Q. ${MnO}_{4}^{-}$ 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 ${MnO}_{4}^{-}$ ions. What volume of solution $Y$ would be required to oxidise 25 mL of a solution $X$ containing ${{Fe}^{2}}^{+}$ ions in neutral condition?
Explanation
(c)
In acidic medium, $MnO_{4}^{-}$ is reduced to $Mn^{2 +}$
$\begin{matrix}
+ 7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
Mn\ O_{4}^{-} \rightarrow {Mn}^{2 +}
\end{matrix}$
Change in oxidation number=7$-$2=5
Solution $X$ Solution $Y$
$N_{1}V_{1} = N_{2}V_{2}$
For $Fe^{2 +}$ For $MnO_{4}^{-}$
$N \times 25 = 5M \times V$ [$\because$ For $MnO_{4}^{-},\ N = 5M$ in acidic medium]
25N= $5M \times 20$
25N=100M …(i)
In neutral medium, $MnO_{4}^{-}$ is reduced to $MnO_{2}$
$\begin{matrix}
+ 7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + 4\ \ \ \\
Mn\ O_{4}^{-} \rightarrow {MnO}_{4}^{-}
\end{matrix}$
Change in oxidation number=7$-$4=3
Solution $X$ Solution $Y$
$N_{1}V_{1} = N_{2}V_{2}$
For $Fe^{2 +}$ For $MnO_{4}^{-}$
$25 \times N = 3M \times V$
[$\because$ For $MnO_{4}^{-},\ N = 3M$ in neutral medium]
25N= $3M \times V$ …(ii)
From Eqs (i) and (ii)
100M=3M$\times V$
$V = \frac{100}{3} = 33.3$ mL
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.