Question

In the following reaction:
$V{O}_3^{2-}+Mn{O}_4^{-}⟶{Mn}^{2+}+V{O}_4^{3-}$, how many moles of $Mn{O}_4$ is required to oxidise $1$ mol of $V{O}_3^{2-}$?

• A. $0.2$ mol
• B. $0.4$ mol
• C. $40.8$ mol
• D. $1.0$ mol

Answer 1

Answer: (A)

$0.2$ mol

$V{O}_3^{2-}+Mn{O}_4^{-}⟶{Mn}^{2+}+V{O}_4^{3-}$
+4 +7 +2 +5
So, change in oxidation number of $Mn=-5$
change in oxidation number of $V=+1$

Charge balance would give us,
$5molV{O}_3^{2-}\equiv 1molMn{O}_4^{-}$
$\therefore 1molV{O}_3^{2-}\equiv 0.2molMn{O}_4^{-}$