Question

For each redox reaction, identify the substance being oxidized, the substance being reduced, the oxidizing agent, and the reducing agent. (a) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{HI}(g)\) (b) \(\mathrm{CO}(g)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(2 \mathrm{Al}(s)+6 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{H}_{2}(g)\)

Short answer

For reaction (a), \(\mathrm{H}_{2}\) is oxidized, \(\mathrm{I}_{2}\) is reduced, \(\mathrm{H}_{2}\) is the reducing agent, and \(\mathrm{I}_{2}\) is the oxidizing agent. For reaction (b), \(\mathrm{CO}\) is reduced, no substance is oxidized, and \(\mathrm{CO}\) is the reducing agent. For reaction (c), \(\mathrm{Al}\) is oxidized, \(\mathrm{H}^{+}\) is reduced, \(\mathrm{Al}\) is the reducing agent, and \(\mathrm{H}^{+}\) is the oxidizing agent.

Step-by-step solution

Determine Oxidation States for Reaction (a)

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{HI}(g)\), identify the oxidation states of each element before and after the reaction: Hydrogen (H) changes from 0 in \(\mathrm{H}_{2}\) to +1 in \(\mathrm{HI}\), and Iodine (I) changes from 0 in \(\mathrm{I}_{2}\) to -1 in \(\mathrm{HI}\).

Identify Changes in Oxidation States for Reaction (a)

Hydrogen is being oxidized because its oxidation state increases, and iodine is being reduced because its oxidation state decreases.

Identify Agents for Reaction (a)

The reducing agent is \(\mathrm{H}_{2}\) since it donates electrons, and the oxidizing agent is \(\mathrm{I}_{2}\) since it accepts electrons.

Determine Oxidation States for Reaction (b)

For the reaction \(\mathrm{CO}(g)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}(s)+\mathrm{H}_{2}\mathrm{O}(g)\), identify the oxidation states: Carbon in \(\mathrm{CO}\) is +2 and changes to 0 in \(\mathrm{C}(s)\), while hydrogen remains at +1 throughout the reaction.

Identify Changes in Oxidation States for Reaction (b)

Carbon is being reduced as its oxidation state decreases, and there is no change in the oxidation state of hydrogen, so it is neither oxidized nor reduced.

Identify Agents for Reaction (b)

The reducing agent is \(\mathrm{CO}\) because it gains electrons, and there is no clear oxidizing agent since no species is being oxidized in this reaction.

Determine Oxidation States for Reaction (c)

For the reaction \(2 \mathrm{Al}(s)+6 \mathrm{H}^{+}(a q) \longrightarrow 2\mathrm{Al}^{3+}(a q)+3 \mathrm{H}_{2}(g)\), aluminum (Al) changes from 0 to +3, while hydrogen (H) changes from +1 in \(\mathrm{H}^{+}\) to 0 in \(\mathrm{H}_{2}\).

Identify Changes in Oxidation States for Reaction (c)

Aluminum is being oxidized as its oxidation state increases, and hydrogen ions are being reduced as their oxidation state decreases.

Identify Agents for Reaction (c)

The reducing agent in this reaction is aluminum (Al), and the oxidizing agent is the hydrogen ion (\(\mathrm{H}^{+}\)).

Key concepts

Oxidation and Reduction

Understanding redox reactions involves a fascinating interplay of elements losing and gaining electrons. When an element loses electrons during a reaction, we say it is oxidized, and the process it undergoes is called oxidation. Conversely, when an element gains electrons, it is reduced, and this gain is termed reduction.

Consider the reaction between hydrogen gas and iodine gas to form hydrogen iodide: \[\begin{equation} \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{HI}(g) \end{equation}\] Here, hydrogen starts in its elemental form and ends up bonded to iodine, going from an oxidation state of 0 to +1. It loses an electron, which means hydrogen is oxidized. In contrast, iodine gains electrons, transitioning from an oxidation state of 0 to -1, making it the reduced species. It's like a dance of electrons: one partner (iodine) takes what the other (hydrogen) gives away.

Oxidizing and Reducing Agents

In the dance of redox reactions, there are key players that facilitate the exchange of electrons: the oxidizing and reducing agents. An oxidizing agent takes electrons from another species, thus enabling oxidation to occur. Conversely, a reducing agent donates electrons, causing reduction.

These agents are essential for a redox reaction to proceed. For instance, in the reaction \[\begin{equation} 2 \mathrm{Al}(s)+6 \mathrm{H}^{+}(a q) \longrightarrow 2\mathrm{Al}^{3+}(a q)+3 \mathrm{H}_{2}(g), \end{equation}\] aluminum is the reducing agent because it donates electrons to the hydrogen ions, which become molecular hydrogen, \[\begin{equation} H_{2}. \end{equation}\] aluminum is oxidized in the process as its oxidation state increases from 0 to +3. Meanwhile, the hydrogen ions act as the oxidizing agent by accepting electrons from aluminum and thus getting reduced from an oxidation state of +1 to 0.

Oxidation States

The concept of oxidation states is a bookkeeping method that helps us keep track of electron movement in redox reactions. Each atom in a compound is assigned an oxidation state, reflecting the potential charge it would have if all shared electrons were granted to the more electronegative element.

Understanding how oxidation states change during a reaction leads to a clearer picture of which atoms are oxidized and which are reduced. A common misunderstanding lies in the reaction involving carbon monoxide and hydrogen gas: \[\begin{equation} \mathrm{CO}(g)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}(s)+\mathrm{H}_{2}\mathrm{O}(g). \end{equation}\] Here, carbon in carbon monoxide has an oxidation state of +2 and becomes elemental carbon with an oxidation state of 0, indicating it has been reduced. Furthermore, correctly identifying changes in oxidation states ensures accurate identification of oxidizing and reducing agents. Using the oxidation state, we understand the movement of electrons and the dynamics of the chemical transformation.