Question

$1$ mole of $Mn{O}_4^{-}$ will oxidise $x$ moles of ferric oxalate in acidic medium. The value of $x$ is :

• A. $\frac{5}{6}$
• B. $\frac{6}{5}$
• C. $5$
• D. $6$

Answer 1

Answer: (A)

$\frac{5}{6}$

$Mn{O}_4^{-}$ is reduced to $M{n}^{2+}$.
$Mn{O}_4^{-}\to M{n}^{+2}$
It will oxidise only oxalate, $(C_2{O}_4^{2-}$) as iron is already in its highest oxidised state.
The reaction is as follows:
${Fe}_2{\left(C_2{O}_4\right)}_3\rightleftharpoons 2{Fe}^{3+}+3C_2{O}_4^{2-}$
At equivalent point, both reactant have same number of equivalents.
Therefore, one mole of $Mn{O}_4^{-}$ will oxidise $\frac{5}{6}$ moles of ferric oxalate.