Question

Copper is purified by electrolysis (see Figure 23.6). A \(5.00-\mathrm{kg}\) anode is used in a cell where the current is \(37.8 \mathrm{~A}\). How long (in hours) must the current run to dissolve this anode and electroplate it onto the cathode?

Short answer

It takes approximately 55.78 hours to dissolve and electroplate the anode.

Step-by-step solution

Identify Key Information

We have a copper anode weighing 5.00 kg and a current of 37.8 A. Our goal is to find out how long it takes to dissolve and electroplate this copper.

Convert Mass to Moles

Determine the molar mass of copper, which is approximately 63.55 g/mol. Convert the mass of the anode to grams: 5.00 kg = 5000 g. Calculate the moles of copper: \[ \text{moles} = \frac{5000 \, \text{g}}{63.55 \, \text{g/mol}} \approx 78.68 \, \text{moles}. \]

Faraday's Laws of Electrolysis

Use the fact that 1 mole of electrons is required to deposit 1 mole of copper. The charge of 1 mole of electrons (Faraday's constant \(F\)) is approximately 96,485 C/mol.

Calculate Total Charge Required

Find the total charge needed to dissolve the copper: \[ Q = 78.68 \, \text{moles} \times 96,485 \, \text{C/mol} \approx 7,591,349 \, \text{C}. \]

Relate Charge, Current, and Time

Use the formula for current \(I = \frac{Q}{t}\) to solve for time: \[ t = \frac{Q}{I} = \frac{7,591,349 \, \text{C}}{37.8 \, \text{A}} \approx 200,822 \, \text{seconds}. \]

Convert Time to Hours

Convert the time from seconds to hours: \[ \text{hours} = \frac{200,822 \, \text{seconds}}{3600 \, \text{seconds/hour}} \approx 55.78 \, \text{hours}. \]

Key concepts

Faraday's laws

Faraday's laws of electrolysis are foundational principles governing the amount of substance that is transformed during electrolysis. They are crucial in predicting and calculating the mass of an element that can be deposited or dissolved at an electrode. These laws are named after Michael Faraday, who laid down the quantitative laws of electrolysis.

• First Law: The amount of substance that undergoes a chemical change at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) that passes through the electrolyte.
• Second Law: When the same quantity of electricity passes through different substances, the masses of these substances are proportional to their chemical equivalent weights.

Faraday's constant, denoted as \( F \), is integral to these laws. It represents the charge of one mole of electrons, roughly 96,485 coulombs. In electrolysis, this constant helps to bridge the relationship between the charge passed through the system and the resulting chemical changes. In other words, to deposit or dissolve one mole of a substance, one mole of electrons is involved, facilitating precise calculations in electrolysis processes.

molar mass of copper

The molar mass of copper is a key factor in determining how much copper you have in a given sample and plays a central role in electrochemical calculations. The molar mass of copper is approximately 63.55 grams per mole. This means that one mole of copper atoms weighs 63.55 grams.To perform electrolysis calculations, we first need to convert the mass of copper into moles using the molar mass. For instance, if you have 5.00 kg of copper (which is 5000 grams), you can find out how many moles of copper this corresponds to by dividing by the molar mass:\[ ext{moles} = \frac{5000 ext{ g}}{63.55 ext{ g/mol}} \approx 78.68 ext{ moles}. \]This conversion is essential because electrochemical reactions function at the molar level, relying on Avogadro's number and the concept of moles to predict and measure chemical changes effectively. Thus, understanding and calculating the molar mass provides a solid foundation for further electrolysis calculations.

anode and cathode

In electrolysis, the anode and cathode are the two electrodes where oxidation and reduction reactions occur, essential to understanding the process of electrolysis.

• Anode: This is the electrode where oxidation occurs. In the context of copper purification, the anode is made of impure copper. During electrolysis, copper atoms lose electrons and dissolve into the electrolyte as copper ions. This contributes to the reduction of the anode's mass.
• Cathode: This is the electrode where reduction occurs. During electrolysis, copper ions in the electrolyte gain electrons at the cathode and are deposited as pure copper. This process allows for the collection of copper in its elemental form, purifying it in the process.

The interaction between the anode and cathode and their roles in oxidation and reduction are fundamental to the success of electrochemical applications like metal plating and purification. The energy and current supplied to the system drive these reactions, allowing for deliberate control over the chemical changes involved. Understanding the distinct reactions at each electrode is key to manipulating and optimizing electrolysis for various practical purposes.