Question

$5\times {10}^{-3}\text{mol}$ of a solution containing an ion $A^{n+}$ require $3\times {10}^{-3}\text{mol}$ of $Mn{O}_4^{-}$ for the oxidation of $A^{n+}$ to $A{O}_3^{-}$ in acidic medium. What is the value of $n$?

Answer 1

In acidic medium, $Mn{O}_4^{-}$, will be reduced into $M{n}^{2+}$
$\stackrel{7+}{Mn}{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}Ox=5$
$A^{n+}\stackrel{\text{oxidation}}{\to }\stackrel{5+}{A{O}_3^{-}}$
As in the oxidation, oxidation state of the elements is increased therefore $5>n$ and $x^{\prime }=(5-n)$.
$\therefore$ Molar ratio at which $A^{n+}$ and $Mn{O}_4^{-}$
react $=5:(5-n)$
$\therefore \frac{5}{(5-n)}=\frac{5\times {10}^{-3}}{3\times {10}^{-3}}=\frac{5}{3}$
$\Rightarrow n=2$.
$\therefore$ The value of $n$ is $2$.