Certainly! To explain how the reaction between iodide ions (I⁻) and peroxydisulfate ions (S₂O₈²⁻) is catalyzed by iron(II) ions (Fe²⁺), we need to look at the individual steps of the catalyzed reaction.
### Step 1:
In the first step, Fe²⁺ reacts with the peroxydisulfate ion (S₂O₈²⁻). The Fe²⁺ gets oxidized to Fe³⁺, while the peroxydisulfate ion (S₂O₈²⁻) gets reduced to sulfate ion (SO₄²⁻).
The equation for this step is:
[tex]\[ 2 \, \text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2 \, \text{Fe}^{3+} + 2 \, \text{SO}_4^{2-} \][/tex]
### Step 2:
In the second step, Fe³⁺ reacts with iodide ions (I⁻). The Fe³⁺ gets reduced back to Fe²⁺, while the iodide ions (I⁻) get oxidized to iodine (I₂).
The equation for this step is:
[tex]\[ 2 \, \text{Fe}^{3+} + 2 \, \text{I}^{-} \rightarrow 2 \, \text{Fe}^{2+} + \text{I}_2 \][/tex]
### Combining the Steps:
These two steps explain the catalytic action of Fe²⁺. The Fe²⁺ ions are not consumed in the overall reaction; instead, they facilitate the process by forming an intermediate (Fe³⁺) that quickly reacts with I⁻ to regenerate Fe²⁺ and produce the final products, iodine (I₂) and sulfate ion (SO₄²⁻).
In summary, the two equations that explain how this reaction is catalyzed are:
1. [tex]\( 2 \, \text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2 \, \text{Fe}^{3+} + 2 \, \text{SO}_4^{2-} \)[/tex]
2. [tex]\( 2 \, \text{Fe}^{3+} + 2 \, \text{I}^{-} \rightarrow 2 \, \text{Fe}^{2+} + \text{I}_2 \)[/tex]
These steps show the role of the Fe²⁺ ions in catalyzing the reaction, making the overall process faster.
