Equivalent weight of KMnO_4 acting as an oxid | vedclass

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

Question

(D) The equivalent weight of an oxidizing agent is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}}$.In an a

Vedclass · Accepted Answer

$\frac{1}{5} 	imes$ Molecular weight of $KMnO_4$

Vedclass · Answer

(D) The equivalent weight of an oxidizing agent is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}}$.In an acidic medium,$KMnO_4$ undergoes the following reduction reaction: $MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$.Here,the change in oxidation state of $Mn$ is from $+7$ to $+2$,which involves a transfer of $5$ electrons.Therefore,the n-factor is $5$.Thus,the equivalent weight of $KMnO_4$ is $\frac{1}{5} 	imes 	ext{Molecular weight}$.

Compound	Oxidation number
$1$. $C_6H_{12}O_6$ (Carbon)	$a$. $+1$
$2$. $CCl_4$ (Carbon)	$b$. $0$
$3$. $NH_4Cl$ (Nitrogen)	$c$. $+4$
$4$. $Ba(H_2PO_2)_2$ (Phosphorus)	$d$. $-3$

Equivalent weight of $KMnO_4$ acting as an oxidant in acidic medium is equal to

Similar Questions

What is the oxidation number of oxygen in a superoxide ion?

Match the following compounds with the oxidation number of the central atom (or specific atom in question):

Compound Oxidation number

$1$. $C_6H_{12}O_6$ (Carbon) $a$. $+1$

$2$. $CCl_4$ (Carbon) $b$. $0$

$3$. $NH_4Cl$ (Nitrogen) $c$. $+4$

$4$. $Ba(H_2PO_2)_2$ (Phosphorus) $d$. $-3$

In the chemical reaction $Cl_2 + H_2S \to 2HCl + S$,the oxidation number of sulphur changes from

What are the oxidation numbers of hydrogen in $KH$,$MgH_2$,and $NaH$ respectively?

The oxidation state of elemental carbon is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module