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प्रश्न
Write redox couples involved in the reactions given.
\[\ce{Cu + Zn^{2+} ->Cu^{2+} + Zn}\]
उत्तर
Redox couple is a reducing element along with its oxidizing form. So for the given example,Redox couples are given as:
\[\ce{Cu^{2+} / Cu and Zn^{2+} / Zn}\].
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संबंधित प्रश्न
Justify that the following reaction is redox reaction:
\[\ce{2K(s) + F2(g) → 2K+F– (s)}\]
Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction:
\[\ce{2AgBr (s) + C6H6O2(aq) → 2Ag(s) + 2HBr (aq) + C6H4O2(aq)}\]
Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction:
\[\ce{HCHO (l) + 2Cu^{2+}(aq) + 5 OH–(aq) → Cu2O(s) + HCOO–(aq) + 3H2O(l)}\]
Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction:
\[\ce{N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l)}\]
What sorts of informations can you draw from the following reaction ?
\[\ce{{(CN)}_{2(g)} + 2OH-_{(aq)} -> CN-_{(aq)} + CNO-_{(aq)} + H_2O_{(l)}}\]
Refer to the periodic table given in your book and now answer the following questions:
Select the possible non-metals that can show disproportionation reaction.
Refer to the periodic table given in your book and now answer the following question:
Select three metals that can show disproportionation reaction.
Which of the following statement(s) is/are not true about the following decomposition reaction.
\[\ce{2KClO3 -> 2KCl + 3O2}\]
(i) Potassium is undergoing oxidation.
(ii) Chlorine is undergoing oxidation.
(iii) Oxygen is reduced.
(iv) None of the species are undergoing oxidation or reduction.
\[\ce{MnO^{2-}4}\] undergoes disproportionation reaction in acidic medium but \[\ce{MnO^{-}4}\] does not. Give reason.
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]
