Question

In the titration between oxalic acid and acidified potassium permanganate, the manganous salt is formed during the reaction and catalyzes the reaction. The manganous salt acts as:
(A) Promoter
(B) Positive catalyst
(C) Autocatalyst
(D) None of the above

Answer 1

To solve this question, we first need to understand the process of titration. A process that is used to determine the concentration of an unknown solution using a solution of known concentration is known as titration.

Complete answer:
Now, the reaction that occurs when acidified potassium permanganate (VII) is titrated against oxalic acid is a redox reaction.
Potassium permanganate (VII) is acidified by adding sulfuric acid.
The reaction takes place as follows.
$2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}+5{{(COOH)}_{2}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+8{{H}_{2}}O+10C{{O}_{2}}\uparrow$
In this reaction, acidified potassium permanganate (VII) acts as a strong oxidizing agent and oxidizes oxalic acid to carbon dioxide.
Whereas, oxalic acid acts as a reducing agent and reduces acidified potassium permanganate (VII) to manganese (I) sulfate.
Now, the ionic reaction that takes place can be written as follows
$2MnO_{4}^{-}+16{{H}^{+}}+5{{C}_{2}}H_{4}^{2-}\to 2M{{n}^{2+}}+10C{{O}_{2}}+8{{H}_{2}}O$
Here we can see that manganese ions are formed and the rate of the reaction depends on the formation of these manganese ions as these catalyze the reaction.

Hence, we can say that the manganese salts formed from the manganese ions act as an option (C) Autocatalyst.

Note:
It should be noted that in this titration reaction, potassium permanganate (VII) acts as a self-indicator as purple-colored $MnO_{4}^{-}$ ions lose color when they are converted into $M{{n}^{2+}}$ ions.
Also, since acids like hydrochloric acid or nitric acid are themselves oxidizing agents, the titration cannot be performed in their presence and hence sulfuric acid is used to acidify potassium permanganate which acts as a strong oxidizing agent.