Question

$MnO_4^-$ ions are reduced in acidic condition to $Mn^{2+}$ ions whereas they are reduced in neutral condition to $MnO_2$. The oxidation of $25\, mL$ of a solution $X$ containing $Fe^{2+}$ ions required in acidic condition $20\, mL$ of a solution $Y$ containing $M\,nO_4^-$ ions. What volume of solution $Y$ would be required to oxidise $25\, mL$ of solution $X$ containing $Fe^{2+}$ ions in neutral condition?

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

Answer 1

Answer: (C)

$MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$

In acidic medium
$5\, vol$. of $Fe^{2+}$ requires $1\, vol$. of $MnO_4^-$ in acidic medium.
$ \therefore \quad25\, vol$. of $Fe^{2+} requires \frac{1}{5} \times 25\, vol$ of $MnO_{4}^{-}$ in acidic medium.
$\frac{1}{5} \times 25\, vol$ of $MnO_{4}^{-} \equiv 20 \,vol$. or $20\, mL$
In neutral medium.
$3 \,vol$. of $Fe^{2+}$ requires $1 \,vol$. of $MnO_{4}^{-}$ in neutral medium Then $25 \,vol$. of $Fe^{2+}$ requires vol. of $\frac{1}{3} \times 25\,MnO^{-}_{4}$ in neutral medium
$\frac{1}{5} \times 25\, vol$ of $MnO_{4}^{-} \equiv 20 \,vol$.
$\therefore \quad\frac{1}{3} \times 25\, vol$ of $MnO_{4}^{-} \approx \frac{20}{5} \times \frac{25}{3} \,vol$ of $MnO_{4}^{-}$
$= 33.3 \,vol$. or $mL$