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 disproportionate to give permanganate as follows.

$2MnO_2+ 4KOH + O_2→2KMnO_4+2H_2O$

$3KMnO_4+ 4H^+→2KMnO_4 + MnO_2 + 2H_2O$

On heating $KMnO_4$ decomposes at 513 K to give $K_2MnO_4$. Permanganate ion is tetrahedral and diamagnetic. Acidified $KMnO_4$ acts 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?

5

The correct answer is Option (1) → 5 **

To solve this, we need to look at the balanced redox reaction between Potassium Permanganate ($KMnO_4$) and Oxalic Acid ($H_2C_2O_4$) in an acidic medium.

1. The Half-Reactions

First, we identify how many electrons are transferred in each part of the reaction:

• Reduction: $MnO_4^-$ is reduced to $Mn^{2+}$.

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

(Each mole of $KMnO_4$ consumes 5 moles of electrons).

• Oxidation: Oxalic acid is oxidized to Carbon Dioxide ($CO_2$).

$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$

(Each mole of oxalate ion releases 2 moles of electrons).

To balance the electrons, we multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5:

$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$

$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$

Adding them together gives the full ionic equation:

$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$

3. Calculating Moles of $CO_2$

From the balanced equation, we can see the stoichiometric ratio:

• 2 moles of $KMnO_4$ produce 10 moles of $CO_2$.

To find how much 1 mole of $KMnO_4$ produces:

$\frac{10 \text{ moles } CO_2}{2 \text{ moles } KMnO_4} = \mathbf{5 \text{ moles of } CO_2}$

Therefore, the correct answer is 5.