Question

Elemental calcium is produced by the electrolysis of molten \(\mathrm{CaCl}_{2}\), (a) What mass of calcium can be preduced 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 this process is about \(1.831 \times 10^5\ g\), and the minimum voltage needed to cause electrolysis is \(2.84\ V\).

Step-by-step solution

Part (a) - Finding the mass of calcium produced

1. Calculate the charge (Q) passed through the electrolyte during the given time period. To find the charge, we can use the formula Q = I * t, where I is the current and t is the time in seconds. First, convert the given time (48 hours) to seconds: \( \) \[ t = 48\ h × 60\ \frac{min}{h} × 60\ \frac{s}{min} = 172800\ s \] \( \) Then, calculate the charge: \( \) \[ Q = (7.5 × 10^{3}\ \mathrm{A}) × (172800\ \mathrm{s}) = 1.296 × 10^{9}\ \mathrm{C} \] 2. Determine the number of moles of electrons using Faraday's constant. Faraday's constant, F, is approximately 96485 C per mole of electrons. Therefore, we can find the number of moles of electrons (n) by dividing the total charge (Q) by the Faraday constant (F): \( \) \[ n = \frac{Q}{F} = \frac{1.296 × 10^{9}\ \mathrm{C}}{96485\ \frac{\mathrm{C}}{\mathrm{mol}}} = 1.343 × 10^{4}\ \mathrm{mol} \] \( \) 3. Calculate the moles of calcium produced. In the electrolysis of CaCl2, one mole of calcium is produced by two moles of electrons due to the reaction: \( \) \[ \mathrm{Ca^{2+}}(l) + 2e^{-} → \mathrm{Ca}(l) \] \( \) Therefore, the number of moles of calcium produced is half the number of moles of electrons: \( \) \[ \text{moles of } \mathrm{Ca} = \frac{1}{2} × 1.343 × 10^{4}\ \mathrm{mol} = 6.715 × 10^{3}\ \mathrm{mol} \] \( \) 4. Adjust for cell efficiency. Considering the 68% cell efficiency, the actual moles of calcium produced will be: \( \) \[ \text{actual moles of }\mathrm{Ca} = 0.68 × 6.715 × 10^{3}\ \mathrm{mol} = 4.5662 × 10^{3}\ \mathrm{mol} \] \( \) 5. Convert moles of calcium to mass. Finally, multiply the number of moles by the molar mass of calcium (40.08 g/mol) to obtain the mass of calcium produced during electrolysis. \( \) \[ \text{mass of }\mathrm{Ca} = (4.5662 × 10^{3}\ \mathrm{mol}) × (40.08\ \frac{\mathrm{g}}{\mathrm{mol}}) = 1.831 × 10^{5}\ \mathrm{g} \] \( \) Therefore, the mass of calcium produced by this process is about 1.831 x 10^5 g.

Part (b) - Finding the minimum voltage needed

1. Determine the change in reduction potential of the reaction. The reduction potential for the reduction of Ca2+ to Ca is -2.84 V: \( \) \[ \mathrm{Ca^{2+}}(l) + 2e^{-} → \mathrm{Ca}(l)\ \ E^\circ = -2.84\ \mathrm{V} \] \( \) 2. Calculate the minimum voltage needed for electrolysis. For the electrolysis of CaCl2 to proceed spontaneously, the cell potential (∆E) must be greater than 0: \( \) \[ \Delta E = E_{cathode} - E_{anode} = -2.84\ \mathrm{V} \] \( \) Since the minimum voltage required is equal to the absolute value of ΔE, the minimum voltage required for the electrolysis of CaCl2 is 2.84 V. In summary, the mass of calcium produced by this process is about 1.831 x 10^5 g, and the minimum voltage needed to cause electrolysis is 2.84 V.

Key concepts

Faraday's Constant

Faraday's constant is crucial in electrolysis calculations, offering a direct link between electrical charge and the amount of substance produced at an electrode. It represents the charge of one mole of electrons, approximately 96485 coulombs per mole. This constant allows us to determine how many electrons are involved in a chemical reaction when we know the total electrical charge.

In the context of electrolysis, to find the moles of electrons, we divide the total charge by Faraday's constant. For example, if you find a charge of 1.296 × 10^9 C, dividing this by 96485 C/mol gives the number of moles of electrons involved. This conversion is foundational for predicting the amount of material produced in an electrolytic cell.

Understanding Faraday's constant helps explain why electrical charge is so critical in chemical processes and allows for precise calculations of electroplated materials or produced elements.

Reduction Potential

Reduction potential is a measurement of a substance's tendency to gain electrons, expressed in volts. In any redox reaction, it determines which species will be reduced. Positive reduction potentials indicate a greater tendency to acquire electrons, whereas negative values suggest a lesser tendency.

In our exercise involving calcium electrolysis, the reaction \(\text{Ca}^{2+}(l) + 2e^{-} → \text{Ca}(l)\) has a reduction potential of -2.84 V. This negative value shows that calcium ions are somewhat reluctant to gain electrons, reflecting why external voltage is needed to drive the reaction.

Understanding reduction potentials is crucial when designing cells and predicting reaction feasibility. It ensures that you calculate the correct voltage required for electrolysis, as it's essential that the applied voltage matches or exceeds the reduction potential to move the reaction forward.

Cell Efficiency

Cell efficiency in electrolysis refers to the effectiveness with which electrical energy is converted into chemical energy to produce the desired substance. This efficiency is usually expressed as a percentage and considers all the energy losses within a cell.

In the provided exercise, the electrolytic cell operates at 68% efficiency. This means out of all the energy supplied, only 68% is effectively used to produce calcium. The remaining 32% is lost due to factors like heat dissipation and other inefficiencies within the cell.

When calculating the actual yield, it is crucial to adjust the theoretical amount produced by the cell efficiency percentage. By doing so, the expected output aligns more closely with real-world outcomes, ensuring more accurate predictions in industrial or laboratory settings.