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Chapter 17: Problem 96

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of 1.00 million \(A\), what mass of aluminum can be produced in \(2.00 \mathrm{h} ?\)

Short Answer

Expert verified
The mass of aluminum produced can be calculated as follows: Moles of electrons = \(\frac{(1.00 \times 10^{6} A)(2.00 \times 3600 \ s)}{96485 \frac{C}{mol}} \approx 74.5 \ mol\) Moles of aluminum = \(\frac{74.5 \ mol}{3} \approx 24.8 \ mol\) Mass of aluminum = \(24.8 \ mol \times 26.98 \frac{g}{mol} \approx 670 \ g\) So, in 2 hours, about 670 g of aluminum can be produced using a plant with a continuous capacity of 1.00 million A.

Step by step solution

01

Write the balanced half-reaction for the electrolysis of aluminum oxide

The electrolysis of aluminum oxide results in the reduction of aluminum ions. The balanced half-reaction is: \[ \mathrm{Al}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Al} \]
02

Calculate the moles of electrons passed through the cell

Using the formula, we can calculate the moles of electrons passed through the electrolytic cell. Moles of electrons = \(\frac{It}{F}\) Where I = current (in amperes), t = time (in seconds), and F = Faraday constant (96,485 C/mol) Convert current and time to appropriate units: I = 1.00 million A = \(1.00 \times 10^{6} A\) t = 2.00 hours = \(2.00 \times 3600\) seconds Calculate the moles of electrons: Moles of electrons = \(\frac{(1.00 \times 10^{6} A)(2.00 \times 3600 \ s)}{96485 \frac{C}{mol}}\)
03

Calculate the moles of aluminum produced

From the balanced half-reaction, 3 moles of electrons are needed to produce 1 mole of aluminum. So, we can calculate the moles of aluminum produced by dividing the moles of electrons calculated in step 2 by 3. Moles of aluminum = \(\frac{\text{moles of electrons}}{3}\)
04

Calculate the mass of aluminum produced

To calculate the mass of aluminum produced, we'll multiply the moles of aluminum by the molar mass of aluminum (26.98 g/mol). Mass of aluminum = moles of aluminum 脳 molar mass of aluminum
05

Evaluate the mass of aluminum produced

Now, we just need to put the values from step 3 and step 4 in the last equation to find the mass of aluminum produced in 2 hours using a plant with a continuous capacity of 1.00 million A.

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Key Concepts

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

Faraday's Laws of Electrolysis
Understanding Faraday's laws of electrolysis is crucial when studying the production of metals through electrolytic processes. These laws quantitatively explain the relationship between the amount of electric charge passed through a substance and the amount of substance that is deposited at an electrode.

The first law states that the mass of a substance altered at an electrode during electrolysis is proportional to the amount of electric charge passed through the electrolyte. Mathematically, it can be represented as \( m = (Q/F) \times M/(n) \), where \( m \) is the mass of the substance, \( Q \) is the total electric charge, \( F \) is Faraday's constant (\( 96,485 C/mol\)), \( M \) is the molar mass of the substance, and \( n \) is the number of moles of electrons required to deposit or dissolve 1 mole of substance.

The second law complements the first by stating that when the same quantity of electric charge is passed through several electrolytes, the masses of substances altered are directly proportional to their chemical equivalent weights. This allows us to predict how much of a substance will be produced or consumed during electrolysis if the amount of electric charge and the identity of the substance are known.
Molar Mass
The molar mass of a substance represents the mass of one mole of that substance, which is a fundamental concept in chemistry and crucial for calculating the outcomes of chemical reactions, including those involving electrolysis.

Molar mass is generally expressed in grams per mole (g/mol) and it's numerically equivalent to the average atomic or molecular mass of the substance. For instance, the molar mass of aluminum (\( Al\)) is \(26.98 g/mol\).

When calculating the mass of aluminum produced in electrolysis, as shown in the given solution, the molar mass allows us to convert moles of aluminum to grams. This ensures that scientists can measure the production of aluminum in terms of weight, a practical commercial quantity.
Molten Salt Electrolysis
Molten salt electrolysis is a method used to extract reactive metals from their naturally occurring compounds, such as aluminum oxide (\( Al_{2}O_{3}\)).

In this process, the compound containing the metal is dissolved in a molten salt, creating an electrolyte that can conduct electricity. When a strong electric current is passed through this electrolyte, it causes the metal cations to migrate to the cathode, where they gain electrons and are reduced to their solid metal form.

This technique is widely used in the production of aluminum because it is one of the few effective methods of reducing aluminum ions to form pure aluminum metal. The high-temperature conditions required to melt aluminum oxide (over 2000掳C) are achieved by dissolving it in another substance, like cryolite, which lowers the melting point of the oxide.

Practical Commercial Significance

Commercially, this process is conducted on a large scale in plants designed to sustain the high-energy demands of maintaining the electrolyte in a molten state, facilitating continuous production of aluminum.
Reduction Half-Reaction
The reduction half-reaction is part of the overall reaction that occurs during electrolysis or any redox process. It specifically involves the gain of electrons by chemical species.

In the context of aluminum electrolysis, the reduction half-reaction can be illustrated by: \[ Al^{3+} + 3e^{-} \rightarrow Al \].

This equation represents the transformation of aluminum ions ( \( Al^{3+}\)) into solid aluminum (\( Al\)) by gaining three electrons. Each aluminum ion requires three electrons for the reduction to occur, so the stoichiometry of this half-reaction determines the amount of aluminum that can be produced per mole of electrons transferred.

Electrodes in Electrolysis

In the electrolysis cell, the cathode is the site where reduction occurs. It's at this electrode that the aluminum ion accepts electrons and is reduced. This understanding of electrolysis, including reduction half-reactions, is central to grasping how electric current can cause chemical change and metal extractions.

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Most popular questions from this chapter

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{C}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are 1.0 atm. a. \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{V}\) \(\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} \quad \quad \mathscr{E}^{\circ}=0.68 \mathrm{V}\) b. \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \quad \mathscr{E}^{\circ}=-1.18 \mathrm{V}\) \(\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{E}^{\circ}=-0.036 \mathrm{V}\)

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M}\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times\) \(10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{V}\). Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co} $$ The overall reaction and equilibrium constant value are $$\begin{aligned} 2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) & \longrightarrow \\ 2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s) & K=2.79 \times 10^{7} \end{aligned}$$ Calculate the cell potential, \(\mathscr{E}\), for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.

In the electrolysis of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what reactions occur at the anode and the cathode (assuming standard conditions)? $$\begin{array}{lr} \mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-} & 80^{\circ} \\ \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow_{2 \mathrm{H}_{2} \mathrm{O}} & 2.01 \mathrm{V} \\ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-} & -0.83 \mathrm{V} \\ \mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na} & -2.71 \mathrm{V} \end{array}$$

Consider a galvanic cell based on the following half-reactions: $$\begin{array}{ll} & \mathscr{E}^{\circ}(\mathrm{V}) \\ \mathrm{La}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{La} & -2.37 \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & -0.44 \\ \hline \end{array}$$ a. What is the expected cell potential with all components in their standard states? b. What is the oxidizing agent in the overall cell reaction? c. What substances make up the anode compartment? d. In the standard cell, in which direction do the electrons flow? e. How many electrons are transferred per unit of cell reaction? f. If this cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Fe}^{2+}\right]=2.00 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{La}^{3+}\right]=3.00 \times 10^{-3} \mathrm{M},\) what is the expected cell potential?
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