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प्रश्न
Consider the reaction:
\[\ce{O3(g) + H2O2(l) → H2O(l) + 2O2(g)}\]
Why it is more appropriate to write these reaction as:
\[\ce{O3(g) + H2O2 (l) → H2O(l) + O2(g) + O2(g)}\]
Also, suggest a technique to investigate the path of the redox reactions.
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उत्तर
O2 is produced from each of the two reactants O3 and H2O2. For this reason, O2 is written twice.
The given reaction involves two steps. First, O3 decomposes to form O2 and O. In the second step, H2O2 reacts with the O produced in the first step, thereby producing H2O and O2.
| \[\ce{O_{3(g)} -> O_{2(g)} + O_{(g)}}\] |
| \[\ce{H_2O_{2(l)} + O_{(g)} -> H_2O_{(l)} + O_{2(g)}}\] |
|
\[\ce{H_2O_{2(l)} + O_{3(g)} -> H_2O_{(l)} + O_{2(g)} + O_{2(g)}}\] |
The path of this reaction can be investigated by using `"H"_2"O"_2^(18)` or `"O"_3^18`.
संबंधित प्रश्न
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
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- \[\ce{H2O2 (aq) + Fe^{2+} (aq) → Fe^{3+} (aq) + H2O (l) (in acidic solution)}\]
- \[\ce{Cr_2O^{2-}_7 + SO2(g) → Cr^{3+} (aq) + SO^{2-}_4 (aq) (in acidic solution)}\]
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\[\ce{N2H4(l) + ClO^-_3 (aq) → NO(g) + Cl–(g)}\]
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\[\ce{xMnO^-_4 + yC2O^{2-}_4 + zH^+ -> xMn^{2+} + 2{y}CO2 + z/2H2O}\]
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