Laboratory Activity 1A – Transfer of Electrons at a Distance:
Aim: To investigate oxidation and reduction reactions in terms of electron transfer at a distance.
Materials: 1.0 mol dm−3 of sulphuric acid, H2SO4, 0.2 mol dm−3 of iron(II) sulphate, FeSO4 solution (freshly prepared), 0.1 mol dm−3 of acidified potassium manganate(VII), KMnO4 solution and 0.5 mol dm−3 of sodium hydroxide, NaOH solution.
Apparatus: U-tube, connecting wires with crocodile clips, galvanometer, retort stand, carbon electrodes, dropper and test tube.
1. Pour 1.0 mol dm−3 of sulphuric acid, H2SO4 into the U-tube until half full and clamp it vertically.
2. Carefully, pour 0.2 mol dm−3 of iron(II) sulphate, FeSO4 solution into arm X of the U-tube using a dropper until the solution reaches the height of 3 cm.
3. Carefully, pour 0.1 mol dm−3 of acidified potassium manganate(VII), KMnO4 solution into arm Y of the U-tube using a dropper until the solution reaches the height of 3 cm.
4. Connect the carbon electrodes to the galvanometer using the connecting wires.
5. Dip one of the carbon electrodes into the iron(II) sulphate, FeSO4 solution while the other carbon electrode into the acidified potassium manganate(VII), KMnO4 solution to complete the circuit as shown in Figure 1.2.
6. Observe the direction of the deflection of the galvanometer needle and determine the positive and negative terminals for each electrode.
7. Leave the apparatus set-up for 30 minutes.
8. Observe the colour change of the iron(II) sulphate, FeSO4 solution and the acidified potassium manganate(VII), KMnO4 solution.
9. Record your observation in the table below.
10. After 30 minutes, draw out a little of the iron(II) sulphate, FeSO4 solution using a dropper and pour it into a test tube.
11. Add sodium hydroxide, NaOH solution into the test tube to verify the product formed.
Data and observation:
Discussion:
1. Based on your observation, write the half equations for electrode X and electrode Y.
2. What are the types of reactions that take place at electrode X and electrode Y?
3. Write the overall ionic equation for the reaction that takes place.
4. Identify the oxidised substance, reduced substance, oxidising agent and reducing agent.
Give reasons for your answers.
5. State the direction of the electron flow in the experiment.
6. State the positive terminal and negative terminal of the experiment.
7. What is the function of sulphuric acid, H2SO4?
8. Other than using the U-tube, draw a labelled set-up of apparatus that can be used for the experiment to investigate the transfer of electrons at a distance between potassium iodide, KI solution and chlorine water, Cl2.
Answer:
1. X : 2I− → I2 + 2e
Y : MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
2. X : Oxidation
Y : Reduction
3. 10I− + 2MnO4− + 16H+ → 5I2 + 2Mn2+ + 8H2O
4.
– I− ions are oxidised because they lose electrons.
– MnO4− ions are reduced because they gain electrons.
– Oxidising agent is MnO4− ions because MnO4− ions are the electron acceptors.
– Reducing agent is I− ions because I− ions are the electron donors.
5. Electrons move from X electrode to Y electrode through the connecting wires.
6. Positive terminal is Y electrode and negative terminal is X electrode.
7. To allow the transfer of ions and to complete the curcuit.
8.
Aim: To investigate oxidation and reduction reactions in terms of electron transfer at a distance.
Materials: 1.0 mol dm−3 of sulphuric acid, H2SO4, 0.2 mol dm−3 of iron(II) sulphate, FeSO4 solution (freshly prepared), 0.1 mol dm−3 of acidified potassium manganate(VII), KMnO4 solution and 0.5 mol dm−3 of sodium hydroxide, NaOH solution.
Apparatus: U-tube, connecting wires with crocodile clips, galvanometer, retort stand, carbon electrodes, dropper and test tube.
1. Pour 1.0 mol dm−3 of sulphuric acid, H2SO4 into the U-tube until half full and clamp it vertically.
2. Carefully, pour 0.2 mol dm−3 of iron(II) sulphate, FeSO4 solution into arm X of the U-tube using a dropper until the solution reaches the height of 3 cm.
3. Carefully, pour 0.1 mol dm−3 of acidified potassium manganate(VII), KMnO4 solution into arm Y of the U-tube using a dropper until the solution reaches the height of 3 cm.
4. Connect the carbon electrodes to the galvanometer using the connecting wires.
5. Dip one of the carbon electrodes into the iron(II) sulphate, FeSO4 solution while the other carbon electrode into the acidified potassium manganate(VII), KMnO4 solution to complete the circuit as shown in Figure 1.2.
6. Observe the direction of the deflection of the galvanometer needle and determine the positive and negative terminals for each electrode.
7. Leave the apparatus set-up for 30 minutes.
8. Observe the colour change of the iron(II) sulphate, FeSO4 solution and the acidified potassium manganate(VII), KMnO4 solution.
9. Record your observation in the table below.
10. After 30 minutes, draw out a little of the iron(II) sulphate, FeSO4 solution using a dropper and pour it into a test tube.
11. Add sodium hydroxide, NaOH solution into the test tube to verify the product formed.
Data and observation:
Discussion:
1. Based on your observation, write the half equations for electrode X and electrode Y.
2. What are the types of reactions that take place at electrode X and electrode Y?
3. Write the overall ionic equation for the reaction that takes place.
4. Identify the oxidised substance, reduced substance, oxidising agent and reducing agent.
Give reasons for your answers.
5. State the direction of the electron flow in the experiment.
6. State the positive terminal and negative terminal of the experiment.
7. What is the function of sulphuric acid, H2SO4?
8. Other than using the U-tube, draw a labelled set-up of apparatus that can be used for the experiment to investigate the transfer of electrons at a distance between potassium iodide, KI solution and chlorine water, Cl2.
Answer:
1. X : 2I− → I2 + 2e
Y : MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
2. X : Oxidation
Y : Reduction
3. 10I− + 2MnO4− + 16H+ → 5I2 + 2Mn2+ + 8H2O
4.
– I− ions are oxidised because they lose electrons.
– MnO4− ions are reduced because they gain electrons.
– Oxidising agent is MnO4− ions because MnO4− ions are the electron acceptors.
– Reducing agent is I− ions because I− ions are the electron donors.
5. Electrons move from X electrode to Y electrode through the connecting wires.
6. Positive terminal is Y electrode and negative terminal is X electrode.
7. To allow the transfer of ions and to complete the curcuit.
8.