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प्रश्न
\[\ce{MnO^{2-}4}\] undergoes disproportionation reaction in acidic medium but \[\ce{MnO^{-}4}\] does not. Give reason.
उत्तर
In \[\ce{MnO^{2-}4}\], Mn is in highest oxidation state that is +7 hence here manganese cannot undergo oxidation that is why disproportionate is not possible whereas in \[\ce{MnO^{2-}4}\] manganese is in +6 oxidation state which can be oxidized as well as reduced.
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संबंधित प्रश्न
Justify that the following reaction is redox reaction:
\[\ce{Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)}\]
Justify that the following reaction is redox reaction:
\[\ce{4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)}\]
Fluorine reacts with ice and results in the change: \[\ce{H2O(s) + F2(g) → HF(g) + HOF(g)}\]
Justify that this reaction is a redox reaction.
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction:
\[\ce{Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)}\]
Consider the reactions:
- \[\ce{H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq)}\]
- \[\ce{H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq)}\]
- \[\ce{C6H5CHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → C6H5COO–(aq) + 2Ag(s) + 4NH3 (aq) + 2 H2O(l)}\]
- \[\ce{C6H5CHO(l) + 2Cu^{2+}(aq) + 5OH–(aq) → No change observed}\]
What inference do you draw about the behaviour of Ag+ and Cu2+ from these reactions?
Which of the following is not an example of redox reaction?
Assertion (A): The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionation reaction.
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 \[\ce{H2O}\].
Why does fluorine not show disporportionation reaction?
In an experiment O3 undergo decomposition as \[\ce{O3 -> O2 + O}\] by the radiations of wavelength 310 Å. The total energy falling on the O3 gas molecules is 2.4 × 1026 eV and quantum yield of the reaction is 0.2.
The volume strength of the H2O2 solution which is obtained from reaction of 1 l H2O and nascent oxygen [O] obtained from the above reactions is (Assuming no change in volume of H2O)
\[\ce{H2O + O -> H2O2}\]
[Given: Na (Avogadro's No.) = 6 × 1023]
