Problem 47 - 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 for approximately 250 hours.

Step 1: Write the balanced chemical equation

The reaction between aluminum (Al) and oxygen (\(\mathrm{O}_{2}\)) is represented as: \[ 4 \mathrm{Al} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{Al}_{2}\mathrm{O}_{3}. \]This equation shows that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.

Step 2: Calculate the number of moles of Al

First, we calculate the number of moles of aluminum (Al) using its molar mass, which is approximately 26.98 g/mol. The formula is:\[ \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{84}{26.98} \approx 3.11 \text{ moles}. \]

Step 3: Determine the total charge produced

Each Al atom releases 3 electrons when it reacts, as can be seen from the unbalanced half-reaction of Al to Al³⁺ ions. Therefore, 3.11 moles of Al will release:\[ 3.11 \times 3 = 9.33 \text{ moles of electrons}. \]Since 1 mole of electrons corresponds to a charge of approximately \( 96,485 \text{ C (Faraday's constant)},\) the total charge will be: \[ 9.33 \times 96,485 \approx 900,376 \text{ C}. \]

Step 4: Calculate the time in hours to produce 1.0 A

To find out how long the cell can produce 1.0 A of current, use the formula: \[ \text{time (s)} = \frac{\text{total charge (C)}}{\text{current (A)}} = \frac{900,376}{1.0} = 900,376 \text{ s}. \]Convert this time from seconds to hours by dividing by 3600:\[ \text{time (h)} = \frac{900,376}{3600} \approx 250.1 \text{ hours}. \]

Key Concepts

Balanced Chemical Equation

Understanding the concept of a balanced chemical equation is crucial in chemistry as it represents a chemical reaction with the same number of atoms for each element on both sides.
This ensures that mass is conserved in the reaction, based on the law of conservation of mass. In the reaction between aluminum (Al) and oxygen (\(\mathrm{O}_{2}\)), the balanced chemical equation is:

• \(4 \mathrm{Al} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{Al}_{2}\mathrm{O}_{3}\).

Here, 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. Balancing involves adjusting coefficients to equalize atoms on both sides. This equation also implies molar relationships that are essential in stoichiometric calculations, such as determining reactant consumption or product formation.

Moles of Aluminum

Calculating the number of moles is a fundamental task in chemistry to understand the amount of substance involved in a reaction. In this scenario, we need to find the moles of aluminum to predict how long it can produce electricity in the voltaic cell.
The amount of aluminum initially given is 84 grams. To convert this mass into moles, we use the molar mass of aluminum, which is approximately 26.98 g/mol. Using the formula:

• \(\text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}}\).

Plugging in the values:

• \(\frac{84}{26.98} \approx 3.11 \text{ moles}\),

we determine there are approximately 3.11 moles of aluminum available for the reaction. This value forms the basis for further calculations related to electron transfer and time duration for current production.

Faraday's Constant

Faraday's constant is a key concept in electrochemistry, named after Michael Faraday.
It is used to calculate the charge carried by one mole of electrons. The constant is approximately 96,485 coulombs per mole of electrons.
For this exercise, knowing that each aluminum atom releases 3 electrons when reacting to form aluminum ions (\(\mathrm{Al}^{3+}\)), and given that there are 3.11 moles of aluminum:

• The total moles of electrons is \(3.11 \times 3 = 9.33 \text{ moles of electrons}\).

The total charge can then be calculated using Faraday’s constant:

• \(9.33 \text{ moles} \times 96,485 = 900,376 \text{ C}\).

This charge value is pivotal in determining how long the voltaic cell can sustain a 1.0 ampere current.

Electric Current Calculation

When considering the production of electricity in a voltaic cell, calculating the duration the cell can sustain a given current is crucial.
The total electric charge calculated previously is 900,376 coulombs. To find out how long the cell can produce a specific electric current—in this case, 1.0 ampere—we use the formula for time:

• \(\text{time (s)} = \frac{\text{total charge (C)}}{\text{current (A)}}\).

Substituting the known values gives:

• \(\frac{900,376}{1.0} = 900,376 \text{ seconds}\).

However, since this is an inconvenient unit for long-term duration, it's practical to convert seconds into hours:

• \(\text{time (h)} = \frac{900,376}{3600} \approx 250.1 \text{ hours}\).

This conversion shows that, with the available aluminum and a constant current of 1.0 A, the voltaic cell can operate for approximately 250.1 hours.