Question

H\(_2\)O\(_2\) reduces KMnO\(_4\) in acidic medium to 'x' and in basic medium to 'y'. What are x and y?

• A. x = MnO\(_2\), y = Mn\(^{2+}\)
• B. x = Mn\(^{2+}\), y = MnO\(_2\)
• C. x = MnO\(_4^{2-}\), y = Mn\(^{2+}\)
• D. x = MnO\(_2\), y = MnO\(_4^{2-}\)

Hint:

Remember the common reduction products of permanganate ion ($\operatorname{MnO}_4^-$) in different media: Acidic Medium}: Mn(+7) \(\rightarrow\) Mn(+2) (e.g., $\operatorname{Mn}^{2+}$ ions, colorless in dilute solution) Neutral or Weakly Basic Medium}: Mn(+7) \(\rightarrow\) Mn(+4) (e.g., $\operatorname{MnO}_2$, a brown precipitate) Strongly Basic Medium}: Mn(+7) \(\rightarrow\) Mn(+6) (e.g., $\operatorname{MnO}_4^{2-}$, manganate ion, green color)

Answer 1

The Correct Option is B

Step 1: Understand the role of H\(_2\)O\(_2\) and the oxidation state of Mn in KMnO\(_4\).
Hydrogen peroxide (H\(_2\)O\(_2\)) can act as a reducing agent (gets oxidized itself to O\(_2\)) or an oxidizing agent. In this problem, it is stated that H\(_2\)O\(_2\) reduces KMnO\(_4\), meaning H\(_2\)O\(_2\) acts as a reducing agent.
In KMnO\(_4\), the manganese (Mn) atom is in the +7 oxidation state (from $\operatorname{MnO}_4^-$). Since it is being reduced, its oxidation state will decrease. The specific product of reduction depends on the reaction medium (acidic or basic). 
Step 2: Determine the product 'x' (reduction of KMnO\(_4\) in acidic medium).
In a strong acidic medium, the permanganate ion ($\operatorname{MnO}_4^-$), which is a very powerful oxidizing agent, is typically reduced to the manganese(II) ion, $\operatorname{Mn}^{2+}$. In this process, the oxidation state of Mn changes from +7 to +2. The half-reaction for the reduction of permanganate in acidic medium is: \[ \operatorname{MnO}_4^- + 8\operatorname{H}^+ + 5\operatorname{e}^- \rightarrow \operatorname{Mn}^{2+} + 4\operatorname{H}_2\operatorname{O} \] Thus, 'x' is $\operatorname{Mn}^{2+}$. 
Step 3: Determine the product 'y' (reduction of KMnO\(_4\) in basic medium).
In a basic (or neutral) medium, the permanganate ion ($\operatorname{MnO}_4^-$) is typically reduced to manganese dioxide, $\operatorname{MnO}_2$. In $\operatorname{MnO}_2$, the oxidation state of Mn is +4. The half-reaction for the reduction of permanganate in basic medium is: \[ \operatorname{MnO}_4^- + 2\operatorname{H}_2\operatorname{O} + 3\operatorname{e}^- \rightarrow \operatorname{MnO}_2 + 4\operatorname{OH}^- \] Thus, 'y' is $\operatorname{MnO}_2$. 
Step 4: Conclude the values of x and y.
Based on the analysis:
x = $\operatorname{Mn}^{2+}$
y = $\operatorname{MnO}_2$
This corresponds to Option (2). The final answer is $\boxed{x = Mn}^{2+}, y = MnO}_2}$.