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

(a) Calculate the mass of Li formed by electrolysis of molten Li.i by a current of \(7.5 \times 10^{4}\) A flowing for a period of \(24 \mathrm{~h}\). Assume the electrolytic cell is \(85 \%\) efficient. (b) What is the minimum voltage required to drive the reaction?

Short Answer

Expert verified
The mass of Li formed by electrolysis is approximately \(395775.26 g\) and the minimum voltage required to drive the reaction is \(3.05 V\).

Step by step solution

01

Calculate the total charge passed through the cell

According to the given information, the current flowing through the cell is \(7.5 \times 10^4 A\) and the process lasts for \(24 h\). To find the total charge passed through the cell, use the formula: Charge (Q) = Current (I) × time (t) First, convert time from hours to seconds: \(t = 24 \mathrm{~h} \times 3600 \frac{s}{\mathrm{hr}} = 86400 s\) Now, calculate the charge: \(Q = (7.5 \times 10^4 A) \times 86400 s = 6.48 \times 10^9 C\)
02

Determine the moles of electrons (F) transferred

According to Faraday's law, the amount of substance produced in an electrolytic cell is directly proportional to the moles of electrons transferred. The Faraday constant (F) represents the charge of one mole of electrons: \(F = 96485 \frac{C}{\mathrm{mol}}\) Calculate the moles of electrons transferred: Moles of electrons \(n = \frac{Q}{F} = \frac{6.48 \times 10^9 C}{96485 \frac{C}{\mathrm{mol}}} = 67132.25 \mathrm{~mol}\) Since the electrolytic cell is 85% efficient, we need to adjust the moles of electrons accordingly: \(n_{effective} = 0.85 \times 67132.25 \mathrm{~mol} = 57062.41 \mathrm{~mol}\)
03

Calculate the moles of lithium produced

In lithium electrolysis, each Li atom requires one electron for its formation. Hence, the moles of Li produced are equal to the moles of effective electrons transferred: Moles of Li \( = 57062.41 \mathrm{~mol} \)
04

Calculate the mass of Li produced

Now that we know the moles of Li produced, we can calculate its mass. The molar mass of Li is 6.939 g/mol. Mass of Li \( = 57062.41 \mathrm{~mol} \times 6.939 \frac{g}{\mathrm{mol}} = 395775.26 g\) #Part (b)#
05

Write the reduction half-reaction

To determine the minimum voltage required, first, write the reduction half-reaction: \(Li^+ + e^- \rightarrow Li\)
06

Determine the minimum voltage

The standard electrode potential for this reaction is -3.05 V. Since the process is carried out in molten Li, it should be equal to the minimum voltage required to drive the reaction: Minimum Voltage \( = -3.05 V\) Since we are only interested in the magnitude of the minimum voltage required, take the absolute value: Minimum Voltage \( = 3.05 V\) So, the mass of Li formed by electrolysis is approximately \(395775.26 g\) and the minimum voltage required to drive the reaction is \(3.05 V\).

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 Vaia!

Key Concepts

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

Faraday's Law
One of the most fundamental concepts in electrochemistry is Faraday's Law. This law helps us understand how electrical energy can cause a chemical change during electrolysis. Essentially, it states that the amount of chemical change is directly proportional to the total electric charge that passes through the solution. In more technical terms, the law tells us that to deposit or dissolve one mole of a substance, we need to pass a specific quantity of electric charge, which is the product of a constant (Faraday's Constant) and the substance's valency.

Faraday's Constant, labeled as F, is universally known as 96,485 Coulombs per mole, which represents the charge carried by one mole of electrons. If we apply this to the exercise regarding the electrolysis of lithium, the mass of lithium formed can be related to the charge passed through the electrolytic cell by knowing that lithium has a valency of +1. Since each lithium ion requires one electron to be reduced to lithium metal, the moles of electrons will be equal to the moles of lithium produced in this scenario.
Electrolytic Cell Efficiency
Electrolytic cell efficiency is a critical factor in the industrial production of elements through electrolysis. It refers to the fraction of the electrical energy that is actually used to produce the desired substance, rather than wasted through heat or side reactions. The efficiency is provided as a percentage and signifies the effectiveness of the electrolytic process.

In our exercise, the given efficiency of the electrolytic cell is 85%. This means that only 85% of the total charge passed through the cell results in the reduction of lithium ions to lithium metal. We must adjust the theoretical number of moles of lithium to account for this efficiency by multiplying the total moles of electrons transferred by 0.85. The result of this multiplication gives us the effective moles of lithium that can be produced in real-world scenarios, factoring in the inevitable losses and inefficiencies that occur during the electrolysis process.
Molar Mass Calculation
The molar mass calculation is essential for converting between the mass of a substance and the amount of substance in moles. The molar mass of an element is the weight of one mole of that element and is expressed in grams per mole (g/mol). For lithium (Li), the molar mass is approximately 6.939 g/mol. Knowing this value allows us to relate the moles of lithium atoms produced in a reaction to the actual mass of lithium formed.

To calculate the mass of lithium from the moles of lithium, as in our exercise example, we multiply the effective moles of lithium by its molar mass. By carrying out this step, we arrive at a quantifiable amount that can be measured and used in practical applications such as battery production or pharmaceutical manufacturing. Understanding how to perform molar mass calculations is a crucial skill in chemistry because it bridges the gap between the microscopic world of atoms and the macroscopic world that we can measure and observe.

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

(a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules?

A common shorthand way to represent a voltaic cell is anode | anode solution || cathode solution |cathode A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such AS from solid to solution. (a) Write the half-reactions and overall cell reaction represented by Fe \(\left|\mathrm{Fe}^{2+}\right|\left|\mathrm{Ag}^{+}\right| \mathrm{Ag}_{\mathrm{g}}\) sketch the cell. (b) Write the half-reactions and overall cell reaction represented by \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}\right|\left|\mathrm{H}^{+}\right| \mathrm{H}_{3}\) s sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q) & \longrightarrow \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Pt is used as an inert electrode in contact with the \(\mathrm{ClO}_{3}^{-}\)and Cl. Sketch the cell.

A voltaic cell that uses the reaction $$ \mathrm{PdCl}_{4}{ }^{2-}(a q)+\mathrm{Cd}(s) \longrightarrow \mathrm{Pd}(s)+4 \mathrm{CT}(a q)+\mathrm{Cd}^{2+}(a q) $$ has a measured standard cell potential of \(+1.03 \mathrm{~V}\). (a) Write the two half-cell reactions. (b) By using data from Appendix \(E\), determine \(E_{\text {rel }}^{0}\) for the reaction involving \(\mathrm{Pd}\). (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

(a) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons appear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent?

By using the data in Appendix E, determine whether each of the following substances is likely to serve as an exidant or a reductant: (a) \(\mathrm{Cl}_{2}\) (g), (b) \(\mathrm{MnO}_{4}^{-}\)(aq, acidic solution), (c) \(\mathrm{Ba}\) (s), (d) \(\mathrm{Zn}(\) s).

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.

Sign-up for free