/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 89 In the commercial preparation of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the commercial preparation of aluminum, aluminum oxide, \(\mathrm{Al}_{2} \mathrm{O}_{3}\), is electrolyzed at \(1000^{\circ} \mathrm{C}\). (The mineral cryolite is added as a solvent.) Assume that the cathode reaction is $$ \mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al} $$ How many coulombs of electricity are required to give \(3.61 \mathrm{~kg}\) of aluminum?

Short Answer

Expert verified
38,706,323 coulombs are required.

Step by step solution

01

Determine Moles of Aluminum

First, calculate the moles of aluminum that corresponds to 3.61 kg. Use the molar mass of aluminum, which is approximately 26.98 g/mol. Convert the mass from kilograms to grams.\[\text{Mass of Al} = 3.61 \text{ kg} \times 1000 \text{ g/kg} = 3610 \text{ g}\]\[\text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}} = \frac{3610 \text{ g}}{26.98 \text{ g/mol}} \approx 133.77 \text{ moles}\]
02

Calculate Total Charge Required

The given reaction shows that one mole of aluminum requires 3 moles of electrons. Therefore, 133.77 moles of aluminum will need three times more electrons in moles.\[\text{Moles of e}^- = 133.77 \text{ moles of Al} \times 3 = 401.31 \text{ moles of e}^-\]Next, use Faraday's constant (approximately 96485 coulombs per mole of electrons) to find the total charge (in coulombs) needed.\[\text{Total charge} = 401.31 \text{ moles of e}^- \times 96485 \text{ C/mol} \approx 38706323.35 \text{ C}\]
03

Round and State the Result

Now, round the number of coulombs to a suitable precision for practical purposes. Generally, rounding to the nearest whole number is appropriate.\[\text{Total charge required} \approx 38,706,323 \text{ coulombs}\]

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Aluminum Production
Aluminum production typically relies on electrolysis, a process in which electricity is used to drive a chemical reaction.In the commercial production of aluminum, alumina (aluminum oxide, \(\text{Al}_2\text{O}_3\)) is dissolved in molten cryolite and subjected to electrolysis.During this process, aluminum ions \((\text{Al}^{3+})\) are reduced to aluminum metal at the cathode by gaining electrons.The reaction can be described by the equation: - \(\text{Al}^{3+} + 3 \text{e}^{-} \rightarrow \text{Al}\)This process is crucial because naturally occurring aluminum compounds are very stable, making the direct extraction of aluminum metal from its ore challenging.Through electrolysis, manufacturers can efficiently produce large quantities of aluminum, which is a lightweight, durable, and highly adaptable metal.
Aluminum is used in various applications, including packaging, transportation, construction, and electrical industries.
Coulombs Calculation
The key aspect of electrolytic processes is the calculation of the charge in coulombs needed to produce a given amount of substance. To compute the required charge, you first need to determine the moles of the substance involved in the reaction. In the example we are exploring, we begin by finding the moles of aluminum from its mass. Once we have the moles of aluminum, the next step is to recognize that each mole of aluminum requires 3 moles of electrons (as derived from the cathodic reaction described earlier). To find the total moles of electrons needed, multiply the moles of aluminum by 3. Finally, convert the moles of electrons to coulombs using Faraday's constant, which provides the linkage between moles and charge. This allows us to find the total electrical charge needed. This calculation is crucial to ensure that enough electricity is provided for the complete reduction of aluminum ions.
Faraday's Constant
A pivotal concept in electrolysis is Faraday's constant.Faraday's constant, approximately \(96485 \text{ C/mol}\), connects the amount of electric charge carried by one mole of electrons.This constant is named after Michael Faraday, a pioneer in the study of electromagnetism and electrochemistry.Understanding Faraday's constant is essential in calculating the charge required in electrolysis processes.By knowing the number of moles of electrons required for a reaction, you can use Faraday's constant to find the total charge in coulombs, making it instrumental in industrial applications like aluminum production.The use of Faraday's constant ensures precise calculation in electrochemical reactions, providing the necessary insights into how much electricity needs to be supplied to achieve the desired chemical transformation.
Chemical Reactions
At the heart of the electrolysis process are chemical reactions facilitated by the passage of electric current. In the case of aluminum production, the primary reaction sees aluminum ions being converted into aluminum metal. This transformation is a reduction reaction, where aluminum ions gain electrons, changing from a positive ion (cation) to a neutral metal. The reaction is crucial in electrochemistry as it shows how electrons can be used to facilitate chemical changes, particularly in breaking down compounds into their elemental form. Beyond aluminum production, electrolysis is a versatile method applied in various chemical processes, from electroplating to producing chlorine and hydrogen gases. These reactions demonstrate the transformative power of electrochemical methods. They showcase technology's ability to convert raw materials into useful products, highlighting the practical applications and importance of chemical reactions driven by electricity.

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

An electrode is prepared from liquid mercury in contact with a saturated solution of mercury(I) chloride, \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\), containing \(1.00 M \mathrm{Cl}^{-}\). The emf of the voltaic cell constructed by connecting this electrode as the cathode to the standard hydrogen half-cell as the anode is \(0.268 \mathrm{~V}\). What is the solubility product of mercury(I) chloride?

A sensitive test for bismuth(III) ion consists of shaking a solution suspected of containing the ion with a basic solution of sodium stannite, \(\mathrm{Na}_{2} \mathrm{SnO}_{2}\). A positive test consists of the formation of a black precipitate of bismuth metal. Stannite ion is oxidized by bismuth(III) ion to stannate ion, \(\mathrm{SnO}_{3}^{2-}\). Write a balanced equation for the reaction.

A voltaic cell has an iron rod in \(0.30 M\) iron(III) chloride solution for the cathode and a zinc rod in \(0.40 M\) zinc sulfate solution for the anode. The half-cells are connected by a salt bridge. Write the notation for this cell.

A solution of copper(II) sulfate is electrolyzed by passing a current through the solution using inert electrodes. Consequently, there is a decrease in the \(\mathrm{Cu}^{2+}\) concentration and an increase in the hydronium ion concentration. Also, one electrode increases in mass and a gas evolves at the other electrode. Write half-reactions that occur at the anode and at the cathode.

Give the mathematical relationships between the members of each possible pair of the three quantities \(\Delta G^{\circ}, E_{\text {cell }}^{\circ}\), and \(K\).

See all solutions

Recommended explanations on Chemistry Textbooks

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.