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?

Short answer

The mass of calcium produced by the electrolysis of molten \(\mathrm{CaCl}_{2}\) with a current of \(7.5 \times 10^{3}\) A applied for \(48\) hours and a cell efficiency of \(68\%\) is approximately \(7.67 \times 10^{4}\) g. The minimum voltage needed to cause electrolysis is equal to the standard reduction potential of calcium, which is \(-2.87\) V.

Step-by-step solution

(Step 1: Calculate the total charge passed through the electrolyte)

Use the formula: charge = current × time. The total charge, \(Q\), passed through the electrolytic cell is given by: $$ Q = I \cdot t $$ Where \(I = 7.5 \times 10^{3} A\) is the current and \(t = 48 h\) is the time. We will convert the time to seconds so that the units are consistent: $$ t = 48 h \cdot \frac{3600 s}{1 h} $$ Hence, calculate the total charge \(Q\).

(Step 2: Calculate the number of moles of electrons passed)

To find the number of moles of electrons we can use Faraday's constant, \(F = 96,485~C/mol\): $$ n(\text{electrons}) = \frac{Q}{F} $$ Calculate the number of moles of electrons by dividing the total charge \(Q\) obtained in Step 1 by Faraday's constant.

(Step 3: Find the number of moles of Calcium produced)

In the electrolysis process, calcium is produced from its ion, \(\mathrm{Ca}^{2+}\). The balanced half-reaction for the process is: $$ \mathrm{Ca}^{2+} + 2e^{-} \rightarrow \mathrm{Ca (s)} $$ This reaction implies that for every 2 moles of electrons, 1 mole of calcium is produced. Therefore, we can calculate the number of moles of calcium produced as: $$ n(\text{Ca}) = \frac{1}{2} n(\text{electrons}) $$ Calculate the number of moles of calcium produced using the number of moles of electrons obtained in Step 2.

(Step 4: Consider cell efficiency and calculate the actual moles of Calcium produced)

The problem states that the efficiency of the electrolytic cell is \(68\%\). To find the actual moles of calcium produced, multiply the number of moles of calcium obtained in Step 3 by the cell efficiency: $$ n_{\text{actual}}(\text{Ca}) = 0.68 \cdot n(\text{Ca}) $$

(Step 5: Calculate the mass of Calcium produced)

The mass of calcium produced can be calculated using the molar mass of calcium (\(M_{\text{Ca}} = 40.08 g/mol\)): $$ m_{\text{actual}}(\text{Ca}) = n_{\text{actual}}(\text{Ca}) \times M_{\text{Ca}} $$ Calculate the actual mass of calcium produced.

(Step 6: Identify the reduction potential for Calcium)

To answer part (b) of the exercise, we need to determine the minimum voltage needed for the electrolysis. Consider the reduction half-reaction for calcium: $$ \mathrm{Ca}^{2+} + 2e^{-} \rightarrow \mathrm{Ca (s)} $$ The standard reduction potential for this half-reaction, \(E^{\circ}\), is \(-2.87~V\). Since voltage is an intensive property, it does not depend on efficiency. Therefore, the minimum voltage needed to cause electrolysis is equal to the standard reduction potential \(E^{\circ}\), which is \(-2.87~V\).

Key concepts

Faraday's Constant

Faraday's constant is a fundamental value that plays a crucial role in electrochemistry, particularly in electrolysis calculations. It represents the magnitude of electric charge carried by one mole of electrons and is approximately equal to 96,485 Coulombs per mole (C/mol). This constant is essential when converting between moles of electrons and charge.

• The constant allows us to calculate the number of moles of electrons, using the relationship: \( n(\text{electrons}) = \frac{Q}{F} \), where \( Q \) is the total charge.
• In electrolysis, knowing the moles of electrons helps determine the amount of substance produced or consumed in the reaction.

For example, during the electrolysis of calcium chloride, calculating the number of moles of electrons transferred is critical for determining how much elemental calcium is produced. The foundation laid by Faraday’s constant simplifies the calculations immensely, enabling a clear understanding of the process at work.

Reduction Potential

The concept of reduction potential is vital in understanding electrolysis processes, as it defines the tendency of a chemical species to acquire electrons and undergo reduction. It is measured in volts (V) and can be found in electrochemical series tables for many elements.

• A reduction potential value tells us how easily a molecule, ion, or atom gains electrons.
• In the context of our exercise, the reduction potential of calcium ion (\( \mathrm{Ca}^{2+} \)) is \(-2.87~V\).
• This negative value indicates that calcium is a strong reducing agent, meaning it is less likely to gain electrons without an external energy input, such as an external voltage in electrolysis.

To perform an electrolysis reaction, the voltage applied to the system must exceed the reduction potential to ensure electron flow. This requirement is known as the threshold voltage. In electrolysis, this is the minimum energy needed to drive the non-spontaneous chemical reaction.

Cell Efficiency

Cell efficiency is a measure of how well an electrolytic cell converts electrical energy into chemical energy during electrolysis. It reflects the effectiveness of the cell in achieving the desired chemical conversion, by accounting for energy losses occurring during the process.

• Efficiency is typically expressed as a percentage, indicating what fraction of the applied energy directly contributes to forming the product.
• For our specific problem, the cell efficiency is 68%, which means that only 68% of the energy is used in actually producing calcium from calcium ions.
• This efficiency factor must be considered when calculating the actual output of calcium, as it reduces the theoretical yield to a practical amount.

To calculate the actual mass of calcium produced, apply the efficiency percentage to the theoretical yield. This way, efficiency provides a more realistic insight into the performance of the electrolytic cell, allowing for more accurate predictions of product formation.