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Fluorine reacts with ice and results in the change: HX2O(s)+FX2(g)⟶HF(g)+HOF(g) Justify that this reaction is a redox reaction.

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प्रश्न

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.

टीपा लिहा
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उत्तर

Let us write the oxidation number of each atom involved in the given reaction above its symbol as:
+1 -2         0              +1 -1        +1 -2 +1
\[\ce{H2O(s) + F2(g) → HF(g) + HOF(g)}\]

Here, we have observed that the oxidation number of F increases from 0 in F2 to +1 in HOF. Also, the oxidation number decreases from 0 in Fto –1 in HF. Thus, in the above reaction, F is both oxidized and reduced.
Hence, the given reaction is a redox reaction.

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पाठ 7: Redox Reactions - EXERCISES [पृष्ठ २८०]

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एनसीईआरटी Chemistry - Part 1 and 2 [English] Class 11
पाठ 7 Redox Reactions
EXERCISES | Q 8.4 | पृष्ठ २८०

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:

\[\ce{4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3(s)}\]


Justify that the following reaction is redox reaction:

\[\ce{2K(s) + F2(g) → 2K+F– (s)}\]


Justify that the following reaction is redox reaction:

\[\ce{4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)}\]


Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.


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

\[\ce{HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → 2Ag(s) + HCOO–(aq) + 4NH3(aq) + 2H2O(l)}\]


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

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


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. \[\ce{C6H5CHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → C6H5COO–(aq) + 2Ag(s) + 4NH3 (aq) + 2 H2O(l)}\]
  4. \[\ce{C6H5CHO(l) + 2Cu^{2+}(aq) + 5OH–(aq) → No change observed}\]

What inference do you draw about the behaviour of Ag+ and Cu2+ from these reactions?


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.


Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg and Zn.


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.


Write redox couples involved in the reactions given.

\[\ce{Cu + Zn^{2+} ->Cu^{2+} + Zn}\]


Write redox couples involved in the reactions given.

\[\ce{Mg + Fe^{2+} -> Mg^{2+} + Fe}\]


Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.

\[\ce{NaClO4, NaClO3, NaClO, KClO2, Cl2O7, ClO3, Cl2O, NaCl, Cl2 , ClO2}\].

Which oxidation state is not present in any of the above compounds?


An acidified manganate solution undergoes disproportionation reaction. The spin-only magnetic moment value of the product having manganese in a higher oxidation state is ______ B.M. (Nearest integer)


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]


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