Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 18: 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 that can be produced in 2.00 hours with a continuous capacity of 1.00 million A is approximately 549022.88 grams.

Step by step solution

01

Identify the information given.

In this problem, we have been given the following information: 1. Current (I) = 1.00 million A (1000000 A) 2. Time (t) = 2.00 hours
02

Convert all given values into base units.

Since we will use the SI unit system, convert the given values to that system: 1. Time (t) = 2.00 hours × 3600 s/hour = 7200 s
03

Recall the Faraday's Law of Electrolysis formula.

Faraday's Law of Electrolysis states that: Mass (m) = (Current * Time * Molecular Weight) / (n * Faraday's constant), where - Current (I) = 1000000 A - Time (t) = 7200 s - Molecular Weight of Aluminum (M) = 26.98 g/mol - n = number of electrons involved in the redox reaction (Al³⁺ + 3e⁻ → Al) - Faraday's constant (F) = 96485 C/mol
04

Determine the number of electrons involved in the redox reaction.

For the given problem, the redox reaction in the electrolysis process is: Al³⁺ + 3e⁻ → Al Clearly, 3 electrons (n = 3) are involved in the reaction.
05

Plug the values into the formula and calculate the mass of aluminum produced.

Now, plug in the given values into the Faraday's Law of Electrolysis formula to calculate the mass of aluminum produced: Mass (m) = (1000000 A * 7200 s * 26.98 g/mol) / (3 * 96485 C/mol) Mass (m) = \( \frac{1000000 \times 7200 \times 26.98}{3\times 96485} \) g ≈ 549022.88 g The mass of aluminum that can be produced in 2.00 hours with a continuous capacity of 1.00 million A is approximately 549022.88 grams.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Current in Electrolysis
Current, in the context of electrolysis, refers to the flow of electric charge. It's measured in amperes (A). In our scenario, we have a plant operating with a continuous capacity of 1,000,000 A (1 million A).
This large amount of current is necessary to drive the electrolysis process, where electrical energy is used to break down compounds.
  • The higher the current, the more chemical reactions occur in a given period.
  • Current is directly proportional to the rate at which products are formed in the electrolysis process.
To calculate the quantity of a substance like aluminum, we multiply the current by the time for which it flows. In this exercise, we converted time into seconds because the SI unit of time is seconds, allowing us to work with current in amperes smoothly.
The Role of Redox Reactions
Redox reactions, short for reduction-oxidation reactions, play a crucial role in electrolysis.
They involve the transfer of electrons between chemical species. In electrolysis, these reactions occur at the electrodes.
  • Reduction occurs at the cathode where ions gain electrons.
  • Oxidation happens at the anode where elements lose electrons.
In our aluminum production, the redox reaction is: \[ \text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al} \]
Here, each aluminum ion gains 3 electrons (reduction) to form aluminum metal. Understanding the number of electrons exchanged is essential as it determines how much aluminum can be produced. In our case, "n", the number of electrons needed, is 3, which fits into Faraday's equation to calculate mass.
Molecular Weight of Aluminum
Molecular weight (also known as molar mass) is the mass of one mole of a substance, expressed in grams/mol.
For aluminum, this is 26.98 g/mol.
  • Molecular weight helps in converting moles of a substance into grams.
  • It's crucial for calculating the physical amount of aluminum produced.
In our exercise, the molecular weight of aluminum is used in Faraday's Law to determine how much aluminum is created when a certain number of moles is deposited.
By combining this with the current, time, and charge information from the electrolysis process, we can accurately determine the mass of aluminum that can be produced.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following half-reactions: $$ \begin{aligned} \mathrm{IrCl}_{6}{ }^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{~V} \\ \mathrm{PtCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{~V} \\ \mathrm{PdCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{~V} \end{aligned} $$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant \(1.0 M\) in chloride ion and \(0.020 M\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

Three electrochemical cells were connected in series so that the same quantity of electrical current passes through all three cells. In the first cell, \(1.15 \mathrm{~g}\) chromium metal was deposited from achromium(III) nitrate solution. In the second cell, \(3.15 \mathrm{~g}\) osmium was deposited from a solution made of \(\mathrm{Os}^{n+}\) and nitrate ions. What is the name of the salt? In the third cell, the electrical charge passed through a solution containing \(\mathrm{X}^{2+}\) ions caused deposition of \(2.11 \mathrm{~g}\) metallic \(\mathrm{X}\). What is the electron configuration of \(\mathrm{X}\) ?

Given the following two standard reduction potentials, $$ \begin{array}{ll} \mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.10 \mathrm{~V} \\ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.50 \mathrm{~V} \end{array} $$ solve for the standard reduction potential of the half-reaction $$ \mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+} $$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)

Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) b. \(\mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)\) c. \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)\)

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of \(2.50\) A passing for \(50.0\) min. If \(2.618 \mathrm{~g}\) Ru is produced at the cathode, what is the charge on the ruthenium ions in solution?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free