$KMnO_4$ in basic medium is reduced to
- A$K_2MnO_4$
- B$MnO_2$
- C$Mn(OH)_2$
- D$Mn^{2+}$
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Similar Questions
Molecular weight of $Cu_2S$ is $M$. In the following reaction its equivalent weight is $M / x$. What is $x$?
$Cu_2S + MnO_4^{-} \rightarrow Cu^{2+} + Mn^{2+} + SO_2$
$Cu_2S + MnO_4^{-} \rightarrow Cu^{2+} + Mn^{2+} + SO_2$
Medium
View SolutionThe set of coefficients of $P_4$,$OH^{\Theta}$ and $H_2PO_2^{\Theta}$ in the balanced reaction $P_4 + OH^{\Theta} \to PH_3 + H_2PO_2^{\Theta}$ is
Medium
View SolutionIn an alkaline solution,$ClO_2$ oxidizes $H_2O_2$ to $O_2$ and is itself reduced to $Cl^-$. How many moles of $H_2O_2$ will be oxidized by $1 \ mol$ of $ClO_2$?
Difficult
View SolutionFor the following balanced equation,$x, y$ and $z$ will be respectively:
$xMnO_4^- + yH^+ + ze^- \to 2Mn^{2+} + 4H_2O$
$xMnO_4^- + yH^+ + ze^- \to 2Mn^{2+} + 4H_2O$
Difficult
View SolutionHow many moles of $H_2O_2$ are required to decolorize $1 \ mol$ of $KMnO_4$ in an acidic medium (in $.5$)?
Difficult
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