NEET Redox Reactions PYQs & Practice Questions

10 free NEET Chemistry questions on Redox Reactions, starting with previous year questions (2017–2026). Tap “Show answer” to check each one. No login needed.

Topics: Oxidation State Determination · Oxidizing and Reducing Agents · Oxidation Number · Balancing Redox Reactions · Disproportionation Reactions · Classical and Electronic Concept · Identifying Redox Reactions · Standard Electrode Potential and Oxidising Strength · Feasibility of Redox Reactions from E° Values · Dual Oxidising/Reducing Behaviour of Thiosulphate · Rules for Assigning Oxidation Numbers · Oxidation States within a Compound

1. Q1NEET 2026Redox Titration - Oxalic Acid vs KMnO4

   In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO4 solution. If the volume of KMnO4 solution required to reach the end point is 10 mL, the strength (molarity) of the KMnO4 solution is:

   • A. 0.20 M
   • B. 0.25 M
   • C. 0.15 M
   • D. 0.10 M
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   Answer: C — 0.15 M

   Oxalic acid has n-factor = 2 (C2O4^2- loses 2 electrons to form 2CO2); its normality = 0.25 × 2 = 0.5 N. KMnO4 in acidic medium has n-factor = 5 (Mn7+ to Mn2+). Using N1V1 = N2V2: 0.5 × 10 = N(KMnO4) × 10, so N(KMnO4) = 0.5 N. Molarity of KMnO4 = Normality/n-factor = 0.5/5 = 0.10 M.

2. Q2NEET 2026Oxidation States of Elements in Compounds

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

   • A. N2O5 > Al2O3 > H2S (N, Al, S underlined respectively)
   • B. PbO2 > N2O3 > SO3 (Pb, N, S underlined respectively)
   • C. P4O6 > Cl2O7 > AlH3 (P, Cl, Al underlined respectively)
   • D. P4O10 > SO3 > H2O (P, S, O underlined respectively)
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   Answer: —

   In N2O5, N is +5. In Al2O3, Al is +3. In H2S, S is -2. The order +5 > +3 > -2 is correctly decreasing, so N2O5 > Al2O3 > H2S represents the correct decreasing order of oxidation states of the underlined atoms.

3. Q3NEET 2025Oxidation State Determination

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

   • A. +1, −1, and +6
   • B. +2, −2, and +6
   • C. +1, −2, and +4
   • D. +4, −4, and +6
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   Answer: —

   In KO2 (potassium superoxide), K is +1. In H2O2 (hydrogen peroxide), O is −1. In H2SO4, S is +6.

4. Q4NEET 2023Balancing Redox Equations

   On balancing the given redox reaction: a Cr2O7^2- + b SO3^2-(aq) + c H+(aq) -> 2a Cr3+(aq) + b SO4^2-(aq) + (c/2) H2O(l), the coefficients a, b and c are found to be, respectively:

   • A. 3, 8, 1
   • B. 1, 8, 3
   • C. 8, 1, 3
   • D. 1, 3, 8
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   Answer: C — 8, 1, 3

   Balancing the redox reaction between dichromate and sulphite ions: Cr2O7^2- + 3SO3^2- + 8H+ -> 2Cr3+ + 3SO4^2- + 4H2O. This gives a=1, b=3, c=8.

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5. Q5NEET 2022Oxidation State Change in Permanganate Reactions

   In the neutral or faintly alkaline medium, KMnO4 oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from:

   • A. +7 to +4
   • B. +6 to +4
   • C. +7 to +3
   • D. +6 to +5
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   Answer: —

   In neutral or faintly alkaline medium, MnO4- (Mn in +7 state) is reduced to MnO2 (Mn in +4 state), while oxidizing iodide (I-) to iodate (IO3-).

6. Q6NEET 2020Change in oxidation number in a halogenation reaction

   What is the change in oxidation number of carbon in the following reaction? CH4(g) + 4Cl2(g) → CCl4(l) + 4HCl(g)

   • A. +4 to +4
   • B. 0 to +4
   • C. -4 to +4
   • D. 0 to -4
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   Answer: B — 0 to +4

   In CH4, carbon has an oxidation state of -4. In CCl4, carbon has an oxidation state of +4 (since Cl is more electronegative). The oxidation number of carbon changes from -4 to +4.

7. Q7NEET 2019Disproportionation Reactions

   Which of the following reactions are disproportionation reactions? (a) 2Cu+ -> Cu2+ + Cu0 (b) 3MnO4^2- + 4H+ -> 2MnO4- + MnO2 + 2H2O (c) 2KMnO4 --heat--> K2MnO4 + MnO2 + O2 (d) 2MnO4- + 3Mn2+ + 2H2O -> 5MnO2 + 4H+. Select the correct option from the following:

   • A. (a), (c) and (d)
   • B. (a) and (d) only
   • C. (a) and (b) only
   • D. (a), (b) and (c)
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   Answer: B — (a) and (d) only

   In (a), Cu+ (oxidation state +1) disproportionates into Cu2+ (+2) and Cu0 (0), both oxidation and reduction of the same element. In (b), Mn in MnO4^2- (+6) disproportionates into MnO4- (+7) and MnO2 (+4). Reactions (c) and (d) are not disproportionation since in (c) oxygen is oxidized not manganese in a self-redox sense matching the pattern, and (d) is a comproportionation (opposite of disproportionation). So (a) and (b) are the correct disproportionation reactions.

8. Q8NEET 2018Disproportionation using standard electrode potential (Latimer) diagrams

   Consider the change in oxidation state of Bromine corresponding to different emf values as shown in a Latimer diagram: BrO4^- (1.82V) BrO3^- (1.5V) HBrO (1.0652V) Br- and Br2 (1.595V) Br-. Then the species undergoing disproportionation is

   • A. BrO3^-
   • B. HBrO
   • C. Br2
   • D. BrO4^-
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   Answer: A — BrO3^-

   A species undergoes disproportionation when the reduction potential for the step converting it to the lower oxidation state species is greater than the reduction potential for the step producing it, i.e. E°(right) > E°(left) around that species in the Latimer diagram. This condition is satisfied for HBrO, identifying it as the species that disproportionates.

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9. Q9NEET 2017Reducing agents and KMnO4

   Name the gas that can readily decolourise acidified KMnO4 solution:

   • A. SO2
   • B. NO2
   • C. P2O5
   • D. CO2
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   Answer: —

   SO2 is a reducing agent that readily decolourises acidified KMnO4 solution by reducing Mn7+ (purple) to Mn2+ (colourless), itself getting oxidised to sulfate/SO3.

10. Q10Oxidation Number

    What is the oxidation number of chromium in K₂Cr₂O₇?

    • A. +3
    • B. +6
    • C. +7
    • D. +4
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    Answer: A — +3

    In K₂Cr₂O₇: K is +1, O is -2. So 2(+1) + 2x + 7(-2) = 0 → 2 + 2x - 14 = 0 → 2x = 12 → x = +6. Chromium has oxidation number +6.