Question

The balanced equation in basic medium is :

$Mn{O}_4+F{e}^{2+}\stackrel{O{H}^{-}}{\to }M{n}^{2+}+F{e}^{3+}$

• A. ${MnO}_4^{-}+5F{e}^{2+}+4H_{2}O\to M{n}^{2+}+5F{e}^{3+}+8O{H}^{-}$
• B. $2{MnO}_4^{-}+10F{e}^{2+}+4H_{2}O\to M{n}^{2+}+5F{e}^{3+}+8O{H}^{-}$
• C. ${MnO}_4^{-}+5F{e}^{2+}+6H_{2}O\to M{n}^{2+}+5F{e}^{3+}+4O{H}^{-}$
• D. none of these

Answer 1

The correct option is A ${MnO}_4^{-}+5F{e}^{2+}+4H_{2}O\to M{n}^{2+}+5F{e}^{3+}+8O{H}^{-}$

The unbalanced redox equation is as follows:
$Mn{O}_4+F{e}^{2+}\stackrel{O{H}^{-}}{\to }M{n}^{2+}+F{e}^{3+}$
All atoms other than H and O are balanced..
The oxidation number of Fe changes from 2 to 3. The change in the oxidation number of Fe is 1.
The oxidation number of Mn changes from 7 to 2. The change in the oxidation number is 5.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $F{e}^{2+}$ and $F{e}^{3+}$ with 5.
$Mn{O}_4+5F{e}^{2+}\stackrel{O{H}^{-}}{\to }M{n}^{2+}+5F{e}^{3+}$
O atoms are balanced by adding 4 water molecules on RHS.
$Mn{O}_4+5F{e}^{2+}\stackrel{O{H}^{-}}{\to }M{n}^{2+}+5F{e}^{3+}+4H_{2}O$
To balance H atoms, add 8 $H^{+}$ on LHS.
$Mn{O}_4+5F{e}^{2+}+8H^{+}\stackrel{O{H}^{-}}{\to }M{n}^{2+}+5F{e}^{3+}+4H_{2}O$
Since the reaction occurs in basic medium, add 8 hydroxide ions on both sides of the equation.
$Mn{O}_4+5F{e}^{2+}+8H^{+}+8O{H}^{-}\stackrel{O{H}^{-}}{\to }M{n}^{2+}+5F{e}^{3+}+4H_{2}O+8O{H}^{-}$
On LHS, 8 $H^{+}$ ions combine with 6 $O{H}^{-}$ ions to form 8 water molecules out of which 4 cancel out with 4 water molecules on RHS.
${MnO}_4^{-}+5F{e}^{2+}+4H_{2}O\to M{n}^{2+}+5F{e}^{3+}+8O{H}^{-}$
This is the balanced chemical equation.