Question

Observe the statements and choose the correct option:

Statement-l: $KMnO_4$ oxidises nitrite to nitrate in acidic medium.

Statement-II: Manganous salt is oxidised to $MnO_2$ by $KMnO_4$ in neutral medium and $ZnSO_4$ or $ZnO$ can catalyse the oxidation.

• A. Statement-l and statement-Il both are correct
• B. Statement-l and statement-Il both are incorrect
• C. Statement-l is correct but statement-ll is incorrect
• D. Statement-l is incorrect but statement-ll is correct

Answer 1

Answer: (A)

Statement-l and statement-Il both are correct

Answer 2

The question asks us to evaluate two statements regarding the reactions of $KMnO_4$.

Statement-I: $KMnO_4$ oxidises nitrite to nitrate in acidic medium.

Let's write the half-reactions for this redox process in acidic medium:

1. Oxidation of nitrite ($NO_2^-$) to nitrate ($NO_3^-$):
   The oxidation state of nitrogen changes from +3 in $NO_2^-$ to +5 in $NO_3^-$.
   $NO_2^- + H_2O \rightarrow NO_3^- + 2H^+ + 2e^-$

2. Reduction of permanganate ($MnO_4^-$) to manganese(II) ion ($Mn^{2+}$) in acidic medium:
   The oxidation state of manganese changes from +7 in $MnO_4^-$ to +2 in $Mn^{2+}$.
   $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

To obtain the overall balanced reaction, we multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to balance the electrons (10 electrons).
$5(NO_2^- + H_2O \rightarrow NO_3^- + 2H^+ + 2e^-)$
$2(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O)$

Adding the two half-reactions:
$5NO_2^- + 5H_2O + 2MnO_4^- + 16H^+ \rightarrow 5NO_3^- + 10H^+ + 2Mn^{2+} + 8H_2O$

Simplifying the $H^+$ and $H_2O$ terms:
$2MnO_4^- + 5NO_2^- + 6H^+ \rightarrow 2Mn^{2+} + 5NO_3^- + 3H_2O$
This reaction is a well-known redox reaction where $KMnO_4$ acts as a strong oxidizing agent.
Therefore, Statement-I is correct.

Statement-II: Manganous salt is oxidised to $MnO_2$ by $KMnO_4$ in neutral medium and $ZnSO_4$ or $ZnO$ can catalyse the oxidation.

This statement describes a comproportionation reaction where manganese in a lower oxidation state ($Mn^{2+}$, +2) reacts with manganese in a higher oxidation state ($MnO_4^-$, +7) to form an intermediate oxidation state ($MnO_2$, +4).
Let's write the half-reactions in neutral/slightly alkaline medium (as $MnO_2$ typically precipitates in such conditions):

1. Oxidation of manganous ion ($Mn^{2+}$) to manganese dioxide ($MnO_2$):
   $Mn^{2+} + 4OH^- \rightarrow MnO_2 + 2H_2O + 2e^-$

2. Reduction of permanganate ($MnO_4^-$) to manganese dioxide ($MnO_2$) in neutral/basic medium:
   $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$

To obtain the overall balanced reaction, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to balance the electrons (6 electrons).
$3(Mn^{2+} + 4OH^- \rightarrow MnO_2 + 2H_2O + 2e^-)$
$2(MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-)$

Adding the two half-reactions:
$3Mn^{2+} + 12OH^- + 2MnO_4^- + 4H_2O \rightarrow 3MnO_2 + 6H_2O + 2MnO_2 + 8OH^-$

Simplifying the $OH^-$ and $H_2O$ terms:
$3Mn^{2+} + 2MnO_4^- + 4OH^- \rightarrow 5MnO_2 + 2H_2O$
This reaction indeed occurs in neutral or slightly alkaline medium, leading to the precipitation of brown $MnO_2$.

Regarding the catalysis by $ZnSO_4$ or $ZnO$:
In the industrial preparation of manganese dioxide, the oxidation of $Mn^{2+}$ by $KMnO_4$ is a known method. It is well-documented that the presence of zinc salts ($ZnSO_4$) or zinc oxide ($ZnO$) can catalyze this reaction. $ZnO$ particularly helps in maintaining a suitable pH by consuming any acid produced and provides nucleation sites for the $MnO_2$ precipitate, thereby enhancing the reaction rate and purity of the product.
Therefore, Statement-II is also correct.

Since both Statement-I and Statement-II are correct, the correct option is A.

Explanation of the solution:

Statement-I is correct because $KMnO_4$ is a strong oxidizing agent that oxidizes nitrite ($NO_2^-$) to nitrate ($NO_3^-$) in acidic medium, while $MnO_4^-$ is reduced to $Mn^{2+}$.
Statement-II is correct because manganous salt ($Mn^{2+}$) reacts with $KMnO_4$ ($MnO_4^-$) in neutral medium to form $MnO_2$ (a comproportionation reaction). This reaction is known to be catalyzed by $ZnSO_4$ or $ZnO$, which help maintain pH and provide nucleation sites.