Problem 112 - A Complete Resource Book in Chemistry for JEE Main

Question

One mole of acidified \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) on reaction with excess KI will liberate ..... moles (s) of \(\mathrm{I}_{2}\). (a) 2 (b) 3 (c) 6 (d) 7

Step-by-Step Solution

Answer

1 mole of acidified \(K_{2}Cr_{2}O_{7}\) produces 3 moles of \(I_{2}\). The correct answer is (b) 3.

Step 1: Write the Balanced Chemical Equation

The reaction between potassium dichromate ({K}_{2}{Cr}_{2}{O}_{7}) and potassium iodide (KI) in an acidic medium is: \[Cr_{2}O_{7}^{2-} + 6I^{-} + 14H^{+} \rightarrow 2Cr^{3+} + 3I_{2} + 7H_{2}O\]. This equation shows that 1 mole of \(Cr_{2}O_{7}^{2-}\) reacts to produce 3 moles of iodine (\(I_{2}\)).

Step 2: Understand the Mole Relationship

From the balanced equation, we see that 1 mole of dichromate ion \(Cr_{2}O_{7}^{2-}\) produces 3 moles of iodine \(I_{2}\). This relationship tells us that if 1 mole of \(K_{2}Cr_{2}O_{7}\) is used, since \(K_{2}Cr_{2}O_{7}\) is the source of the \(Cr_{2}O_{7}^{2-}\), this will lead to the production of 3 moles of \(I_{2}\).

Step 3: Verify the Result

Double-check the stoichiometry of the balanced equation. Since we confirmed that 1 mole of \(Cr_{2}O_{7}^{2-}\) indeed leads to the production of 3 moles of \(I_{2}\), we can verify that the stoichiometry supports the production of exactly 3 moles of \(I_{2}\) from 1 mole of \(K_{2}Cr_{2}O_{7}\).

Key Concepts

Balanced Chemical Equations

In chemistry, a balanced chemical equation is crucial because it represents the conservation of mass. It shows that the number of atoms for each element is the same on both sides of the equation. This ensures everything is accounted for during the reaction. For example, in the reaction of potassium dichromate and potassium iodide, the equation is \[Cr_{2}O_{7}^{2-} + 6I^{-} + 14H^{+} \rightarrow 2Cr^{3+} + 3I_{2} + 7H_{2}O \].

• This balanced equation tells us what happens: 1 mole of \(Cr_{2}O_{7}^{2-}\) reacts with 6 moles of iodide ions \(I^{-}\) in the presence of 14 hydrogen ions \(H^{+}\).
• As a result, the products formed include 2 moles of \(Cr^{3+}\), 3 moles of iodine \(I_{2}\), and 7 moles of water \(H_{2}O\).

Without balancing the equation, there would be confusion about how many moles of each substance react or form. It's crucial for determining the quantities in stoichiometry.

Mole Concept

The mole concept is fundamental to chemical calculations. It provides a bridge between the atomic scale and the macroscopic quantities we work with in the lab. A mole is defined as \(6.022 \times 10^{23}\) particles of a substance, which could be atoms, molecules, or ions.For the problem in focus, using the mole concept we relate to the balanced equation as follows:

• 1 mole of \(Cr_{2}O_{7}^{2-}\) ion from the balanced equation contributes to the reaction.
• It results in the formation of 3 moles of \(I_{2}\) based on the stoichiometry of the reaction.

By understanding this concept, you can calculate the volumes, masses, and quantities of different reactants and products formed in a chemical reaction. It's truly a vital part of learning and performing stoichiometry.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between substances. These reactions are crucial in various chemical processes, including the one demonstrated here.In the reaction between \(K_2Cr_2O_7\) and \(KI\), the following occurs:

• The dichromate ion \(Cr_{2}O_{7}^{2-}\) is reduced to \(Cr^{3+}\). Reduction is gain of electrons, marked by the decrease in oxidation state from +6 in \(Cr_{2}O_{7}^{2-}\) to +3 in \(Cr^{3+}\).
• The iodide ion \(I^{-}\) is oxidized to \(I_{2}\). Oxidation is loss of electrons, with the oxidation state changing from -1 in \(I^{-}\) to 0 in molecular iodine \(I_{2}\).

Understanding redox reactions helps in identifying how electrons are exchanged during these processes, determining energy changes, and predicting product formation in varied reactions.