Question

The number of moles of $KMn{O_4}$ reduced by one mole of KI in alkaline medium is:
A.One-fifth
B.Five
C.One
D.Two

Answer 1

Hint: The balanced equation will be $2KMn{O_4} + {H_2}O + KI \to2Mn{O_2} + 2KOH + KI{O_3}$
The increase in the oxidation number of iodine is balanced with a decrease in the oxidation number of Manganese.

Complete step by step answer:
Alkaline solution is a base and water mixture and the solution shows the midpoint (7) in pH scale.
So, the reaction is:
 \[KMn{O_4} + KI + {H_2}O \to Mn{O_2} + KI{O_3} + KOH\]
In the presence of an alkaline medium, $2KMn{O_4}$ will react with KI to form Potassium Iodate, Potassium hydroxide and manganese dioxide.
So, first we need to balance the above equation so that the number of all the atoms in the equation on the right side is equal to the left side of the reaction.
The oxidation number of iodine will increase from -1 in KI to +5 in $KI{O_3}$ . So, the total increase in the oxidation number of iodine is 6. The oxidation number of Manganese (Mn) will decrease from +7 in $KMn{O_4}$ to +4 in \[Mn{O_2}\] . So, there is a decrease of 6 in oxidation number for 2 Manganese (Mn) atoms.
Thus, the increase in the oxidation number of iodine is balanced with a decrease in the oxidation number of Mn if the mole ratio $KMn{O_4} = KI = 2:1$
The balanced chemical reaction is:
 \[2KMn{O_4} + KI + {H_2}O \to 2Mn{O_2} + KI{O_3} + 2KOH\]
Hence, 2 moles of $KMn{O_4}$ will be reduced by one mole of KI in an alkaline medium.
Therefore, the correct answer is option (D).

Note:$KMn{O_4}$ acts as an oxidising agent in alkaline medium. When alkaline $KMn{O_4}$ ​ is treated with KI, iodide ion is oxidised to $I{O_3}^ - $