Experiment 1B (The Effects of Concentrations of Ions in Solutions on the Selective Discharge of Ions) :
Aim: To study the effects of the concentrations of ions in hydrochloric acid, HCl on the selective discharge of ions at the electrode.
Problem statement: Does the concentration of hydrochloric acid, HCl affect the selection of ions to be discharged at the anode?
Hypothesis: When 1.0 mol dm−3 of hydrochloric acid, HCl is used, chloride ions, Cl− are discharged at the anode; when 0.0001 mol dm−3 of hydrochloric acid, HCl is used, hydroxide ions, OH− are discharged at the anode.
Variables:
(a) Manipulated variable : Concentration of hydrochloric acid, HCl.
(b) Responding variable : Ions discharged at the anode.
(c) Fixed variable : Hydrochloric acid, HCl and carbon electrodes.
Materials: 1.0 mol dm−3 of hydrochloric acid, HCl and 0.0001 mol dm−3 of hydrochloric acid, HCl.
Apparatus: Electrolytic cell, battery, carbon electrode, connecting wires with crocodile clips, switch, ammeter, test tube, wooden splinter and blue litmus paper.
Procedure:
1. Pour 1.0 mol dm−3 of hydrochloric acid, HCl into the electrolytic cell until half full.
2. Fill two test tubes with 1.0 mol dm−3 of hydrochloric acid, HCl until full and invert the test tubes onto the carbon electrodes in the electrolytic cell.
3. Connect the electrodes to the switch, ammeter and battery with connecting wires as shown in Figure 1.30.
4. Turn on the switch to complete the circuit and leave aside for a few minutes.
5. Observe and record any changes that occur at the anode and cathode.
6. Collect the gas produced at the anode.
7. Carry out a confirmatory test on the gas collected. Insert a damp blue litmus paper into the test tube at the anode.
8. Repeat steps 1 to 6 using 0.0001 mol dm−3 of hydrochloric acid, HCl.
9. Carry out a confirmatory test for the gas collected. Insert a glowing wooden splinter inside the test tube at the anode.
10. Record your observations.
Results:
Record all observations in the table below.
Discussion:
1. For each 1.0 mol dm−3 and 0.0001 mol dm−3 of hydrochloric acid, HCl:
(a) name the products formed at the anode during electrolysis. Explain your answer.
(b) write the half equations to show the formation of the products at the anode during electrolysis.
2. State the operational definition of electrolysis in this experiment.
3. After electrolysis of 1.0 mol dm−3 of hydrochloric acid, HCl is carried out for 1 hour, state the observation at the anode. Explain why.
Answer:
1.
Hydrochloric acid, HCl 1.0 mol dm−3
(a) Chlorine.
Chloride ions are discharged because the concentration of Cl− ions is higher than OH− ions.
Two chloride ions lose two electrons to form one chlorine molecule.
(b) 2Cl− → Cl2 + 2e−
Hydrochloric acid, HCl 0.0001 mol dm−3
(a) Oxygen
Hydroxide ions are discharged because the E0 value of OH− ions are less positive compared to the E0 value of Cl− ions.
Four OH− ions lose four electrons to form one oxygen molecule and two water molecules.
(b) 4OH− → O2 + 2H2O + 4e−
2. The electrolyte decomposition process produces gas bubbles at the anode when carbon electrodes that are connected to the batteries are dipped into hydrochloric acid.
3. Colourless gas is released. Hydroxide ions are discharged because the E0 value of OH− ions is less positive compared to the E0 value of Cl− ions. Four OH− ions lose four electrons to form one oxygen molecule and two water molecules.
Aim: To study the effects of the concentrations of ions in hydrochloric acid, HCl on the selective discharge of ions at the electrode.
Problem statement: Does the concentration of hydrochloric acid, HCl affect the selection of ions to be discharged at the anode?
Hypothesis: When 1.0 mol dm−3 of hydrochloric acid, HCl is used, chloride ions, Cl− are discharged at the anode; when 0.0001 mol dm−3 of hydrochloric acid, HCl is used, hydroxide ions, OH− are discharged at the anode.
Variables:
(a) Manipulated variable : Concentration of hydrochloric acid, HCl.
(b) Responding variable : Ions discharged at the anode.
(c) Fixed variable : Hydrochloric acid, HCl and carbon electrodes.
Materials: 1.0 mol dm−3 of hydrochloric acid, HCl and 0.0001 mol dm−3 of hydrochloric acid, HCl.
Apparatus: Electrolytic cell, battery, carbon electrode, connecting wires with crocodile clips, switch, ammeter, test tube, wooden splinter and blue litmus paper.
Procedure:
1. Pour 1.0 mol dm−3 of hydrochloric acid, HCl into the electrolytic cell until half full.
2. Fill two test tubes with 1.0 mol dm−3 of hydrochloric acid, HCl until full and invert the test tubes onto the carbon electrodes in the electrolytic cell.
3. Connect the electrodes to the switch, ammeter and battery with connecting wires as shown in Figure 1.30.
4. Turn on the switch to complete the circuit and leave aside for a few minutes.
5. Observe and record any changes that occur at the anode and cathode.
6. Collect the gas produced at the anode.
7. Carry out a confirmatory test on the gas collected. Insert a damp blue litmus paper into the test tube at the anode.
8. Repeat steps 1 to 6 using 0.0001 mol dm−3 of hydrochloric acid, HCl.
9. Carry out a confirmatory test for the gas collected. Insert a glowing wooden splinter inside the test tube at the anode.
10. Record your observations.
Results:
Record all observations in the table below.
Discussion:
1. For each 1.0 mol dm−3 and 0.0001 mol dm−3 of hydrochloric acid, HCl:
(a) name the products formed at the anode during electrolysis. Explain your answer.
(b) write the half equations to show the formation of the products at the anode during electrolysis.
2. State the operational definition of electrolysis in this experiment.
3. After electrolysis of 1.0 mol dm−3 of hydrochloric acid, HCl is carried out for 1 hour, state the observation at the anode. Explain why.
Answer:
1.
Hydrochloric acid, HCl 1.0 mol dm−3
(a) Chlorine.
Chloride ions are discharged because the concentration of Cl− ions is higher than OH− ions.
Two chloride ions lose two electrons to form one chlorine molecule.
(b) 2Cl− → Cl2 + 2e−
Hydrochloric acid, HCl 0.0001 mol dm−3
(a) Oxygen
Hydroxide ions are discharged because the E0 value of OH− ions are less positive compared to the E0 value of Cl− ions.
Four OH− ions lose four electrons to form one oxygen molecule and two water molecules.
(b) 4OH− → O2 + 2H2O + 4e−
2. The electrolyte decomposition process produces gas bubbles at the anode when carbon electrodes that are connected to the batteries are dipped into hydrochloric acid.
3. Colourless gas is released. Hydroxide ions are discharged because the E0 value of OH− ions is less positive compared to the E0 value of Cl− ions. Four OH− ions lose four electrons to form one oxygen molecule and two water molecules.