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 h. Assume the electrolytic cell is \(85 \%\) efficient. (b) What is the minimum voltage required to drive the reaction?

Short answer

After applying Faraday's laws and accounting for the 85% efficiency, the mass of Li formed is calculated as \(1.98 \times 10^3\) g. The minimum voltage required to drive the reaction is 2.32 V.

Step-by-step solution

1. Calculate the total charge passed through the cell

To calculate the total charge passed through the cell, we will use the formula: Total charge (Q) = Current (I) x Time (t) The given current is \(7.5 \times 10^4\) A and the time is 24 hours. We need to convert the time into seconds before plugging the values into the formula. 1 hour = 3600 seconds So, 24 hours = 24 x 3600 = 86,400 seconds Now, we can calculate the total charge: Q = \(7.5 \times 10^4\) A x 86,400 s

2. Calculate the number of moles of electrons

Once we have the total charge, we can calculate the number of moles of electrons using Faraday's constant (F = 96,485 C/mol): Moles of electrons (n) = Total charge (Q) / Faraday's constant (F) n = Q / 96,485 C/mol

3. Apply the electrolytic cell efficiency

Given that the cell is 85% efficient, we now need to account for the efficiency in our calculations: Moles of electrons (n') = Moles of electrons (n) x 0.85

4. Calculate the moles of Lithium formed

Lithium is formed by the reduction of Li+ ions, which requires one mole of electrons for each mole of Li+ ions: Moles of Li formed = Moles of electrons (n')

5. Calculate the mass of Lithium formed

Now that we have the moles of Li formed, we can calculate the mass by using the molar mass of Li (6.94 g/mol): Mass of Li = Moles of Li x Molar mass of Li Mass of Li = Moles of Li x 6.94 g/mol

6. Calculate the minimum voltage required

For part (b), we're asked to find the minimum voltage required to drive the reaction. We can find this by using the standard electrode potentials of the half-reactions which are: Li+ + e- -> Li (E° = -3.04 V) 2Cl- -> Cl2 + 2e- (E° = +1.36 V) To calculate the minimum voltage required for the cell, we can use the formula: E°cell = E°cathode - E°anode E°cell = E°(Li) - E°(Cl) Now we can plug in the values to find the minimum voltage required for the electrolysis of molten LiCl.

Key concepts

Electrolysis

Electrolysis is a chemical process where electrical energy is used to drive a non-spontaneous chemical reaction. It's like charging a battery, but instead, we are causing a chemical change. Specifically, for the electrolysis of molten lithium chloride (LiCl), electricity is used to break it down into lithium metal (Li) and chlorine gas (Cl₂). This process happens in a special device called an electrolytic cell.
In the electrolytic cell, positive lithium ions (Li⁺) gain electrons at the cathode (negative electrode) to become lithium metal. Chloride ions (Cl⁻) lose electrons at the anode (positive electrode) to form chlorine gas. It's essential for us to understand these roles, as they highlight the movement and transformation of ions under the effect of an external electric current. Thus, electrolysis offers an insightful look into how electrical and chemical energies can be interconverted.

Faraday's constant

Faraday's constant is a crucial part of electrochemistry and is represented by the symbol F. It's approximately equal to 96,485 coulombs per mole. This number tells us how much charge, in terms of electrons, is needed to reduce or oxidize one mole of ions. Think of it as a bridge between electricity and chemistry because it links the amount of electrical charge to the number of particles involved in a chemical reaction.
When we calculate the number of moles of electrons that participated in the reaction, we use the formula:

• Number of moles of electrons = Total charge / Faraday's constant

This fundamental relationship helps us understand how chemical changes in electrolysis equate to measurable electrical changes. With Faraday's constant, we decode how much electricity is needed to produce a certain amount of substance, such as lithium in our exercise example.

Standard electrode potential

The standard electrode potential (E°) indicates how likely a species is to gain or lose electrons. It's a measure of the driving force behind a redox reaction and is expressed in volts (V). Each substance involved in our reaction has its own E° value, calculated under standard conditions (1M concentration, 1 atm, at 25°C).
In electrochemistry, the potential influences the direction and magnitude of electronic flow. In our exercise, the equation Li⁺ + e⁻ → Li has a standard potential of -3.04 V, indicating it's a reduction process that occurs at the cathode. Conversely, the oxidation process at the anode, 2Cl⁻ → Cl₂ + 2e⁻, has a potential of +1.36 V.
Knowing these values helps determine the overall cell potential needed to keep the reaction going. We calculate the minimum voltage required for the reaction by finding the difference between these potentials:

• E°cell = E°(cathode) - E°(anode)

This calculation provides the minimum voltage necessary for the electrolysis process to proceed.

Current and charge calculations

In electrochemistry, understanding the relationship between current, time, and charge is essential for quantifying reactions. Current, denoted as I, measures the flow of electric charge in amperes (A). Charge, represented by Q, measures the total electricity passed, in coulombs (C).
Using the formula

• Total charge (Q) = Current (I) x Time (t)

we can determine how much charge has passed through an electrolytic cell. Since time is often given in hours, it's crucial to convert it to seconds (1 hour = 3600 seconds) for calculations involving coulombs, as seen in the exercise where 24 hours equals 86,400 seconds.
Understanding this formula is pivotal. It allows us to compute how much electrical energy drives the electrochemical reactions. By knowing the total charge, alongside other parameters like efficiency and Faraday's constant, we can determine the amounts of products like lithium formed in electrolysis.