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

01

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.

02

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.

03

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.

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