Question

$ MnO_{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 $ 25\, mL $ 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 $ 25\, mL $ of a solution $ X $ containing $ F{{e}^{2+}} $ ions in neutral condition?

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

Answer 1

Answer: (C)

In acidic medium, $ MnO_{4}^{-} $ is reduced to $ M{{n}^{2+}} $
$ \overset{+7}{\mathop{Mn}}\,O_{4}^{-}\xrightarrow{{}}M{{n}^{2+}} $
Change in oxidation number $ =7-2=5
$ Solution X Solution Y
$ {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} $ For $ F{{e}^{2+}} $ For $ MnO_{4}^{-} $
$ N\times 25=5\,M\times V $ $ [\because For\,MnO_{4}^{-},N=5\,M] $ in acidic medium]
$ 25N=5M\times 20 $ $ 25N=100M $ ...(i)
In neutral medium, $ MnO_{4}^{-} $ is reduced to $ Mn{{O}_{2}} $
$ \overset{+7}{\mathop{MnO_{4}^{-}}}\,\xrightarrow{{}}\overset{+4}{\mathop{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 $ MnO_{4}^{-} $
$ 25\times N=3\text{ }M\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 $