Question:
In acid medium, \(\mathrm{H_2O_2}\) reacts with aqueous \(\mathrm{KMnO_4}\) to form \(\mathrm{Mn^{2+}}\), \(\mathrm{H_2O}\) and \(X\). In basic medium, \(\mathrm{H_2O_2}\) reacts with aqueous \(\mathrm{KMnO_4}\) to form \(\mathrm{MnO_2}\), \(\mathrm{H_2O}\), \(\mathrm{OH^-}\) and \(Y\). What are \(X\) and \(Y\) respectively?
In acid medium, \(\mathrm{H_2O_2}\) reacts with aqueous \(\mathrm{KMnO_4}\) to form \(\mathrm{Mn^{2+}}\), \(\mathrm{H_2O}\) and \(X\). In basic medium, \(\mathrm{H_2O_2}\) reacts with aqueous \(\mathrm{KMnO_4}\) to form \(\mathrm{MnO_2}\), \(\mathrm{H_2O}\), \(\mathrm{OH^-}\) and \(Y\). What are \(X\) and \(Y\) respectively?
Show Hint
Hydrogen peroxide acts as a reducing agent towards \(\mathrm{KMnO_4}\) in both acidic and basic media and gets oxidized to
\[
\boxed{\mathrm{O_2}.}
\]
Updated On: Jul 15, 2026
- \(\mathrm{O_2,\ H_2}\)
- \(\mathrm{H_2,\ H_2}\)
- \(\mathrm{O_2,\ O_2}\)
- \(\mathrm{H_2,\ O_2}\)
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The Correct Option is C
Solution and Explanation
Step 1: Reaction in acidic medium.
In acidic medium,
\[
\mathrm{MnO_4^-}
\]
is reduced to
\[
\mathrm{Mn^{2+}},
\]
while hydrogen peroxide is oxidized to oxygen.
\[
\boxed{
2\mathrm{MnO_4^-}+5\mathrm{H_2O_2}+6\mathrm{H^+}
\rightarrow
2\mathrm{Mn^{2+}}+5\mathrm{O_2}+8\mathrm{H_2O}
}
\]
Hence,
\[
\boxed{X=\mathrm{O_2}}.
\]
Step 2: Reaction in basic medium. In basic medium, \[ \mathrm{MnO_4^-} \] is reduced to \[ \mathrm{MnO_2}, \] and hydrogen peroxide is again oxidized to oxygen. Hence, \[ \boxed{Y=\mathrm{O_2}}. \] Therefore, \[ \boxed{X=\mathrm{O_2},\qquad Y=\mathrm{O_2}} \] Thus, \[ \boxed{(C)} \] is the correct answer.
Step 2: Reaction in basic medium. In basic medium, \[ \mathrm{MnO_4^-} \] is reduced to \[ \mathrm{MnO_2}, \] and hydrogen peroxide is again oxidized to oxygen. Hence, \[ \boxed{Y=\mathrm{O_2}}. \] Therefore, \[ \boxed{X=\mathrm{O_2},\qquad Y=\mathrm{O_2}} \] Thus, \[ \boxed{(C)} \] is the correct answer.
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