Laboratory Activity 1C ( Displacement of Metal from Its Salt Solution):
Aim : To investigate a redox reaction in the displacement of metal from its salt solution.
Materials : Magnesium ribbon, Mg, lead plate, Pb, copper plate, Cu, 0.5 mol dm−3 of lead(II)nitrate, Pb(NO3)2, solution, 0.5 mol dm−3 of magnesium nitrate, Mg(NO3)2, solution and 0.5 mol dm−3 of copper(II) nitrate, Cu(NO3)2 solution.
Apparatus : Test tube, test tube rack and sandpaper.
Procedure:
1. Using the metal plates and salt solutions provided, plan an experiment to investigate a redox reaction in the displacement of metal from its salt solution.
2. Write clear steps of the investigation.
3. Record all observations.
Results:
Construct a suitable table to record all observations and inferences.
Discussion:
1. For each set of experiment:
(a) Write the oxidation half equation, reduction half equation and overall ionic equation.
(b) Identify:
(i) Oxidised substance.
(ii) Reduced substance.
(iii) Oxidising agent.
(iv) Reducing agent.
(c) Give reasons for your answers in (b)(i), (ii), (iii) and (iv).
Answer:
Procedure:
1. Using the sand paper, clean the magnesium ribbons, lead plates and copper plates.
2. Put the magnesium ribbons into two difference test tubes.
3. Pour 0.5 mol dm−3 lead(II) nitrate solution into the first test tube and 0.5 mol dm−3 copper(II) nitrate solution into the second test tube until each magnesium ribbon is immersed.
4. Repeat step 2 using lead plates and copper plates.
5. Pour 0.5 mol dm−3 magnesium(II) nitrate solution into the third test tube and 0.5 mol dm−3 copper(II) nitrate solution into the fourth test tube until each lead plate is immersed.
6. Pour 0.5 mol dm−3 magnesium(II) nitrate solution into the fifth test tube and 0.5 mol dm−3 lead(II) nitrate solution into the sixth test tube until each copper plate is immersed.
7. Put all the test tubes on the test tube rack.
8. Record all observations.
Result:
Table to record all observations and inferences of metal displacement reactions.
Experiment: Mg + Pb(NO3)2
(a) Oxidation half equation: Mg → Mg2+ + 2e−
Reduction half equation: Pb2+ + 2e− → Pb
Overall ionic equation: Mg + Pb2+ → Mg2+ + Pb
(b)(i) Magnesium
(ii) PLead(II) nitrate
(iii) Lead(II) nitrate
(iv) Magnesium
(c)(i) Magnesium atoms lose electrons to form Mg2+ ions.
(ii) Pb2+ ions gain electrons to form lead, Pb atoms.
(iii) Ion Pb2+ ions are the electron acceptors
(iv) Mg is the electron donor
Experiment: Mg + Cu(NO3)2
(a) Oxidation half equation: Mg → Mg2+ + 2e−
Reduction half equation: Cu2+ + 2e− → Cu
Overall ionic equation: Mg + Cu2+ → Mg2+ + Cu
(b)(i) Magnesium
(ii) Copper(II) nitrate
(iii) Copper(II) nitrate
(iv) Magnesium
(c)(i) Magnesium atoms lose electrons to form Mg2+ ions.
(ii) Cu2+ ions gain electrons to form copper, Cu atoms.
(iii) Cu2+ ions are the electron acceptors
(iv) Mg is the electron donor
Experiment: Pb + Cu(NO3)2
(a) Oxidation half equation:: Pb → Pb2+ + 2e−
Reduction half equation: Cu2+ + 2e− → Cu
Overall ionic equation: Pb + Cu2+ → Pb2+ + Cu
(b)(i) Lead
(ii) Copper(II) nitrate
(iii) Copper(II) nitrate
(iv) Lead
(c)(i) Pb atoms lose electrons to form Pb2+ ions.
(ii) Cu2+ ions gain electrons to form copper, Cu atoms.
(iii) Cu2+ ions are the electron acceptors.
(iv) Pb is the electron donor
Aim : To investigate a redox reaction in the displacement of metal from its salt solution.
Materials : Magnesium ribbon, Mg, lead plate, Pb, copper plate, Cu, 0.5 mol dm−3 of lead(II)nitrate, Pb(NO3)2, solution, 0.5 mol dm−3 of magnesium nitrate, Mg(NO3)2, solution and 0.5 mol dm−3 of copper(II) nitrate, Cu(NO3)2 solution.
Apparatus : Test tube, test tube rack and sandpaper.
Procedure:
1. Using the metal plates and salt solutions provided, plan an experiment to investigate a redox reaction in the displacement of metal from its salt solution.
2. Write clear steps of the investigation.
3. Record all observations.
Results:
Construct a suitable table to record all observations and inferences.
Discussion:
1. For each set of experiment:
(a) Write the oxidation half equation, reduction half equation and overall ionic equation.
(b) Identify:
(i) Oxidised substance.
(ii) Reduced substance.
(iii) Oxidising agent.
(iv) Reducing agent.
(c) Give reasons for your answers in (b)(i), (ii), (iii) and (iv).
Answer:
Procedure:
1. Using the sand paper, clean the magnesium ribbons, lead plates and copper plates.
2. Put the magnesium ribbons into two difference test tubes.
3. Pour 0.5 mol dm−3 lead(II) nitrate solution into the first test tube and 0.5 mol dm−3 copper(II) nitrate solution into the second test tube until each magnesium ribbon is immersed.
4. Repeat step 2 using lead plates and copper plates.
5. Pour 0.5 mol dm−3 magnesium(II) nitrate solution into the third test tube and 0.5 mol dm−3 copper(II) nitrate solution into the fourth test tube until each lead plate is immersed.
6. Pour 0.5 mol dm−3 magnesium(II) nitrate solution into the fifth test tube and 0.5 mol dm−3 lead(II) nitrate solution into the sixth test tube until each copper plate is immersed.
7. Put all the test tubes on the test tube rack.
8. Record all observations.
Result:
Table to record all observations and inferences of metal displacement reactions.
Experiment: Mg + Pb(NO3)2
(a) Oxidation half equation: Mg → Mg2+ + 2e−
Reduction half equation: Pb2+ + 2e− → Pb
Overall ionic equation: Mg + Pb2+ → Mg2+ + Pb
(b)(i) Magnesium
(ii) PLead(II) nitrate
(iii) Lead(II) nitrate
(iv) Magnesium
(c)(i) Magnesium atoms lose electrons to form Mg2+ ions.
(ii) Pb2+ ions gain electrons to form lead, Pb atoms.
(iii) Ion Pb2+ ions are the electron acceptors
(iv) Mg is the electron donor
Experiment: Mg + Cu(NO3)2
(a) Oxidation half equation: Mg → Mg2+ + 2e−
Reduction half equation: Cu2+ + 2e− → Cu
Overall ionic equation: Mg + Cu2+ → Mg2+ + Cu
(b)(i) Magnesium
(ii) Copper(II) nitrate
(iii) Copper(II) nitrate
(iv) Magnesium
(c)(i) Magnesium atoms lose electrons to form Mg2+ ions.
(ii) Cu2+ ions gain electrons to form copper, Cu atoms.
(iii) Cu2+ ions are the electron acceptors
(iv) Mg is the electron donor
Experiment: Pb + Cu(NO3)2
(a) Oxidation half equation:: Pb → Pb2+ + 2e−
Reduction half equation: Cu2+ + 2e− → Cu
Overall ionic equation: Pb + Cu2+ → Pb2+ + Cu
(b)(i) Lead
(ii) Copper(II) nitrate
(iii) Copper(II) nitrate
(iv) Lead
(c)(i) Pb atoms lose electrons to form Pb2+ ions.
(ii) Cu2+ ions gain electrons to form copper, Cu atoms.
(iii) Cu2+ ions are the electron acceptors.
(iv) Pb is the electron donor