
The oxidation of oxalic acid by acidified \[KMn{O_4}\] is an example of autocatalysis. It is due which of the following?
A) \[S{O^{2-}_4}\]
B) \[Mn{O^{2-}_4}\]
C) \[M{n^{2 + }}\]
D) \[{K^ + }\]
Answer
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Hint: Autocatalysis or autocatalysis reaction is a type of chemical reaction in which one of the reaction products acts as catalyst for further reaction. The oxidation of oxalic acid by potassium permanganate (\[KMn{O_4}\]) is an example of autocatalysis.
Complete answer:
Before answering the question, we need to understand the term ‘autocatalysis’.
Autocatalysis or autocatalytic reaction is a type of chemical reaction in which one of the reactant products acts as a catalyst for the same or a coupled reaction. The reaction is generally slow at the initial phase but with the formation of the product which acts as catalyst the reaction rate is increased.
In oxidation of oxalic acid by potassium permanganate (\[KMn{O_4}\]) permanganate in acidic condition is a very strong oxidizing agent. In permanganate (\[Mn{O_4}\] ) manganese is in + 7 oxidation state meaning it is devoid of electrons. So, when permanganate ion reacts with oxalic acid \[{\left( {COOH} \right)_2}\], its oxides oxalic acid \[{\left( {COOH} \right)_2}\] to give carbon dioxide (\[C{O_2}\] ) and water reducing itself to \[M{n^{2 + }}\] .
Half reaction involved is as follows -
Oxidation half reaction
\[{\left( {COOH} \right)_2} \to \,2C{O_2} + 2{H^ + } + \,2{e^ - }\]
Reaction half reaction
\[Mn{O_4}^ - + 8{H^ + } + 5{e^ - }\, \to \,M{n^{2 + }} + 4{H_2}O\]
Overall reaction is as follows –
\[5{\left( {COOH} \right)_2} + 2Mn{O_4}^ - + 6{H^ + }\, \to \,10C{O_2} + 8{H_2}O\]
The reaction is autocatalytic due to the formation of $Mn^{2+}$ where it further catalysis the overall reaction.
Thus option (C) is the correct option.
Note: Student should note that the start of the reaction is relatively slow as expected from an autocatalytic reaction, due to the absence of the catalyst \[M{n^{2 + }}\]. As \[M{n^{2 + }}\] starts forming in the reaction, the rate of the reaction increases. So usually in order to make the reaction fast, the oxalic acid solution is warmed. Also, alternative to this one can also add \[MnS{O_4}\] (which contains \[M{n^{2 + }}\] ions) before the start of reaction.
Complete answer:
Before answering the question, we need to understand the term ‘autocatalysis’.
Autocatalysis or autocatalytic reaction is a type of chemical reaction in which one of the reactant products acts as a catalyst for the same or a coupled reaction. The reaction is generally slow at the initial phase but with the formation of the product which acts as catalyst the reaction rate is increased.
In oxidation of oxalic acid by potassium permanganate (\[KMn{O_4}\]) permanganate in acidic condition is a very strong oxidizing agent. In permanganate (\[Mn{O_4}\] ) manganese is in + 7 oxidation state meaning it is devoid of electrons. So, when permanganate ion reacts with oxalic acid \[{\left( {COOH} \right)_2}\], its oxides oxalic acid \[{\left( {COOH} \right)_2}\] to give carbon dioxide (\[C{O_2}\] ) and water reducing itself to \[M{n^{2 + }}\] .
Half reaction involved is as follows -
Oxidation half reaction
\[{\left( {COOH} \right)_2} \to \,2C{O_2} + 2{H^ + } + \,2{e^ - }\]
Reaction half reaction
\[Mn{O_4}^ - + 8{H^ + } + 5{e^ - }\, \to \,M{n^{2 + }} + 4{H_2}O\]
Overall reaction is as follows –
\[5{\left( {COOH} \right)_2} + 2Mn{O_4}^ - + 6{H^ + }\, \to \,10C{O_2} + 8{H_2}O\]
The reaction is autocatalytic due to the formation of $Mn^{2+}$ where it further catalysis the overall reaction.
Thus option (C) is the correct option.
Note: Student should note that the start of the reaction is relatively slow as expected from an autocatalytic reaction, due to the absence of the catalyst \[M{n^{2 + }}\]. As \[M{n^{2 + }}\] starts forming in the reaction, the rate of the reaction increases. So usually in order to make the reaction fast, the oxalic acid solution is warmed. Also, alternative to this one can also add \[MnS{O_4}\] (which contains \[M{n^{2 + }}\] ions) before the start of reaction.
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