Chemistry Essay Cluster 2 by IFFAH ADANI BINTI KAMARUL HISYAM Moe

Question 1

m4494869 — 2021-08-29 16:48:35 UTC

(a)(i) - Ag, Y, X

- X can displace silver nitrate

- Because X is more electropositive than Ag

- Y can displace silver nitrate

- Because Y is more electropositive than Ag

- Y can’t displace X nitrate

- Because Y is less electropositive than X

(ii) - Product formed: Copper (II) nitrate

- Ionic equation: Cu + 2Ag+ —> Cu2+ + 2Ag

(b) - Balanced chemical equation: CuO + H2 —> Cu + H2O

- Copper (II) oxide undergoes reduction while hydrogen undergoes oxidation

- The oxidation number of Cu2+ decrease from +2 to 0

- The oxidation number of H2increase from 0 to +2

- Cu2+ ion is oxidising agent

- H2 is reducing agent

- When iron is in contact with more electropositive metal like magnesium, rusting of iron will become slower

- Magnesium atom releases electrons to form magnesium ion

- Magnesium corrodes or undergoes oxidation instead of iron

- Observation: High intensity of pink spot and no blue spots found in the test tube

- Oxidation half equation: Mg → Mg2+ + 2e-

- Reduction half equation: 2H2O + O2 + 4e- → 4OH

Question 2

m4494869 — 2021-08-29 16:53:53 UTC

(a)(i) - Oxidation number of aluminium: +3

- Oxidation number of iron: +3

(ii) - Al2O3, aluminium oxide

- Fe2O3, iron (II) oxide

(iii) - Aluminium has only 1 oxidation number while iron has 2 oxidation number

- The roman numeral III indicates the oxidation number of iron (III) is +3 whereas the position of aluminium in group 13 indicates the oxidation number of aluminium is +3

- Iron is a transition element which normally have more than one oxidation number while aluminium is not a transition element that has only one oxidation number.

(b)(i) - Experiment I: Act as a reducing agent

- Experiment II: Act as an oxidising agent

(ii) - Experiment II

* Half equation of oxidation: Mg → Mg2+ + 2e-

* Half equation of reduction: Fe2+ +2e- → Fe

- Experiment III

* Half equation of oxidation: Fe → Fe2+ + 2e-

(c) Experiment I
- Iron (II) ions, Fe2+ is a reducing agent while manganate (VII) ions, MnO4- is an 2+ oxidising agent
- Iron (II) ions, Fe2+ undergoes oxidation process while manganate (VII) ions, MnO4- undergoes reduction process
- Oxidation half equation: Fe2+ → Fe3+ +e-
- Reduction half equation: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
- Enode value of iron (II) ions, Fe2+ is lower than manganate (VII) ions , MnO4-
- Iron (II) ions, Fe2+ releases 2 electrons and are oxidised to iron (III) ions, Fe3+
- Manganate (VII) ions, MnO4-, receive 7 electrons and are reduced to manganese (II) ions, Mn2+

Experiment II
- Magnesium, Mg is a reducing agent while iron, Fe is an oxidizing agent
- Magnesium, Mg undergoes oxidation process while iron, Fe undergoes reduction
- Oxidation half equation: Mg → Mg2+ + 2e-
- Reduction half equation: Fe2+ + 2e- → Fe
- Enode value of magnesium ion, Mg2+ is lower than iron (II) ion, Fe2+
- Magnesium releases 2 electrons to produce magnesium ion, Mg2+
- Iron (II) ion, Fe2+ receives 2 electrons to produce iron.

Experiment III
- Iron atom is an reducing agent
- Iron atom undergoes oxidation process
- Oxidation half equation: Fe → Fe2+ + 2e-
- Enode value of iron metal, Fe is lower than copper metal, Cu
- Iron atom releases 2 electrons to form iron (II) ion, Fe2+

Question 3

m4494869 — 2021-08-29 16:54:32 UTC

(a) - Halogen X: Bromine water

- Iron (II) sulphate undergoes oxidation because the oxidation number of iron in iron (II) sulphate increase from +2 to +3

- Bromine undergoes reduction because the oxidation number decreases from 0 to -1

- Reducing agent: Iron (II) sulphate

- Oxidising agent: Bromine water

- Half equation at L: 2Fe2+ → 2Fe3+ + 2e

- Half equation at M: Br2 +2e →2Br-

- Overall ionic equation: 2Fe2+ + Br2 → 2Fe3+ + 2Br-

(b)

Diagram 1

E° value is more positive

X ion on the left side is a stronger oxidising agent
It is easier for X ion to receive electron and undergo reduction
Conversely, X atom on the right side is difficult to release electron

E° value is more negative

X atom on the right side is the stronger reducing agent
It is easier for X atom to release electrons and undergo oxidation
Conversely, X on the left side is difficult to accept electrons

Diagram 2

Half equation of hydrogen
2H+ + 2e- <=> H2
E° value of standard hydrogen electrode potential
E° = 0.00V

If E° value is more positive and less negative
The strength of oxidising agent increases
If E° value is more negative and less positive
The strength of reducing agent increase

Question 4

m4494869 — 2021-08-29 16:57:45 UTC

(a) Set I
- Equation: Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
- Substance undergo oxidation: Cu
- Substance undergo reduction: Ag+
- Half equation oxidation: Cu → Cu2+ + 2e-
- Half equation reduction: 2Ag+ + 2e- → 2Ag
- Observation:
- AgNO3 solution changes colour from colourless to blue
- Copper ribbon become thinner

Set II
- Equation: Fe2(SO4)3 + 3Zn → 3ZnSO4 + Fe
- Substance undergo oxidation: Zn
- Substance undergo reduction: Fe3+
- Half equation oxidation: Zn → Zn2+ + 2e-
- Half equation reduction: Fe3+ + e- → Fe2+
- Observation:
- Fe2(SO4)3 solution changes colour from brown to green
- Some zinc powder dissolved

(b)

Metals used: Tin, Sn, magnesium, Mg and copper, Cu
Materials: Agar solution, phenolphthalein, potassium hexacyanoferrate (III), KFe(CN)6 solution, iron nails, magnesium ribbon, Mg, tin strip, Sn and copper strip, Cu
Procedure:

1. Label three test tube with Set I (Tin), Set II (Magnesium) and Set III (Copper)
2. Clean all three iron nails, magnesium ribbon, strips of copper and tin with sand papers
3. Coil three iron nail with magnesium ribbon, strips of copper and tin respectively
4. Place all three iron nails in three separate test tubes
5. Pour the same amount of hot jelly solution containing of potassium hexacyanoferrate(II) and phenolphthalein into the test tubes to complete cover all the nails
6. Keep the test tubes in a test tube rack and leave aside for a day
7. Observe and record any changes

Conclusion:

- Set I- Iron undergoes oxidation or corrosion instead of tin.
- Set Il - Magnesium undergoes oxidation or corrosion instead of iron.
- Set III - Iron undergoes oxidation or corrosion instead of copper.

Question 5

m4494869 — 2021-08-29 17:00:33 UTC

(a)

The surface of the iron in the middle of the water droplet with lower oxygen concentration becomes anode
Oxidation occurs at anode
Oxidation half equation at anode: 4Fe → Fe2+ + 4e-
Iron atom releases electrons to form Fe2+
Electrons flow to the end of water droplet with higher oxygen concentration
The end becomes the cathode
Reduction occurs at cathode
Reduction half equation at cathode: O2 + H2O + 4e- → 4OH-
Oxygen molecules receive electrons to form hydroxide ions
Fe2+ ions combine with OH- ions to form Fe(OH)2
Fe(OH)2 undergoes further oxidation to form Fe2O3 + H2O

(b)

Extraction of metals using electrolysis for metals more reactive than carbon
At anode, oxide ions donate electrons and undergoes oxidation
Oxidation half equation at anode: 2O2- → O2 + 4e-
At cathode, Al2+ is reduced by receiving electrons to form molten aluminium, Al
Reduction half equation at cathode: Al3+ + 3e- → Al
Ionic equation: 6O2- + 4Al3+ → 3O2 + 4Al

Question 6

m4494869 — 2021-08-29 17:01:03 UTC

(a) - Reaction X is not a redox reaction

- Because there’s no change in oxidation number or any ions in reaction X

- Reaction Y is a redox reaction

- Because Zn atom had an increase in oxidation number from 0 to +2

- Copper (II) ion had a decrease in oxidation number from +2 to 0

(b)(i) - Oxidation number of iron in compound R is +2

- Iron (II) chloride

- Oxidation number of iron in compound S is +3

- Iron (III) chloride

(ii) Oxidising agent: Bromine water

Observation: Green colour of solution turns brown

Procedure:

1. Fill in the U-tube half full with dilute sulphuric acid and clamp it vertically.

2. Fill one arm of the U-tube using dropper with solution of Br2 water and another with KI solution.

3. Dip the carbon electrodes into the solutions and connect the galvanometer using connecting wire as shown in diagram.

4. Observe the galvanometer pointer and the colour change of Br2 water and KI solution.

Observation: Brown colour of Br2 water turns colourless

Conclusion: Br2 molecules are reduced to Br- ions

Question 7

m4494869 — 2021-08-29 17:06:39 UTC

(a)

SIMILARITIES

- Electrodes are dipped into electrolyte
- Oxidation reaction at the anode
- Reduction reaction at the cathode
- Electron flow from the anodes to the cathodes through the connecting wire

DIFFERENCES

Cell P
- Half equation copper A: Cu → Cu2+ + 2e-
- Half equation copper B: Cu2+ + 2e- → Cu
- Reaction at copper A: Oxidation
- Reaction at copper B: Reduction
- Observations:
- Copper A becomes thinner
- Copper B becomes thicker
- Blue solution remain unchanged

Cell Q
- Half equation copper: Cu2+ + 2e- → Cu
- Half equation magnesium: Mg → Mg2+ + 2e-
- Reaction at copper A: Oxidation
- Reaction at magnesium: Reduction
- Observations:
- Magnesium electrode becomes thinner
- Copper electrode becomes thicker
- Blue solution becomes paler

(b)

Materials: 1 mol dm-3 sulphuric acid, H2SO4, 0.1 mol dm-3 iron (III) sulphate solution, Fe2(SO4)3, 0.1 mol dm-3 potassium bromide solution, KBr and sodium hydroxide solution, NaOH

Apparatus : U-tube, galvanometer, connecting wires with crocodile clips, galvanometer, retort stand, carbon electrodes, dropper and test tube

Procedure :

Measure and pour 1 mol dm-3 of sulphuric acid, H2SO4 into the 7-tube until half full and clamp it vertically.
Measure and pour 0.5 mol dm-3 of potassium bromide, KBr solution into one arm using dropper.
Measure and pour 0.5 mol dm-3 of iron (II) sulphate, Fe2(SO4)3 solution into other arm of U-tube.
Dip the carbon electrodes into the solutions and connect to the galvanometer using connecting wires.
Observe the galvanometer pointer and the colour changes of the solutions of potassium bromide, KBr solution and iron (II) sulphate, Fe2(SO4)3 solution.

Observations:

Oxidising agent: Iron (III) sulphate
Half equation for oxidising agent: Fe3+ + 2e- → Fe2+
The brown iron (III) sulphate solution turns green

Confirmatory test for Fe3+:

Sodium hydroxide solution is added to the mixture until excess
Green precipitate is formed insoluble in excess
Reducing agent: Potassium bromide solution
Half equation of reduction agent: 2Br- → Br2 + 2e-
The colourless potassium bromide solution turns brown