Problem 110 - 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

(b) (molecular weight)/6

Step 1: Understanding the Reaction

In the iodometric titration involving potassium dichromate (K2Cr2O7) and sodium thiosulfate (Na2S2O3), the dichromate ion (Cr2O7^{2-}) oxidizes iodide ions (I^{-}) to iodine (I_2). It is important to first understand the balanced chemical reaction:\[ ext{Cr}_2 ext{O}_7^{2-} + 6 ext{I}^- + 14 ext{H}^+ ightarrow 2 ext{Cr}^{3+} + 3 ext{I}_2 + 7 ext{H}_2 ext{O}\]This equation shows that one mole of dichromate oxidizes iodide to form three moles of iodine.

Step 2: Determine the Change in Oxidation State

For the equivalent weight, we need to determine the change in oxidation state. The oxidation state of Cr in Cr2O7^{2-} is +6, while in Cr^{3+} it is +3. Thus, for two chromium atoms, the total change is:\[2 imes (6 - 3) = 6\]This means that 6 electrons are involved in the reduction of two chromium atoms.

Step 3: Calculating the Equivalent Weight Formula

The equivalent weight of a substance in a redox reaction is defined as the molecular weight of the substance divided by the number of electrons involved per molecule in the redox change. For K2Cr2O7, it accepts 6 electrons, derived from the balanced half-reaction equation.Therefore, the equivalent weight can be expressed as:\[\frac{\text{Molecular weight of } K2Cr2O7}{6}\]

Step 4: Identifying the Correct Answer

Given the formula for calculating the equivalent weight of K2Cr2O7 from the oxidation reaction and options provided, we conclude that the correct choice which matches the computation of molecular weight divided by number of electrons involved is option (b):\[(\text{molecular weight})/6\]

Key Concepts

Standardization

Standardization is a critical process in quantitative analysis to determine the exact concentration of a solution. It's like calibrating a scale so that you know exactly against what you are measuring.

In the context of iodometry, standardization involves using a solution with a known concentration to titrate another solution, which we want to determine the concentration of. For instance, potassium dichromate (\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)) can be used as a primary standard to determine the concentration of a sodium thiosulfate (\(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\)) solution in an iodometric titration. This ensures that the sodium thiosulfate we use has a known concentration, allowing it to be used accurately in further analyses.

The steps often involve:

• Preparing the standard solution.
• Performing a titration with the solution needing standardization.
• Calculating the unknown concentration with the help of stoichiometry from the balanced equation.

Equivalent Weight

The equivalent weight of a substance in a chemical reaction is the mass of that substance which will either gain or lose one mole of electrons. It’s a bit like dividing the substance’s efforts in a reaction by how much it achieves.

For redox reactions like the titration involving \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), the calculation of equivalent weight is crucial. To find it, you take the molecular weight of the compound and divide it by the number of electrons transferred per formula unit in the reaction.

In our exercise:

• Potassium dichromate accepts six electrons in its redox reaction, thus the equivalent weight is the molecular weight divided by 6.
• These calculations are fundamental to ensuring accurate stoichiometric measurements in various analytical procedures.

Redox Reaction

Redox reactions are chemical reactions where oxidation and reduction processes occur simultaneously. It's like a chemical seesaw where if something goes up (oxidation), something else comes down (reduction).

In our iodine titration:

• Potassium dichromate (\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)) serves as the oxidizing agent. It takes in electrons from iodide ion,i.e., it gets reduced itself.
• Iodide ions (\(\mathrm{I}^-\)) lose electrons to become iodine (\(\mathrm{I}_{2}\)). This change reflects the reduction of potassium dichromate and oxidation of iodide.

Redox reactions are important for their use in energy transformations, batteries, and metabolic processes.

Oxidation State Change

The change in oxidation state is the backbone of identifying redox reactions and therefore defining what gets oxidized and what gets reduced. It’s like using a scorecard to tell what's happening to each element in a reaction.

In the reaction involving potassium dichromate:

• Chromium's oxidation state changes from +6 in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) to +3 in \(\mathrm{Cr}^{3+}\).
• The combined oxidation state change for two chromiums is a change of 6 (from +12 to +6).
• Iodide changes from -1 to 0 as it forms iodine.

Recognizing and calculating these changes in oxidation states is crucial because it allows chemists to balance redox reactions by ensuring the electron gain is balanced by electron loss.