Question

Potassium iodide in solution reacts readily with many reagents. In the following reactions, identify the atoms that are being oxidized and reduced, and specify the oxidizing and reducing agents. a. \(\mathrm{Cl}_{2}(g)+\mathrm{KI}(a q) \rightarrow \mathrm{KCl}(a q)+\mathrm{I}_{2}(s)\) b. \(2 \mathrm{FeCl}_{3}(a q)+2 \mathrm{KI}(a q) \rightarrow 2 \mathrm{FeCl}_{2}(a q)+2 \mathrm{KCl}(a q)+\mathrm{I}_{2}(s)\) c. \(2 \mathrm{CuCl}_{2}(a q)+4 \mathrm{KI}(a q) \rightarrow 2 \mathrm{CuI}(s)+4 \mathrm{KCl}(a q)+\mathrm{I}_{2}(s)\)

Short answer

a. Iodine is oxidized from -1 to 0; chlorine is reduced from 0 to -1. Oxidizing agent: \(Cl_2\), Reducing agent: \(KI\) b. Iodine is oxidized from -1 to 0; iron is reduced from +3 to +2. Oxidizing agent: \(FeCl_3\), Reducing agent: \(KI\) c. Iodine is oxidized from -1 to 0; copper is reduced from +2 to +1. Oxidizing agent: \(CuCl_2\), Reducing agent: \(KI\)

Step-by-step solution

a. Identify oxidation and reduction for: \(\mathrm{Cl}_{2}(g)+\mathrm{KI}(a q) \rightarrow \mathrm{KCl}(a q)+\mathrm{I}_{2}(s)\)

: Let's write the half reactions for oxidation and reduction: Oxidation: \(2\mathrm{I}^- \rightarrow \mathrm{I}_{2} + 2e^-\) Reduction: \(\mathrm{Cl}_{2} + 2e^- \rightarrow 2\mathrm{Cl}^-\) In this case, iodine (I) is being oxidized from an oxidation number of -1 to 0, while chlorine (Cl) is being reduced from 0 to -1. Oxidizing agent: \(\mathrm{Cl}_{2}\) Reducing agent: \(\mathrm{KI}\)

b. Identify oxidation and reduction for: \(2 \mathrm{FeCl}_{3}(a q)+2 \mathrm{KI}(a q) \rightarrow 2 \mathrm{FeCl}_{2}(a q)+2 \mathrm{KCl}(a q)+\mathrm{I}_{2}(s)\)

: Let's write the half reactions for oxidation and reduction: Oxidation: \(2\mathrm{I}^- \rightarrow \mathrm{I}_{2} + 2e^-\) Reduction: \(2\mathrm{Fe}^{3+} + 2e^- \rightarrow 2\mathrm{Fe}^{2+}\) In this case, iodine (I) is being oxidized from an oxidation number of -1 to 0, while iron (Fe) is being reduced from +3 to +2. Oxidizing agent: \(\mathrm{FeCl}_{3}\) Reducing agent: \(\mathrm{KI}\)

c. Identify oxidation and reduction for: \(2 \mathrm{CuCl}_{2}(a q)+4 \mathrm{KI}(a q) \rightarrow 2 \mathrm{CuI}(s)+4 \mathrm{KCl}(a q)+\mathrm{I}_{2}(s)\)

: Let's write the half reactions for oxidation and reduction: Oxidation: \(4\mathrm{I}^- \rightarrow 2\mathrm{I}_{2} + 4e^-\) Reduction: \(2\mathrm{Cu}^{2+} + 2e^- \rightarrow 2\mathrm{Cu}^+\) In this case, iodine (I) is being oxidized from an oxidation number of -1 to 0, while copper (Cu) is being reduced from +2 to +1. Oxidizing agent: \(\mathrm{CuCl}_{2}\) Reducing agent: \(\mathrm{KI}\)

Key concepts

Oxidation and Reduction Agents

Understanding the roles of oxidation and reduction agents is critical in redox reactions, which are types of chemical reactions involving the transfer of electrons between two species. The reducing agent, or reductant, loses electrons and thereby gets oxidized. Conversely, the oxidizing agent, or oxidant, gains electrons and is reduced.

In the provided examples, potassium iodide (KI) acts as a reducing agent because it loses electrons as iodine's oxidation state changes from -1 to 0. In each example, the other reactant is the oxidizing agent; chlorine (Cl₂), iron(III) chloride (FeCl₃), and copper(II) chloride (CuCl₂) gain electrons, thus being reduced.

To empower student comprehension, imagine the reducing agent as 'giving away' electrons, while the oxidizing agent is 'receiving' them. Like generous donors and grateful recipients in a transaction, these agents help us track the flow of electrons in a reaction.

Oxidation Numbers

Oxidation numbers, often referred to as oxidation states, play a crucial role in identifying how electrons are transferred in redox reactions. These are hypothetical charges that an atom would have if the compound was composed of ions. In redox processes, oxidation numbers help determine which atoms are oxidized and which are reduced.

In the textbook examples, the oxidation number of iodine goes from -1 in KI to 0 in I₂, indicating it is oxidized. For chlorine, iron, and copper, their oxidation numbers decrease, showing reduction. When teaching this concept, use relatable analogies like 'scores in a game' that can either 'go up' (oxidation) or 'go down' (reduction) to clarify the changes in electron ownership.

Chemical Reactions

Chemical reactions are the processes in which substances, known as reactants, transform into new substances, called products. In the context of redox reactions, we witness a special transformation where the change involves the transfer of electrons. The reactions in our exercise exemplify chemical processes where the reactants undergo a change in their oxidation states, thus forming new compounds with different properties.

By comparing a chemical reaction to a recipe, where the reactants are 'ingredients' and the products are the 'finished dish', students can better visualize the transformation. Just as ingredients combine in particular ways to yield a dish, reactants combine and change to yield products in a chemical reaction, illustrating the dynamic and fascinating nature of chemistry.