Problem 12 - Chemistry and Chemical Reactivity

Question

For each of the following electrochemical cells, write equations for the oxidation and reduction half-reactions and for the overall reaction. (a) \(\mathrm{Pb}(\mathrm{s})\left|\mathrm{Pb}^{2+}(\mathrm{aq}) \| \mathrm{Sn}^{4+}(\mathrm{aq}), \mathrm{Sn}^{2+}(\mathrm{aq})\right| \mathrm{C}(\mathrm{s})\) (b) \(\mathrm{Hg}(\ell)\left|\mathrm{Hg}_{2} \mathrm{Cl}_{2}(\mathrm{s})\right| \mathrm{Cl}^{-}(\mathrm{aq}) \| \mathrm{Ag}^{+}(\mathrm{aq}) | \mathrm{Ag}(\mathrm{s})\)

Step-by-Step Solution

Answer

For cell (a): Overall reaction is \( \mathrm{Pb}(\mathrm{s}) + \mathrm{Sn}^{4+}(\mathrm{aq}) \rightarrow \mathrm{Pb}^{2+}(\mathrm{aq}) + \mathrm{Sn}^{2+}(\mathrm{aq}) \). For cell (b): Overall reaction is \( \mathrm{Hg}_2\mathrm{Cl}_2(\mathrm{s}) + 2\mathrm{Ag}^+(\mathrm{aq}) \rightarrow 2\mathrm{Hg}(\ell) + 2\mathrm{Ag}(\mathrm{s}) + 2\mathrm{Cl}^-(\mathrm{aq}) \).

Step 1: Identify the Half-Reactions for Cell (a)

For cell (a), we have the system: \[ \mathrm{Pb}(\mathrm{s}) \mid \mathrm{Pb}^{2+}(\mathrm{aq}) \parallel \mathrm{Sn}^{4+}(\mathrm{aq}), \mathrm{Sn}^{2+}(\mathrm{aq}) \mid \mathrm{C}(\mathrm{s}) \] - The oxidation half-reaction occurs at the anode: \[ \mathrm{Pb}(\mathrm{s}) \rightarrow \mathrm{Pb}^{2+}(\mathrm{aq}) + 2e^- \] - The reduction half-reaction occurs at the cathode: \[ \mathrm{Sn}^{4+}(\mathrm{aq}) + 2e^- \rightarrow \mathrm{Sn}^{2+}(\mathrm{aq}) \]

Step 2: Write the Overall Reaction for Cell (a)

Combine the oxidation and reduction reactions from Step 1: - Add the two half-reactions: \[ \mathrm{Pb}(\mathrm{s}) + \mathrm{Sn}^{4+}(\mathrm{aq}) \rightarrow \mathrm{Pb}^{2+}(\mathrm{aq}) + \mathrm{Sn}^{2+}(\mathrm{aq}) \] - This is the overall redox reaction for cell (a).

Step 3: Identify the Half-Reactions for Cell (b)

For cell (b), we have the system: \[ \mathrm{Hg}(\ell) \mid \mathrm{Hg}_{2} \mathrm{Cl}_{2}(\mathrm{s}) \parallel \mathrm{Cl}^-(\mathrm{aq}) \mid \mathrm{Ag}^+(\mathrm{aq}) \mid \mathrm{Ag}(\mathrm{s}) \] - The oxidation half-reaction is: \[ \mathrm{Hg}_2\mathrm{Cl}_2(\mathrm{s}) + 2e^- \rightarrow 2\mathrm{Hg}(\ell) + 2\mathrm{Cl}^-(\mathrm{aq}) \] - The reduction half-reaction is: \[ \mathrm{Ag}^+(\mathrm{aq}) + e^- \rightarrow \mathrm{Ag}(\mathrm{s}) \] Note: Multiply the reduction half-reaction by 2 to balance the electrons.

Step 4: Write the Overall Reaction for Cell (b)

Combine the oxidation and adjusted reduction reactions from Step 3: - Multiply the reduction half-reaction by 2: \[ 2\mathrm{Ag}^+(\mathrm{aq}) + 2e^- \rightarrow 2\mathrm{Ag}(\mathrm{s}) \] - Add the oxidation and the adjusted reduction reactions: \[ \mathrm{Hg}_2\mathrm{Cl}_2(\mathrm{s}) + 2\mathrm{Ag}^+(\mathrm{aq}) \rightarrow 2\mathrm{Hg}(\ell) + 2\mathrm{Ag}(\mathrm{s}) + 2\mathrm{Cl}^-(\mathrm{aq}) \] - This is the overall redox reaction for cell (b).

Key Concepts

Oxidation Half-Reaction

The oxidation half-reaction is a fundamental process in electrochemical cells where a substance loses electrons. This reaction takes place at the anode, which is one of the key components in electrochemical setups. When a substance undergoes oxidation, it increases its oxidation state while releasing electrons.
For example, in the electrochemical cell \( a \) from the original exercise, lead (Pb) at the anode becomes lead ions \( \text{Pb}^{2+} \), and releases two electrons: \[ \text{Pb} (\text{s}) \rightarrow \text{Pb}^{2+} (\text{aq}) + 2e^- \] This process of losing electrons marks the oxidation step.
In general, recognizing an oxidation half-reaction involves:

• Identifying the species at the anode that loses electrons.
• Observing an increase in the oxidation state of this species.
• Recording the electrons released during this process.

Such reactions are vital as they provide the electrons necessary for the coupled reduction half-reaction in the system.

Reduction Half-Reaction

In contrast to oxidation, the reduction half-reaction involves the gain of electrons by a substance, reducing its oxidation state. This process occurs at the cathode, another crucial part of electrochemical cells. When a substance is reduced, it typically becomes more negative or less positive, signifying electron gain.
Taking the cell \( a \) example again, the reduction half-reaction happens with the transformation of \( \text{Sn}^{4+} \) ions to \( \text{Sn}^{2+} \) at the cathode via electron gain, as shown in the equation: \[ \text{Sn}^{4+} (\text{aq}) + 2e^- \rightarrow \text{Sn}^{2+} (\text{aq}) \]
Some key points to identify a reduction half-reaction include:

• Locating the species at the cathode that gains electrons.
• Observing a decrease in the oxidation state of this species.
• Including the electrons that are accepted during this process.

Reduction half-reactions are crucial for completing the electron transaction started by oxidation, thus sustaining the flow of electrical current.

Redox Reaction

Redox reactions, short for reduction-oxidation reactions, are foundational to the operations of electrochemical cells. These reactions simultaneously involve the processes of oxidation, where electrons are lost, and reduction, where electrons are gained. As a result, there is a flow of electrons from the substance oxidized to the substance reduced.
In essence, a redox reaction integrates both half-reactions occurring in a cell, creating a complete equation representative of the chemical changes. For example, in electrochemical cell \( a \), the comprehensive redox reaction is: \[ \text{Pb} (\text{s}) + \text{Sn}^{4+} (\text{aq}) \rightarrow \text{Pb}^{2+} (\text{aq}) + \text{Sn}^{2+} (\text{aq}) \]
Key aspects of redox reactions include:

• Incorporating separate equations for both oxidation and reduction processes.
• Balancing electrons between the transferred and received states.
• Combining the balanced half-reactions to showcase the overall chemical proportion.

These reactions are not only important for educational purposes but are also critical for real-world applications such as batteries and electrolysis.