Problem 94 - Chemistry The Central Science In Si Units

Question

Elemental calcium is produced by the electrolysis of molten \(\mathrm{CaCl}_{2}\). (a) What mass of calcium can be produced by this process if a current of \(7.5 \times 10^{3} \mathrm{~A}\) is applied for \(48 \mathrm{~h}\) ? Assume that the electrolytic cell is \(68 \%\) efficient. (b) What is the minimum voltage needed to cause the electrolysis?

Step-by-Step Solution

Answer

(a) Approximately 183.18 kg of calcium can be produced. (b) Minimum voltage needed is slightly above 2.87 V.

Step 1: Determine Charge Passed

To find the total charge, multiply the current by the time for which it is applied. The time, 48 hours, must be converted to seconds.\[\text{Current } (I) = 7.5 \times 10^{3} \text{ A}\]\[\text{Time } (t) = 48 \text{ hours} \times 3600 \text{ s/hr} = 1.728 \times 10^{5} \text{ s}\]Charge (Q) is given by:\[Q = I \times t = 7.5 \times 10^{3} \text{ A} \times 1.728 \times 10^{5} \text{ s} = 1.296 \times 10^{9} \text{ C}\]

Step 2: Calculate Moles of Electrons Transferred

Convert the total charge into moles of electrons using Faraday's constant (approximately 96485 C/mol).\[n_e (\text{moles of electrons}) = \frac{Q}{F} = \frac{1.296 \times 10^9 \text{ C}}{96485 \text{ C/mol}} \approx 13434 \text{ mol}\]

Step 3: Relate Moles of Electrons to Moles of Calcium

According to the reaction: \[\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}\]Two moles of electrons are needed for each mole of calcium produced. Therefore, calculate the moles of calcium:\[n_{\text{Ca}} = \frac{n_e}{2} = \frac{13434}{2} \approx 6717 \text{ mol}\]

Step 4: Calculate Mass of Calcium Produced

Convert moles of calcium to mass, using the molar mass of calcium (40.08 g/mol). Also account for the efficiency.\[\text{Mass of } \mathrm{Ca} = 6717 \text{ mol} \times 40.08 \text{ g/mol} \times 0.68 \approx 183,176 \text{ g} \approx 183.18 \text{ kg}\]

Step 5: Determine Minimum Voltage for Electrolysis

The minimum voltage required for electrolysis is slightly more than the standard reduction potential, which is approximately -2.87 V for the reduction of \(\text{Ca}^{2+}\) to \(\text{Ca}\). This value may need to be adjusted for overpotentials and other conditions beyond the scope of this calculation.Therefore, the minimum theoretical voltage needed to cause electrolysis is slightly above 2.87 V.

Key Concepts

Faraday's Laws of Electrolysis

Faraday's laws of electrolysis are key to understanding the relationship between the electric charge passed through the electrolyte and the amount of substance that undergoes chemical change. These laws are vital in predicting the outcomes of electrolysis:

1. **First Law**: The amount of substance that is altered or deposited at an electrode is directly proportional to the quantity of electric charge passed through the electrolyte. In simpler terms, the more electricity you pass, the more of the substance you change.

2. **Second Law**: The amounts of different substances liberated by the same quantity of electricity passing through the electrolyte are proportional to their equivalent weights. This means each element will react based on its specific electronic requirements.

In the calcium production example, Faraday's laws help us translate the current and duration of electrolysis into a specific amount of deposited calcium. Here, we used these principles to relate the charge passed (from the current and time) to the moles of electrons transferred and, eventually, the moles of calcium produced.

Current and Charge

Current and charge are essential concepts in electrolysis, playing a critical role in how much chemical change occurs during the process. Current, measured in amperes (A), is the rate of flow of electric charge in a circuit. Charge, measured in coulombs (C), is the total quantity of electricity transported.

Using the formula for charge \(Q = I \times t\), where \(Q\) is the charge, \(I\) is the current, and \(t\) is the time in seconds, we can calculate the total charge passed during electrolysis. This calculation forms the foundation for determining how much of a compound will react or be deposited through electrolysis. For the task at hand, a significant charge is passed over 48 hours to produce elemental calcium from \(\mathrm{CaCl}_{2}\).

Moles and Molar Mass

In electrolysis calculations, converting between charge and moles of electrons using Faraday's constant is crucial. One mole of electrons ( _{e^-} ) equates approximately to 96485 coulombs of charge—this is Faraday's constant.

From the charge calculated by the current and time, we determine the moles of electrons, which is critical to finding how much of a particular substance is produced or consumed. The relationship further extends to moles of the substance formed, guided by the stoichiometry of the reaction.

For example, in this case, each mole of calcium requires two moles of electrons. Once we have moles of calcium, we multiply by its molar mass (40.08 g/mol) to convert this into mass. This direct proportionality helps in precise calculation needed for the accurate amounts of production in electrolysis processes.

Efficiency in Electrolysis

Efficiency is a measure of how much of the electrical energy is converted into the expected chemical change. No electrolysis process is perfectly efficient due to energy losses, side reactions, or incomplete reactions.

In the example provided, we apply an efficiency rate of 68%, meaning that only that portion of the calculated theoretical production actually materializes as usable product. This efficiency impacts the final mass of calcium we can expect to produce compared to the theoretical calculations.

A key aspect to improve efficiency includes optimizing the electrode material and conditions to minimize overriding losses and ensure maximum product formation from the input current.

Standard Reduction Potential

Standard reduction potential is an indicator of the tendency of ions to be reduced or gain electrons, which is fundamental for predicting the feasibility of the electrolysis.

Each electrochemical reaction has a standard reduction potential measured in volts (V), which represents the voltage required under standard conditions to drive the reaction.

For calcium, the standard reduction potential is around -2.87 V for the reduction of \(\text{Ca}^{2+}\) to \(\text{Ca}\). However, practical electrolysis requires a bit more voltage than this theoretical minimum to overcome barriers and losses, known as overpotentials. This added voltage ensures the electrolysis proceeds efficiently under the specific operating conditions of the electrolytic cell.