Problem 23 - Chemistry The Molecular Science

Question

Draw a diagram of each cell. Label the anode, the cathode, the species in each half-cell solution, the direction of electron movement in an external circuit, and the direction of movement of ions within the cell. (a) \(\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq}) \| \mathrm{Fe}^{2+}(\mathrm{aq})\right| \mathrm{Fe}(\mathrm{s})\) (b) \(\mathrm{Pt}(\mathrm{s})\left|\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}), \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{Fe}^{2+}(\mathrm{aq}), \mathrm{Fe}^{3+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})\)

Step-by-Step Solution

Answer

For (a), Cu is the anode, and Fe is the cathode. For (b), H2O2 is the anode, and Fe2+/Fe3+ is the cathode.

Step 1: Identify the Half-Cells for Exercise (a)

The given electrochemical cell notation is \( \mathrm{Cu} (\mathrm{s}) \left| \mathrm{Cu}^{2+} (\mathrm{aq}) \| \mathrm{Fe}^{2+} (\mathrm{aq}) \right| \mathrm{Fe} (\mathrm{s}) \). This cell consists of two half-cells:1. Copper half-cell: \( \mathrm{Cu} (\mathrm{s}) \left| \mathrm{Cu}^{2+} (\mathrm{aq}) \right. \)2. Iron half-cell: \( \left. \mathrm{Fe}^{2+} (\mathrm{aq}) \right| \mathrm{Fe} (\mathrm{s}) \)

Step 2: Determine the Anode and Cathode for Exercise (a)

The cell notation shows the anode on the left and the cathode on the right. In this setup:- Anode (Oxidation): \( \mathrm{Cu} (\mathrm{s}) \to \mathrm{Cu}^{2+} (\mathrm{aq}) + 2e^- \)- Cathode (Reduction): \( \mathrm{Fe}^{2+} (\mathrm{aq}) + 2e^- \to \mathrm{Fe} (\mathrm{s}) \)

Step 3: Diagram for Exercise (a)

1. Draw two beakers (half-cells) connected by a salt bridge.2. Label the left beaker as the Copper half-cell and the right beaker as the Iron half-cell.3. In the Copper half-cell, label the solid copper as the anode. Label the \( \mathrm{Cu}^{2+} \) ions in solution.4. In the Iron half-cell, label the solid iron as the cathode. Label the \( \mathrm{Fe}^{2+} \) ions in solution.5. Draw an external wire connecting the anode to the cathode and indicate electron flow from Anode to Cathode.6. Draw the direction of ion flow in the salt bridge to complete the circuit.

Step 4: Identify the Half-Cells for Exercise (b)

The given electrochemical cell notation is \( \mathrm{Pt} (\mathrm{s}) \left| \mathrm{H}_{2} \mathrm{O}_{2} (\mathrm{aq}), \mathrm{H}^{+} (\mathrm{aq}) \| \mathrm{Fe}^{2+} (\mathrm{aq}), \mathrm{Fe}^{3+} (\mathrm{aq}) \right| \mathrm{Pt} (\mathrm{s}) \).This cell consists of two half-cells:1. Platinum electrode in hydrogen peroxide and hydrogen ion solution2. Iron (II)/Iron (III) solution with platinum electrode

Step 5: Determine the Anode and Cathode for Exercise (b)

In this exercise, the anode and cathode need to be determined based on the reactions:- At the anode, Hydrogen peroxide is reduced: \( \mathrm{H}_2 \mathrm{O}_2 (\mathrm{aq}) + 2 \mathrm{H}^+ (\mathrm{aq}) \to \mathrm{O}_2 (\mathrm{g}) + 2e^- \)- At the cathode, Iron is oxidized: \( \mathrm{Fe}^{2+} (\mathrm{aq}) \to \mathrm{Fe}^{3+} (\mathrm{aq}) + e^- \)

Step 6: Diagram for Exercise (b)

1. Draw two beakers (half-cells) connected by a salt bridge.2. Label the left beaker with platinum in a \( \mathrm{H}_2 \mathrm{O}_2 \) and \( \mathrm{H}^+ \) solution and label this side as the anode.3. Label the right beaker with iron ions in solution as the cathode.4. Indicate the platinum electrode in each half-cell.5. Draw the external wire connecting the two platinum electrodes, indicating electron flow from anode to cathode.6. Illustrate ion flow in the salt bridge to complete the circuit.

Key Concepts

Anode and Cathode Identification

Every electrochemical cell comprises two electrodes: an anode and a cathode. Identifying these is fundamental in understanding how the cell operates. The anode is where oxidation occurs, and the cathode is where reduction takes place. You can remember this by using the mnemonic "An Ox" (Anode: Oxidation) and "Red Cat" (Reduction: Cathode). In cell notations, the anode is listed on the left and the cathode on the right.

For Exercise (a), the reactions are:

• At the anode: \( \mathrm{Cu} (\mathrm{s}) \rightarrow \mathrm{Cu}^{2+} (\mathrm{aq}) + 2e^- \)
• At the cathode: \( \mathrm{Fe}^{2+} (\mathrm{aq}) + 2e^- \rightarrow \mathrm{Fe} (\mathrm{s}) \)

For Exercise (b), identification involves recognizing that \(\mathrm{H}_2\mathrm{O}_2\) is reduced, acting as the anode, while \(\mathrm{Fe}^{2+}\) oxidizes at the cathode:

• Anode: \( \mathrm{H}_2 \mathrm{O}_2 (\mathrm{aq}) + 2 \mathrm{H}^+ (\mathrm{aq}) + 2e^- \rightarrow \mathrm{O}_2 (\mathrm{g}) \)
• Cathode: \( \mathrm{Fe}^{2+} (\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+} (\mathrm{aq}) + e^- \)

Electron Flow Direction

Understanding the direction of electron flow in an electrochemical cell is as simple as remembering that electrons naturally flow from the anode to the cathode. This flow is crucial as it generates an electric current that can be harnessed for work.

For Exercise (a), electrons move from the copper electrode (anode) where oxidation occurs, to the iron electrode (cathode) where reduction takes place. Similarly, in Exercise (b), electrons flow from the \(\mathrm{H}_2\mathrm{O}_2\) at the anode to the \(\mathrm{Fe}^{3+}\) ions at the cathode.

In both scenarios, this happens through an external wire connecting the two electrodes. Visualizing this can aid in understanding how batteries and other electrochemical cells generate power.

Ion Movement

Ion movement within an electrochemical cell completes the electrical circuit and maintains charge balance. This movement typically occurs through a salt bridge or porous barrier connecting the two half-cell solutions.

In Exercise (a), as electrons move from the copper anode to the iron cathode:

• \(\mathrm{Cu}^{2+}\) ions are created at the anode, moving into the solution.
• \(\mathrm{Fe}^{2+}\) ions gain electrons at the cathode, forming solid iron.
• To balance the charge, negatively charged ions from the salt bridge migrate towards the copper half-cell, while positively charged ions move towards the iron half-cell.

For Exercise (b), \(\mathrm{H}^{+}\) ions at the anode react with \(\mathrm{H}_2\mathrm{O}_2\), whereas \(\mathrm{Fe}^{2+}\) ions are oxidized to \(\mathrm{Fe}^{3+}\). Similarly, the salt bridge allows counter ions to migrate, preventing charge buildup and allowing the redox reactions to proceed.

Half-Cell Diagram

Drawing half-cell diagrams is an effective way to visualize the working of electrochemical cells. These diagrams often consist of two separate containers (the half-cells) linked by a salt bridge or another ion-conducting medium. Each half-cell consists of an electrode immersed in a solution.

In Exercise (a), the copper half-cell acts as the anode, and the iron half-cell as the cathode. In Exercise (b), the anode is the \(\mathrm{H}_2\mathrm{O}_2\) with a platinum electrode, and the cathode is the \(\mathrm{Fe}^{2+}/\mathrm{Fe}^{3+}\) iron solution, also with platinum electrodes.

• The oxidation and reduction reactions are labeled within each half-cell to illustrate the transformations occurring on the electrodes.
• Direction of electron flow is shown from anode to cathode.
• A salt bridge is depicted with ions moving appropriately to balance the cell’s charge, completing the circuit through ion flow.

Such diagrams help in understanding the entire setup and process clearly, highlighting the roles of different components in conducting electricity.