Question

Write a balanced net ionic equation for (a) the oxidation of iodide to iodine by sulfate ion in acidic solution. Sulfur dioxide gas is also produced. (b) The preparation of iodine from an iodide salt and chlorine gas.

Short answer

Question: Write balanced net ionic equations for the following reactions: (a) oxidation of iodide to iodine by sulfate ion in acidic solution, with sulfur dioxide as a byproduct, and (b) preparation of iodine from an iodide salt and chlorine gas. Answer: (a) 2 I⁻ (aq) + SO₄²⁻ (aq) + 4 H⁺ (aq) → I₂ (s) + SO₂ (g) + 2 H₂O (l), (b) 2 I⁻ (aq) + Cl₂ (g) → I₂ (s) + 2 Cl⁻ (aq)

Step-by-step solution

Part (a): Oxidation of iodide to iodine by sulfate ion in an acidic solution

To form the balanced net ionic equation for this reaction, we first need to identify the overall reaction and separate the species involved. The overall reaction can be written as: 2 I⁻ (aq) + SO₄²⁻ (aq) + 4 H⁺ (aq) → I₂ (s) + SO₂ (g) + 2 H₂O (l) Next, we can break this reaction down into its half-reactions. The oxidation half-reaction is that of iodide converting to iodine. The reduction half-reaction is sulfate reducing to sulfur dioxide while gaining electrons. Oxidation half-reaction: 2 I⁻ (aq) → I₂ (s) + 2 e⁻ Reduction half-reaction: SO₄²⁻ (aq) + 4 H⁺ (aq) + 2 e⁻ → SO₂ (g) + 2 H₂O (l) Now that we have both half-reactions, we can add them together to form the balanced net ionic equation: 2 I⁻ (aq) + SO₄²⁻ (aq) + 4 H⁺ (aq) → I₂ (s) + SO₂ (g) + 2 H₂O (l)

Part (b): Preparation of iodine from an iodide salt and chlorine gas

Similar to part (a), we need to first identify the overall reaction and separate the species involved. The overall reaction for the preparation of iodine from an iodide salt and chlorine gas can be written as: 2 I⁻ (aq) + Cl₂ (g) → I₂ (s) + 2 Cl⁻ (aq) As before, we can divide the overall reaction into its half-reactions. The oxidation half-reaction is iodide converting to iodine, and the reduction half-reaction is chlorine gas converting to chloride ions. Oxidation half-reaction: 2 I⁻ (aq) → I₂ (s) + 2 e⁻ Reduction half-reaction: Cl₂ (g) + 2 e⁻ → 2 Cl⁻ (aq) Now, we can combine the two half-reactions to form the balanced net ionic equation for the preparation of iodine from an iodide salt and chlorine gas: 2 I⁻ (aq) + Cl₂ (g) → I₂ (s) + 2 Cl⁻ (aq)