The number of moles of $KMn{{O}_{4}}$ reduced by 1 mol of KI in alkaline medium is:
a.) 1
b.) 2
c.) 5
d.) $\dfrac{1}{5}$

Hint: Oxidation is defined as a procedure that involves gain of oxygen or a loss of hydrogen. The reduction is defined as a procedure that involves gain of hydrogen or a loss of oxygen.

Complete Solution :
In alkaline medium the reduction of $KMn{{O}_{4}}$ with KI will take place as:
\[\begin{align}
 & 2KMn{{O}_{4}}+{{H}_{2}}O\to 2KOH+2Mn{{O}_{2}} \\
 & KI+3[O]\to KI{{O}_{3}} \\
\end{align}\]

Hence, the overall reaction will be:
\[KI+2KMn{{O}_{4}}+{{H}_{2}}O\to KI{{O}_{3}}+2Mn{{O}_{2}}\]
The number of atoms at the reactant side must be equal to the number of atoms at the product side in a balanced chemical equation.
There are some steps to balance a chemical reaction as :

1. Assign oxidation numbers to every atom in the equation and compose the numbers above the atom as:
\[\overset{+7}{\mathop{KMn{{O}_{4}}}}\,+\overset{-1}{\mathop{KI}}\,\to \overset{+4}{\mathop{Mn{{O}_{2}}}}\,+\overset{+5}{\mathop{KI{{O}_{3}}}}\,\]

2) Distinguish the molecules that are reduced and those that are oxidized as:
Reduction:
\[\overset{+7}{\mathop{KMn{{O}_{4}}}}\,\to \overset{+4}{\mathop{Mn{{O}_{2}}}}\,\]
Oxidation:
\[\overset{-1}{\mathop{KI}}\,\to \overset{+5}{\mathop{KI{{O}_{3}}}}\,\]

3) The change in oxidation number is:
Reduction: Gain of a total of 3 electrons
Oxidation: Loss of a total 6 electrons

4) Now, balance the total change in oxidation number as:
Reduction:
\[\overset{+7}{\mathop{KMn{{O}_{4}}}}\,\to \overset{+4}{\mathop{Mn{{O}_{2}}}}\,\times 2\]
Gain of total six electrons
Oxidation:
\[\overset{-1}{\mathop{KI}}\,\to \overset{+5}{\mathop{KI{{O}_{3}}}}\,\times 1\]
Loss of total six electrons
Therefore, reduction =$2KMn{{O}_{4}}\to 2Mn{{O}_{2}}$
Oxidation: $KI\to KI{{O}_{2}}$

5) Balance O atoms in reduction reaction by adding water and then balance H by ${{H}^{+}}$.

6) Add $O{{H}^{-}}$ to both the sides to neutralize ${{H}^{+}}$ for a base-catalyzed reaction as:
Hence, the balanced chemical reaction is:
\[2KMn{{O}_{4}}+{{H}_{2}}O+KI\to 2Mn{{O}_{2}}+2KOH+KI{{O}_{3}}\]
Thus, the number of moles of $KMn{{O}_{4}}$ reduced by 1 mol of KI in the alkaline medium is 2.
So, the correct answer is “Option B”.

Note: The possibility to make a mistake is that all alkalis are soluble in water but all bases are not soluble in water they may or may not be soluble. Don’t get confused between these two.