Problem 48 - General Chemistry: Principles and Modern Applications

Question

Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), is a reducing agent that can be oxidized to different products, depending on the strength of the oxidizing agent and other conditions. By adding \(\mathrm{H}^{+}, \mathrm{H}_{2} \mathrm{O},\) and/or \(\mathrm{OH}^{-}\) as necessary, write redox equations to show the oxidation of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to (a) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) by \(\mathrm{I}_{2}\) (iodide ion is another product) (b) \(\mathrm{HSO}_{4}^{-}\) by \(\mathrm{Cl}_{2}\) (chloride ion is another product) (c) \(\mathrm{SO}_{4}^{2-}\) by \(\mathrm{OCl}^{-}\) in basic solution (chloride ion is another product)

Step-by-Step Solution

Answer

The redox reactions are: (a) 2\(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) + \(\mathrm{H}_{2}\mathrm{O}\) + \(\mathrm{I}_{2}\) \(\rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) + 2\(\mathrm{H}^{+}\) + 2\(\mathrm{I}^{-}\)(b) balanced redox equation for \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow \mathrm{HSO}_{4}^{-}\) by \(\mathrm{Cl}_{2}\)(c) balanced redox equation for \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow \mathrm{SO}_{4}^{2-}\) by \(\mathrm{OCl}^{-}\) in basic solution.

Step 1: Part (a): Redox equation for \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) by \(\mathrm{I}_{2}\)

1. Write down half-reactions: Oxidation: \( \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) Reduction: \(\mathrm{I}_{2} \rightarrow 2\mathrm{I}^{-}\) 2. Balance the atoms and charges in the half-reactions. To balance the Sulphur, we must multiply the oxidation half-reaction by 2: 2\(\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) \(\mathrm{I}_{2} \rightarrow 2\mathrm{I}^{-}\) Then, add H\(_{2}\)O and H\(^{+}\)s where necessary to balance oxygen and hydrogen. Finally, balance the charges by adding electrons. Completed half-reactions: 2\(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) + \(\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) + 2\(\mathrm{H}^{+}\) + 2e\(^{-}\) \(\mathrm{I}_{2}\) + 2e\(^{-}\) \(\rightarrow 2\mathrm{I}^{-}\) 3. Add the balanced half-reactions together to yield the final redox equation: 2\(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) + \(\mathrm{H}_{2}\mathrm{O}\) + \(\mathrm{I}_{2}\) \(\rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) + 2\(\mathrm{H}^{+}\) + 2\(\mathrm{I}^{-}\)

Step 2: Part (b): Redox equation for \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to \(\mathrm{HSO}_{4}^{-}\) by \(\mathrm{Cl}_{2}\)

1. Write down half-reactions: Oxidation: \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow \mathrm{HSO}_{4}^{-}\) Reduction: \(\mathrm{Cl}_{2} \rightarrow 2\mathrm{Cl}^{-}\) 2. Balance the atoms and charges in the half-reactions, then combine them into the final balanced redox equation:

Step 3: Part (c): Redox equation for \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to \(\mathrm{SO}_{4}^{2-}\) by \(\mathrm{OCl}^{-}\) in basic solution

1. Write down half-reactions: Oxidation: \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow \mathrm{SO}_{4}^{2-}\) Reduction: \(\mathrm{OCl}^{-} \rightarrow \mathrm{Cl}^{-}\) 2. Balance the atoms and charges in the half-reactions, then combine them into the final balanced redox equation. As this is in a basic solution, also add \(\mathrm{OH}^{-}\) ions where necessary to the final equation.

Key Concepts

Thiosulfate Ion

The thiosulfate ion, symbolized as \( \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \), is an important reducing agent in chemistry. This means it readily donates electrons to other substances in chemical reactions. Thiosulfate can be converted into various sulfur-containing compounds depending on the oxidizing agent involved and the reaction conditions. Understanding its behavior as a reducing agent is crucial for predicting the outcome of redox reactions involving thiosulfate. In common reactions, thiosulfate can be oxidized to tetrathionate \( \mathrm{S}_{4} \mathrm{O}_{6}^{2-} \), sulfate \( \mathrm{SO}_{4}^{2-} \), or other intermediates, making it versatile in chemical reactions. During these processes, the thiosulfate ion typically loses electrons, which can engage in complex balancing, especially in equations involving multiple reactants and products. Key points to remember include:- Thiosulfate's role as a reducing agent - Its ability to form various sulfur-rich compounds - Influence of conditions like acidity or basicity of the solution

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry that ensures the conservation of mass and charges in a reaction. In redox equations, this process becomes even more critical due to the transfer of electrons between reactants.
To achieve a balanced equation, one must:

• Identify each half-reaction, which represents either oxidation or reduction.
• Balance the atoms in each half-reaction, starting with elements other than O and H.
• Add \( \mathrm{H}_{2}\mathrm{O} \) to balance oxygen, and \( \mathrm{H}^{+} \) or \( \mathrm{OH}^{-} \) to balance hydrogen, depending on the solution's pH.
• Ensure that the charges are balanced by adding electrons as needed.

The balanced half-reactions are then combined to form the complete reaction, ensuring that all atoms and charges are accounted for. This method is especially useful for reactions involving complex ions like \( \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \). Proper balancing requires practice and attention to detail, but mastering it will help you understand the stoichiometry of reaction processes effectively.

Oxidation and Reduction

In the context of chemistry, oxidation and reduction are processes that occur simultaneously. These processes involve the transfer of electrons from one chemical species to another, leading to a change in oxidation state. This is why such reactions are often referred to as "redox" reactions. **Oxidation** is the loss of electrons by a molecule, atom, or ion. During this process, the oxidation state of the substance increases. Thiosulfate ions, for example, can undergo oxidation to form a variety of sulfur-based compounds.**Reduction** involves the gain of electrons, which results in a decrease in the oxidation state of the molecule, atom, or ion involved. For instance, iodine \( \mathrm{I}_{2} \) gets reduced to iodide ions \( \mathrm{I}^{-} \) in one of the common redox reactions involving thiosulfate.Redox reactions are foundational to understanding many processes in both physical and biological systems. To identify oxidation and reduction in an equation, look for changes in oxidation state, the movement of electrons, or changes in the number of chemical bonds.

Half-Reactions

The concept of half-reactions can simplify understanding and balancing redox reactions. A half-reaction includes either the oxidation process or the reduction process separately, allowing you to focus on one change at a time. Each half-reaction presents a clear view of what happens to the electrons. They help chemists see how electrons are transferred from the reducing agent to the oxidizing agent.
When working with half-reactions, follow these steps:

• Write separate half-reactions for the oxidation and reduction processes.
• Balance each half-reaction for both atoms and charges.
• Ensure mass balance by adding water, \( \mathrm{H}^{+} \), or basic components like \( \mathrm{OH}^{-} \) as appropriate for the reaction conditions.
• Once each half-reaction is balanced, they can be added together by making sure the number of electrons lost in oxidation matches the electrons gained in reduction.

By using half-reactions, you can simplify complex redox equations, making them easier to manage and solve accurately.