What is the change in the oxidation state of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In an acidic medium,the permanganate ion $(MnO_4^{-})$ acts as a strong oxidizing agent and reacts with hydrogen peroxide $(H_2O_2)$.The balanced

Vedclass · Accepted Answer

$7 \rightarrow 2$

Vedclass · Answer

(C) In an acidic medium,the permanganate ion $(MnO_4^{-})$ acts as a strong oxidizing agent and reacts with hydrogen peroxide $(H_2O_2)$.The balanced chemical equation for this reaction is: $2MnO_4^{-} + 5H_2O_2 + 6H^{+} \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$.In $MnO_4^{-}$,the oxidation state of $Mn$ is calculated as: $x + 4(-2) = -1$,which gives $x = +7$.In the product $Mn^{2+}$,the oxidation state of $Mn$ is $+2$.Therefore,the oxidation state of $Mn$ changes from $+7$ to $+2$.

What is the change in the oxidation state of $Mn$ in the reaction of $MnO_4^{-}$ with $H_2O_2$ in acidic medium?

Similar Questions

Arrange $ICl$,$HI$,$I_2$,and $HIO_4$ in the decreasing order of the oxidation number of iodine.

The oxidation number and covalency of sulphur in the sulphur molecule $S_8$ are respectively

In alkaline condition,$KMnO_4$ reacts as $2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + [O]$. The equivalent weight of $KMnO_4$ would be (Atomic mass of $K = 39$,$Mn = 55$,$O = 16$).

When zinc reacts with very dilute $HNO_3$,the oxidation state of nitrogen changes from

Calculate the oxidation number of $Fe$ in $Fe_{0.94}O$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module