Question

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

Answer 1

Answer: (D)

two

Answer 2

A base and water combination produces an alkaline solution, which displays the pH scale's middle number (7.0).
Thus, the process is as follows: KMnO4+KI+H2OMnO2+KIO3+KOH
When an alkaline media is present, the reaction between KI and 2KMnO4 produces potassium iodate, potassium hydroxide, and manganese dioxide.
Therefore, the first step is to balance the equation above such that the number of atoms on the right side of the equation equals the number on the left of the reaction.
Iodine's oxidation number will go from -1 in KI to +5 in KIO3. Iodine's oxidation number has increased by 6 in total. Manganese's (Mn) oxidation number will drop from +7 in KMnO4 to +4 in MnO2.
Thus, the oxidation number of 2 Manganese (Mn) atoms has decreased by 6.
If the mole ratio of KMnO4 = KI = 2:1 is used, the rise in iodine's oxidation number is counterbalanced by a reduction in Mn's oxidation number.
The chemical process is as follows: 2KMnO4+KI+H2O2MnO2+KIO3+2KOH.
As a result, in an alkaline media, one mole of KI will decrease two moles of KMnO4.

Therefore, the correct option is (D): two