Question
Question: \(Mn{O_2}\)is fused with\(KOH\) in the presence of air, a colored compound is formed, the product an...
MnO2is fused withKOH in the presence of air, a colored compound is formed, the product and its color is:
A. K2MnO4,dark green
B. KMnO4, purple
C. Mn2O3, brown
D. Mn3O4,black
Solution
Redox reactions are one of those reactions in which oxidation and Reduction takes place simultaneously.
Increase in oxidation Number during reaction is known as oxidation.
Decrease in Reduction Number during ration is called Reduction.
During Redox reaction increase in oxidation number and decrease in oxidation number takes place simultaneously.
Complete step by step answer:
Let us discuss a given reaction MnO2 fused withKOH in presence of air A green coloredK2MnO4 is obtained.
2MnO2+4KOH+O2→2K2MnO4+2H2O
Manages Potassium Potassium
Dioxide hydroxide magnate
Therefore, from the above explanation the correct option is (A) K2MnO4, dark green.
This is the first step of preparation of potassium permanganate [KMnO4].
In this reaction 2MnO2 Oxidizes to K2MnO4.
The fused mass obtained in the reaction contains K2MnO4. It is Heated with water and men convert it into KMnO4either by oxidation or by electrolysis.
Oxidation of K2MnO4 to KMnO4 carried out by H2SO4 or Cl2 or CO2 thorough solution.
3KMnO4+6+2H2SO4→2K2SO4+2K4MnO+7+2H2O+MnO2+4
2K2MnO4+Cl2→2KMnO4+2KCl
First reaction is disproportion reaction in which K2MnO4 Oxidizes to KMnO4 and reduces to MnO2 In electrolytic oxidation magnate solution is electrolytic. between ion electrodes.
At Ca mode 2H++2e−→H2↑ Reduction
The oxygen evolved at anode converts manganate to permanganate:
2K2MnO4+H2O+[O]→2KMnO4+2KOH.
Note:
Potassium permanganate[KMnO4] is. An oxidizing agent. It is used in the industry and laboratory.
It is used as Baeyer’s reagent for defecting unsaturation in organic compound