Question

Iodide ion, I\(^{-}\), is one of the most easily oxidized species. Balance each of the following oxidationreduction reactions, which take place in acidic \(50^{-}\) lution, by using the "half-reaction" method. a. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{I}_{2}(a q)\) b. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{C} \mathrm{r}^{3+}(a q)+\mathrm{I}_{2}(a q)\) c. \(\mathrm{Cu}^{2+}(a q)+\mathrm{I}^{-}(a q) \rightarrow \operatorname{CuI}(s)+\mathrm{I}_{2}(a q)\)

Short answer

a. \(4I^{-}(aq) + 2IO_{3}^{-}(aq) + 12H^{+}(aq) \rightarrow 5I_{2}(aq) + 6H_{2}O(l)\) b. \(14H^{+}(aq) + Cr_{2}O_{7}^{2-}(aq) + 6I^{-}(aq) \rightarrow 2Cr^{3+}(aq) + 3I_{2}(aq) + 7H_{2}O(l)\) c. \(2Cu^{2+}(aq) + 4I^{-}(aq) \rightarrow 2CuI(s) + I_{2}(aq)\)

Step-by-step solution

Identify the half-reactions

Oxidation half-reaction: I⁻(aq) → I₂(aq) Reduction half-reaction: IO₃⁻(aq) → I₂(aq)

Balance each half-reaction

Oxidation half-reaction: 2I⁻(aq) → I₂(aq) + 2e⁻ Reduction half-reaction: IO₃⁻(aq) + 6H⁺(aq) + 6e⁻ → 3H₂O(l) + I⁻(aq)

Multiply the half-reactions

We multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the number of electrons: 6I⁻(aq) → 3I₂(aq) + 6e⁻ 2(IO₃⁻(aq) + 6H⁺(aq) + 6e⁻) → 2(3H₂O(l) + I⁻(aq))

Add the half-reactions and simplify

6I⁻(aq) + 2IO₃⁻(aq) + 12H⁺(aq) → 3I₂(aq) + 6H₂O(l) + 2I⁻(aq) Now, we can simplify: 6I⁻(aq) + 2IO₃⁻(aq) + 12H⁺(aq) → 3I₂(aq) + 6H₂O(l) + 2I⁻(aq) 4I⁻(aq) + 2IO₃⁻(aq) + 12H⁺(aq) → 5I₂(aq) + 6H₂O(l)

Check the final balanced equation

Ensure the equation is balanced for mass and charge: 4I⁻(aq) + 2IO₃⁻(aq) + 12H⁺(aq) → 5I₂(aq) + 6H₂O(l) Mass is balanced with 10 iodine atoms and 12 hydrogen atoms on each side. Charges are balanced with a total charge of 0 on both sides. b. Cr₂O₇²⁻(aq) + I⁻(aq) → Cr³⁺(aq) + I₂(aq) We will follow the same steps as demonstrated in part (a), showing only the final balanced equation: 14H⁺(aq) + Cr₂O₇²⁻(aq) + 6I⁻(aq) → 2Cr³⁺(aq) + 3I₂(aq) + 7H₂O(l) c. Cu²⁺(aq) + I⁻(aq) → CuI(s) + I₂(aq) We will follow the same steps as demonstrated in part (a), showing only the final balanced equation: Copper (II) ions are reduced to copper (I) ions in this reaction. 2Cu²⁺(aq) + 4I⁻(aq) → 2CuI(s) + I₂(aq) These are the balanced redox reactions for the given problems under acidic conditions using the half-reaction method.

Key concepts

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox reactions. It splits the overall reaction into two separate half-reactions: one for oxidation and another for reduction. Each half-reaction is balanced separately for mass and charge.

The key steps are: identifying the half-reactions, balancing them for atoms and charge, making sure the same number of electrons are involved in both the oxidation and reduction processes, and then combining the half-reactions. This ensures mass and charge conservation, a fundamental principle in chemistry. Chemical equations balanced using this method show a clear electron transfer process, underlying the redox concept.

Acidic Solution Balancing

Many redox reactions occur under acidic conditions. Balancing redox reactions in acidic solutions involves additional steps. After using the half-reaction method to balance the atoms and electrons, it is usually necessary to balance for hydrogen and oxygen atoms present in an acidic medium.

This is typically done by adding hydrogen ions (H+) to balance hydrogen atoms, and water molecules to balance oxygen atoms. For instance, in the provided exercise, hydrogen ions and water molecules are added to balance the reduction half-reaction in acidic medium. The complete balanced equation not only includes the reactants and products but also accounts for the conditions under which the reaction takes place.