Problem 8 - Chemistry for Allied Health

Question

Identify each process as an oxidation or a reduction. a. \(\mathrm{Rb} \rightarrow \mathrm{Rb}^{+}\) b. \(\mathrm{Te} \rightarrow \mathrm{Te}^{2-}\) c. \(2 \mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\) d. \(\mathrm{P}^{3-} \rightarrow \mathrm{P}\) e. \(2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2}\) f. \(\mathrm{Sn}^{4+} \rightarrow \mathrm{Sn}^{2+}\) g. \(\mathrm{Br}_{2} \rightarrow \mathrm{Br}^{-}\) h. \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}\)

Step-by-Step Solution

Answer

a: Oxidation, b: Reduction, c: Reduction, d: Oxidation, e: Oxidation, f: Reduction, g: Reduction, h: Oxidation.

Step 1: Introduction to Oxidation and Reduction

Before we begin, recall that oxidation involves the loss of electrons, resulting in an increase in the oxidation state, whereas reduction involves the gain of electrons, resulting in a decrease in the oxidation state.

Step 2: Evaluating Process a

In a. \( \mathrm{Rb} \rightarrow \mathrm{Rb}^{+} \), Rb loses one electron. The oxidation state increases from 0 to +1, indicating oxidation.

Step 3: Evaluating Process b

In b. \( \mathrm{Te} \rightarrow \mathrm{Te}^{2-} \), Te gains two electrons. The oxidation state decreases from 0 to -2, indicating reduction.

Step 4: Evaluating Process c

In c. \( 2 \mathrm{H}^{+} \rightarrow \mathrm{H}_{2} \), \( \mathrm{H^{+}} \) ions gain electrons (each \( \mathrm{H^{+}} \) gains one electron). The oxidation state decreases from +1 to 0, indicating reduction.

Step 5: Evaluating Process d

In d. \( \mathrm{P}^{3-} \rightarrow \mathrm{P} \), P loses electrons (3 electrons are lost). The oxidation state increases from -3 to 0, indicating oxidation.

Step 6: Evaluating Process e

In e. \( 2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2} \), Cl loses electrons (each Cl gains one electron), moving from -1 to 0 each. This indicates oxidation.

Step 7: Evaluating Process f

In f. \( \mathrm{Sn}^{4+} \rightarrow \mathrm{Sn}^{2+} \), Sn gains electrons (specifically, 2 electrons are gained) as the charge decreases. The oxidation state decreases, indicating reduction.

Step 8: Evaluating Process g

In g. \( \mathrm{Br}_{2} \rightarrow \mathrm{Br}^{-} \), br gains electrons (each Br gains one electron). The oxidation state decreases from 0 to -1, indicating reduction.

Step 9: Evaluating Process h

In h. \( \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} \), Fe loses one electron. The oxidation state increases from +2 to +3, indicating oxidation.

Key Concepts

Oxidation

Oxidation is a fundamental concept in chemistry, particularly in the study of redox reactions. It refers to the process whereby an atom, ion, or compound loses electrons.
When an electron is lost, the oxidation state (or oxidation number) of the element increases. Think of oxidation as the element becoming more positive. This can happen when an atom loses one or more electrons to another atom.

• For example, in the scenario \(\mathrm{Rb} \rightarrow \mathrm{Rb}^{+}\), rubidium loses an electron. This loss results in an increase of the oxidation state from 0 to +1, signaling oxidation.
• Keep in mind that oxidation does not always involve a reaction with oxygen, even though the name suggests it might.

The concept of oxidation is central to many chemical reactions, especially those involving metals and their interactions with other elements.

Reduction

Reduction is the counterpart to oxidation and is equally important in redox reactions. In a reduction process, an atom, ion, or compound gains electrons. This gain results in a decrease in its oxidation state, becoming more negative.

• For instance, in the process \(\mathrm{Te} \rightarrow \mathrm{Te}^{2-}\), tellurium gains two electrons. This change causes a decrease in oxidation state from 0 to -2, indicating reduction.
• A common mnemonic to remember reduction is "OIL RIG": Oxidation Is Loss, Reduction Is Gain, regarding electrons.

Reduction often occurs simultaneously with oxidation in reactions, since electrons move from one substance to another, resulting in one being oxidized and the other reduced.

Electron Transfer

Electron transfer is at the heart of redox reactions. These reactions involve the movement of electrons from one element to another.

• During oxidation, electrons are lost from an element, as in \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}\), where iron loses an electron.
• Conversely, during reduction, electrons are gained, as seen in \(\mathrm{Br}_{2} \rightarrow \mathrm{Br}^{-}\), where bromine gains electrons.

Understanding electron transfer is key to mastering redox reactions. It's the flow of electrons that drives the changes in oxidation states and powers many chemical processes, from metal rusting to energy production in fuel cells.

Oxidation State Changes

Oxidation states (or numbers) are useful tools to track the electron transfers in redox reactions. They indicate "electron management" of an element — whether it's richer or poorer in electrons compared to a neutral state.

• An increase in oxidation state, like in \(\mathrm{P}^{3-} \rightarrow \mathrm{P}\), means oxidation, as phosphorus loses three electrons.
• A decrease in oxidation state, such as in \(\mathrm{Sn}^{4+} \rightarrow \mathrm{Sn}^{2+}\), indicates reduction with tin gaining two electrons.

Tracking changes in oxidation state is fundamental in determining whether oxidation or reduction has occurred. It forms the backbone of interpreting reactions, helping chemists understand both the framework of reaction mechanisms and practical applications.