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) \(\frac{\text { mol } w t}{6}\)

Short answer

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

Step-by-step solution

Understanding the Reaction

In the iodometry process, \[ \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \] is used to oxidize iodide ions \( (\mathrm{I}^-) \) into iodine \( (\mathrm{I}_2) \). The reaction can be represented as follows: \[ \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} + 6\mathrm{I}^- + 14\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} + 3\mathrm{I}_2 + 2\mathrm{K}^+ \] This indicates that one mole of \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) is responsible for liberating three moles of iodine \( (\mathrm{I}_2) \).

Define Equivalent Weight

Equivalent weight is the molecular weight divided by the change in moles of electrons per formula unit. In this reaction, 6 moles of electrons are transferred because each iodine ion releases one electron: \[ 6 \text{ moles of } \mathrm{I}^- \rightarrow 3 \text{ moles of } \mathrm{I}_2 \]. Therefore, the n-factor (electron change per mole) for \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) is 6.

Calculate the Equivalent Weight

To find the equivalent weight of \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \), divide its molar mass by the n-factor found above:\[ \text{Equivalent weight} = \frac{\text{Molar mass of } \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}}{6} \]. Thus, the equivalent weight is \( \frac{\text{mol. wt}}{6} \).

Key concepts

Equivalent Weight

In chemistry, the concept of equivalent weight is crucial, especially when dealing with reactions like titrations. The equivalent weight of a substance is defined as the mass that will combine with or displace a fixed amount of another substance. It's calculated as the molar mass divided by the n-factor, which represents the number of moles of electrons lost or gained in the reaction.
For the compound \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) in redox reactions, it's not enough to know just its molar mass to determine equivalent weight. You need to identify how many electrons are involved in the change during the chemical reaction. With the given reaction, six moles of electrons are transferred based on the stoichiometry of idoine formation, so the n-factor is 6.
This means that the equivalent weight of \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) is its molecular weight divided by 6. This is very useful to understand when you are working with solutions in titrations, making this an essential concept to master.

Redox Reactions

Redox reactions stand for reduction-oxidation reactions. They encompass all reactions where there's a transfer of electrons between species. In such reactions, one species gets oxidized (loses electrons) while another gets reduced (gains electrons).
In the reaction involving \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \), the potassium dichromate acts as an oxidizing agent by accepting electrons from iodide ions \((\mathrm{I}^-)\), converting them to iodine \((\mathrm{I}_2)\).
Redox reactions are characterized by changes in the oxidation states of the molecules involved. By knowing how oxidation states change, you can easily identify the agents of oxidation and reduction. For example, in the given reaction, potassium dichromate gains electrons and reduces chromium's oxidation state, demonstrating the dynamic nature of redox processes.

Oxidation-Reduction Titration

An oxidation-reduction titration, often called a redox titration, involves a chemical reaction where a known concentration of a titrant oxidizes or reduces the analyte. These titrations rely on the transfer of electrons from one reactant to another.
In iodometry, which is a type of redox titration, reagents like \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) are used to titrate iodine's concentration. During this titration, the reaction is indirect, with iodide ions being first oxidized to iodine, which is then titrated.
These titrations are valuable for analyzing compounds that can be easily oxidized or reduced. By precisely quantifying the oxidizing agent like potassium dichromate, it becomes possible to determine an unknown quantity of a substance via the difference in electron numbers, emphasizing how powerful and versatile redox titrations are in analytical chemistry.