Advertisements
Advertisements
Question
Justify that the following reaction is redox reaction:
\[\ce{2K(s) + F2(g) → 2K+F– (s)}\]
Solution
The oxidation number of each element in the given reaction can be represented as:
0 0 +1 -1
\[\ce{2K(s) + F2(g) → 2K+F– (s)}\]
In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F2to –1 in KF i.e., F2 is reduced to KF.
Hence, the above reaction is a redox reaction.
APPEARS IN
RELATED QUESTIONS
Justify that the following reaction is redox reaction:
\[\ce{4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3(s)}\]
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?
How do you count for the following observations?
When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent-smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?
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:
\[\ce{2S_2O_3^{(2-)}(aq) + l_2(S) -> S_4O_6^{(2-)}(aq) + 2l-(aq)}\]
\[\ce{S_2O_3^{(2-)}(aq) + 2Br_2(l) + 5H_2O(l) -> 2SO_4^{2-} (aq) + 4Br-(aq) + 10H+ (aq)}\]
Why does the same reductant, thiosulphate react differently with iodine and bromine?
Why does the following reaction occur?
\[\ce{XeO^{4-}_6 (aq) + 2F- (aq) + 6H+ (aq) -> XeO3(g) + F_2(g) + 3H_2O(l)}\]
What conclusion about the compound Na4XeO6 (of which `"XeO"_6^(4+)` is a part) can be drawn from the reaction.
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.
Which of the following is not an example of redox reaction?
Identify 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.
Identify the correct statements with reference to the given reaction.
\[\ce{P4 + 3OH- + 3H2O -> PH3 + 3H2PO^{-}2}\]
(i) Phosphorus is undergoing reduction only.
(ii) Phosphorus is undergoing oxidation only.
(iii) Phosphorus is undergoing oxidation as well as reduction.
(iv) Hydrogen is undergoing neither oxidation nor 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?
Which of the following examples does not represent disproportionation?
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)
\[\ce{H2O2 -> H2O + O2}\]
This represents ______.
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]
For the decomposition reaction \[\ce{NH2COONH4 (s) <=> 2NH3 (g) + CO2 (g)}\] the Kp = 2.9 × 10-5 atm3. The total pressure of gases at equilibrium when 1 mol of \[\ce{NH2COONH4 (s)}\] was taken initially could be ______.
