Question

Consider the reactions : \(2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{~s}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})\) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+2 \mathrm{Br}_{2}(\mathrm{l})+5 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+4 \mathrm{Br}^{-}(\mathrm{aq})+10 \mathrm{H}^{+}(\mathrm{aq})\) Why does the same reductant, thiosulphate react differently with iodine and bromine?

Short answer

Thiosulphate reacts differently due to iodine's weaker oxidizing strength compared to bromine.

Step-by-step solution

Understanding the Concept of Redox Reactions

In redox (reduction-oxidation) reactions, one substance is reduced while the other is oxidized. The substance that donates electrons is called the reductant (reducing agent), and the one that accepts electrons is the oxidant (oxidizing agent). In this problem, thiosulphate (\(\mathrm{S}_{2}\mathrm{O}_{3}^{2-}\)) is the reductant.

Analyzing the Reaction with Iodine

In the reaction with iodine, \(2 \mathrm{~S}_{2}\mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{~s})\) turns into \(\mathrm{S}_{4}\mathrm{O}_{6}^{2-}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})\). Here, the thiosulphate ions donate electrons to iodine, reducing it to iodide ions, while thiosulphate is transformed into tetrathionate.

Analyzing the Reaction with Bromine

Consider the reaction where \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+2 \mathrm{Br}_{2}(\mathrm{l})+5 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) forms \(2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+4 \mathrm{Br}^{-}(\mathrm{aq})+10 \mathrm{H}^{+}(\mathrm{aq})\). Thiosulphate's electrons reduce bromine to bromide ions, and thiosulphate is oxidized to sulphate ions.

Comparing the Products

When reacting with iodine, thiosulphate forms tetrathionate, but with bromine, it forms sulphate. This difference occurs because iodine is a less powerful oxidizing agent than bromine due to their respective positions in the halogen group.

Explaining the Different Reactions

Thiosulphate reacts differently with iodine and bromine because the strength of the oxidizing agent (iodine vs. bromine) affects which products are formed. Bromine, a stronger oxidizing agent, completely oxidizes thiosulphate to sulfate, while iodine, being weaker, only partially oxidizes it to tetrathionate.

Key concepts

Thiosulphate Reactions

Thiosulphate ions, represented by \(\mathrm{S}_{2}\mathrm{O}_{3}^{2-}\), are nifty players in chemistry, often acting as reducing agents in various redox reactions. When a reducing agent like thiosulphate. interacts in these reactions, it tends to donate electrons. It undergoes a transformation as it assists in reducing another element.
For example, in its reaction with iodine, thiosulphate donates electrons and reduces iodine to iodide ions. This process keeps the thiosulphate relatively intact, resulting in the formation of tetrathionate \((\mathrm{S}_{4}\mathrm{O}_{6}^{2-})\). However, when it reacts with stronger oxidizing agents like bromine, the scenario changes. Here, the thiosulphate ends up being oxidized to sulfate \((\mathrm{SO}_{4}^{2-})\), showing its versatility in reactions. Understanding the behavior of thiosulphate can be crucial for predicting outcomes in chemical reactions involving multiple reactants.

• In iodine reactions: Thiosulphate \(\rightarrow \) Tetrathionate \((\mathrm{S}_{4}\mathrm{O}_{6}^{2-})\)
• In bromine reactions: Thiosulphate \(\rightarrow \) Sulfate \((\mathrm{SO}_{4}^{2-})\)

Oxidizing Agents

Oxidizing agents are substances that receive electrons in a chemical reaction, inducing the oxidation of another substance. They are central to understanding how and why thiosulphate reacts differently with iodine and bromine. Oxidizing agents vary in strength, impacting how completely they can oxidize a substance like thiosulphate.
Bromine, a stronger oxidizing agent compared to iodine, has the capability to fully oxidize thiosulphate ions into sulfate ions. This is a stark difference from iodine, which, due to its weaker oxidizing power, leads to the formation of tetrathionate ions instead. This difference can be attributed to the elements’ positions on the periodic table where bromine is placed above iodine in the halogen group, indicating greater electronegativity and oxidizing power.
This behavior suggests that:

• Strong oxidizing agents (like bromine) can facilitate more complete reactions, turning \(\mathrm{S}_{2}\mathrm{O}_{3}^{2-}\) to \(\mathrm{SO}_{4}^{2-}\).
• Weaker oxidizing agents (like iodine) result in partial oxidation, yielding \(\mathrm{S}_{4}\mathrm{O}_{6}^{2-}\).

Halogens Chemistry

Halogens, encompassing elements like iodine and bromine, are essential components in various chemical reactions. Known for their reactivity, these elements are often involved in oxidation reactions. The chemistry of halogens can be quite intriguing, as these elements show gradual changes in reactivity and strength as oxidizing agents down the group.
Bromine and iodine, while both halogens, display varying degrees of oxidizing power. Bromine, being higher on the periodic table, acts as a stronger oxidizing agent compared to iodine. This difference in their chemical behavior is why a substance like thiosulphate will react differently with each.
Understanding halogen chemistry is crucial for anticipating the products formed in reactions involving them. This knowledge is especially useful when predicting the behavior of halogens in various redox reactions, which often involve the transfer of electrons.
Key points about halogens include:

• Bromine is more reactive and a stronger oxidizer than iodine.
• Halogens readily participate in redox reactions, showcasing their versatility.
• The reactivity trends of halogens are influenced by their electron configurations and positions in the periodic table.