Question

In dilute alkaline solution, $ \text{MnO}_4^- $ changes to

• A. \( \text{MnO}_2 \)
• B. \( \text{MnO}_4^{2-} \)
• C. \( \text{MnO} \)
• D. \( \text{Mn}_2\text{O}_3 \)

Hint:

In alkaline solutions, potassium permanganate (\( \text{MnO}_4^- \)) is reduced to manganate ions (\( \text{MnO}_4^{2-} \)), which results in a color change from purple to green.

Answer 1

The Correct Option is B

In a dilute alkaline solution, the permanganate ion \( \text{MnO}_4^- \) undergoes a chemical reduction reaction to form the manganate ion \( \text{MnO}_4^{2-} \). This process can be explained by analyzing the changes in oxidation states and balancing the resultant half-reaction in a basic environment.

When potassium permanganate \( \text{KMnO}_4 \) is dissolved in a dilute alkaline solution, the \(\text{MnO}_4^-\) ion is reduced. The reaction can be represented as follows:

Reduction half-reaction:

\( \text{MnO}_4^- + e^- \rightarrow \text{MnO}_4^{2-} \)

In this reaction, the oxidation state of manganese changes from +7 in \( \text{MnO}_4^- \) to +6 in \( \text{MnO}_4^{2-} \). This change indicates a gain of one electron, demonstrating a reduction process. In an alkaline medium, the addition of hydroxide ions \( \text{OH}^- \) assists in maintaining charge balance:

\( \text{MnO}_4^- + \text{e}^- + \text{OH}^- \rightarrow \text{MnO}_4^{2-} + \text{OH}^- \)

Thus, in dilute alkaline conditions, \( \text{MnO}_4^- \) will most commonly convert into \( \text{MnO}_4^{2-} \), which is known as the manganate ion.

Answer 2

In a dilute alkaline solution, permanganate ion (\(\text{MnO}_4^-\)) undergoes a reduction reaction. The permanganate ion is a strong oxidizing agent, and in such a basic environment, it will be reduced to manganate ion (\(\text{MnO}_4^{2-}\)). This can be represented by the following half-reaction:

\[\text{MnO}_4^- + e^- \rightarrow \text{MnO}_4^{2-}\]

This reduction occurs because the alkaline environment provides the necessary conditions for the transformation of the \(\text{MnO}_4^-\) ion into the \(\text{MnO}_4^{2-}\) ion. Both ions have different oxidation states of manganese, \(\text{Mn}\) in \(\text{MnO}_4^-\) is in the +7 oxidation state, while in \(\text{MnO}_4^{2-}\), it is in the +6 oxidation state, indicating a reduction process by the gain of an electron.