Question

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

Short answer

The answer is (b) 3 moles of \( \mathrm{I}_2 \).

Step-by-step solution

Understanding the Reaction

The reaction in question involves acidified potassium dichromate, \( \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \), reacting with excess potassium iodide, \( \mathrm{KI} \), to liberate iodine, \( \mathrm{I}_2 \). We need to calculate the moles of \( \mathrm{I}_2 \) produced from one mole of \( \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \).

Write the Balanced Chemical Equation

The balanced redox reaction for this can be written as:\[\mathrm{Cr}_2 \mathrm{O}_7^{2-} + 6\mathrm{I}^- + 14\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 3\mathrm{I}_2 + 7\mathrm{H}_2\mathrm{O}\]This shows us how the iodide ions are oxidized and the dichromate ions are reduced.

Analyze the Stoichiometry

From the above equation, 1 mole of \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \) reacts with 6 moles of iodide ions \( \mathrm{I}^- \) to liberate 3 moles of iodine \( \mathrm{I}_2 \). This tells us that for every mole of \( \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \), 3 moles of \( \mathrm{I}_2 \) are produced.

Key concepts

Stoichiometry

Stoichiometry is a core concept in chemistry that helps us understand the quantitative relationships between substances involved in a chemical reaction. By using stoichiometry, we can predict the amount of products and reactants needed or produced in a reaction. This involves using the balanced chemical equation, which shows the proportions in which substances react.

To delve deeper, stoichiometry relies on the concept of moles, which is a unit of measurement that indicates a quantity of a substance. In our example, we start with 1 mole of potassium dichromate, \( \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \), and need to understand how many moles of iodine, \( \mathrm{I}_2 \), are created. Through stoichiometric calculations, we see that the reaction will produce 3 moles of iodine for every mole of potassium dichromate that reacts.

Understanding stoichiometry is crucial for solving problems such as this, as it allows us to scale the reaction up or down quantitatively.

Balanced Chemical Equations

A balanced chemical equation is fundamental in representing a chemical reaction. Balancing these equations involves ensuring that the number of atoms for each element is equal on both the reactant and product sides of the equation. This reflects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

In our specific reaction, the balanced equation is:\[\mathrm{Cr}_2 \mathrm{O}_7^{2-} + 6\mathrm{I}^- + 14\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 3\mathrm{I}_2 + 7\mathrm{H}_2\mathrm{O}\]This equation is balanced because it shows the correct number of each atom on both sides. Six iodide ions \(\mathrm{I}^-\) react with one mole of dichromate ions \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) to produce three moles of \(\mathrm{I}_2\).

Balancing chemical equations is vital for understanding how substances interact in a chemical reaction and ensuring the accurate calculation of reactants and products.

Oxidation-Reduction (Redox) Reactions

Redox reactions are chemical processes in which electrons are transferred between species. These reactions consist of two main components: oxidation and reduction. In oxidation, a species loses electrons, while in reduction, a species gains electrons.

The reaction in our example is a classic redox reaction. The iodide ions \(\mathrm{I}^-\) undergo oxidation as they lose electrons and form iodine \(\mathrm{I}_2\). Simultaneously, the dichromate ions \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) are reduced as they gain electrons to form \(\mathrm{Cr}^{3+}\).

Understanding redox reactions is essential for explaining many chemical processes, including those in the environment, industrial applications, and biological systems. Recognizing oxidizing and reducing agents in a reaction helps predict the course and products of the chemical reaction.