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 halfreactions 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) Ox: \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \), Red: \( \text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O} \), Overall: \( 4\text{Fe}^{2+} + \text{O}_2 + 4\text{H}^+ \rightarrow 4\text{Fe}^{3+} + 2\text{H}_2\text{O} \). (b) Anode: oxidation of Fe, Cathode: reduction of O₂. (c) Electrons flow from Fe to O₂ electrode, negative ions move cathode to anode half-cell.

Step 1: Determine Oxidation and Reduction Half-Reactions

For the voltaic cell, identify which species is oxidized and which is reduced. - The half-reaction at the anode involves the oxidation of Fe: \[ \text{Fe}^{2+} (\text{aq}) \rightarrow \text{Fe}^{3+} (\text{aq}) + e^- \]- The reduction half-reaction at the cathode involves O₂ and H₂O in an acidic solution: \[ \text{O}_2 (\text{g}) + 4\text{H}^+ (\text{aq}) + 4e^- \rightarrow 2\text{H}_2\text{O} (\text{l}) \]

Step 2: Write the Overall Cell Reaction

Sum the balanced oxidation and reduction half-reactions to give the overall cell reaction, making sure the electrons cancel out:- Balance electrons by multiplying the iron oxidation by 4: \[ 4 \text{Fe}^{2+} (\text{aq}) \rightarrow 4 \text{Fe}^{3+} (\text{aq}) + 4e^- \]- Add it to the reduction half-reaction: \[ 4 \text{Fe}^{2+} (\text{aq}) + \text{O}_2 (\text{g}) + 4 \text{H}^+ (\text{aq}) \rightarrow 4 \text{Fe}^{3+} (\text{aq}) + 2 \text{H}_2 \text{O} (\text{l}) \]

Step 3: Determine Anode and Cathode Reactions

Identify which reactions occur at each electrode:- The oxidation reaction occurs at the anode: \( \text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + e^- \)- The reduction reaction occurs at the cathode: \( \text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O} \).

Step 4: Understand Electron and Ion Flow

In the external circuit, electrons flow from the anode to the cathode. Since oxidation (loss of electrons) occurs at the anode, electrons must travel to the cathode where reduction occurs. - Electrons flow from the **Fe electrode (anode)** to the **O₂ electrode (cathode)**. - Negative ions in the salt bridge move from the **cathode half-cell** to the **anode half-cell** to maintain charge neutrality.

Key Concepts

Oxidation-Reduction Reactions

In a voltaic cell, oxidation-reduction reactions are the key processes that drive the generation of electric current. These reactions are composed of two complementary processes: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons. Together, these reactions are known as redox reactions. Oxidation always occurs at the anode, and reduction always occurs at the cathode.
For instance, in our voltaic cell exercise, the oxidation reaction involves iron ions ( Fe^{2+} ) losing electrons to become iron ions ( Fe^{3+} ), while the reduction reaction involves oxygen ( O_2 ) gaining electrons to form water ( H_2O ). Redox reactions are fundamental to energy conversion in electrochemical cells.

Half-Reaction Equations

Half-reaction equations are used to represent the separate processes of oxidation and reduction. They are termed 'half-reactions' because they each represent one half of the complete redox process. Each half-reaction describes the change in oxidation state of a particular species and helps determine which species is being oxidized or reduced.
For example, the oxidation half-reaction for our voltaic cell is:

• Fe^{2+} ( ext{aq}) ightarrow Fe^{3+} ( ext{aq}) + e^-

The reduction half-reaction is:

• O_2 ( ext{g}) + 4 ext{H}^+ ( ext{aq}) + 4e^- ightarrow 2 ext{H}_2O ( ext{l})

These equations must be balanced not only in terms of mass but also in terms of charge to ensure that the number of electrons lost in oxidation equals the number gained in reduction.

Electrode Reactions

Electrode reactions describe the processes occurring at the electrodes of a voltaic cell, which are an integral part of oxidation-reduction reactions. The electrode at which oxidation occurs is termed the anode, and the electrode at which reduction occurs is termed the cathode.
In our voltaic cell example:

• The anode reaction is: Fe^{2+}( ext{aq}) ightarrow Fe^{3+}( ext{aq}) + e^-.
• The cathode reaction is: O_2 + 4 ext{H}^+ + 4e^- ightarrow 2 ext{H}_2O.

It's vital to understand these reactions as they highlight the flow of electrons that creates electrical energy within the cell.

Electron Flow

The flow of electrons in a voltaic cell is essential for generating an electric current. Electrons are produced at the anode during oxidation and travel through an external circuit to the cathode, where they participate in the reduction reaction.
In our example, electrons flow from the iron electrode, which serves as the anode, to the oxygen electrode, which is the cathode. This movement of electrons from anode to cathode occurs because electrons naturally move towards areas of lower potential energy, facilitating the redox reactions at the respective electrodes. Always keep in mind that electron flow is from anode to cathode in any voltaic cell.

Anode and Cathode Compartments

In a voltaic cell, the anode and cathode are located in separate compartments, known as half-cells. These compartments are connected by a salt bridge, which allows for the flow of ions and maintains electrical neutrality.
The anode compartment houses the oxidation reaction, which in our example is: Fe^{2+} ightarrow Fe^{3+} + e^-.
The cathode compartment contains the reduction reaction: O_2 + 4 ext{H}^+ + 4e^- ightarrow 2 ext{H}_2O.
The salt bridge enables negative ions to move from the cathode half-cell to the anode half-cell, balancing the charge that results from the migration of electrons in the external circuit. Understanding the role of these compartments and the salt bridge is crucial for comprehending how voltaic cells function to produce electricity.