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Question: \(Mn{O_2}\)is fused with\(KOH\) in the presence of air, a colored compound is formed, the product an...

MnO2Mn{O_2}is fused withKOHKOH in the presence of air, a colored compound is formed, the product and its color is:
A. K2MnO4,{K_2}Mn{O_4},dark green
B. KMnO4KMn{O_4}, purple
C. Mn2O3M{n_2}{O_3}, brown
D. Mn3O4M{n_3}{O_4},black

Explanation

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 MnO2Mn{O_2} fused withKOHKOH in presence of air A green coloredK2MnO4{K_2}Mn{O_4} is obtained.
2MnO2+4KOH+O22K2MnO4+2H2O2Mn{O_2} + 4KOH + {O_2} \to 2{K_2}Mn{O_4} + 2{H_2}O
Manages Potassium Potassium
Dioxide hydroxide magnate
Therefore, from the above explanation the correct option is (A) K2MnO4,{K_2}Mn{O_4}, dark green.
This is the first step of preparation of potassium permanganate [KMnO4]\left[ {KMn{O_4}} \right].
In this reaction 2MnO22Mn{O_2} Oxidizes to K2MnO4{K_2}Mn{O_4}.

The fused mass obtained in the reaction contains K2MnO4{K_2}Mn{O_4}. It is Heated with water and men convert it into KMnO4KMn{O_4} either by oxidation or by electrolysis.
Oxidation of K2MnO4{K_2}Mn{O_4} to KMnO4KMn{O_4} carried out by H2SO4{H_2}S{O_4} or Cl2C{l_2} or CO2C{O_2} thorough solution.
3KMnO4+6+2H2SO42K2SO4+2KMnO4+7+2H2O+MnO2+43K\mathop {Mn{O_4}}\limits^{ + 6} + 2{H_2}S{O_4} \to 2{K_2}S{O_4} + 2K{\mathop {MnO}\limits^{ + 7} _4} + 2{H_2}O + \mathop {Mn{O_2}}\limits^{ + 4}
2K2MnO4+Cl22KMnO4+2KCl2{K_2}Mn{O_4} + C{l_2} \to 2KMn{O_4} + 2KCl
First reaction is disproportion reaction in which K2MnO4{K_2}Mn{O_4} Oxidizes to KMnO4KMn{O_4} and reduces to MnO2Mn{O_2} In electrolytic oxidation magnate solution is electrolytic. between ion electrodes.
At CaCa mode 2H++2eH22{H^ + } + 2{e^ - } \to {H_2} \uparrow Reduction
The oxygen evolved at anode converts manganate to permanganate:
2K2MnO4+H2O+[O]2KMnO4+2KOH.2{K_2}Mn{O_4} + {H_2}O + \left[ O \right] \to 2KMn{O_4} + 2KOH.

Note:
Potassium permanganate[KMnO4]\left[ {KMn{O_4}} \right] 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