement-II is incorrect
   4. DOption D: Statement-I is incorrect but Statement-II is correct

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   Correct answer: (D) Statement-I is incorrect but Statement-II is correct

   Explanation

   Statement-I: $$\mathrm{4NaCl + MnO_2 + 4H_2SO_4 \rightarrow MnCl_2 + 4NaHSO_4 + 2H_2O + Cl_2}$$ $\mathrm{Cl^-}$is oxidised to$\mathrm{Cl_2}$and Mn goes from$+4$(in$\mathrm{MnO_2}$) to$+2$(in$\mathrm{MnCl_2}$). Mn is reduced, not oxidised, so Statement-I is incorrect. Statement-II:$$\mathrm{2NaI + MnO_2 + 3H_2SO_4 \rightarrow 2NaHSO_4 + MnSO_4 + 2H_2O + I_2}$$$\mathrm{I^-}$is oxidised to$\mathrm{I_2}$and Mn is reduced from$+4$to$+2$. Statement-II is correct. So the answer is option (D).

3. Question 3 (NEET 2026)

   NEET 2026EasyOxidation number calculation

   The correct decreasing order of oxidation state of the underlined atom in each molecule is

   1. AOption A: $\mathrm{\underline{P}_4O_{10} > \underline{S}O_3 > H_2\underline{O}}$
   2. BOption B: $\mathrm{\underline{N}_2O_5 > \underline{Al}_2O_3 > H_2\underline{S}}$
   3. COption C: $\mathrm{\underline{Pb}O_2 > \underline{N}_2O_3 > \underline{S}O_3}$
   4. DOption D: $\mathrm{\underline{P}_4O_6 > \underline{Cl}_2O_7 > \underline{Al}H_3}$

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   Correct answer: (B) $\mathrm{\underline{N}_2O_5 > \underline{Al}_2O_3 > H_2\underline{S}}$

   Explanation

   Find the oxidation state of each underlined atom. (A) P in $\mathrm{P_4O_{10}}$: $+5$; S in $\mathrm{SO_3}$: $+6$; O in $\mathrm{H_2O}$: $-2$. Since $+5 < +6$, the order is wrong. (B) N in $\mathrm{N_2O_5}$: $+5$; Al in $\mathrm{Al_2O_3}$: $+3$; S in $\mathrm{H_2S}$: $-2$. $+5 > +3 > -2$ is correct. (C) Pb in $\mathrm{PbO_2}$: $+4$; N in $\mathrm{N_2O_3}$: $+3$; S in $\mathrm{SO_3}$: $+6$. Wrong. (D) P in $\mathrm{P_4O_6}$: $+3$; Cl in $\mathrm{Cl_2O_7}$: $+7$; Al in $\mathrm{AlH_3}$: $+3$. Wrong. So the answer is option (B).

4. Question 4 (NEET 2025)

   NEET 2025MediumOxidation Number

   Consider the following compounds: KO₂, H₂O₂ and H₂SO₄. The oxidation states of the underlined elements in them are, respectively,

   1. AOption A: +1, −1, and +6
   2. BOption B: +2, −2, and +6
   3. COption C: +1, −2, and +4
   4. DOption D: +4, −4, and +6

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   Correct answer: (A) +1, −1, and +6

   Explanation

   In KO₂, potassium has oxidation state +1. In H₂O₂, oxygen has oxidation state −1 due to peroxide linkage. In H₂SO₄, sulfur has oxidation state +6.

5. Question 5 (NEET 2024)

   NEET 2024EasyOxidation & Reduction Concepts

   Which reaction is NOT a redox reaction?

   1. AOption A: H₂ + Cl₂ → 2 HCl
   2. BOption B: BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl
   3. COption C: Zn + CuSO₄ → ZnSO₄ + Cu
   4. DOption D: 2 KClO₃ + I₂ → 2 KIO₃ + Cl₂

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   Correct answer: (B) BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl

   Explanation

   A redox reaction involves change in oxidation states. In: $$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$$ there is no change in oxidation states of any element. It is only a double displacement reaction. Hence, option (B) is correct.

6. Question 6 (NEET 2023)

   NEET 2023MediumBalancing redox reactions

   On balancing the given redox reaction, a Cr₂O₇²⁻ + b SO₃²⁻(aq) + c H⁺(aq) → 2a Cr³⁺(aq) + b SO₄²⁻(aq) + c/2 H₂O(l) the coefficients a, b and c are found to be, respectively -

   1. AOption A: 1, 3, 8
   2. BOption B: 3, 8, 1
   3. COption C: 1, 8, 3
   4. DOption D: 8, 1, 3

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   Correct answer: (A) 1, 3, 8

   Explanation

   Balancing the reaction in acidic medium gives: Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O Hence, a = 1, b = 3 and c = 8.

7. Question 7 (NEET 2021)

   NEET 2021EasyOxidation & Reduction Concepts

   Which of the following reactions is the metal displacement reaction? Choose the right option.

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 4

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   Correct answer: (C) Option 3

   Explanation

   A metal displacement reaction occurs when a more reactive metal displaces a less reactive metal from its compound. In reaction (3), aluminium displaces chromium from chromium oxide: Cr₂O₃ + 2Al → Al₂O₃ + 2Cr This is an example of metal displacement (thermite-type reaction). Reaction (1) and reaction (2) are decomposition reactions. Reaction (4) is a hydrogen displacement reaction where iron displaces hydrogen from HCl. Therefore, option C is correct.

8. Question 8 (NEET 2020)

   NEET 2020EasyOxidation Number

   What is the change in oxidation number of carbon in the following reaction? $CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(l) + 4HCl(g)$

   1. AOption A: $+4$ to $+4$
   2. BOption B: $0$ to $+4$
   3. COption C: $-4$ to $+4$
   4. DOption D: $0$ to $-4$

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   Correct answer: (C) $-4$ to $+4$

   Explanation

   In $$CH_4$$, hydrogen has oxidation number $$+1$$. Let the oxidation number of carbon be $$x$$: $$x + 4(+1) = 0$$ $$x = -4$$ In $$CCl_4$$, chlorine has oxidation number $$-1$$. Let the oxidation number of carbon be $$x$$: $$x + 4(-1) = 0$$ $$x = +4$$ Thus, the oxidation number of carbon changes from $$-4$$ to $$+4$$. Hence, option (C) is correct.

9. Question 9 (NEET 2018)

   NEET 2018MediumBalancing redox reactions

   For the redox reaction: $$\mathrm{MnO_4^- + C_2O_4^{2-} + H^+ \rightarrow Mn^{2+} + CO_2 + H_2O}$$ The correct coefficients of the reactants for the balanced equation are

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (B) Option (2)

   Explanation

   The balanced redox reaction in acidic medium is: $$\mathrm{2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O}$$ Thus, the coefficients of: - MnO₄⁻ = 2 - C₂O₄²⁻ = 5 - H⁺ = 16 Hence, option (B) is correct.

10. Question 10 (NEET 2018)

    NEET 2018HardOxidation & Reduction Concepts

    Consider the change in oxidation state of Bromine corresponding to different E° values as shown in the diagram below. Then the species undergoing disproportionation is

    1. AOption A: BrO₃⁻
    2. BOption B: BrO₄⁻
    3. COption C: HBrO
    4. DOption D: Br₂

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    Correct answer: (C) HBrO

    Explanation

    For disproportionation of HBrO: Reduction half reaction: $$\mathrm{HBrO \rightarrow Br_2} \qquad E^\circ = 1.595\ V$$ Oxidation half reaction: $$\mathrm{HBrO \rightarrow BrO_3^-} \qquad E^\circ = 1.5\ V$$ Therefore, $$E^\circ_{cell} = 1.595 - 1.5 = 0.095\ V$$ Since the value is positive, HBrO undergoes disproportionation.