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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 \(\mathrm{O}_{2} ?\)

Short Answer

Expert verified
The cell can produce 1.0 A for approximately 250 hours.

Step by step solution

01

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.
02

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:\[ \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{84}{26.98} \approx 3.11 \text{ moles}. \]
03

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 Al鲁鈦 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: \[ 9.33 \times 96,485 \approx 900,376 \text{ C}. \]
04

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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鈥攊n this case, 1.0 ampere鈥攚e 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.

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Most popular questions from this chapter

An aqueous solution of \(\mathrm{Na}_{2} \mathrm{S}\) is placed in a beaker with two inert platinum electrodes. When the cell is attached to an external battery, electrolysis occurs. (a) Hydrogen gas and hydroxide ion form at the cathode. Write an equation for the half-reaction that occurs at this electrode. (b) Sulfur is the primary product at the anode. Write an equation for its formation.

One half-cell in a voltaic cell is constructed from a copper wire electrode in a \(4.8 \times 10^{-3} \mathrm{M}\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} .\) The other half-cell consists of a zinc electrode in a \(0.40 \mathrm{M}\) solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} .\) Calculate the cell potential.

Balance the following redox equations. All occur in basic solution. (a) \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{Cr}(\mathrm{s}) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s})\) (b) \(\mathrm{NiO}_{2}(\mathrm{s})+\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Ni}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{s})\) $$\begin{aligned}&\text { (c) } \mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{CrO}_{4}^{2-}(\mathrm{aq}) \rightarrow\\\&&\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\left[\mathrm{Cr}(\mathrm{OH})_{4}\right]^{-}(\mathrm{aq})\end{aligned}$$ $$\text { (d) } \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{Ag}(\mathrm{s})$$

Which product, \(\mathrm{O}_{2}\) or \(\mathrm{F}_{2}\), is more likely to form at the anode in the electrolysis of an aqueous solution of KF? Explain your reasoning.

Two \(\mathrm{Ag}^{+}\) (aq) \(|\) Ag half-cells are constructed. The first has \(\left[\mathrm{Ag}^{+}\right]=1.0 \mathrm{M},\) the second has \(\left[\mathrm{Ag}^{+}\right]=1.0 \times 10^{-5} \mathrm{M}\) When linked together with a salt bridge and external circuit, a cell potential is observed. (This kind of voltaic cell is referred to as a concentration cell.) (a) Draw a picture of this cell, labeling all components. Indicate the cathode and the anode, and indicate in which direction electrons flow in the external circuit. (b) Calculate the cell potential at \(298 \mathrm{K}\)

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