Keywords: Electrolysis, Copper Sulfate, Copper Metal, IGCSE, Prescribed Practical, Electrolytic cell, Anode, Cathode

Introduction: In this practical investigation, we will be using electrolysis to extract copper metal from a copper sulfate solution. Electrolysis is a process in which an electric current is passed through an electrolytic cell, causing a chemical reaction to occur at the anode and cathode. In this case, the copper ions in the copper sulfate solution will be reduced at the cathode to form copper metal, while the sulfate ions will be oxidized at the anode to form oxygen gas.

Equipment:

• Copper sulfate solution (0.1M)
• Copper electrodes
• Power supply
• Beaker (250mL)
• Voltmeter
• Ammeter
• Crocodile clips
• Filter funnel and paper
• Balance
• Distilled water

Method:

1. Weigh two copper electrodes and record their masses.
2. Pour the copper sulfate solution into a beaker and insert the two copper electrodes into the solution.
3. Connect the positive end of the power supply to one of the electrodes and the negative end to the other electrode.
4. Attach the ammeter and voltmeter to the circuit.
5. Turn on the power supply and adjust the voltage to 3V.
6. Observe the electrodes and the solution.
7. Record the time and current readings every 5 minutes.
8. After 30 minutes, turn off the power supply and remove the electrodes from the solution.
9. Wash the electrodes with distilled water and dry them using a filter funnel and paper.
10. Weigh the electrodes and record their masses.
11. Calculate the mass of copper deposited on the cathode using the difference in masses of the electrodes before and after electrolysis.

Calculations and Expected Findings: The expected findings from this investigation are that copper ions will be reduced at the cathode to form copper metal, while the sulfate ions will be oxidized at the anode to form oxygen gas. The mass of copper deposited on the cathode can be calculated using the difference in masses of the electrodes before and after electrolysis.

Conclusion: The results of this investigation show that electrolysis can be used to extract copper metal from a copper sulfate solution. The amount of copper extracted can be calculated using the mass of the electrodes before and after electrolysis. This process has important industrial applications, such as in the production of pure copper for electrical wiring.

Questions:

1. What is electrolysis?
2. What happens at the anode and cathode during electrolysis?
3. What is the purpose of using copper electrodes in this experiment?
4. What is the expected product of the oxidation reaction?
5. How can the mass of copper deposited on the cathode be calculated?

Answers:

1. Electrolysis is a process in which an electric current is passed through an electrolytic cell, causing a chemical reaction to occur at the anode and cathode.
2. At the cathode, copper ions are reduced to form copper metal, while at the anode, sulfate ions are oxidized to form oxygen gas.
3. Copper electrodes are used because they are good conductors of electricity and are not affected by the electrolysis process.
4. The expected product of the oxidation reaction is oxygen gas.
5. The mass of copper deposited on the cathode can be calculated using the difference in masses of the electrodes before and after electrolysis.

Extension: Measuring the mass of electrolysis products.

To calculate the number of coulombs of charge in a circuit, you need to know the current flowing through the circuit and the time for which the current flows. The formula to calculate the number of coulombs of charge (Q) is:

Q = I × t

where Q is the charge in coulombs (C), I is the current in amperes (A), and t is the time in seconds (s).

For example, if a current of 2 amperes flows through a circuit for 2500 seconds, the number of coulombs of charge that passes through the circuit can be calculated as:

Q = I × t

= 2 A × 2500 s

= 5000 C

Therefore, the number of coulombs of charge that passes through the circuit in 10 seconds at a current of 2 amperes is 20 C.

Suppose during electrolysis, a total charge of 5000 C passed through the circuit. Given that the number of moles of electrons transferred is equal to the total charge passed divided by the Faraday constant (F = 96500 C/mol), we can calculate the number of moles of electrons transferred:

Number of moles of electrons transferred = 5000 C ÷ 96500 C/mol = 0.0518 mol

Since the reduction of each copper ion requires the transfer of two electrons, the number of moles of copper deposited can be calculated by dividing the number of moles of electrons transferred by 2:

Number of moles of copper deposited = 0.0518 mol ÷ 2 = 0.0259 mol

Finally, we can use the molar mass of copper (63.55 g/mol) to calculate the mass of copper deposited:

Mass of copper deposited = Number of moles of copper deposited × Molar mass of copper = 0.0259 mol × 63.55 g/mol = 1.65 g

Therefore, the mass of copper deposited on the cathode during electrolysis is 1.65 grams.

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