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Chapter 13: Problem 81

The reaction \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{I}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}_{2}\) proceeds slowly in aqueous solution, but it can be catalyzed by the \(\mathrm{Fe}^{3+}\) ion. Given that \(\mathrm{Fe}^{3+}\) can oxidize \(\mathrm{I}^{-}\) and \(\mathrm{Fe}^{2+}\) can reduce \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-},\) write a plausible two-step mechanism for this reaction. Explain why the uncatalyzed reaction is slow.

Short Answer

Expert verified
The two-step reaction mechanism is: \(Fe^{3+} + I^- \longrightarrow Fe^{2+} + I\) and \(Fe^{2+} + S_{2}O_{8}^{2-} \longrightarrow Fe^{3+} + 2SO_{4}^{2-}\). The uncatalyzed reaction is slow because it requires \(S_{2}O_{8}^{2-}\) and \(I^-\) to collide directly with enough energy, which has a high activation energy. The catalyst \(Fe^{3+}\) provides an alternative reaction pathway with a lower activation energy.

Step by step solution

01

Construct the first half of the reaction

The first half of the reaction consists of the oxidation of \(I^-\) by \(\mathrm{Fe}^{3+}\). This could be written as \(Fe^{3+} + I^{-} \longrightarrow Fe^{2+} + I\). Here, \(Fe^{3+}\) has been reduced to \(Fe^{2+}\) and \(I^{-}\) has been oxidized to atomic iodine.
02

Construct the second half of the reaction

The second half of the reaction consists of the reduction of \(S_{2}O_{8}^{2-}\) by \(Fe^{2+}\), which could be written as \(Fe^{2+} + S_{2}O_{8}^{2-} \longrightarrow Fe^{3+} + 2SO_{4}^{2-}\). Here, \(Fe^{2+}\) gets oxidized to \(Fe^{3+}\) and \(S_{2}O_{8}^{2-}\) gets reduced to \(2SO_{4}^{2-}\).
03

Evaluate the overall mechanism

Combining both halves of the reaction gives us \(Fe^{3+}+ 2I^- \longrightarrow 2SO_{4}^{2-} + I_{2}\). The fact that \(Fe^{3+}\) is regenerated shows it acting as a catalyst, it is not consumed in the overall reaction.
04

Explain why the uncatalyzed process is slow

In the uncatalyzed reaction, \(S_{2}O_{8}^{2-}\) and \(I^-\) have to collide with the correct orientation and sufficient energy to lead to product formation. This requires a high activation energy. When \(Fe^{3+}\) is present, it lowers the activation energy by providing an alternative pathway where \(S_{2}O_{8}^{2-}\) and \(I^-\) do not need to collide with each other directly, making the reaction faster.

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Most popular questions from this chapter

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