Problem 135 - General Chemistry: Principles and Modern Applications

Question

Use the following electrode potential diagram for basic solutions to classify each of the statements below as true or false. Assume standard conditions. $$\begin{aligned}\mathrm{SO}_{4}^{2-} \stackrel{-0.936 \mathrm{V}}{\longrightarrow} \mathrm{SO}_{3}^{2-} & \stackrel{-0.576 \mathrm{V}}{\longrightarrow} \\\& \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \stackrel{-0.74 \mathrm{V}}{\longrightarrow} \mathrm{S} \stackrel{-0.476 \mathrm{V}}{\longrightarrow} \mathrm{S}^{2-}\end{aligned}$$ (a) Sulfate \(\left(\mathrm{SO}_{4}^{2-}\right)\) is a stronger oxidant than thiosulfate \(\left(\mathrm{S}_{2} \mathrm{O}_{3}^{2}\right)\) in basic solution. (b) \(S^{2-}\) can be used as a reducing agent in basic solutions. (c) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is stable with respect to disproportionation to \(\mathrm{SO}_{3}^{2-}\) and \(\mathrm{S}\) in basic solution.

Step-by-Step Solution

Answer

(a) True, (b) True, (c) True

Step 1: Analyze Statement (a)

To determine whether \(\mathrm{SO}_{4}^{2-}\) is a stronger oxidant than \(\mathrm{S}_{2} \mathrm{O}_{3}^{2}\), compare their reduction potentials. The reduction potential of \(\mathrm{SO}_{4}^{2-}\) is -0.936 V, while that of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is -0.74 V. A stronger oxidant has a more negative reduction potential, meaning it's more likely to get reduced, i.e., gain electrons. Therefore, \(\mathrm{SO}_{4}^{2-}\) with a reduction potential of -0.936 V is a stronger oxidant than \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) with a reduction potential of -0.74 V. Therefore, statement (a) is true.

Step 2: Analyze Statement (b)

\(\mathrm{S}^{2-}\) is formed from the reduction of \(\mathrm{S}\) with a potential of -0.476 V. As a reducing agent, it would go through oxidation, meaning losing electrons and producing \(\mathrm{S}\). This is the reverse of the equation given and would have a potential of +0.476 V in standard conditions. Since the potential is positive, it means that this reaction is spontaneous, and therefore \(\mathrm{S}^{2-}\) can act as a reducing agent. Thus, statement (b) is true.

Step 3: Analyze Statement (c)

Disproportionation refers to a reaction where a substance is simultaneously oxidized and reduced. For \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to disproportionate into \(\mathrm{SO}_{3}^{2-}\) and \(\mathrm{S}\), it means that part of the \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) gets oxidized and part gets reduced. Looking at the electrode potentials, to form \(\mathrm{SO}_{3}^{2-}\) from \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) would need to get oxidized (reverse process of the given reaction) which is non-spontaneous as it has a positive potential. Therefore, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is stable with respect to disproportionation to \(\mathrm{SO}_{3}^{2-}\) and \(\mathrm{S}\) in basic solution. Thus, statement (c) is true.

Key Concepts

Oxidizing Agent

In chemistry, an oxidizing agent is a substance that gains electrons during a chemical reaction and, in the process, becomes reduced. This means it helps to oxidize other substances. The electrode potential provides a measure for this tendency. In the context of the provided exercise, we compared two species, sulfate \((SO_4^{2-})\) and thiosulfate \((S_2O_3^{2-})\).

An oxidizing agent is stronger when it possesses a more negative reduction potential. The reduction potential is crucial as it indicates the likelihood of a substance gaining electrons. In our example, \(SO_4^{2-}\) has a reduction potential of -0.936 V, while \(S_2O_3^{2-}\) has a potential of -0.74 V.

Here are the steps to determine which is a stronger oxidizing agent:

• Compare the values of reduction potential.
• The more negative the value, the greater the tendency to gain electrons and oxidize other substances.
• Thus, since \(SO_4^{2-}\) has a more negative value than \(S_2O_3^{2-}\), it is the stronger oxidizing agent.

Recognizing these potentials helps one predict how substances will behave in redox reactions.

Reducing Agent

A reducing agent is a chemical species that donates electrons to another species in a redox reaction. As it gives away electrons, it becomes oxidized itself. In the provided solution, we encounter \(S^{2-}\) serving as a reducing agent when it gets oxidized back to sulfur \((S)\).

To determine whether a substance can act as a reducing agent, you should examine the electrode potential when it undergoes oxidation, not reduction. For \(S^{2-}\), this reaction is spontaneous, characterized by a positive oxidation potential. Here are some insights:

• Reducing agents have negative reduction potentials but possess a positive oxidation potential.
• They effectively lose electrons in reactions, facilitating the reduction of other substances.
• In our example, \(S^{2-}\) has a reduction potential of -0.476 V when considered in reverse (oxidation), it shows a positive potential, suggesting spontaneous electron donation in reaction.

Recognizing which species are reducing agents in reactions is key to understanding their role in driving chemical changes.

Disproportionation Reaction

A disproportionation reaction is a type of redox reaction where a single substance undergoes simultaneous oxidation and reduction, producing at least two different products. In the context of the exercise, we examined \(S_2O_3^{2-}\) to check if it undergoes disproportionation into \(SO_3^{2-}\) and sulfur \((S)\).

This stability assessment relies heavily on electrode potentials. A material being oxidized shows non-spontaneity when it needs a reaction involving a positive potential for the reverse direction. Let's break down the disproportionation consideration:

• To consider if a disproportionation is feasible, look at both the oxidation and reduction potentials of the involved transformation.
• If either of these potentials suggests non-spontaneous transformation, the parent compound is stable against disproportionation.
• In \(S_2O_3^{2-}\), oxidizing to \(SO_3^{2-}\) involves a non-spontaneous potential, leading to stability in basic solutions.

Understanding disproportionation is essential when exploring reactions that involve a single species transforming in multiple pathways simultaneously.