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?
Answer & explanation
Correct answer: option 1
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.