Question
Question: The number of moles of \(KMn{O_4}\) reduced by one mole of KI in alkaline medium is: A.One-fifth ...
The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is:
A.One-fifth
B.Five
C.One
D.Two
Solution
The balanced equation will be 2KMnO4+H2O+KI→2MnO2+2KOH+KIO3
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:
KMnO4+KI+H2O→MnO2+KIO3+KOH
In the presence of an alkaline medium, 2KMnO4 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 KIO3 . So, the total increase in the oxidation number of iodine is 6. The oxidation number of Manganese (Mn) will decrease from +7 in KMnO4 to +4 in MnO2 . 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 KMnO4=KI=2:1
The balanced chemical reaction is:
2KMnO4+KI+H2O→2MnO2+KIO3+2KOH
Hence, 2 moles of KMnO4 will be reduced by one mole of KI in an alkaline medium.
Therefore, the correct answer is option (D).
Note: KMnO4 acts as an oxidising agent in alkaline medium. When alkaline KMnO4 is treated with KI, iodide ion is oxidised to IO3−