Problem 30 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In the chemical reaction between stoichiometric quantities of \(\mathrm{KMnO}_{4}\) and KI in weakly basic solution, what is the number of moles of \(\mathrm{I}_{2}\) released for 4 moles of \(\mathrm{KMnO}_{4}\) consumed? [Adv. 2020]

Step-by-Step Solution

Answer

10 moles of \( \mathrm{I}_{2} \) are formed.

Step 1: Understand the reaction

In a weakly basic solution, the reaction between potassium permanganate (\( \mathrm{KMnO}_{4} \)) and potassium iodide (\( \mathrm{KI} \)) is a redox reaction. In this reaction, \( \mathrm{KMnO}_{4} \) is reduced, and \( \mathrm{KI} \) is oxidized.

Step 2: Write the balanced equation

The chemical equation for the reaction is: \[ 2 \mathrm{KMnO}_{4} + 10 \mathrm{KI} + 8 \mathrm{H_{2}O} \rightarrow 5 \mathrm{I}_{2} + 2 \mathrm{MnO}_{2} + 6 \mathrm{KOH} \] This equation shows that 2 moles of \( \mathrm{KMnO}_{4} \) produce 5 moles of \( \mathrm{I}_{2} \).

Step 3: Determine the moles of \( \mathrm{I}_{2} \) produced

Given that 2 moles of \( \mathrm{KMnO}_{4} \) produce 5 moles of \( \mathrm{I}_{2} \), for 4 moles of \( \mathrm{KMnO}_{4} \), we use the ratio: \[ \text{moles of } \mathrm{I}_{2} = \left(\frac{5}{2}\right) \times 4 = 10 \] Therefore, 10 moles of \( \mathrm{I}_{2} \) are produced.

Key Concepts

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It allows us to predict how much product will form from a given amount of reactant. In the given exercise involving potassium permanganate ( KMnO_4 ) and potassium iodide ( KI ), stoichiometry helps us calculate the amount of iodine ( I_2 ) produced when a specific number of moles of KMnO_4 is used. The key takeaway is the use of molar ratios derived from a balanced chemical equation. This ratio provides the proportionate quantities of reactants and products. By using stoichiometry, we can apply this ratio to determine the amount of products generated. In our exercise, the molar ratio between KMnO_4 and I_2 is obtained from the balanced equation as 2:5. Thus, for every 2 moles of KMnO_4 , 5 moles of I_2 are produced. With this in mind, if 4 moles of KMnO_4 are reacted, the moles of I_2 can be calculated by recognizing that doubling the moles of KMnO_4 effectively doubles the moles of I_2 , resulting in 10 moles of I_2 .

Chemical Equation Balancing

Balancing a chemical equation is crucial because it ensures that the law of conservation of mass is adhered to—where matter is neither created nor destroyed in a chemical reaction. Every chemical equation must have the same number of each type of atom on both sides of the equation.In redox reactions, such as the one between KMnO_4 and KI, balancing becomes more interesting due to the changes in oxidation states. This process involves a few extra steps like assigning oxidation numbers and ensuring both charge and mass balance. For our exercise, the reaction is balanced as follows: \[ 2 \mathrm{KMnO_{4}} + 10 \mathrm{KI} + 8 \mathrm{H_{2}O} \rightarrow 5 \mathrm{I_{2}} + 2 \mathrm{MnO_{2}} + 6 \mathrm{KOH} \]Each side reflects the same number of atoms for each element, ensuring it accurately represents what occurs in the reaction. For example, both sides contain 2 manganese (Mn) atoms, 10 iodine (I) atoms, and 8 oxygen (O) atoms.

Oxidation-Reduction Reactions

Oxidation-reduction, or redox, reactions are chemical processes where the oxidation state of atoms changes. They involve two key simultaneous processes: oxidation (loss of electrons) and reduction (gain of electrons). In a redox reaction, the substance that donates electrons is oxidized, while the substance receiving electrons is reduced. Understanding this electron transfer is central for predicting the products of the reaction and its balance. Focusing on the reaction of KMnO_4 with KI , the permanganate ion is reduced as manganese goes from a +7 to a +4 oxidation state, producing manganese dioxide. On the other hand, iodide ions are oxidized to iodine, undergoing a change from -1 to 0 oxidation state. Managing these transformations is essential for determining not only the stoichiometric coefficients but also the proportionate electron transfer needed to balance the equation. Recognizing these changes highlights the interplay of oxidation and reduction within the equation, making redox reactions a distinct and fascinating type of chemical reaction.