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?

Answer 1

Hint: First we will write a balanced redox reaction of permanganate and iron in both acidic and neutral mediums. From acidic medium reaction using unitary method we will find permanganate volume and using this we will find required volume from neutral medium again using unitary method.

Complete step by step Answer:
Given that:
$Mn{{O}_{4}}^{-}$$+{{H}^{+}}\to M{{n}^{2+}}$: ACIDIC MEDIUM
Balancing the above reactions as:
\[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O\]
The redox reaction of $Mn{{O}_{4}}^{-}$​  and $F{{e}^{2+}}$in acidic condition can be represented and balanced as:
\[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O\]
\[\dfrac{[F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}]\times 5}{MnO_{4}^{-}+5F{{e}^{2+}}+8{{H}^{+}}\to M{{n}^{2+}}+5F{{e}^{3+}}+4{{H}_{2}}O}\]
From balanced reaction, $5$ volume of $F{{e}^{2+}}$  requires $1$ volume of $Mn{{O}_{4}}^{-}$​  in acidic medium.
Therefore, $25$ Volume of $F{{e}^{2+}}$ requires $\dfrac{1}{5}\times 25$volume of  $Mn{{O}_{4}}^{-}$
Or, $\dfrac{1}{5}\times 25$$=5$ volume of  $Mn{{O}_{4}}^{-}$
Given that $20ml$of $Mn{{O}_{4}}^{-}$ is required
Thus $5$ volume of $Mn{{O}_{4}}^{-}$ $=20ml$
Given: \[MnO_{4}^{-}-\to Mn{{O}_{2}}\] neutral condition
Balancing the above reactions as:
\[MnO_{4}^{-}+4{{H}^{+}}+3{{e}^{-}}\to Mn{{O}_{2}}+2{{H}_{2}}O\]
The redox reaction of $Mn{{O}_{4}}^{-}$​ and $F{{e}^{2+}}$in neutral condition can be represented and balanced as:
\[MnO_{4}^{-}+4{{H}^{+}}+3{{e}^{-}}\to Mn{{O}_{2}}+2{{H}_{2}}O\]
\[\dfrac{[F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}]\times 3}{MnO_{4}^{-}+3F{{e}^{2+}}+4{{H}^{+}}\to M{{n}^{2+}}+3F{{e}^{3+}}+2{{H}_{2}}O}\]
In neutral medium. From balanced reaction, $3$ volume of $F{{e}^{2+}}$ requires $1$ volume of  $Mn{{O}_{4}}^{-}$
Then $25$ Volume of $F{{e}^{2+}}$ requires $\dfrac{1}{3}\times 25=\dfrac{25}{3}$volume of  $Mn{{O}_{4}}^{-}$
Since from acidic medium we know $5$ volume of $Mn{{O}_{4}}^{-}$$=20ml$
$1$ Volume of $Mn{{O}_{4}}^{-}$$=\dfrac{20}{5}=4ml$
Thus, $\dfrac{25}{3}$volume of $Mn{{O}_{4}}^{-}$$=4\times \dfrac{25}{3}ml$ of $Mn{{O}_{4}}^{-}$$=33.3ml$

Additional information: n factor of potassium permanganate in acidic medium is $5$.
The factor of potassium permanganate in basic medium is $1$.
The factor of potassium permanganate in neutral medium is $3$.
a factor for any ion will be the charge that ion is containing including the sign of the charge. This n factor of ion can be one, two, etc anything.
Here in the case of $F{{e}^{2+}}$ the n factor will be $+2$.

Note: Always write the reactions first in whichever medium it is asked in these types of questions. There are high chances of mistakes in these types of questions whenever we try to solve them without writing reactions. After writing reactions these types of questions become very easy and can be solved in one or two lines using a unitary method as shown above in the solution part.