Question

$2.68\times {10}^{-3}mol$ of a solution containing an ion $A^{n+}$ required $1.6\times {10}^{-3}$ mol of ${Mn{O}_4}^{-1}$ for oxidation of $A^{n+}$ to ${A{O}_3}^{-1}$ ion in acid medium. What is the value of $n$?

Answer 1

$A^{n+}$ is oxidised to $A{O}_3^{-}$
Change in oxidation number $=5\left(inA{O}_3^{-}\right)-n\left(in{A}^{n+}\right)$
$=5-n.....\left(i\right)$
$2.68\times {10}^{-3}$ mol of $A^{n+}$ ion react with $1.6\times {10}^{-3}$ mol of $Mn{O}_4^{-}$ ions
$\therefore 1mol$ of $A^{n+}$ ion will react with $\frac{1.6\times {10}^{-3}}{2.6\times {10}^{-3}}$ mol of $Mn{O}_4^{-}$ ions
$2K\stackrel{+7}{M}n{O}_4+3H_{2}S{O}_4\to K_{2}S{O}_4+2\stackrel{+2}{M}nS{O}_4+3H_{2}O+5\left[O\right]$
Number of equivalents of $Mn{O}_4^{-}$ used in oxidation of $A^{n+}$ to $A{O}_3^{-}=0.597\times 5=2.985\approx 3$
Thus, from equation (i), $5-n=3$
$n=2$.