In alkaline condition KMnO_4 reacts as follow | 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{Molar mass}}{	ext{n-factor}}$In the given re

Vedclass · Accepted Answer

$158.0$

Vedclass · Answer

(D) The equivalent weight of an oxidizing agent is calculated as:$	ext{Equivalent weight} = \frac{	ext{Molar mass}}{	ext{n-factor}}$In the given reaction,the reduction half-reaction is:$MnO_4^- + e^- 	o MnO_4^{2-}$The change in oxidation state of $Mn$ is from $+7$ to $+6$,which involves the gain of $1$ electron per $KMnO_4$ molecule.Thus,the n-factor is $1$.The molar mass of $KMnO_4 = 39.1 + 54.9 + 4 	imes 16 = 158 \ g/mol$.$	ext{Equivalent weight} = \frac{158}{1} = 158$.

In alkaline condition $KMnO_4$ reacts as follows:
$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$
Therefore,its equivalent weight will be:

Similar Questions

In neutral or alkaline solution,$MnO_{4}^{-}$ oxidises thiosulphate to.

Which of the following reactions is a redox reaction?

The equivalent weight of ${S_2O_3^{2-}}$ in the reaction where it acts as a reductant is ...... . ${2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-}$

$H^{+} + Cr_2O_7^{2-} + SO_3^{2-} \to Cr^{3+} + SO_4^{2-} + H_2O$. In the balanced reaction,the coefficients of $H^{+}$ and $SO_4^{2-}$ will be respectively:

Which of the following equations is a balanced one?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

In alkaline condition $KMnO_4$ reacts as follows:$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$Therefore,its equivalent weight will be: