Laboratory Activity 1E (Electrolysis of Molten Lead(II) bromide, PbBr2):
Aim: To study the electrolysis of molten lead(II) bromide, PbBr2.
Material: Lead(II) bromide powder, PbBr2.
Apparatus: Beaker, crucible, light bulb, battery, carbon electrodes, connecting wire with crocodile clip, switch, tripod stand, pipe clay triangle and Bunsen burner.
Procedure:
1. Fill a crucible with lead(II) bromide powder, PbBr2 until half full.
2. Place the crucible on top of the pipe clay triangle at the tripod stand.
3. Connect the carbon electrodes to the battery, light bulb and switch using the connecting wires as shown in Figure 1.24.
4. Heat the lead(II) bromide powder, PbBr2 until it melts.
5. Dip the carbon electrodes, C into the molten lead(II) bromide, PbBr2 and turn on the switch to complete the circuit.
6. Observe and record any changes that occur at the anode.
7. After 5 minutes, turn off the switch and carefully pour the molten lead(II) bromide, PbBr2 into a beaker.
8. Observe the substances that are formed at the cathode and anode and record your observation.
Results:
Record all observations in the following table.
Discussion:
1. State the name of the ions that move to the cathode and anode during electrolysis.
2. Write the half equations for the reactions at the:
(a) cathode.
(b) anode.
3. Identify the products formed at the cathode and anode.
4. Explain how the products at the cathode and anode are formed.
5. Write the overall ionic equation that represents the electrolysis of molten lead(II) bromide, PbBr2.
Answer:
1. Lead(II) ions move to the cathode and bromide ions move to the anode.
2.
(a) Pb2+ + 2e− → Pb
(b) 2Br− → Br2 + 2e−
3.
Cathode : Lead
Anode : Bromine
4.
Cathode : A lead(II) ion gains two electrons to form a lead atom.
Anode : A bromide ion loses one electron to form a bromine atom. Two bromine atoms combine to form a bromine molecule. (OR two bromide ions lose two electrons to form a bromine molecule)
5. Pb2+ + 2Br− → Pb + Br2
Aim: To study the electrolysis of molten lead(II) bromide, PbBr2.
Material: Lead(II) bromide powder, PbBr2.
Apparatus: Beaker, crucible, light bulb, battery, carbon electrodes, connecting wire with crocodile clip, switch, tripod stand, pipe clay triangle and Bunsen burner.
Procedure:
1. Fill a crucible with lead(II) bromide powder, PbBr2 until half full.
2. Place the crucible on top of the pipe clay triangle at the tripod stand.
3. Connect the carbon electrodes to the battery, light bulb and switch using the connecting wires as shown in Figure 1.24.
4. Heat the lead(II) bromide powder, PbBr2 until it melts.
5. Dip the carbon electrodes, C into the molten lead(II) bromide, PbBr2 and turn on the switch to complete the circuit.
6. Observe and record any changes that occur at the anode.
7. After 5 minutes, turn off the switch and carefully pour the molten lead(II) bromide, PbBr2 into a beaker.
8. Observe the substances that are formed at the cathode and anode and record your observation.
Results:
Record all observations in the following table.
Discussion:
1. State the name of the ions that move to the cathode and anode during electrolysis.
2. Write the half equations for the reactions at the:
(a) cathode.
(b) anode.
3. Identify the products formed at the cathode and anode.
4. Explain how the products at the cathode and anode are formed.
5. Write the overall ionic equation that represents the electrolysis of molten lead(II) bromide, PbBr2.
Answer:
1. Lead(II) ions move to the cathode and bromide ions move to the anode.
2.
(a) Pb2+ + 2e− → Pb
(b) 2Br− → Br2 + 2e−
3.
Cathode : Lead
Anode : Bromine
4.
Cathode : A lead(II) ion gains two electrons to form a lead atom.
Anode : A bromide ion loses one electron to form a bromine atom. Two bromine atoms combine to form a bromine molecule. (OR two bromide ions lose two electrons to form a bromine molecule)
5. Pb2+ + 2Br− → Pb + Br2