Problem 76 - 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) same as mol. wt (b) \(\frac{\text { mol. wt }}{2}\) (c) \(\frac{\text { mol. wt }}{4}\) (d) \(\frac{\text { mol. wt }}{6}\)

Step-by-Step Solution

Answer

The equivalent weight of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is \(\frac{\text{mol. wt}}{6}\).

Step 1: Understand the Reaction

In the iodometry process, the reaction involves potassium dichromate (\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)) reacting with potassium iodide (\(\mathrm{KI}\)) in an acidic medium to liberate iodine (\(I_2\)). The iodine is then titrated with sodium thiosulfate (\(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\)) solution. The balanced reaction for \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is: \[\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6\mathrm{e}^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}\]

Step 2: Calculate the Number of Electrons Transferred

In the reduction half-reaction of \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) to \(\mathrm{Cr}^{3+}\), 6 electrons are transferred. This indicates that the equivalent weight of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) will be connected to the transfer of these electrons.

Step 3: Determine Equivalent Weight Formula

The equivalent weight is calculated by dividing the molecular weight by the number of electrons transferred per formula unit. Thus, the equivalent weight of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is given by: \[\text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n}\] where \(n\) is the number of electrons transferred (6 in this case).

Step 4: Calculate Final Equivalent Weight

Using the formula, the equivalent weight of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is \(\frac{\text{Molecular Weight}}{6}\). Hence, the correct choice is:\ (d) \frac{\text{mol. wt}}{6}\.

Key Concepts

Iodometry

Iodometry is a titration method that involves the reaction of iodine with a titrant. It is commonly used in redox (reduction-oxidation) titrations to determine the concentration of oxidizing agents in a solution. The process involves a few crucial steps:

• A known amount of the oxidizing agent, such as potassium dichromate (\( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \)), is allowed to react and liberate iodine from an iodide source.
• The liberated iodine (\( \text{I}_2 \)) is then titrated with a standard solution of sodium thiosulfate (\( \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \)).
• The endpoint of this titration is usually indicated by the disappearance of the blue color of the iodine-starch complex.

This technique is reliable due to its precise and clear endpoint detection, which makes it suitable for quantitative analysis in various chemical applications. Understanding iodometry is essential for mastering redox titrations as it helps in determining the exact amount of analyte in a sample.

Potassium Dichromate

Potassium dichromate (\( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \)) is an important chemical compound, often used as an oxidizing agent in laboratory settings. Its role in iodometry is pivotal as:

• It reacts with potassium iodide to release iodine, which is then used in the titration step.
• The compound itself provides a stable source of chromium ions involved in the redox reactions.
• \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) has a characteristic orange color that can aid in visualization during reactions before the iodine is liberated.

Due to its strong oxidative properties, potassium dichromate is often chosen for experiments where a reliable oxidizing substance is required. It also acts as a primary standard for preparing solutions due to its high purity and stability, making it an essential reagent in iodometric methods.

Reduction Half-Reaction

In any redox process, a reduction half-reaction plays a critical role. It involves the gain of electrons by a substance. In the context of using \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) within iodometric titrations:

• The dichromate ion (\( \mathrm{Cr}_2\mathrm{O}_7^{2-} \)) undergoes reduction, gaining electrons to form chromium ions (\( \mathrm{Cr}^{3+} \)).
• The balanced equation for this reduction half-reaction is:\[\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6\mathrm{e}^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}\]This equation shows the transfer of six electrons.
• The equivalent weight of the oxidizing agent is found by considering this electron transfer.

Understanding this concept is crucial for calculating the equivalent weight, which is essential for accurate titration results. This reduction process provides insight into how substances interact at the molecular level during chemical transformations.