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Chapter 20: Problem 93

(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) 脳 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 \(3.05 V\).

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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.

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Most popular questions from this chapter

A student designs an ammeter (a device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the device for \(2.00 \mathrm{~min}, 12.3 \mathrm{~mL}\) of water-saturated \(\mathrm{H}_{2}(g)\) is collected. The temperature of the system is \(25.5^{\circ} \mathrm{C}\), and the atmospheric pressure is 768 torr. What is the magnitude of the current in amperes?

A \(1 \mathrm{M}\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a beaker with a strip of \(\mathrm{Cu}\) metal. A \(1 \mathrm{M}\) solution of \(\mathrm{SnSO}_{4}\) is placed in a second beaker with a strip of \(\mathrm{Sn}\) metal. A salt bridge connects the two beakers, and wires to a voltmeter link the two metal electrodes. (a) Which electrode serves as the anode and which as the cathode? (b) Which electrode gains mass and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

Hydrogen gas has the potential for use as a clean fuel in reaction with oxygen. The relevant reaction is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I) $$ Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at \(85^{\circ} \mathrm{C}\). (a) Use data in Appendix C to calculate \(\Delta H^{t}\) and \(\Delta S^{\circ}\) for the reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of \(\Delta G\) for the reaction as the temperature increases? (c) What is the significance of the change in the magnitude of \(\Delta G\) with temperature with respect to the utility of hydrogen as a fuel? (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen?

s each of the following substances likely to serve as an oxidant or a reductant: (a) \(\mathrm{Ce}^{3+}(\mathrm{aq})\), (b) \(\mathrm{Ca}(\mathrm{s})\), (c) \(\mathrm{CO}_{3}^{-}(\mathrm{aq})\), (d) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) ?

(a) How many coulombs are required to plate a layer of chromium metal \(0.25 \mathrm{~mm}\) thick on an auto bumper with a total area of \(0.32 \mathrm{~m}^{2}\) from a solution containing \(\mathrm{CrO}_{4}{ }^{2-}\) ? The density of chromium metal is \(7.20 \mathrm{~g} / \mathrm{cm}^{3}\), (b) What current flow is required for this electroplating if the bumper is to be plated in \(10.0 \mathrm{~s}\) ? (c) If the external source has an emf of \(+6.0 \mathrm{~V}\) and the electrolytic cell is \(65 \%\) efficient, hew much electrical power is expended to electroplate the bumper?
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