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 \(\quad\) half-cell.

Short answer

(a) Oxidation: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\), Reduction: \(\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}\), Overall: \(4\text{Fe}^{2+} + \text{O}_2 + 4\text{H}^+ \rightarrow 4\text{Fe}^{3+} + 2\text{H}_2\text{O}\). (b) Anode: Fe reaction, Cathode: \(\text{O}_2\) reaction. (c) Electrons: anode to cathode, Ions: cathode to anode.

Step-by-step solution

Identify Oxidation and Reduction

The reaction involves the half-cell reactions of iron and oxygen in acid solution. The \(\text{Fe}^{2+}\) is oxidized to \(\text{Fe}^{3+}\) and releases electrons, while \(\text{O}_2\) is reduced to \(\text{H}_2\text{O}\) in the presence of \(\text{H}^+\) ions, gaining electrons.

Write Oxidation Half-Reaction

The oxidation half-reaction involves \(\text{Fe}^{2+}\) losing electrons and converting into \(\text{Fe}^{3+}\). The equation: \[\text{Fe}^{2+} (\text{aq}) \rightarrow \text{Fe}^{3+} (\text{aq}) + e^-\]

Write Reduction Half-Reaction

The reduction half-reaction involves \(\text{O}_2\) reduced to \(\text{H}_2\text{O}\) in acidic conditions, consuming electrons and protons:\[\text{O}_2 (\text{g}) + 4\text{H}^+ (\text{aq}) + 4e^- \rightarrow 2\text{H}_2\text{O} (\text{l})\]

Balance Electrons and Write Overall Reaction

To balance the electrons, multiply the oxidation half-reaction by 4 so both half-reactions exchange the same number of electrons. Then add the balanced half-reactions.The balanced equations:\[4(\text{Fe}^{2+} (\text{aq}) \rightarrow \text{Fe}^{3+} (\text{aq}) + e^-)\]Combine both half-reactions:\[4\text{Fe}^{2+} (\text{aq}) + \text{O}_2 (\text{g}) + 4\text{H}^+ (\text{aq}) \rightarrow 4\text{Fe}^{3+} (\text{aq}) + 2\text{H}_2\text{O} (\text{l})\]

Identify Anode and Cathode

The oxidation half-reaction occurs at the anode, whereas the reduction half-reaction occurs at the cathode. Hence, \(\text{Fe}^{2+}\) oxidizes at the anode and \(\text{O}_2\) reduces at the cathode.

Complete Electron and Ion Movement Sentences

- Electrons flow in the external circuit from the anode (where oxidation occurs) to the cathode (where reduction occurs). - Negative ions move in the salt bridge from the cathode half-cell to the anode half-cell to maintain electrical neutrality.

Key concepts

Half-Reaction

In electrochemistry, a half-reaction is a simplified way to consider either the oxidation or reduction processes taking place separately in a redox reaction. These reactions are crucial as they allow us to understand and balance complex redox reactions by breaking them down into two separate equations.
In the context of the voltaic cell involving iron and oxygen in an acidic solution, the oxidation half-reaction details the process where iron (\( ext{Fe}^{2+} \)), loses an electron. The reduction half-reaction, on the other hand, involves oxygen (\( ext{O}_2 \)) gaining electrons to form water.

• Oxidation: Loss of electrons.
• Reduction: Gain of electrons.
• Half-reactions help balance the overall equation by ensuring that the number of electrons lost and gained are equal.

Oxidation and Reduction

Oxidation and reduction are chemical processes that always occur together in a redox reaction. In these reactions, one substance loses electrons (oxidation), while another substance gains them (reduction).
For the given voltaic cell example:

• Oxidation occurs when \( ext{Fe}^{2+} \) is transformed into \( ext{Fe}^{3+} \) through the loss of an electron.
• Reduction takes place when \( ext{O}_2 \) gains electrons and protons to form \( ext{H}_2 ext{O} \).

These processes are essential for the transfer of electrons in the voltaic cell, driving the overall chemical reaction that generates electricity.

Voltaic Cell

A voltaic cell, also known as a galvanic cell, is an electrochemical cell that uses a spontaneous redox reaction to generate electricity. It consists of two half-cells connected by a wire and a salt bridge. Each half-cell contains a metal electrode submerged in an electrolyte solution.
In the described cell:

• The anode is where oxidation takes place (\( ext{Fe}^{2+} \) to \( ext{Fe}^{3+} \)) and is the source of electrons in the circuit.
• The cathode is where reduction occurs (\( ext{O}_2 \) to \( ext{H}_2 ext{O} \)) and consumes electrons delivered through the circuit.
• The salt bridge maintains electrical neutrality by allowing ion flow between the solutions.

Voltaic cells are foundational in understanding how chemical reactions can be harnessed to produce electrical energy.

Electron Flow

Electron flow in a voltaic cell is essential for generating electricity. Electrons are the carriers of electrical charge, moving from the anode to the cathode through an external circuit.
For the iron-oxygen cell:

• Electrons are released at the anode during oxidation.
• These electrons travel through the circuit to the cathode where they are used in the reduction of oxygen.
• This flow creates an electrical current that can be harnessed for practical uses, such as powering a device.
• The direction is always from the site of oxidation (anode) to the site of reduction (cathode).

Understanding electron flow is vital to appreciate how energy is transferred in a voltaic cell.

Cell Reaction Balance

Balancing a cell reaction involves ensuring that the number of electrons lost in the oxidation equals the number of electrons gained in the reduction. This balance is crucial in maintaining the cell's functionality and ensuring it can continuously produce electricity.
For the discussed cell reaction:

• The oxidation of \( ext{Fe}^{2+} \) is balanced by multiplying its half-reaction by 4 to match the electrons utilized in the reduction of \( ext{O}_2 \).
• The overall balanced reaction becomes: \[ 4 ext{Fe}^{2+} ( ext{aq}) + ext{O}_2 ( ext{g}) + 4 ext{H}^+ ( ext{aq}) ightarrow 4 ext{Fe}^{3+} ( ext{aq}) + 2 ext{H}_2 ext{O} ( ext{l}) \].

Balancing ensures electron conservation in the cell, allowing the reaction to proceed smoothly and steadily.