
$MnO_4^{2 - }$ changes to $Mn{O_2}$ into $MnO_4^ - $ in acidic medium. Calculate equivalent weight of $MnO_4^{2 - }$.
A) $1M$
B) $\dfrac{1}{2}M$
C) $\dfrac{3}{2}M$
D) $3M$
Answer
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Hint:At first we will write what is given in the question. Then we will write what $MnO_4^{2 - }$ is. Then we will write the reaction of $MnO_4^{2 - }$ with water. We will know what is the type of the reaction. Then we will find the equivalent weight of $MnO_4^{2 - }$ . We will also find where reduction and oxidation reactions happen.
Complete solution:
Step1. We are given with $MnO_4^{2 - }$. It will be disproportionate to $Mn{O_2}$ and $MnO_4^ - $. We need to calculate the equivalent weight of $MnO_4^{2 - }$.
Step2. $MnO_4^{2 - }$ is called manganite . Its molecular weight is 118.936 g/mol. It is an inorganic compound which is a divalent anion formed by removal of both the protons from manganic acid.
Step3. The $MnO_4^{2 - }$ breaks into two different compounds. While it water it breaks and releases hydrogen ions. It is the disproportionation reaction. In the disproportionation reaction the equivalent weight of the species undergoing the disproportionation. While disproportionation one will be oxidizing asn another is reducing it can also be called redox reaction.
Step4. $MnO_4^{2 - }$$ \to $$Mn{O_2}$
The oxidation number of Mn in $MnO_4^{2 - }$ is six and the oxidation number of $Mn{O_2}$is four. It is reducing. The difference is two then the molecular weight is $mol.wt/2$.
$MnO_4^{2 - }$$ \to $$MnO_4^ - $
The oxidation number of Mn in $MnO_4^{2 - }$is six and the oxidation number of $MnO_4^ - $ is seven so the equivalent weight is $\dfrac{{mol.wt}}{1}$ .
So total equivalent weight is $\dfrac{M}{2} + \dfrac{M}{1} = \dfrac{3}{2}M$
Hence the correct answer is $\dfrac{3}{2}M$.
Note:To solve such questions, we must know what redox reactions are and how they work, A redox reaction can be defined as a chemical reaction in which electrons are transferred between two reactants participating in it. The oxidation reaction can be identified as the oxidation state changes.
Complete solution:
Step1. We are given with $MnO_4^{2 - }$. It will be disproportionate to $Mn{O_2}$ and $MnO_4^ - $. We need to calculate the equivalent weight of $MnO_4^{2 - }$.
Step2. $MnO_4^{2 - }$ is called manganite . Its molecular weight is 118.936 g/mol. It is an inorganic compound which is a divalent anion formed by removal of both the protons from manganic acid.
Step3. The $MnO_4^{2 - }$ breaks into two different compounds. While it water it breaks and releases hydrogen ions. It is the disproportionation reaction. In the disproportionation reaction the equivalent weight of the species undergoing the disproportionation. While disproportionation one will be oxidizing asn another is reducing it can also be called redox reaction.
Step4. $MnO_4^{2 - }$$ \to $$Mn{O_2}$
The oxidation number of Mn in $MnO_4^{2 - }$ is six and the oxidation number of $Mn{O_2}$is four. It is reducing. The difference is two then the molecular weight is $mol.wt/2$.
$MnO_4^{2 - }$$ \to $$MnO_4^ - $
The oxidation number of Mn in $MnO_4^{2 - }$is six and the oxidation number of $MnO_4^ - $ is seven so the equivalent weight is $\dfrac{{mol.wt}}{1}$ .
So total equivalent weight is $\dfrac{M}{2} + \dfrac{M}{1} = \dfrac{3}{2}M$
Hence the correct answer is $\dfrac{3}{2}M$.
Note:To solve such questions, we must know what redox reactions are and how they work, A redox reaction can be defined as a chemical reaction in which electrons are transferred between two reactants participating in it. The oxidation reaction can be identified as the oxidation state changes.
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