Question

Electrolysis of a solution of \(\mathrm{CuSO}_{4}(\mathrm{aq})\) to give copper metal is carried out using a current of 0.66 A. How long should electrolysis continue to produce \(0.50 \mathrm{g}\) of copper?

Short answer

Electrolysis should continue for about 38.34 minutes to produce 0.50 g of copper.

Step-by-step solution

Determine Molar Mass of Copper

Copper's molar mass is required for the conversion of mass to moles. The atomic mass of copper (Cu) is approximately 63.55 g/mol.

Convert Mass of Copper to Moles

Use the formula for moles: \[moles = \frac{\text{mass of copper}}{\text{molar mass of copper}} = \frac{0.50 \text{ g}}{63.55 \text{ g/mol}} \approx 0.00787 \text{ mol} \]

Calculate Number of Moles of Electrons Required

In the reduction process, each copper ion (Cu²⁺) gains 2 electrons to become copper metal (Cu). So, the moles of electrons required will be twice the moles of copper.\[moles\ of\ electrons = 2 \times 0.00787 \text{ mol} = 0.01574 \text{ mol} \]

Calculate Total Charge Needed

Use the relation between moles of electrons and charge: Faraday's constant (F) is approximately 96485 C/mol. Multiply the moles of electrons by Faraday's constant to get the total charge:\[total\ charge = 0.01574 \text{ mol} \times 96485 \text{ C/mol} \approx 1518.85 \text{ C} \]

Use Current to Find Time

Use the formula relating current, charge, and time. Current (I) is 0.66 A, and charge (Q) is 1518.85 C:\[I = \frac{Q}{t} \Rightarrow t = \frac{Q}{I} = \frac{1518.85 \text{ C}}{0.66 \text{ A}} \approx 2300.68 \text{ seconds} \]

Convert Time to Minutes

Convert the time from seconds to minutes:\[time\ in\ minutes = \frac{2300.68\ seconds}{60\ seconds/minute} \approx 38.34\ minutes \]

Key concepts

Copper Molar Mass

Understanding the molar mass of copper is crucial when working with electrolysis. Copper, symbolized as Cu, has an atomic mass of about 63.55 grams per mole. This means that every mole of copper atoms has a mass of 63.55 grams. - Molar mass helps convert the mass of a substance to the amount of substance in moles. - In electrolysis, knowing the molar mass allows us to determine how many moles of copper are formed or used. When performing calculations like converting 0.50 grams of copper to moles, we use the formula:\[ \text{moles of copper} = \frac{\text{mass of copper}}{\text{molar mass of copper}} \]In this case, the calculation is \( \frac{0.50 \text{ g}}{63.55 \text{ g/mol}} \), yielding approximately 0.00787 moles of copper. Grasping this concept ensures the correct usage of copper in electrolysis processes.

Faraday's Constant

Faraday's constant is a fundamental element in electrochemistry, representing the total electric charge carried by one mole of electrons. It is approximately 96485 coulombs per mole. This constant is named after Michael Faraday, who played a significant role in studying electromagnetism and electrochemistry. - Faraday's constant is used to calculate the charge needed to deposit or dissolve a specific amount of a substance during electrolysis.- It relates the moles of electrons exchanged in a chemical reaction to the total electric charge involved.In our specific exercise, we calculated that copper requires two electrons per copper ion to reduce to copper metal:\[ \text{moles of electrons} = 2 \times 0.00787 \text{ mol} = 0.01574 \text{ mol} \]Then, by multiplying the moles of electrons by Faraday's constant, we find the total charge needed:\[ \text{total charge} = 0.01574 \text{ mol} \times 96485 \text{ C/mol} \approx 1518.85 \text{ C} \] Understanding Faraday's constant allows us to connect the world of chemical reactions to practical electrical applications.

Current and Charge Relationship

In electrolysis, the relationship between current, charge, and time is fundamental. Current, measured in amperes (A), represents the flow of electric charge. Charge, measured in coulombs (C), is the total amount of electricity that passes through a circuit. - The formula that relates them is: \(I = \frac{Q}{t}\).- Here, \(I\) is the current, \(Q\) is the charge, and \(t\) is the time for which the current flows.In our scenario, we determined the total charge required as 1518.85 C. Given the current of 0.66 A, we placed these values into the formula to solve for time:\[ t = \frac{Q}{I} = \frac{1518.85 \text{ C}}{0.66 \text{ A}} \approx 2300.68 \text{ seconds} \]Finally, converting seconds to minutes provides a more digestible time frame:\[ \text{time in minutes} = \frac{2300.68 \text{ seconds}}{60 \text{ seconds/minute}} \approx 38.34 \text{ minutes} \] Grasping this relationship aids in understanding how much time is needed to complete an electrolysis process, making it a crucial concept in electrochemistry education.