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Identify disproportionation reaction - Chemistry

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प्रश्न

Identify disproportionation reaction

विकल्प

  • \[\ce{CH4 + 2O2 -> CO2 + 2H2O}\]

  • \[\ce{CH4 + 4Cl2 -> CCl4 + 4HCl}\]

  • \[\ce{2F2 + 2OH- -> 2F- + OF2 + H2O}\]

  • \[\ce{2NO2 + 2OH- -> NO^{-}2 + NO^{-}3 + H2O}\]

MCQ

उत्तर

\[\ce{2NO2 + 2OH- -> NO^{-}2 + NO^{-}3 + H2O}\]

Explanation:

Disproportionate reactions are defined as the reactions in which the same substance is oxidized as well as reduced. Here, the below reaction is given as-

\[\ce{2NO2 + 2OH- -> NO^{-}2 + NO^{-}3 + H2O}\]

In this reaction, N is both oxidized as well as reduced since O.N. of N increases from +4 in \[\ce{NO^{-}3}\] to +5 in \[\ce{NO2}\] and decreases from +4 in \[\ce{NO}\] to +3 in \[\ce{NO^{-}2}\].

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Oxidation Number - Types of Redox Reactions
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अध्याय 8: Redox Reactions - Multiple Choice Questions (Type - I) [पृष्ठ १०६]

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एनसीईआरटी एक्झांप्लर Chemistry [English] Class 11
अध्याय 8 Redox Reactions
Multiple Choice Questions (Type - I) | Q 10 | पृष्ठ १०६

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\[\ce{HCHO (l) + 2Cu^{2+}(aq) + 5 OH–(aq) → Cu2O(s) + HCOO–(aq) + 3H2O(l)}\]


Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction:

\[\ce{N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l)}\]


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\[\ce{Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)}\]


Consider the reactions:

  1. \[\ce{H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq)}\]
  2. \[\ce{H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq)}\]
  3. \[\ce{C6H5CHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → C6H5COO–(aq) + 2Ag(s) + 4NH3 (aq) + 2 H2O(l)}\]
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\[\ce{P4 + 3OH- + 3H2O -> PH3 + 3H2PO^{-}2}\]

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(iv) Hydrogen is undergoing neither oxidation nor reduction.


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Reason (R): The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in \[\ce{O2}\] and –2 oxidation state in \[\ce{H2O}\].


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\[\ce{S8(s) + {a} OH^-(aq) -> {b} S^{2-}(aq) + {c} S2O^{2-}3(aq) + {d} H2O(l)}\]

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