
Name the gas that readily decolourises the acidified $KMn{O_4}$ solution.
A. $C{O_2} $
B. $S{O_2} $
C. $N{O_2} $
D. ${P_2}{O_5} $
Answer
574.5k+ views
Hint: One needs to understand the qualitative analysis of potassium permanganate to identify the gas that would decolourise the acidified potassium permanganate solution.
Complete step by step solution:
Permanganate and manganate solutions can be prepared by methods such as manganate can be obtained by boiling potassium permanganate with conc.KOH.
$4KMn{O_4} + 4KOH\xrightarrow{{boiling}}4{K_2}Mn{O_4} + 2{H_2}O + 2{O_2}$
The above green solution is quite stable in alkali, but decomposes in pure water and in the presence of acids, depositing manganate oxide and giving a purple solution of permanganate.
$3{K_2}Mn{O_4} + 2{H_2}O \rightleftharpoons 2KMn{O_4} + Mn{O_2} \downarrow + 4KOH$
This reaction is also a method of preparation of $KMn{O_4}$ which is purple in colour.
For use as an oxidizing agent, $KMn{O_4}$ should be stored in dark bottles and standardized just before use due to the following reasons:
1. It is not a primary standard since it is difficult to get it in a high degree of purity and free form traces of $Mn{O_2}$.
2. It is slowly reduced to $Mn{O_2}$ especially in the presence of light or acid.
All available manganate and permanganate salts are soluble in water. There are several tests of $MnO_4^ - $ which are not shown by $MnO_4^{2 - }$. Because $MnO_4^{2 - }$ is stable in solution with pH=12 only. For example, the purple colour of $MnO_4^ - $ is discharged by several reducing agents as is evident from the reaction given below:
$MnO_4^ - + S{O_2}\xrightarrow{{{H^ + }}}M{n^{2 + }} + SO_4^{2 - } + {H_2}O$
In this test permanganate radical gives a green solution on treatment with hot and concentrated NaOH. The purple colour of permanganate solution is restored by acidification or by dilution with large amounts of water.
Hence we can conclude that the correct option is B.
Note:
These solutions need to be handled with care and should be performed under expert supervision. Since the heating effect of potassium permanganate gives black residue.
Complete step by step solution:
Permanganate and manganate solutions can be prepared by methods such as manganate can be obtained by boiling potassium permanganate with conc.KOH.
$4KMn{O_4} + 4KOH\xrightarrow{{boiling}}4{K_2}Mn{O_4} + 2{H_2}O + 2{O_2}$
The above green solution is quite stable in alkali, but decomposes in pure water and in the presence of acids, depositing manganate oxide and giving a purple solution of permanganate.
$3{K_2}Mn{O_4} + 2{H_2}O \rightleftharpoons 2KMn{O_4} + Mn{O_2} \downarrow + 4KOH$
This reaction is also a method of preparation of $KMn{O_4}$ which is purple in colour.
For use as an oxidizing agent, $KMn{O_4}$ should be stored in dark bottles and standardized just before use due to the following reasons:
1. It is not a primary standard since it is difficult to get it in a high degree of purity and free form traces of $Mn{O_2}$.
2. It is slowly reduced to $Mn{O_2}$ especially in the presence of light or acid.
All available manganate and permanganate salts are soluble in water. There are several tests of $MnO_4^ - $ which are not shown by $MnO_4^{2 - }$. Because $MnO_4^{2 - }$ is stable in solution with pH=12 only. For example, the purple colour of $MnO_4^ - $ is discharged by several reducing agents as is evident from the reaction given below:
$MnO_4^ - + S{O_2}\xrightarrow{{{H^ + }}}M{n^{2 + }} + SO_4^{2 - } + {H_2}O$
In this test permanganate radical gives a green solution on treatment with hot and concentrated NaOH. The purple colour of permanganate solution is restored by acidification or by dilution with large amounts of water.
Hence we can conclude that the correct option is B.
Note:
These solutions need to be handled with care and should be performed under expert supervision. Since the heating effect of potassium permanganate gives black residue.
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