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 20 mL of a solution $Y$ containing $Mn{O}_4$ ions. What volume of solution $Y$ would be required to oxidise $25mL$ of a solution $X$ containing $F{e}^{2+}$ ions in neutral condition?

• A. $11.4mL$
• B. $12.0mL$
• C. $33.3mL$
• D. $35.0mL$
• E. $25.0mL$

Answer 1

Answer: (C)

In acidic medium, $Mn{O}_4^{-}$ is reduced to $M{n}^{2+}$
$\stackrel{+7}{\mathrm{Mn}}{O}_4^{-}\to M{n}^{2+}$
Change in oxidation number $=7-2=5<br/>$ Solution X Solution Y
$N_1{V}_1=N_2{V}_2$ For $F{e}^{2+}$ For $Mn{O}_4^{-}$
$N\times 25=5M\times V$ $[\because ForMn{O}_4^{-},N=5M]$ in acidic medium]
$25N=5M\times 20$ $25N=100M$ ...(i)
In neutral medium, $Mn{O}_4^{-}$ is reduced to $Mn{O}_2$
$\stackrel{+7}{Mn{O}_4^{-}}\to \stackrel{+4}{\mathrm{Mn}}{O}_2$
Change in oxidation number $=7-4=3$
Solution X Solution Y
$N_1{V}_1=N_2{V}_2$
For $F{e}^{2+}$ For $Mn{O}_4^{-}$
$25\times N=3M\times V$ [ $\because$ For $Mn{O}_4^{-},N=3M$ in neutral medium
$25N=3M\times V$ ...(ii) From Eqs (i) and (ii) $100M=3M\times V$ $V=\frac{100}{3}=33.3mL$