Problem 16 - A Complete Resource Book in Chemistry for JEE Main

Question

The equivalent weight of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) in the reaction \(2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+\mathrm{I}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}+2 \mathrm{NaI}\) is \((\mathrm{M}=\) molecular weight \()\) (a) \(\mathrm{M}\) (b) \(\mathrm{M} / 4\) (c) \(\mathrm{M} / 0.5\) (d) \(\mathrm{M} / 2\)

Step-by-Step Solution

Answer

The equivalent weight is \(\mathrm{M} / 4\).

Step 1: Understand the Reaction

The reaction given is \(2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+\mathrm{I}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}+2 \mathrm{NaI}\). Here, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) reacts with iodine to form \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) and \(2 \text{NaI}\).

Step 2: Determine Changes in Oxidation States

Thiosulphate \((\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3})\) is oxidized to tetrathionate \((\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6})\). In the process, 2 electrons are lost per molecule of thiosulphate. Since 2 moles of thiosulphate are used, a total of 4 electrons are exchanged.

Step 3: Calculate Equivalent Weight

Equivalent weight of a substance \( = \frac{\text{Molecular weight}}{\text{Number of electrons exchanged per mole}}\). Given that one reaction involves 2 moles of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) and exchanges 4 electrons in total, the equivalent weight is \(\frac{\mathrm{M}}{4}\) for one mole.

Key Concepts

Oxidation States

Oxidation states are a fundamental concept in chemistry that help describe the transfer of electrons in chemical reactions. They represent the effective charge that an atom would have if all bonds were ionic, giving insight into the electron distribution in compounds. In our reaction, thiosulphate \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) is transitioning to tetrathionate \(\mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}\), showing a clear change in oxidation state for sulfur atoms.
\[\text{Thiosulphate: } (\text{Sulfur oxidation state in } \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \text{ is } +2)\]
This indicates that sulfur atoms within the thiosulphate ion are experiencing a change in oxidation state as they change their bonding environment to form the tetrathionate ion. As sulfur atoms lose electrons, their oxidation state increases, signifying oxidation. Understanding these changes is critical for calculating the number of electrons transferred in redox reactions.

Thiosulphate to Tetrathionate Reaction

The transformation from thiosulphate to tetrathionate is a classic redox reaction where iodine acts as the oxidizing agent. In this specific reaction:

• Two moles of thiosulphate ions \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) combine to form one mole of tetrathionate ion \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\).
• The sulfur atoms in thiosulphate are in a \(+2\) oxidation state and are converted into a higher oxidation state in the product.
• This transformation indicates oxidation, as sulfur loses electrons to iodine.

The partner reaction with iodine \(\mathrm{I}_{2}\) determines the reduction aspect, as iodine is reduced while oxidizing thiosulphate. An important characteristic of thiosulphate is that it forms a larger sulfur-oxygen framework in tetrathionate, allowing it to stabilize by sharing electron density among more sulfur atoms.
This reaction is essential in analytical chemistry for titration purposes because it is predictable and occurs with distinct chemical changes.

Electron Exchange in Redox Reactions

In redox reactions, electron exchange is the core process. It's what drives one substance to oxidize while another reduces. This exchange is quantified by understanding the number of electrons lost or gained.
In the thiosulphate to tetrathionate reaction, 2 electrons per thiosulphate molecule are exchanged, totaling 4 electrons for the entire reaction.
Knowing the electrons exchanged is crucial when calculating the equivalent weight of a substance, as shown here:

• Equivalent weight \( = \frac{\text{Molecular weight}}{\text{Number of electrons exchanged per mole}}\)

The equivalent weight helps chemists determine how much of a compound partakes in reactions. For example, in our exercise this value is \(\frac{\mathrm{M}}{4}\), given the 4 electrons involved.
This concept is vital in stoichiometry, offering insight into reaction yields and reagent quantities needed, thus forming a backbone for chemical quantification and analysis.