Question

$2.68\times {10}^{-3}$ moles of a solution containing an ion $A^{n+}$ require $1.61\times {10}^{-3}$ moles of $Mn{O}_4^{-}$ for the oxidation of $A^{n+}$ to $A{O}_3^{-}$ in acidic medium. What is the value of $n$?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

$A^{n+}+\stackrel{+7}{M}n{O}_4^{-}\stackrel{H^{+}}{⟶}\stackrel{+2}{M}{n}^{2+}+\stackrel{+5}{A}{O}_3^{-}$

$n_f=\left(|\text{Change in O.S.}|\right)\times \text{Number of atoms}$
$n_{f_{Mn{O}_4^{-}}}=|+2-(+7)|\times 1=5$

From the law of equivalence,
$\text{Equivalents of}{A}^{n+}=\text{Equivalents of}Mn{O}_4^{-}$
$2.68\times {10}^{-3}\times n_{f_{A^{n+}}}=1.61\times {10}^{-3}\times 5$
$\therefore n_{f_{A^{n+}}}\approx 3$
From the formula for n-factor,
$n_{f_{A^{n+}}}=|+5-(+n)|\times 1=3$
$\therefore n=2$