Problem 61 - General Chemistry

Question

In the following reactions, label the oxidizing agent and the reducing agent. a. \(\mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)\) b. \(\mathrm{Co}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CoCl}_{2}(s)\)

Step-by-Step Solution

Answer

Reaction a: \(\mathrm{P}_{4}\) is the reducing agent and \(\mathrm{O}_2\) is the oxidizing agent. Reaction b: \(\mathrm{Co}\) is the reducing agent and \(\mathrm{Cl}_2\) is the oxidizing agent.

Step 1: Identify oxidation states - Reaction a

In the reaction \(\mathrm{P}_{4}(s)+5\mathrm{O}_{2}(g) \to \mathrm{P}_{4}\mathrm{O}_{10}(s)\), start by identifying the oxidation states of each element. Phosphorus (\(\mathrm{P}\)) in \(\mathrm{P}_{4}\) is 0, as it's in its elemental form. In \(\mathrm{O}_2\), oxygen is also 0. In \(\mathrm{P}_{4}\mathrm{O}_{10}\), phosphorus is \(+5\) and oxygen is \(-2\).

Step 2: Determine the oxidized and reduced elements - Reaction a

Phosphorus is going from 0 to \(+5\), meaning it is losing electrons, thus being oxidized. Oxygen is going from 0 to \(-2\), meaning it is gaining electrons, thus being reduced.

Step 3: Identify agents in Reaction a

An element that is oxidized is a reducing agent, so \(\mathrm{P}_{4}\) (or phosphorus) is the reducing agent. An element that is reduced is an oxidizing agent, so \(\mathrm{O}_2\) (or oxygen) is the oxidizing agent.

Step 4: Identify oxidation states - Reaction b

In the reaction \(\mathrm{Co}(s)+\mathrm{Cl}_{2}(g) \to \mathrm{CoCl}_{2}(s)\), identify oxidation states. Cobalt (\(\mathrm{Co}\)) in its elemental form is 0. Chlorine (\(\mathrm{Cl}_2\)) in its elemental form is 0 as well. In \(\mathrm{CoCl}_2\), cobalt is \(+2\) and chlorine is \(-1\).

Step 5: Determine the oxidized and reduced elements - Reaction b

Cobalt is going from 0 to \(+2\), meaning it is losing electrons, thus being oxidized. Chlorine is going from 0 to \(-1\), meaning it is gaining electrons, thus being reduced.

Step 6: Identify agents in Reaction b

Since cobalt is oxidized, \(\mathrm{Co}\) is the reducing agent. Since chlorine is reduced, \(\mathrm{Cl}_2\) is the oxidizing agent.

Key Concepts

Oxidizing Agent

In a redox reaction, an oxidizing agent plays a crucial role by gaining electrons and getting reduced in the process. This is key to the reaction proceeding as it helps another element to lose electrons. Let's break it down further:

• In the reaction \[\mathrm{P}_4(s) + 5\mathrm{O}_2(g) \longrightarrow \mathrm{P}_4\mathrm{O}_{10}(s)\]oxygen goes from an oxidation state of 0 in \(\mathrm{O}_2\) to -2 in \(\mathrm{P}_4\mathrm{O}_{10}\).
• This change indicates that oxygen is gaining electrons.
• Hence, \(\mathrm{O}_2\) acts as the oxidizing agent.

In general, substances such as oxygen, chlorine, and fluorine are common oxidizing agents because they have a high affinity for electrons. Keep an eye out for elements or compounds being reduced in a reaction; they will likely be your oxidizing agents.

Reducing Agent

The reducing agent is essentially the opposite of the oxidizing agent. It is the element or compound that donates electrons and is oxidized during the reaction. Uncovering the reducing agent involves checking which element increases in oxidation state:

• Consider the reaction \[\mathrm{Co}(s) + \mathrm{Cl}_2(g) \longrightarrow \mathrm{CoCl}_2(s) \]Cobalt starts with an oxidation state of 0 and moves to +2.
• This transition signals it is losing electrons.
• Thus, \(\mathrm{Co}\) is the reducing agent as it supports the reduction of \(\mathrm{Cl}_2\).

Reducing agents are typically metals or hydrides as they tend to lose electrons easily. Remember, the reducing agent's main role is to provide electrons to the oxidizing agent in a redox reaction.

Oxidation State

The oxidation state, or oxidation number, indicates the degree of oxidation of an atom in a compound, also used to determine the electron transfer between atoms. Understanding how to identify and calculate these states can clarify many redox reactions:

• If an element is uncombined, like \(\mathrm{P}_4\) or \(\mathrm{Cl}_2\), it has an oxidation state of 0.
• For a compound like \(\mathrm{P}_4\mathrm{O}_{10}\), phosphorus in its compound form has an oxidation state of +5 and oxygen -2.
• The change in these values helps identify which elements are oxidized or reduced.

Assigning oxidation states correctly is crucial to determine the electron transfer occurring in the reaction. Look at periodic table trends and known compounds for guidance with oxidation states.

Electron Transfer

Electron transfer is the foundation of redox reactions, driving the process of oxidation and reduction. By following the movement of electrons, we gain insight into why these reactions occur:

• Oxidation involves losing electrons, while reduction involves gaining them.
• In \(\mathrm{Co} + \mathrm{Cl}_2 \to \mathrm{CoCl}_2\), \(\mathrm{Co}\) loses two electrons becoming \(\mathrm{Co}^{2+}\).
• Conversely, each \(\mathrm{Cl}\) atom gains one of these electrons, producing two \(\mathrm{Cl}^{-}\) ions.

Tracking who gives and receives electrons illustrates major changes in a reaction. Electron transfer is what balances the redox equation and demonstrates the chemical transformation taking place. Investigating the directions of electron flow clarifies the roles of elements in a reaction.