Problem 9 - Chemistry and Chemical Reactivity

Question

The half-cells \(\mathrm{Fe}^{2+}(\text { aq }) | \mathrm{Fe}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}) | \mathrm{H}_{2} \mathrm{O}\) (in acid solution) are linked to create a voltaic cell. (a) Write equations for the oxidation and reduction half-reactions and for the overall (cell) reaction. (b) Which half-reaction occurs in the anode compartment, and which occurs in the cathode compartment? (c) Complete the following sentences: Electrons in the external circuit flow from the ________ electrode to the ___________ electrode. Negative ions move in the salt bridge from the _________ half-cell to the__________ half-cell.

Step-by-Step Solution

Answer

(a) Oxidation: \( \mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+} + 2e^- \), Reduction: \( \mathrm{O}_{2}(g) + 4H^{+} + 4e^- \rightarrow 2\mathrm{H}_{2} \mathrm{O}(l) \). Overall: \( 4\mathrm{Fe}(s) + \mathrm{O}_{2}(g) + 8H^{+} \rightarrow 4\mathrm{Fe}^{2+} + 4\mathrm{H}_{2} \mathrm{O}(l) \). (b) Oxidation at anode, reduction at cathode. (c) Electrons: anode to cathode. Negative ions: cathode to anode.

Step 1: Identify the Half-Reactions

The half-reaction for \( \mathrm{Fe}^{2+} \) to \( \mathrm{Fe} \) is \( \mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe}(s) \), which is reduction. The half-reaction for \( \mathrm{O}_{2} \) in acidic solution is \( \mathrm{O}_{2}(g) + 4H^{+} + 4e^- \rightarrow 2\mathrm{H}_{2} \mathrm{O}(l) \), also a reduction reaction.

Step 2: Determine the Oxidation Half-Reaction

Typically, the reaction with lower standard electrode potential will proceed as an oxidation. For the iron reaction: \( \mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+} + 2e^- \).

Step 3: Write the Overall Cell Reaction

Combine the two half-reactions: \( \mathrm{4Fe}(s) + \mathrm{O}_{2}(g) + 8H^{+} \rightarrow 4\mathrm{Fe}^{2+} + 4\mathrm{H}_{2} \mathrm{O}(l) \). Adjust coefficients to balance electrons.

Step 4: Identify Anode and Cathode Reactions

The oxidation reaction (\( \mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+} + 2e^- \)) occurs at the anode. The reduction reaction (\( \mathrm{O}_{2}(g) + 4H^{+} + 4e^- \rightarrow 2\mathrm{H}_{2} \mathrm{O}(l) \)) occurs at the cathode.

Step 5: Describe Electron Flow and Ion Movement

Electrons flow in the external circuit from the anode (\( \mathrm{Fe} \)) to the cathode (\( \mathrm{O}_{2} \)). Negative ions move in the salt bridge from the cathode half-cell to the anode half-cell.

Key Concepts

Half-Reaction Equations

In a voltaic cell, it is crucial to understand the nature and purpose of half-reaction equations. These equations help us visualize how electrons are transferred during chemical reactions. By breaking down a full redox reaction into its half-reactions, we manage two separate processes: oxidation and reduction.

- **Oxidation** refers to the loss of electrons. For the reaction involving iron, the equation is written as \( \mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+} + 2e^- \).
- **Reduction** refers to the gain of electrons. In the case of oxygen in an acidic solution, the equation is \( \mathrm{O}_{2}(g) + 4H^{+} + 4e^- \rightarrow 2\mathrm{H}_{2} \mathrm{O}(l) \).

By combining these half-reactions, we derive the overall cell reaction: \( \mathrm{4Fe}(s) + \mathrm{O}_{2}(g) + 8H^{+} \rightarrow 4\mathrm{Fe}^{2+} + 4\mathrm{H}_{2} \mathrm{O}(l) \). Balancing is crucial to ensure that the electrons lost in the oxidation half-reaction match those gained in the reduction half-reaction.

Anode and Cathode Identification

Understanding which part of a voltaic cell acts as either the anode or the cathode ensures we recognize where oxidation and reduction occur.

- **Anode**: This is where oxidation takes place. It loses electrons, which are then transferred to the circuit. In our exercise, the anode reaction is: \( \mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+} + 2e^- \).
- **Cathode**: This is where reduction happens. It receives electrons from the circuit. In the given exercise, the cathode reaction is: \( \mathrm{O}_{2}(g) + 4H^{+} + 4e^- \rightarrow 2\mathrm{H}_{2} \mathrm{O}(l) \).

By pinpointing the anode and cathode, we can map the flow of electrons and better understand how each half-cell contributes to the full voltaic cell operation.

Electron Flow Direction

In a voltaic cell, the direction of electron flow is what generates electricity. Understanding this flow helps explain how the cell converts chemical energy into electrical energy.

- Electrons are emitted from the anode, where oxidation takes place.
- These electrons travel through an external circuit to reach the cathode.
- At the cathode, received electrons participate in the reduction process.

In our exercise problem, electrons depart from the \( \mathrm{Fe} \) anode, traverse the external circuit, and are eventually received by \( \mathrm{O}_{2} \) at the cathode. This movement is crucial for the cell's ability to do work and maintain electrical potential.

Salt Bridge Function

The salt bridge is a critical component in a voltaic cell, balancing charge and allowing the cell to continue operating without interruption. It allows ions to flow between the two half-cells, which compensates for the charge buildup that would otherwise stop the flow of electrons.

- Negative ions traverse the salt bridge from the cathode compartment to the anode compartment.
- This movement prevents positive charge accumulation at the anode and negative charge at the cathode, which would otherwise halt electron flow.

In the exercise example, negative ions flow in the salt bridge from the \( \mathrm{O}_{2} \) half-cell to the \( \mathrm{Fe} \) half-cell. This ensures that the voltaic cell can continue its operation, maintaining a steady flow of electrons and producing a consistent electric current.