The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is:
one fifth
five
one
two
The Correct Option is D
Solution and Explanation
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
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Notes on Redox Reactions
Concepts Used:
Balancing a redox reaction
The method that is based on the difference in oxidation number of oxidizing agent and the reducing agent is known as oxidation number. The half-reaction method entirely depends on the division of the redox reactions into oxidation half and reduction half. It entirely depends on the individual which method to choose and use.
Oxidation Number Method:
Like various other reactions, it is very important to write the correct compositions and formulas. A very important thing to keep in mind while writing oxidation-reduction reactions is to write the compositions and formulas of the substances and the products present in the chemical reaction in a very correct manner.
Half-Reaction Method:
In this procedure, we decouple the equation into two halves. After that, we balance both the parts of the reaction separately. Finally, we add them together to get a balanced equation.