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To analyze the given chemical reaction [tex]\( \text{IO}_3^{-}(aq) + \text{I}^-(aq) \rightarrow \text{I}_2(s) \)[/tex], we need to balance the reaction and understand the underlying processes. However, considering the instructions, let's proceed to detail the steps that's usually taken for solving such reactions, although the exact output of the reaction might not be simplicity inferred by calculation alone.
### Step-by-Step Solution:
1. Identify the Reactants and Products:
- Reactants: Iodate ion ([tex]\(\text{IO}_3^-\)[/tex])(aqueous) and Iodide ion ([tex]\(\text{I}^-\)[/tex])(aqueous)
- Product: Iodine ([tex]\(\text{I}_2\)[/tex])(solid)
2. Oxidation States:
- Determine the oxidation states of iodine in the reactants and products.
- In [tex]\(\text{IO}_3^-\)[/tex], iodate ion: oxidation state of Iodine (I) is +5.
- In [tex]\(\text{I}^-\)[/tex], iodide ion: oxidation state of Iodine (I) is -1.
- In [tex]\(\text{I}_2\)[/tex], elemental iodine: oxidation state of Iodine (I) is 0.
3. Redox Reaction Analysis:
- [tex]\(\text{IO}_3^-\)[/tex] (Iodate ion) is reduced.
- [tex]\(\text{I}^-\)[/tex] (Iodide ion) is oxidized.
4. Balancing the Reaction:
- Write separate half-reactions for oxidation and reduction.
- Oxidation: [tex]\( \text{I}^- \rightarrow \text{I}_2\)[/tex]
- Reduction: [tex]\( \text{IO}_3^- \rightarrow \text{I}_2 \)[/tex]
Balance Oxygen by adding [tex]\( \text{H}_2\text{O} \)[/tex]:
- [tex]\(\text{IO}_3^- \rightarrow \text{I}_2 + 3\text{H}_2\text{O} \)[/tex]
Balance Hydrogen by adding [tex]\( \text{H}^+\)[/tex]:
- [tex]\(6\text{H}^+ + \text{IO}_3^- \rightarrow \text{I}_2 + 3\text{H}_2\text{O} \)[/tex]
Balance charges by adding electrons:
- [tex]\(6\text{H}^+ + \text{IO}_3^- + 5e^- \rightarrow \text{I}_2 + 3\text{H}_2\text{O} \)[/tex]
- [tex]\(\text{I}^- \rightarrow \text{I}_2 + 2e^- \)[/tex]
5. Combining the Half Reactions:
- Ensure the electrons lost in oxidation equal the electrons gained in reduction. You may have to multiply the half-reactions by appropriate coefficients to equalize electron exchange:
- Multiply the oxidation reaction by 5 and the reduction reaction by 2:
- Oxidation: [tex]\(5\text{I}^- \rightarrow \frac{5}{2}\text{I}_2 + 10e^- \)[/tex]
- Reduction: [tex]\(2\text{IO}_3^- + 12\text{H}^+ + 10e^- \rightarrow \text{I}_2 + 6\text{H}_2\text{O} \)[/tex]
- Adding the balanced half-reactions together, we get:
[tex]\(2\text{IO}_3^- + 12\text{H}^+ + 10e^- + 5\text{I}^- \rightarrow 2\text{I}_2 + 6\text{H}_2\text{O} \)[/tex]
- Cancel out identical species to simplify:
[tex]\(2\text{IO}_3^- + 10\text{I}^- + 6\text{H}^+ \rightarrow 4\text{I}_2 + 3\text{H}_2\text{O} \)[/tex]
### Conclusion
Considering the balanced equation, the reaction is complete and balanced. The result indicates the complex nature of the redox reaction leading to the formation of iodine. Remember, the numerical validity or exact quantity might infer experimental outcomes to establish exact figures for laboratory scale reactions.
The final balanced chemical reaction should be:
[tex]\[ 2\text{IO}_3^- + 10\text{I}^- + 12\text{H}^+ \rightarrow 6\text{H}_2\text{O} + 4\text{I}_2 \][/tex]
Therefore, the final answer derived from our resources is detailed but the summarized output converges correctly to analyzing the chemical reaction's implication. This representation showcases a balanced expression theoretically.
### Step-by-Step Solution:
1. Identify the Reactants and Products:
- Reactants: Iodate ion ([tex]\(\text{IO}_3^-\)[/tex])(aqueous) and Iodide ion ([tex]\(\text{I}^-\)[/tex])(aqueous)
- Product: Iodine ([tex]\(\text{I}_2\)[/tex])(solid)
2. Oxidation States:
- Determine the oxidation states of iodine in the reactants and products.
- In [tex]\(\text{IO}_3^-\)[/tex], iodate ion: oxidation state of Iodine (I) is +5.
- In [tex]\(\text{I}^-\)[/tex], iodide ion: oxidation state of Iodine (I) is -1.
- In [tex]\(\text{I}_2\)[/tex], elemental iodine: oxidation state of Iodine (I) is 0.
3. Redox Reaction Analysis:
- [tex]\(\text{IO}_3^-\)[/tex] (Iodate ion) is reduced.
- [tex]\(\text{I}^-\)[/tex] (Iodide ion) is oxidized.
4. Balancing the Reaction:
- Write separate half-reactions for oxidation and reduction.
- Oxidation: [tex]\( \text{I}^- \rightarrow \text{I}_2\)[/tex]
- Reduction: [tex]\( \text{IO}_3^- \rightarrow \text{I}_2 \)[/tex]
Balance Oxygen by adding [tex]\( \text{H}_2\text{O} \)[/tex]:
- [tex]\(\text{IO}_3^- \rightarrow \text{I}_2 + 3\text{H}_2\text{O} \)[/tex]
Balance Hydrogen by adding [tex]\( \text{H}^+\)[/tex]:
- [tex]\(6\text{H}^+ + \text{IO}_3^- \rightarrow \text{I}_2 + 3\text{H}_2\text{O} \)[/tex]
Balance charges by adding electrons:
- [tex]\(6\text{H}^+ + \text{IO}_3^- + 5e^- \rightarrow \text{I}_2 + 3\text{H}_2\text{O} \)[/tex]
- [tex]\(\text{I}^- \rightarrow \text{I}_2 + 2e^- \)[/tex]
5. Combining the Half Reactions:
- Ensure the electrons lost in oxidation equal the electrons gained in reduction. You may have to multiply the half-reactions by appropriate coefficients to equalize electron exchange:
- Multiply the oxidation reaction by 5 and the reduction reaction by 2:
- Oxidation: [tex]\(5\text{I}^- \rightarrow \frac{5}{2}\text{I}_2 + 10e^- \)[/tex]
- Reduction: [tex]\(2\text{IO}_3^- + 12\text{H}^+ + 10e^- \rightarrow \text{I}_2 + 6\text{H}_2\text{O} \)[/tex]
- Adding the balanced half-reactions together, we get:
[tex]\(2\text{IO}_3^- + 12\text{H}^+ + 10e^- + 5\text{I}^- \rightarrow 2\text{I}_2 + 6\text{H}_2\text{O} \)[/tex]
- Cancel out identical species to simplify:
[tex]\(2\text{IO}_3^- + 10\text{I}^- + 6\text{H}^+ \rightarrow 4\text{I}_2 + 3\text{H}_2\text{O} \)[/tex]
### Conclusion
Considering the balanced equation, the reaction is complete and balanced. The result indicates the complex nature of the redox reaction leading to the formation of iodine. Remember, the numerical validity or exact quantity might infer experimental outcomes to establish exact figures for laboratory scale reactions.
The final balanced chemical reaction should be:
[tex]\[ 2\text{IO}_3^- + 10\text{I}^- + 12\text{H}^+ \rightarrow 6\text{H}_2\text{O} + 4\text{I}_2 \][/tex]
Therefore, the final answer derived from our resources is detailed but the summarized output converges correctly to analyzing the chemical reaction's implication. This representation showcases a balanced expression theoretically.
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