Question:
Observe the following unbalanced reactions
\[
KO_2 \xrightarrow{\;HOH\;} X + Y\uparrow + KOH
\]
\[
KMnO_4 \xrightarrow[\text{basic medium}]{\;X\;}
Y + Z + KOH + H_2O
\]
Y and Z are respectively
Observe the following unbalanced reactions
\[
KO_2 \xrightarrow{\;HOH\;} X + Y\uparrow + KOH
\]
\[
KMnO_4 \xrightarrow[\text{basic medium}]{\;X\;}
Y + Z + KOH + H_2O
\]
Y and Z are respectively
Show Hint
Important reactions:
\[
2KO_2+2H_2O
\rightarrow
2KOH+H_2O_2+O_2
\]
and in alkaline medium,
\[
KMnO_4 + H_2O_2
\rightarrow
MnO_2 + O_2.
\]
Thus, \(H_2O_2\) acts as a reducing agent and converts purple permanganate into brown \(MnO_2\).
Updated On: Jul 29, 2026
- \(O_2,\ MnO\)
- \(O_2,\ MnO_2\)
- \(O_3,\ MnO_2\)
- \(O_3,\ MnO\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Identify \(X\) and \(Y\) from the first reaction.
Potassium superoxide reacts with water as
\[
2KO_2+2H_2O
\rightarrow
2KOH+H_2O_2+O_2.
\]
Therefore,
\[
X=H_2O_2
\]
and
\[
Y=O_2.
\]
Step 2: Use \(X\) in the second reaction. In alkaline medium, hydrogen peroxide reduces permanganate to manganese dioxide. \[ 2KMnO_4+3H_2O_2 \rightarrow 2MnO_2+3O_2+2KOH+2H_2O. \] Hence, \[ Y=O_2 \] and \[ Z=MnO_2. \]
Final Answer: \[ \boxed{Y=O_2,\qquad Z=MnO_2} \] \[ \boxed{\text{Answer = (B)}} \]
Step 2: Use \(X\) in the second reaction. In alkaline medium, hydrogen peroxide reduces permanganate to manganese dioxide. \[ 2KMnO_4+3H_2O_2 \rightarrow 2MnO_2+3O_2+2KOH+2H_2O. \] Hence, \[ Y=O_2 \] and \[ Z=MnO_2. \]
Final Answer: \[ \boxed{Y=O_2,\qquad Z=MnO_2} \] \[ \boxed{\text{Answer = (B)}} \]
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