Problem 49 - Chemistry and Chemical Reactivity

Question

A voltaic cell can be built using the reaction between Al metal and \(\mathrm{O}_{2}\) from the air. If the Al anode of this cell consists of \(84 \mathrm{g}\) of aluminum, how many hours can the cell produce 1.0 A of electricity, assuming an unlimited supply of \(\mathrm{O}_{2} ?\)

Step-by-Step Solution

Answer

The cell can produce 1.0 A of electricity for about 250 hours.

Step 1: Determine the Moles of Aluminum

To find the moles of aluminum, use the formula for moles \( n = \frac{mass}{molar\ mass} \). The molar mass of aluminum, \( Al \), is \( 26.98 \text{ g/mol} \). Therefore, the moles of aluminum are \( n = \frac{84 \text{ g}}{26.98 \text{ g/mol}} \approx 3.11 \text{ moles of Al} \).

Step 2: Write the Half-Reaction for Aluminum

The half-reaction for aluminum in the voltaic cell is \( 2Al \rightarrow 2Al^{3+} + 6e^- \). This equation shows that 2 moles of aluminum produce 6 moles of electrons.

Step 3: Calculate Electrons Needed

From the half-reaction, we know that 2 moles of \( Al \) will release 6 moles of electrons. For \( 3.11 \) moles of \( Al \), the moles of electrons are \( 3.11 \times \frac{6}{2} = 9.33 \text{ moles of electrons} \).

Step 4: Convert Moles of Electrons to Charge

To find the total charge, use the formula \( Q = n \times F \), where \( F = 96485 \text{ C/mol} \) (Faraday's constant). Thus, the total charge is \( Q = 9.33 \times 96485 \approx 900168 \text{ C} \).

Step 5: Find Time in Seconds

Using the formula \( Q = I \times t \), where \( I = 1.0 \text{ A} \), solve for time \( t = \frac{Q}{I} \). Substituting in the charge, \( t = \frac{900168}{1} = 900168 \text{ seconds} \).

Step 6: Convert Time to Hours

Convert the time from seconds to hours: \( 1 \text{ hour} = 3600 \text{ seconds} \). Therefore, \( t = \frac{900168}{3600} \approx 250 \text{ hours} \).

Key Concepts

Aluminum Half-Reaction

In the context of a voltaic cell, a half-reaction refers to the process where one part of the redox reaction occurs. Here we focus on what's happening with aluminum in the cell. The half-reaction for aluminum is expressed as: \[ 2Al \rightarrow 2Al^{3+} + 6e^- \]This reaction describes the oxidation half of the overall cell reaction. **Oxidation** is the loss of electrons, and for each 2 moles of aluminum, 6 electrons are released. This movement of electrons is crucial as it facilitates the flow of electricity in the cell. To visualize it:

• Two aluminum atoms lose a total of six electrons, which they release to the external circuit.
• The electrons then travel through the circuit, doing electrical work.
• This process allows the aluminum metal to transform into aluminum ions, \(Al^{3+}\).

This conversion and electron flow drive the electric current that the cell produces.

Faraday's Constant

Understanding **Faraday's Constant** is essential for calculating the charge involved in a reaction. It represents the total charge of one mole of electrons, specifically\[ F = 96485 \text{ C/mol} \] Here's what you need to know:

• Faraday's Constant is used to relate the amount of substance (in moles) and the total electric charge it represents.
• It allows conversion from moles of electrons to coulombs. So, in any electrochemical context, you multiply the moles of electrons by Faraday’s constant to obtain the charge.
• In the original exercise and solution, this conversion was essential to determine the total charge produced by oxidizing the aluminum.

By understanding and applying Faraday's constant, you can bridge the gap between the atomic world and macroscopic phenomena like electric current.

Electrochemistry

**Electrochemistry** deals with the study of chemical reactions that involve an exchange of electricity. In voltaic cells, like the one in the exercise with aluminum, these reactions are harnessed to produce electrical energy. Here’s a snapshot of the process:

• In a typical voltaic cell, chemical energy is transformed into electrical energy through a spontaneous redox reaction.
• This involves two main components: oxidation at the anode and reduction at the cathode.
• Electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs), generating an electrical current.

Electrochemistry allows us to design and understand devices that can convert chemical energy into usable electrical energy, making it a fundamental concept to grasp in fields like battery and fuel cell technology. By understanding these concepts, we gain insight into how chemical reactions can efficiently power electronic devices and industrial processes.