Question

\[{A_2}{O_n}\] is oxidized to $A{O^{3 - }}$ by $KMn{O_4}$ in acidic medium If $1.34mmol$ of ${A_2}{O_n}$ require $32.2ml$ of $0.5M$ acidified solution for complete oxidation. Find the value of n.
$A.$ $4$
$B.$ $2$
$C.$ $6$
$D.$ $5$

Answer 1

Hint: In the given question we have to find the value of n which let the oxidation state of $A$ . The medium is acidic. Also given that ${A_2}{O_n}$ is oxidized to $AO_3^ - $ by $KMn{O_4}$ Oxidized means to make something combine with atom that is increase in oxidation number.

Complete step by step answer:
As per the question the medium is acidic, therefore, $MnO_4^ - $ from $KMn{O_4}$ will change to $M{n^{2 + }}$ . The oxidation state on $M{n^{2 + }}$ in $MnO_4^ - $ is $ + 7$ and the oxidation state of $Mn$ in $M{n^{2 + }}$ is $ + 2$ .
So, $MnO_4^ - $ is reduced to $M{n^{2 + }}$ as oxidation state is decreased. Thus the net reaction would be
${A_2}{O_n} + KMn{O_4} \to AO_3^ - + M{n^{2 + }}$
Let oxidation state of $ = n$ .
The above reaction is a redox reaction. In a chemical reaction in which oxidation and reduction occur simultaneously it is called redox reaction. The number of electrons lost in oxidation is always equal to the number of electrons lost gained in reduction.
Number of electron in reduction of $M{n^{7 + }}$ to $M{n^{2 + }}$ $ = 7 - 2 = 5$
For $0.05M$ $KMn{O_4}$ ,
$ \Rightarrow $ Molarity $\left( M \right)$ $ = $ Number of moles/Volume in liter
$ \Rightarrow $ $0.05 = $ Number of moles/$\dfrac{{32.2}}{{1000}}$
$ \Rightarrow $ Number of moles$ = \dfrac{{32.2 \times 0.05}}{{1000}}$
So, number of for $\left( {\dfrac{{32.2 \times 0.05}}{{1000}}} \right)$ moles of $M{n^{7 + }}$ being converted to $M{n^{2 + }}$$ = \dfrac{{5 \times 32.2 \times 0.05}}{{1000}}$
For ${A_2}{O_n}$ , let oxidation state of $A = n$ and oxidation state $A$ in $AO_3^ - $ is $5$
$\therefore $ Number of electron for conversion of $1$ molecules of ${A_2}{O_n}$ to $AO_3^ - $ $ = $ $5 - n$
Given moles of ${A_2}{O_n}$ $ = 1.34mmol = 1.34 \times {10^{ - 3}}mol$
$\therefore $ For conversion of $1.34 \times {10^{ - 3}}$ moles of ${A_2}{O_n}$ to $AO_3^ - $ , number of electron $ = 1.34 \times {10^{ - 3}} \times \left( {5 - n} \right)$
Now, Number of electron in reduction $ = $ Number of electron is oxidation
$ \Rightarrow $ $\dfrac{{5 \times 32.2 \times 0.05}}{{1000}} = 1.34 \times {10^{ - 3}} \times \left( {5 - n} \right)$
By solving above equation, we get
$ \Rightarrow $ $n = 2$

Thus option $B$ is correct.

Note:
Number of electrons lost in oxidation is always equal to the number of electrons gained in reduction. The term oxidation is defined in many ways which summarizes in given table below:

Oxidation	Reduction
Increase in oxidation state	Decrease in oxidation state
Lose of electron	Gain of electron
Addition of oxygen atom	Addition of Hydrogen atom
Removal of Hydrogen atom	Removal of oxygen atom