PEKA FORM 4

PEKA 1

ELECTROPLATING


PREPARE THE FORMAT OF REPORT AS BELOW:

Aim                           : To study the electroplating of an object with copper.

Problem statement  

Hypothesis               

Variables                

                                  

Materials                  

Apparatus               

Procedure

Observation           

Set	Electrode		Observation
	Anode	Cathode	Anode	Cathode
I	Iron spoon	Copper
II	Copper	Iron spoon

                                  

                 

Discussion

     

Conclusion    

EXAMPLE OF PEKA

Aim                           :To investigate the effect of concentration of ions on selective discharge of ions at the electrodes

Problem statement   :How does the concentration of ions in hydrochloric acid, HCl affect the discharge of ions at the anode?

Hypothesis               :When the concentration of chloride ion is higher, then the chloride ion will be selectively discharged at anode.

                                    

Variables                  :Manipulated variable  :  Concentration of chloride ion

                                   Responding variable   :    Ion discharged

                                   Constant variable        :    Type of electrolyte, type of electrode, duration of electrolysis

                                  

Materials                   1.0 mol dm^-3 hydrochloric acid, 0.001 mol dm^-3 hydrochloric acid

Apparatus                batteries, carbon electrodes, connecting wires with crocodile clips, ammeter, electrolytic cell, test tubes, litmus paper and splinter

Procedure                 1. A electrolytic cell is filled with 1.0 mol dm^-3 hydrochloric acid until it is half full.

                                   2. The switch is turned on.

                                   3. Gas produced at anode is collected and tested with moist litmus paper and glowing splinter.

                                   4. Observations are recorded.

                                   5. Steps 1 to 4 are repeated by using 0.001 mol dm^-3 hydrochloric acid to replace 1.0 mol dm^-3 hydrochloric acid.

 Observation

Electrolyte	Observation at anode
1.0  mol dm-3 hydrochloric acid	A greenish-yellow gas with pungent smell is released. The gas turns the blue litmus paper to red then to white.
0.001 dm-3 hydrochloric acid	Gas bubbles are released. A colourless gas relight a glowing splinter.

                  
                                  

                                   

Discussion                  :

1.      the aqueous solution of hydrochloric acid consists of hydrogen ions, H+, chloride ions, Cl^- and hydroxide ions, OH^- tat move freely.

2.      During the electrolysis, the Cl^- ions and OH^- ions move to the anode.

a)      Electrolysis of 0.001 dm^-3 hydrochloric acid

The OH^- ions are selectively discharged at anode to form oxygen and water. This is because OH^-ion is lower than Cl^- ion in the electrochemical series.

4OH^-     →  O₂  +  2H₂O  +  4e

b)      Electrolysis of 1.0 mol dm^-3 hydrochloric acid

The Cl^- ions are selectively discharged at anode to form chlorine gas. This is because the concentration of Cl^- ions are higher than OH^- ions.

2Cl^-    → Cl₂  +  2e

Conclusion     : During electrolysis of 1.0 mol dm^-3 hydrochloric acid, the Cl^- ions are selectively discharged at anode instead of OH-.to form chlorine gas due tohigher concetration of Cl^-.The hypothesis is accepted.

ANSWER FOR PAPER 1 FORM 4 CHEMISTRY (AUG 2012)

 

1	B		26	C /D
2	D		27	B
3	B		28	C
4	C		29	C
5	D		30	A
6	A		31	C
7	B		32	C
8	B		33	B
9	B		34	C
10	D		35	A
11	D		36	B
12	C		37	D
13	C/D		38	A
14	A		39	B
15	C		40	A
16	B		41	C
17	C		42	C
18	B		43	D
19	D		44	D
20	C		45	C
21	A		46	B
22	D		47	D
23	B		48	C
24	C		49	B
25	D		50	B

Electrolysis of aqueous solution

Electrolysis of aqueous solution

1.     Electrolysis of potassium nitrate solution, KNO₃ using carbon electrodes.

	Anode (+)	Cathode (-)
Ion presents	NO-3, OH-	K+, H+
Half equation	4OH- → 2H2O + O2 +4e	2H+ + 2e → H2
Observation	Colourless gas bubbles are released	Colourless gas bubbles are released
Product	Oxygen gas	Hydrogen gas

2.     Electrolysis of silver nitrate solution, AgNO₃ using carbon electrodes.

	Anode (+)	Cathode (-)
Ion presents	NO-3, OH-	Ag+, H+
Half equation	4OH- → 2H2O + O2 +4e	Ag+ + e → Ag
Observation	Colourless gas bubbles are released	A grey solid is formed
Product	Oxygen gas	Silver metal

3.     Electrolysis of copper(II) sulphate solution, CuSO₄ using carbon electrodes.

	Anode (+)	Cathode (-)
Ion presents	SO2-4, OH-	Cu2+, H+
Half equation	4OH- → 2H2O + O2 +4e	Cu2+ + 2e → Cu
Observation	Colourless gas bubbles are released	Brown solid is formed
Product	Oxygen gas	Copper metal

4.     Electrolysis of concentrated potassium chloride solution, KCl using carbon electrodes.

	Anode (+)	Cathode (-)
Ion presents	Cl- , OH-	K+, H+
Half equation	Cl- → Cl2 + 2e	2H+ + 2e → H2
Observation	Greenish-yellow gas bubbles are released	Colourless gas bubbles are released
Product	Chlorine gas	Hydrogen gas

5.     Electrolysis of silver sulphate solution, Ag₂SO₄ using silver electrode as anode and carbon electrode as cathode.

	Anode (+)	Cathode (-)
Ion presents	-
Half equation	Ag→ Ag+ +e	Ag+ + e → Ag
Observation	Silver electrode corrodes and become thinner	A grey solid is deposited and electrode becomes thicker.
Product	Silver ion	Silver metal

6.     Electrolysis of copper(II) nitrate solution, Cu(NO₃)₂ using copper electrode as anode and carbon electrode as cathode.

	Anode (+)	Cathode (-)
Ion presents	-	-
Half equation	Cu → Cu2+ +2e	Cu2+ + 2e → Cu
Observation	Copper electrode corrodes and become thinner	Brown solid is deposited and electrode becomes thicker.
Product	Copper(II) ion	Copper metal

Formation of covalent bond in chlorine molecule

Formation of  hydrogen chloride molecule, HCl ( covalent compound)

The formation of covalent bonds in hydrogen chloride molecule is explained as follows:

§  Each hydrogen atom has an electron arrangement of 1.

§  The outermost shell needs one electron in order to achieve a stable duplet electron arrangement.

§  Clorine atom has an electron arrangement of 2.8.7.

§  The outermost shell needs one electron in order to achieve a stable octet electron arrangement.

§  Clorine atom contributes one valence electron while hydrogen atom contributes one valence electron for sharing.

§  A single covalent bond is formed.

Remarks:

§  The electrons must be drawn in the overlap of two shells.

§  Students may use Lewis structure in drawing the formation of covalent compound. ( Lewis structure only show the valence electrons)

Formation of sodium chloride, NaCl ( ionic compound)

The formation of ionic bonds in sodium chloride, NaCl is explained as follows:

•    The electron arrangement of a sodium atom is 2.8.1.
•     Sodium atom releases one valence electron to form sodium ion, Na⁺

                             Na → Na⁺ + e

• The electron arrangement of sodium ion is 2.8.
• Sodium ion achieves a stable octet electron arrangement.
• The electron arrangement of a chlorine atom is 2.8.7.
• Chlorine atom receives one electron to form chloride ion, Cl^-

                             Cl + e → Cl^-

• The electron arrangement of chloride ion is 2.8.
•  Chloride ion achieves a stable octet electron arrangement.
•  Strong electrostatic force pulls the sodium ion and chloride ion together.
• An ionic bond is formed.
• Sodium chloride is an ionic compound.

Remark:
Students always confuse following terms.
chlorine is atom
choride is ion


When answer this question, please draw the diagram of formation of ionic bond in sodium chloride because some time the diagram will contribute marks for you.