Given these balanced half-reactions: 3H2O + I2O3 + 6 Cl- -> 2 IO3- + 6 HCl Enter the overall balanced redox reaction in an acidic solution. Do not include phases in your answer. Enter Your answer: 45 Chemical equation: 3 H2O + I2O3 + 6 Cl- -> 2 IO3- + 6 HCl
The Correct Answer and Explanation is:
The correct overall balanced redox reaction in acidic solution is:
3 H₂O + I₂O₃ + 6 Cl⁻ → 2 IO₃⁻ + 6 H⁺ + 6 Cl⁻
Let’s unpack this.
Step-by-step explanation:
This problem involves a redox reaction between iodine trioxide (I₂O₃) and chloride ions (Cl⁻) in acidic medium.
1. Break the overall process into half-reactions:
We identify two half-reactions:
- Oxidation half-reaction (Cl⁻ to Cl₂): But here, Cl⁻ is not oxidized, it remains Cl⁻ in the product. So it is a spectator.
- Reduction half-reaction (I in I₂O₃ to IO₃⁻): To determine how iodine is transformed, consider the change in oxidation states.
Each iodine in I₂O₃ has an average oxidation state of +3. It is converted to IO₃⁻, where iodine is +5. Thus, iodine is oxidized, which might seem counterintuitive since it usually acts as an oxidizing agent. But since Cl⁻ does not change, this process is simply:
I₂O₃ + 3 H₂O → 2 IO₃⁻ + 6 H⁺ + 6 e⁻
We’re missing an explicit reduction half-reaction because Cl⁻ appears on both sides. It’s not undergoing oxidation or reduction in this case but is present to balance charge as part of HCl.
2. Balance charges and combine:
To provide chloride ions for 6 HCl molecules, we include 6 Cl⁻ on the left-hand side. HCl is written as a product to represent hydrogen ions pairing with Cl⁻:
3 H₂O + I₂O₃ + 6 Cl⁻ → 2 IO₃⁻ + 6 HCl
However, this representation combines neutralization directly. To clarify acidic conditions and charge balance separately, one might instead write:
3 H₂O + I₂O₃ + 6 Cl⁻ → 2 IO₃⁻ + 6 H⁺ + 6 Cl⁻
Either format is acceptable depending on whether the goal is to show HCl as product or to focus on the redox process under acidic conditions.
