Read the passage carefully and answer the Questions.
\( KMnO_4 \) is prepared by the fusion of \( MnO_2 \) with an alkali metal hydroxide and an oxidizing agent like \( KNO_3 \) to give a dark- green manganate ion which disproportionates to give permanganate as follows.
\( 2MnO_2+ 4KOH + O_2→2K_2MnO_4+2H_2O \)
\( 3K_2MnO_4+ 4H^+→2KMnO_4 + MnO_2 + 2H_2O \)
On heating \( KMnO_4 \) decomposes at 513 K to give \( K_2MnO_4 \). Permanganate ion is tetrahedral and diamagnetic. Acidified \( KMnO_4 \) acts as a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.
When one mole of \( KMnO_4 \) is used for the oxidation of oxalic acid, how many moles of \( CO_2 \) will be produced?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5
In acidic solution the permanganate ion is a strong oxidising agent. Manganese goes from +7 in \( \ce{MnO4^-} \) to +2 in \( \ce{Mn^2+} \), so each permanganate ion takes up 5 electrons. Oxalic acid (as the oxalate ion) is oxidised to carbon dioxide; carbon goes from +3 to +4, and each oxalate ion gives up 2 electrons while forming 2 molecules of \( \ce{CO2} \).
The two half-reactions are:
\( \displaystyle \ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O} \)
\( \displaystyle \ce{C2O4^2- -> 2CO2 + 2e^-} \)
To balance the electrons, the reduction is multiplied by 2 and the oxidation by 5 (10 electrons each). Adding them gives the full ionic equation:
\( \displaystyle \ce{2MnO4^- + 5C2O4^2- + 16H^+ -> 2Mn^2+ + 10CO2 + 8H2O} \)
So 2 mol of \( \ce{KMnO4} \) produce 10 mol of \( \ce{CO2} \):
\( \displaystyle \text{mol of } \ce{CO2} \text{ per mol of } \ce{KMnO4} = \frac{10}{2} = 5 \)
One mole of \( \ce{KMnO4} \) therefore produces 5 mol of carbon dioxide.
Shortcut: Count electrons. One \( \ce{MnO4^-} \) accepts 5 electrons, and every electron given up by oxalate comes with one \( \ce{CO2} \) (2 electrons per 2 \( \ce{CO2} \)). So 5 electrons mean 5 mol of \( \ce{CO2} \).
Note: It seems that there is a mistake in the original NTA paper. The second equation of the passage is printed as 3K2MnO4 + 4H+ → 2KMnO4 + MnO2 + 2H2O, in which the charges do not balance; it should be 3MnO42− + 4H+ → 2MnO4− + MnO2 + 2H2O.