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Redox Reactions question

2022 · 26 Jul · Shift 1 · Q4
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Redox Reactions question

2022 · 26 Jul · Shift 1 · Q4

JEE MainChemistryRedox ReactionsMCQ+4 / −1
Which of the given reactions is not an example of disproportionation reaction?
  1. A
    2H2O2→2H2O+O22 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}2H2​O2​→2H2​O+O2​
  2. B
    2NO2+H2O→HNO3+HNO22 \mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HNO}_{3}+\mathrm{HNO}_{2}2NO2​+H2​O→HNO3​+HNO2​
  3. C
    MnO4−+4H++3e−→MnO2+2H2O\mathrm{MnO}_{4}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}MnO4−​+4H++3e−→MnO2​+2H2​O
  4. D
    3MnO42−+4H+→2MnO4−+MnO2+2H2O3 \mathrm{MnO}_{4}^{2-}+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{MnO}_{4}^{-}+\mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}3MnO42−​+4H+→2MnO4−​+MnO2​+2H2​O
View written solutionFree

Correct answer: C

  1. Concept of disproportionation

A disproportionation reaction is one in which the same species undergoes both:

  • oxidation (increase in oxidation number), and
  • reduction (decrease in oxidation number).

So, we check whether the same element in the same reactant ends up in two different oxidation states in the products.


  1. Option A: 2H2O2→2H2O+O22\mathrm{H}_2\mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}+\mathrm{O}_22H2​O2​→2H2​O+O2​

Find oxidation state of oxygen:

  • In H2O2\mathrm{H}_2\mathrm{O}_2H2​O2​, oxygen is −1-1−1
  • In H2O\mathrm{H}_2\mathrm{O}H2​O, oxygen is −2-2−2
  • In O2\mathrm{O}_2O2​, oxygen is 000

Thus, oxygen from −1-1−1 goes to:

  • −2-2−2 (reduction)
  • 000 (oxidation)

Hence, this is a disproportionation reaction.


  1. Option B: 2NO2+H2O→HNO3+HNO22\mathrm{NO}_2+\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HNO}_3+\mathrm{HNO}_22NO2​+H2​O→HNO3​+HNO2​

Oxidation state of nitrogen:

  • In NO2\mathrm{NO}_2NO2​: let oxidation state of N be xxx x+2(−2)=0⇒x=+4x+2(-2)=0 \Rightarrow x=+4x+2(−2)=0⇒x=+4
  • In HNO3\mathrm{HNO}_3HNO3​: N is +5+5+5
  • In HNO2\mathrm{HNO}_2HNO2​: N is +3+3+3

Thus, nitrogen from +4+4+4 goes to:

  • +5+5+5 (oxidation)
  • +3+3+3 (reduction)

So, this is also a disproportionation reaction.


  1. Option C: MnO4−+4H++3e−→MnO2+2H2O\mathrm{MnO}_4^-+4\mathrm{H}^+ +3\mathrm{e}^- \rightarrow \mathrm{MnO}_2+2\mathrm{H}_2\mathrm{O}MnO4−​+4H++3e−→MnO2​+2H2​O

Oxidation state of Mn:

  • In MnO4−\mathrm{MnO}_4^-MnO4−​: x+4(−2)=−1⇒x=+7x+4(-2)=-1 \Rightarrow x=+7x+4(−2)=−1⇒x=+7
  • In MnO2\mathrm{MnO}_2MnO2​: x+2(−2)=0⇒x=+4x+2(-2)=0 \Rightarrow x=+4x+2(−2)=0⇒x=+4

Mn changes from +7+7+7 to +4+4+4 only. This is only reduction, not simultaneous oxidation and reduction of the same species.

Hence, this is not a disproportionation reaction.


  1. Option D: 3MnO42−+4H+→2MnO4−+MnO2+2H2O3\mathrm{MnO}_4^{2-}+4\mathrm{H}^+ \rightarrow 2\mathrm{MnO}_4^-+\mathrm{MnO}_2+2\mathrm{H}_2\mathrm{O}3MnO42−​+4H+→2MnO4−​+MnO2​+2H2​O

Oxidation state of Mn:

  • In MnO42−\mathrm{MnO}_4^{2-}MnO42−​: x+4(−2)=−2⇒x=+6x+4(-2)=-2 \Rightarrow x=+6x+4(−2)=−2⇒x=+6
  • In MnO4−\mathrm{MnO}_4^-MnO4−​: Mn is +7+7+7
  • In MnO2\mathrm{MnO}_2MnO2​: Mn is +4+4+4

Thus, Mn from +6+6+6 goes to:

  • +7+7+7 (oxidation)
  • +4+4+4 (reduction)

So, this is a disproportionation reaction.


  1. Final conclusion

The reaction which is not an example of disproportionation is: C\boxed{\text{C}}C​


  1. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

They match.

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