Question

In the reaction,
$KMn{O_4} + {H_2}S{O_4} + {H_2}{C_2}{O_4}\xrightarrow{{M{n^{2 + }}}}\text{Products}$,$M{n^{2 + }}$ ion act as:
A. positive catalyst
B. negative catalyst
C. autocatalyst
D. enzyme catalyst

Answer 1

The compound potassium permanganate is a strong oxidizing agent and on reacting with sulfuric acid it behaves as a powerful oxidizing agent. In an acidic medium, the permanganate ion reduces itself to form manganese ion and decolorizes the purple color solution.

Complete step by step answer:
The standardization of potassium permanganate is done using oxalic acid. The reaction of potassium permanganate and oxalic acid in the presence of sulfuric acid is a type of redox reaction.
The reaction between oxalic acid and potassium permanganate is shown below.
Reduction half reaction:
$2KMn{O_4} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + 5[O]$
$KMn{O_4}$ is a strong oxidizing agent which itself gets reduced to $MnS{O_4}$.
Oxidation half reaction:
$5{(COOH)_2} + 5[O] \to 5{H_2}O + 10C{O_2}$
In this reaction, oxalic acid is oxidized to carbon dioxide by potassium permanganate.
The overall reaction is shown below.
$2KMn{O_4} + 3{H_2}S{O_4} + 5{(COOH)_2} \to {K_2}S{O_4} + 2MnS{O_4} + 8{H_2}O + 10C{O_2}$
In this reaction, two mole of potassium permanganate reacts with three mole of sulfuric acid and five mole of oxalic acid to form one mole of potassium sulfate, two mole of manganese sulfate, eight mole of water and ten mole of carbon dioxide.
The ionic equation of this reaction is shown below.
Reduction half reaction:
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$
Oxidation half reaction:
${C_2}H_4^{2 - } \to 2C{O_2} + 2{e^ - }$
The overall ionic reaction is shown below.
$2MnO_4^ - + 16{H^ + } + 5{C_2}H_4^{2 - } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$
As the manganese ion .. is one of the by-products formed in reaction which itself then act as the catalyst in the reaction. Therefore, it is said as an autocatalyst.

Therefore, the correct option is C

Note:
In the given reaction, even if the extra $M{n^{2 + }}$catalyst is not added in the reaction, then also the reaction speed will be increased as the $M{n^{2 + }}$ ion is already formed in the reaction.