Question

$Mn{O}_4^{-}$ ions are reduced in acidic condition to $M{n}^{2+}$ ions whereas they are reduced in neutral condition to $Mn{O}_2$. The oxidation of $25mL$ of a solution $X$ containing $F{e}^{2+}$ ions required in acidic condition $20mL$ of a solution $Y$ containing $Mn{O}_4^{-}$ ions. What volume of solution $Y$ would be required to oxidise $25mL$ of solution $X$ containing $F{e}^{2+}$ ions in neutral condition?

• A. $11.4mL$
• B. $12.0mL$
• C. $33.3mL$
• D. $35.0mL$

Answer 1

Answer: (C)

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

In acidic medium
$5vol$. of $F{e}^{2+}$ requires $1vol$. of $Mn{O}_4^{-}$ in acidic medium.
$\therefore 25vol$. of $F{e}^{2+}requires\frac{1}{5}\times 25vol$ of $Mn{O}_4^{-}$ in acidic medium.
$\frac{1}{5}\times 25vol$ of $Mn{O}_4^{-}\equiv 20vol$. or $20mL$
In neutral medium.
$3vol$. of $F{e}^{2+}$ requires $1vol$. of $Mn{O}_4^{-}$ in neutral medium Then $25vol$. of $F{e}^{2+}$ requires vol. of $\frac{1}{3}\times 25Mn{O}_4^{-}$ in neutral medium
$\frac{1}{5}\times 25vol$ of $Mn{O}_4^{-}\equiv 20vol$.
$\therefore \frac{1}{3}\times 25vol$ of $Mn{O}_4^{-}\approx \frac{20}{5}\times \frac{25}{3}vol$ of $Mn{O}_4^{-}$
$=33.3vol$. or $mL$