Chemistry report on the first 4 experiments in chapter 1

1. PERKARA TP
1
TP
2
TP
3
TP
4
TP
5
TP
6
Planning an
experiment
Aim of the investigation √ √ √ √ √ √
Hypothesis √ √ √ √ √ √
All the variable involved √ √ √ √ √
Apparatus and materials √ √ √ √ √
Method or technique √ √ √ √ √
Conducting an
experiment
Using apparatus according to plans made √ √ √ √ √
Cleaning and keeping away the apparatus √ √ √ √ √
Collecting and
recording
data/observatio
n
Construct table with manipulated and responding variables √ √ √ √
Completing table with correct data or observation of
responding variable
√ √ √ √
Interpret and
making
conclusion
Writing discussion or drawing graph that supports the
collected data
√ √ √ √
Stating weather the hypothesis is accepted or not √ √ √ √
Making correct conclusion from the experiment √ √ √ √
Justify the results of the investigation relating with theory,
principles and science law when writing report
√ √ √
Evaluate and suggest improvement to the investigation
method and further inquiry method when needed
√ √
Discuss data validity and suggest improvement on data
collection method
√
SUMMARY OF SCIENTIFIC
INVESTIGATION/EXPERIMENT
RUBRIC
Name: __________________
Date :___________
Class: ____________________
Experiment:
__________________________

2. Day: Date:
Title:
Aim:
Problem statement:
Hypothesis: relationship between manipulated and responding variable.
Not all experiments have a hypothesis
Manipulated Variable:
Responding Variable:
Fixed Variable (s) :
Operational Defination : what you do and what you see
Materials:
Apparatus:
Diagrams : Labelled and functional diagrams
Procedure: (use past tense)
Data and Observation: (labelled table with units and to the correct
decimals, equations and graphs)
Burette reading – 2 decimal places
Temperature, voltage, - 1 decimal place
Interpreting data:
Discussion:
Conclusion:
Format of a
Chemistry
Report

5. Problem Statement : How can the oxidation and reduction reaction for the
transfer of electron at a distance be determined?
Operational Definition: When iron (II) sulphate and acidified potassium
manganate (VII) solution are put in a U tube and separated with sulphuric acid
and connected to a galvanometer, there is a deflection of the needle of the
galvanometer and a change in the colour of the two solutions.

6. light green colour of
iron(II) sulphate
solution turns brown
iron(II) ions Fe2+ are oxidised
to become iron(III) ion, Fe3+.
purple solution
becomes colourless.
MnO4- is reduced
from Mn7+ to Mn2+
Discussion
1. Half equations at the electrode X
Electrode Y
2. Electrode X : Oxidation reaction , Electrode Y : Reduction reaction
3. Overall ionic equation:

7. 4. 0xidised substance
Iron (II) sulphate // Iron (II) ion undergoes oxidation by releasing electrons to form iron (III) ion.
Reduced substance
Managante (VII) ion undergoes reduction by receiving electrons to form manganase (II) ion.
Oxidising agent
Acidified Potassium Managanate (VII) solution//Managante (VII) ion acts as oxidizing agent because it
oxidises Iron (II) sulphate //iron (II) ion to iron(III) sulphate // iron(III) ion.
Reducing agent
Iron(II) sulphate // iron (II) ion acts as reducing agent because it reduces Potassium Managanate (VII)
solution//Managante (VII) ion to manganase (II) ion
5. The electron flow is from the electron X through the external circuit/connecting wires to the electrode Y
6. Positive terminal : Electrode Y
Negative terminal : Electrode X
7. The sulphuric acid is an electrolyte and allows the flow of ion and separates the two solutions.
8. **Set –up of apparatus to investigate the transfer of electrons at a distance between KI and chlorine water
– diagram on simple voltaic cell with salt bridge
Conclusion:

10. Experiment Observation Inference
Iron (II) sulphate and Bromine The brown colour of bromine
turns colourless.
The colour of the solution changes
from light green to yellowish
brown.
iron(II) ions Fe2+ is oxidised
to become iron(III) ion, Fe3+.
bromine molecules are
to become bromide ions
Iron (III) chloride and zinc powder Mass of Zinc powder becomes
The brown coloured of iron(III)
sulphate solution turn light green.
iron(III) ions, Fe3+ are
reduced to iron(II) ion, Fe2+
Zinc powder added is oxidised to
form zinc ion, Zn2+
Iron (III) chloride
Data and Observation

12. Discussion:
1. Section A ; Substance oxidised : Iron (II) sulphate
Substance reduced : Bromine water
a. change in oxidation number: Iron (II) sulphate is oxidised because the oxidation number of iron
ion increases from +2 to +3. Bromine water is reduced because the oxidation number of bromine decreases
from 0 to -1
b. Transfer of electrons : iron (II) sulphate is oxidised because iron (II) ion loses 1 electron to form
iron (III) ion. Bromine water is reduced because bromine receive 1 electron to form bromide ion.
Section B ; Substance oxidized : zinc powder
Substance reduced : Iron (III) chloride
a. change in oxidation number: zinc is oxidised because the oxidation number of zinc increases
0 to +2. Iron (III) chloride is reduced because the oxidation number of iron (III) decreases from +3 to +2
b. Transfer of electrons : zinc is oxidised because zinc loses 2 electron to form zinc ion. Iron (III)
chloride is reduced because Iron (III) ion receives 1 electron to form Iron (II) ion.

13. 2. Role of bromine water in section A: acts as a oxidizing agent because it oxidises
iron (II) ion to iron (III) ion. Zinc powder in section B acts as a reducing agent
because is reduces iron (III) ion to iron (II) ion.
3. A freshly prepared iron (II) sulphate solution is used for the experiment in section
A. This is because iron (II) ion is easily oxidised to iron (III) ion when exposed to air.
4. Reagent to verify the presence of iron (III) ion in section A and iron (II) ion in
section B is sodium hydroxide solution// ammonia aqueous (NH3) // potassium
thiocyanate.
5. (a) Another substance that can replace bromine water in section A is : chlorine
water
Another substance that can replace zinc powder in section B is :
Aluminium//Magnesium.

14. Q4 Confirmation test Fe3+
Test 1: Dilute sodium hydroxide solution (NaOH) is
then added to the test tube until excess.
Result: Brown precipitate formed. The precipitate
does not dissolve in excess sodium hydroxide
solution.
Test 2: Dilute ammonium hydroxide solution
(NH4OH)/ammonia aqueous (NH3) is then added into
the test tube until excess
Result: Brown precipitate formed. The precipitate
does not dissolve in excess ammonium hydroxide
solution /ammonia aqueous.
Test 3: 2cm³ of potassium thiocyanate is added to the
test tube.
Result: Red blood solution formed.
Q4 Confirmation test for Fe2+
2 cm³ of the solution of the product is filled into a test tube.
Test 1: Dilute sodium hydroxide solution (NaOH) is then
added to the test tube until excess.
Result: Dirty green precipitate formed. The precipitate
does not dissolve in excess sodium hydroxide solution.
Test 2: Dilute ammonium hydroxide solution
(NH4OH)/ammonia aqueous (NH3) is then added into the
test tube until excess
Result: Green precipitate formed. The precipitate does not
dissolve in excess ammonium hydroxide solution
/ammonia aqueous.
Test 3: 2cm³ of potassium thiocyanate is added to the test
tube.
Result: No change observed

15. Q5 (a)Section A
Other oxidation agents that get to replace bromine water
1.Chlorine water
Half Equations
Cl2 + 2e → 2Cl–
2.Acidic potassium manganate (VII)
Half Equations
MnO4
– + 8H+ + 5e → Mn2+ + 4H2O
3.Potassium dichromate (VI)
Half Equations
Cr2O7
2- + 14H+ + 6e → 2Cr3+ + 7H2O
4.Hydrogen peroxide
Half Equations
H2O2 + 2H+ + 2e → 2H2O
5.Concentrated nitric acid
Half Equations
NO3
– + 4H+ + 3e → NO + 2H2O

16. Conclusion:

18. Procedure:
1. 2 cm3 of 0.5 mol dm-3 2 Lead (II) nitrate, 2 cm3 of 0.5 mol dm-3 magnesium nitrate
solution and 2 cm3 of 0.5 mol dm-3 copper (II) nitrate solution are poured into three separate
test tubes.
2. Three strips of magnesium is cleaned with sandpaper. The strips are then dropped into
the test tubes as shown in Figure .
3.Steps 1-2 is repeated with lead plate and copper plate
4.Any change in colour and whether any metal is deposited is observed.
5.Record all observations
Results and Observations:
A. 0.5 mol dm-3 lead (II) nitrate,
B. 0.5 mol dm-3 magnesium nitrate solution
C. 0.5 mol dm-3 copper (II) nitrate solution
Test tube Test tube Test tube
A B C
Metal Observations
Lead (II) nitrate solution Magnesium (II) nitrate
solution
Copper (II) nitrate solution
Magnesium Silver colour deposit No reaction Brown deposit , blue colour of the solution
turns pale blue
Lead No reaction No reaction Brown deposit , blue colour of the solution
turns pale blue
Copper No reaction No reaction No reaction

19. 1 a,b,c Mg + lead (II) nitrate Mg + Copper(II) nitrate Lead + Copper (II) nitrate
Oxidation half
equation
Reduction half
equation
Overall ionic
equation
Oxidised substance Magnesium because its
oxidation number is increased
from 0 to +2
Magnesium because its
oxidation number is
increased from 0 to +2
Lead because its oxidation
number is increased from 0
to +2
Reduced substance Lead (II) nitrate because lead (II)
ion’s oxidation number is
reduced from +2 to 0
Copper (II) nitrate because
oxidation number for
Copper (II) ion is reduced
from +2 to 0
Copper (II) ion because its
oxidation number is
reduced from +2 to 0
Oxidising agent Lead (II) nitrate// lead (II) ion
oxidises magnesium to
magnesium ion
Copper (II) nitrate oxidises
magnesium to magnesium
ion
Copper (II) ion because it
oxidises lead to lead (II) ion
Reducing agents Magnesium because it reduces
lead (II) ion to lead
Magnesium because it
reduces copper (II) ion to
copper
Lead because it reduces
copper (II) nitrate /Copper
(II) ion to Copper
Pb2+ + 2e- → Pb Cu2+ + 2e- → Cu
Pb → Pb2+ + 2e-
Cu2+ + 2e- → Cu
Mg + Pb2+ → Pb + Mg2+
Discussion
Mg + Cu2+ → Cu + Mg2+
Pb + Cu2+ → Cu + Pb2+

20. •Generally, metals are good electron donors and therefore are good reducing agents. However, different metals
have different strength as reducing agents.
•The strength of metals as reducing agents can be compared by using the electrochemical series.
•The electrochemical series lists metals according to their electropositivity, that is, according to their ability to lose
electrons to form positive ions.
The higher the position of a metal in the electrochemical series, the more electropositive the metal is, the easier it
is for the metal to lose its electrons. Thus, the better reducing agent the metal is.
•On the other hand, the ability of a metal ion to accept electrons increases down the series. Thus, the strength of a
metal ion as an oxidising agent increases down the electrochemical series.

21. Conclusion:
In a displacement of metal, a more electropositive metal will displace a less electropositive
metal from its salt solution.
(a) The more electropositive metal acts as the reducing agent. It loses electrons and undergoes
oxidation to form positive ions.
(b) The ions of the less electropositive metal act as an oxidising agent by accepting the electrons.
While doing so, the ions are reduced to metallic atoms.
(c) In short, there is an electron transfer from the more electropositive metal to the ions of the less
electropositive metal.

24. Halide
Solution
Halogen Observation
Colour of aqueous solution Colour of halogen in 1,1,1 -
trichloroethane
Potassium
Chloride
Bromine water Brown or yellowish brown or yellow
depending on concentration
Brown or reddish yellow or yellow
depending on concentration
Iodine solution Brown or yellowish brown or yellow
depending on concentration
Purple
Potassium
Bromide
Chlorine water Brown or yellowish brown or yellow
depending on concentration
Brown or reddish yellow or yellow
depending on concentration
Iodine solution Brown or yellowish brown or yellow
depending on concentration
Purple
Potassium
Iodide
Chlorine water Brown or yellowish brown or yellow
depending on concentration
Purple
Bromine water Brown or yellowish brown or yellow
depending on concentration
Purple
DATA AND OBSERVATION:
Halogen Halogen in aqueous solution Halogen in 1,1,1,-trichloroethane
Chlorine Pale yellow and almost colourless Pale yellow and almost colourless
Bromine Brown or yellowish brown or yellow
depending on concentration
Brown or reddish yellow or yellow
depending on concentration
Iodine Brown or yellowish brown or yellow
depending on concentration
Purple
A
B

27. Conclusion:
Halogen which placed higher in group 17 are more electronegative and act as strong oxidising
agent and can be reduced easily
The halogen which is at a higher position in group 17 ( more electronegative and reactive) can
displace a halogen that is below it (less electronegative and less reactive) from its solution of halide
ions.
When the displacement reaction of the halogen occurs:
a.transfer of electrons from the halide ions which are positioned further down in group 17 to
halogens which are positioned further up occurs.
b.Halogens which are positioned further up in group 17 act as oxidation agents.
c.These halogens undergo reduction and are reduced to halide ions. The halide ions which are
positioned further down in group 17 act as reducing agents. These ions undergo oxidation and
are oxidised to halogens.