Problem 86 - A Complete Resource Book in Chemistry for JEE Main

Question

In the standardization of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) using \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) by iodometry, the equivalent weight of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is (a) (molecular weight) \(/ 2\) (b) (molecular weight)/ 6 (c) (molecular weight)/3 (d) same as molecular weight

Step-by-Step Solution

Answer

The equivalent weight of \(K_2 Cr_2 O_7\) is (molecular weight)/6.

Step 1: Understanding the Reaction

The given problem relates to iodometry, a method involving iodine and redox reactions. Potassium dichromate ( K_2 Cr_2 O_7 ) acts as an oxidizing agent and reacts in a medium to consume iodide ions, releasing iodine.

Step 2: Determining Change in Oxidation State

In the reaction, chromium in K_2 Cr_2 O_7 changes from oxidation state +6 to +3 . Since there are two chromium atoms, the total change in oxidation number is 6 for K_2 Cr_2 O_7 .

Step 3: Calculating Equivalent Weight

The equivalent weight of a compound in a redox reaction is the molar mass divided by the change in valency or electrons exchanged per molecule. Since 6 electrons are involved in the reduction of K_2 Cr_2 O_7 , the equivalent weight is its molecular weight divided by 6 .

Step 4: Conclusion

Thus, the equivalent weight of K_2 Cr_2 O_7 in the reaction is (molecular weight)/6. This corresponds to option (b).

Key Concepts

Redox Reactions

Redox reactions, which stand for reduction-oxidation reactions, are essential chemical processes where the oxidation state of molecules changes due to electron transfer. These reactions occur ubiquitously in nature and laboratories. Here, there are always two components: one gains electrons and the other loses electrons, making one a reducing agent and the other an oxidizing agent.

• Oxidation: This is when a molecule loses electrons. Its oxidation state increases.
• Reduction: This is when a molecule gains electrons. Its oxidation state decreases.

In many applications, such as iodometry, understanding redox reactions is crucial, as titration principles are often based on these electron exchanges. In the exercise, potassium dichromate acts as an excellent oxidizing agent in the redox process, facilitating the standardization.

Equivalent Weight

Equivalent weight is an important concept in chemistry, especially in reactions and titrations. Essentially, it refers to the amount of a substance that reacts with a standard amount of another substance. This can also involve releasing or consuming a set number of electrons during a reaction.
Calculating the equivalent weight requires an understanding of the change in the oxidation number during a reaction.

• Formula: Equivalent Weight = \( \frac{\text{Molecular weight}}{\text{n}} \)
  where \( n \) is the number of electrons exchanged or the number of hydrogen ions replaced.

For potassium dichromate in the exercise, since it exchanges 6 electrons, its equivalent weight is calculated by dividing its molecular weight by 6.

Potassium Dichromate

Potassium dichromate, represented as \( \text{K}_2 \text{Cr}_2 \text{O}_7 \), is a bright orange crystalline substance frequently used in laboratory settings as an oxidizing agent. This compound is well-regarded for its stable and strong oxidative properties.

• Properties: It acts vigorously to accept electrons, which means it can readily oxidize other substances.
• Uses: Besides its use in redox reactions, potassium dichromate finds applications in cleaning laboratory glassware and as a colorant because of its vivid color.

In the provided exercise, it assists in the conversion of iodide ions to iodine, helping to establish its role in the iodometry process.

Oxidation State

Understanding oxidation states is a cornerstone of grasping redox chemistry. An oxidation state, or oxidation number, is a hypothetical charge assigned to an atom in a molecule assuming the transfer of electrons.

• Positive Values: Reflects loss of electrons.
• Negative Values: Indicates gain of electrons.

In the exercise, the oxidation state of chromium in potassium dichromate changes from +6 to +3 as the reaction proceeds. The change in these values helps in the calculation of the equivalent weight by determining the number of electrons exchanged.

Standardization

Standardization is a critical process in analytical chemistry to determine the exact concentration of a solution, which is crucial for accurate titrations. During standardization, a solution's concentration, such as sodium thiosulfate, is fixed using a primary standard like potassium dichromate, which is known for its purity and stability.

• Purpose: Ensures reproducibility and precision in quantitative analysis.
• Methods: Often achieved through titration with a standard, which facilitates a known reaction, as seen with redox titration.

In iodometry, standardization aids in establishing the amount of the reagent needed for complete reaction, demonstrating its role in reliable quantitative analysis.