Question

Balance each of the following half-reactions. a. \(\mathrm{I}^{-}(a q) \rightarrow \mathrm{I}_{2}(s)\) b. \(\mathrm{O}_{2}(g) \rightarrow \mathrm{O}^{2-}(s)\) c. \(\mathrm{P}_{4}(s) \rightarrow \mathrm{P}^{3-}(s)\) d. \(\mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl}^{-}(a q)\)

Short answer

a. \( \mathrm{2I^{-}(aq) \rightarrow I_{2}(s)} \) b. \( \mathrm{O_{2}(g) \rightarrow 2O^{2-}(s)} \) c. \( \mathrm{P_{4}(s) \rightarrow 4P^{3-}(s)} \) d. \( \mathrm{Cl_{2}(g) \rightarrow 2Cl^{-}(aq)} \)

Step-by-step solution

a. Balancing the half-reaction: I⁻(aq) → I₂(s)

First, identify the change in oxidation numbers by finding the oxidized and reduced forms of iodine. The I⁻ ion has an oxidation number of -1 while I₂ molecule has an oxidation number of 0. We need two I⁻ ions to form one I₂ molecule so that the total change in oxidation numbers is 0: \[\rm 2I^{-}(aq) \rightarrow I_{2}(s)\] This balanced half-reaction shows the conversion of two I⁻ ions into one I₂ molecule.

b. Balancing the half-reaction: O₂(g) → O²⁻(s)

Determine the change in oxidation numbers by finding the oxidized and reduced forms of oxygen. Oxygen in O₂ molecule has an oxidation number of 0, while O²⁻ ion has an oxidation number of -2. In order to balance this half-reaction, we need two O²⁻ ions: \[\rm O_{2}(g) \rightarrow 2O^{2-}(s)\] The balanced half-reaction represents the reduction of one O₂ molecule into two O²⁻ ions.

c. Balancing the half-reaction: P₄(s) → P³⁻(s)

Determine the change in oxidation numbers by looking at the oxidized and reduced forms of phosphorus. Phosphorus in P₄ molecule has an oxidation number of 0, while P³⁻ ion has an oxidation number of -3. In order to balance this half-reaction, we need four P³⁻ ions: \[\rm P_{4}(s) \rightarrow 4P^{3-}(s)\] This balanced half-reaction represents the reduction of one P₄ molecule into four P³⁻ ions.

d. Balancing the half-reaction: Cl₂(g) → Cl⁻(aq)

Determine the change in oxidation numbers by finding the oxidized and reduced forms of chlorine. Chlorine in Cl₂ molecule has an oxidation number of 0 while Cl⁻ ion has an oxidation number of -1. In order to balance this half-reaction, we need two Cl⁻ ions: \[\rm Cl_{2}(g) \rightarrow 2Cl^{-}(aq)\] This balanced half-reaction represents the reduction of one Cl₂ molecule into two Cl⁻ ions.

Key concepts

Oxidation Numbers

Oxidation numbers are crucial for understanding how elements participate in chemical reactions, especially redox reactions. They help us determine the loss and gain of electrons during chemical changes. Let's break it down:

• Each element in its pure form has an oxidation number of 0. For example, \( \text{O}_2 \) and \( \text{Cl}_2 \) are in their elemental states and thus have oxidation numbers of 0.
• Common ions have oxidation numbers equal to their charges. So, the \( \text{I}^- \) ion has an oxidation number of -1.

Understanding oxidation numbers helps in identifying which species get oxidized and which ones get reduced in redox reactions. When an oxidation number increases, it means that the species is oxidized since it loses electrons. Conversely, a decrease in oxidation number signifies reduction, where electrons are gained.

Chemical Equations

In chemistry, chemical equations are a symbolic way to represent a chemical reaction. These equations show the reactants on the left and the products on the right. In redox reactions, balancing these equations is crucial. Let's dig deeper into how this is done:

• Identify the reactants and products in the reaction. For redox reactions, also note the oxidation states.
• Use the half-reaction method to separate the oxidation process from the reduction process.
• Each half-reaction is balanced separately with respect to mass and charge.
• In our example, for \( \text{I}^- \) to \( \text{I}_2 \), we notice the need to balance the number of iodine atoms. Similarly, charge balance is achieved by ensuring the sum of oxidation numbers is the same on both sides.

Combined, these steps ensure that the law of conservation of mass and charge is respected, as no atoms or charges are lost in a balanced chemical equation.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are characterized by the transfer of electrons between chemical species. Here's how they work:

• Oxidation involves the loss of electrons and an increase in oxidation number. For instance, in the reaction \( \text{O}_2(g) \rightarrow 2\text{O}^{2-}(s) \), \( \text{O}_2 \) gains electrons, making it a reduction process.
• Reduction is the gain of electrons and a decrease in oxidation number.
• In each redox process, the substance that loses electrons gets oxidized, and the substance that gains electrons gets reduced.

This dual process ensures that electrical neutrality is maintained as electrons are exchanged. Balancing redox reactions requires balancing both the electron transfer and the atoms involved in each half-reaction. This is achieved using the oxidation numbers, making it easier to see which electrons are gained and lost.