Question

(a) Calculate the mass of Li formed by electrolysis of molten LiCl 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

The mass of Li formed by electrolysis of molten LiCl by a current of \(7.5 \times 10^{4} \mathrm{A}\) flowing for a period of 24 hours, assuming the electrolytic cell is 85% efficient, is approximately \(3.96 \times 10^{5} \mathrm{g}\). The minimum voltage required to drive the reaction is \(3.04 \mathrm{V}\).

Step-by-step solution

Find the total charge passed during electrolysis

The total charge (Q) passed during electrolysis can be found using the formula: \[ Q = I \times t \] where \( I \) is the current in amperes and \( t \) is the time in seconds. Given current \( I = 7.5 \times 10^{4} \mathrm{~A} \) and time \( t = 24 \mathrm{~hours} \). We need to convert the time into seconds: \[ t = 24 \mathrm{~h} \times \frac{3600 \mathrm{~s}}{1 \mathrm{~h}} = 86400 \mathrm{~s} \] Now, calculate the total charge: \[ Q = (7.5 \times 10^{4} \mathrm{~A}) \times (86400 \mathrm{~s}) = 6.48 \times 10^{9} \mathrm{~C} \]

Calculate the charge passed during Li formation considering cell efficiency

The cell is 85% efficient, so we need to find the effective charge (Q') that is used in Li formation by multiplying the total charge by the cell efficiency: \[Q' = Q \times \frac{85}{100} \] \[Q' = (6.48 \times 10^{9} \mathrm{~C}) \times 0.85 = 5.508 \times 10^{9} \mathrm{~C} \]

Use Faraday's laws of electrolysis to calculate the amount of substance of Li formed

From Faraday's laws of electrolysis, the amount of substance (in moles) of Li formed can be calculated using: \[ n = \frac{Q'}{zF} \] Where: - \( n \) is the amount of substance of lithium formed (in moles) - \( Q' \) is the effective charge in coulombs - \( z \) is the number of electrons transferred in the reduction half-reaction (for Li, \( z = 1 \) since the reaction is \( \mathrm{Li^{+} + e^{-} \rightarrow Li} \)) - \( F \) is the Faraday constant, which is \( 9.6485 \times 10^{4} \mathrm{~C/mol} \) Now, calculate the amount of substance of Li formed: \[ n = \frac{5.508 \times 10^{9} \mathrm{~C}}{(1)(9.6485 \times 10^{4} \mathrm{~C/mol})} = 5.703 \times 10^{4} \mathrm{~mol} \]

Calculate the mass of Li formed

The mass of lithium (m) formed can be calculated using: \[ m = n \times M \] Where: - \( m \) is the mass of lithium in grams - \( n \) is the amount of substance of lithium formed (from Step 3) - \( M \) is the molar mass of lithium, which is \( 6.94 \mathrm{~g/mol} \) Now, calculate the mass of Li formed: \[ m = (5.703 \times 10^{4} \mathrm{~mol}) \times (6.94 \mathrm{~g/mol}) = 3.96 \times 10^{5} \mathrm{~g} \] Thus, the mass of Li formed by electrolysis is \( 3.96 \times 10^{5} \mathrm{~g} \).

Calculate the minimum voltage required

The reduction half-reaction of Li is: \[ \mathrm{Li^{+} + e^{-} \rightarrow Li} \] The standard electrode potential for the reduction of Li is \( -3.04 \mathrm{~V} \) (data from the standard reduction potential table). However, since the given reaction is an electrolysis (non-spontaneous) reaction, the minimum voltage required is at least the positive value of the standard electrode potential, which is: \[ V_{min} = |-3.04 \mathrm{~V}| = 3.04 \mathrm{~V} \] So, the minimum voltage required to drive the reaction is \( 3.04 \mathrm{~V} \).

Key concepts

Faraday's Laws of Electrolysis

Faraday's Laws of Electrolysis are two fundamental laws in electrochemistry that explain the relationship between the amount of electric charge passed through a solution and the amount of substance that undergoes a chemical change. The first law states that the amount of chemical change or the mass of a substance altered at an electrode during electrolysis is directly proportional to the total electric charge passed through the solution.

In our exercise, we use this principle to find out how much lithium forms when we know the charge and efficiency of the electrolytic cell. The second law reveals that the masses of different substances produced by the same quantity of electricity are proportional to their equivalent weights or molar masses divided by the number of electrons involved in their half-reactions. For lithium, since only one electron is transferred per ion, it allows us to simplify calculations and make accurate predictions about substance formation.

Applying Faraday's laws helps students understand how electricity and matter interact during electrolysis. It's an essential cornerstone for problems involving the calculation of mass and moles from current and time.

Electrode Potential

The concept of electrode potential is central to understanding electrolysis. Electrode potential refers to the ability of an electrode in an electrochemical cell to lose or gain electrons. It is a vital measure in determining the voltage needed to make a non-spontaneous reaction proceed.

This potential depends on the specific materials involved in the reaction as well as the conditions such as concentration and temperature. In the given exercise, we encounter a standard electrode potential of \(-3.04 \text{ V}\) for the lithium reduction reaction. This means that under standard conditions, lithium ions need an additional voltage input to gain electrons and form metallic lithium.

• Standard conditions typically refer to 25°C, 1M concentration for all ions, and 1 atm pressure.
• Electrode potentials provided in tables are standard values and can be used to evaluate the feasibility of electrochemical reactions.
• An understanding of electrode potential allows us to adjust our electrolysis setup for maximum efficiency and effectiveness.

Remember that, to drive the reaction, an external voltage greater than or equal to the electrode potential is required, making it vital for experimental setups.

Efficiency in Electrolysis

Efficiency in electrolysis quantifies how effectively the electrical energy is used to drive the desired chemical reaction. Inefficiencies can arise due to side reactions, heat loss, or operational inefficiencies inherent within the cell.

In our problem, the electrolytic cell is operated at an 85% efficiency, which means that 15% of the electrical energy is lost to factors other than forming lithium. Efficiency is calculated by comparing the actual charged used to the theoretical charge required to deposit the desired amount of substance.

• Real plants strive to maximize efficiency by optimizing cell design and operating conditions.
• Understanding efficiency helps reduce operational costs and increase the economic viability of industrial electrolysis.
• For calculation, we adjusted the total charge by multiplying it with the efficiency (85%). This gives the effective charge used for lithium formation.

Calculating efficiency offers insights into how electrolysis can be improved, making technical adjustments very relevant to practical applications.

Charge Calculation

Charge calculation is a fundamental step to determine how much material will be deposited on the electrode during electrolysis. The charge, denoted as \(Q\), is calculated by multiplying current \(I\) flowing through the electrolysis cell by the time \(t\) for which it flows.

In our case, we calculated charge using \(Q = I \times t\). It's important to convert the time into seconds for correct calculations, since current is typically given in amperes (C/s). Given a massive current of \(7.5 \times 10^{4}\) amps over 24 hours, it results in a substantial charge being passed through the cell.

• This charge determines the amount of lithium produced as it corresponds to the number of electrons available to partake in the redox reaction.
• Accurate charge calculation ensures the precision of predictions regarding the mass of lithium.

Effective charge is an important basis to further determine factors such as cell efficiency and material yield.

Moles and Molar Mass

Moles and molar mass are foundational concepts that describe the quantity of atoms or molecules in chemistry and their mass. In this electrolysis problem, these concepts help us determine how much lithium will form at the cathode.

A mole is a unit that measures the amount of substance and is defined as containing as many particles (atoms, molecules) as there are in 12 grams of carbon-12. Molar mass is the mass of one mole of a substance in grams. For lithium, its molar mass is \(6.94 \text{ g/mol}\), making it easy to convert moles of lithium to mass using the formula \( m = n \times M \).

• This relationship lets us predict the practical yield expected from electrolysis operations.
• Knowing the number of moles involved allows for conversions between atomic scale and bulk measurements.

Ultimately, understanding moles and molar mass enables precise calculations for any chemical process, ensuring a coherent link between theoretical concepts and real-world applications.