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b) Metals used: Tin, Sn, Magnesium, Mg and Copper, Cu

Materials : Agar solution, phenolphthalein, potassium hexacyanoferrate(III), K3Fe(CN)6 solution, iron nails, magnesium ribbon, Mg, tin strip, Sn and copper strip, Cu
Procedure:

1. Label three test tubes 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 nails 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 potassium hexacyanoferrate(III) and phenolphthalein into the test tubes to completely cover all the nails

6. Keep the test tubes in a test tube rack and leave them aside for a day.

7. Observe and record any changes.

Conclusion:

Set I - Iron undergoes oxidation or corrosion instead of Tin.

Set II - Magnesium undergoes oxidation or corrosion instead of iron.

Set III - Iron undergoes oxidation or corrosion instead of copper.

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a) ii - Copper (II) nitrate
Cu -> Cu2+ + 2e-
2Ag+ + 2e- -> 2Ag
Ionic equation: Cu + 2Ag+ -> Cu2+ + 2Ag

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a) i - Ag, Y, X
- In set 1, metal X can displace silver to form X nitrate solution
- Metal X is more electropositive than silver metal
- In set 2, metal Y can displace silver to form Y nitrate solution
- Metal Y is more electropositive than silver metal
- In set 3, metal Y cannot displace metal X to form Y nitrate solution
- Metal X is less electropositive than metal X

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b)  equation
- Copper(II) Oxide undergoes reduction while hydrogen undergoes   oxidation
- The oxidation number of Cu2+ decreases from +2 to 0
- The oxidation number of H2 increase from 0 to +1
- Cu2+ ion is oxidising agent
- H2 is reducing agent

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c) - Magnesium
- When iron is in contact with more electropositive metal, rusting of iron is slower
- Magnesium atom releases electrons to form magnesium ion
- Magnesium corrodes or undergoes oxidation instead of iron
- The observation is high intensity of pink spot and no blue spots in the test tube

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a) i- Al=x
2x + 3(-2)=0
2x= 6
x= +3

Fe=x
2x + 3(-2)=0
2x= 6
x= +3

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a) Halogen X = Bromine water

- Iron (II) Sulphate undergoes oxidation because the oxidation number of iron in iron

(II) sulphate increase form +2 to +3

- Iron (II) sulphate is the reducing agent

- Bromine undergoes reduction because the oxidation number of bromine decreases

from 0 to -1

- Bromine water is the oxidising agent

Half equation

At electrode L:

- 2Fe 2+ → 2Fe3+ + 2e-

At electrode M:

- Br2 +2e -> 2Br-

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

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b) i- Exp 1: Reducing Agent
Exp 2: Oxidising Agent

ii- Exp 2:-
H/E oxidation: Mg→ Mg2+ +2e-
H/E reduction: Fe2+ +2e- → Fe
Exp 3:-
H/E oxidation:  Fe→Fe2+ +2e-

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c) Experiment I

- Iron (II) ions, Fe is a reducing agent while manganate (VII) ions, MnO4- , is an oxidizing 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+ + 4H20

- 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 (II) ion, Fe2+ an oxidizing agent

- Magnesium, Mg undergoes oxidation process while iron (II) ion, Fe2+ undergoes reduction process

- 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 → Fe 2+ + 2e-

- Enode value of iron metal, Fe is lower than copper metal, Cu

- Iron atom releases 2 electrons to form iron (II)ion, Fe2+

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a) ii- Al2O3: Aluminium Oxide
Fe2O3: Iron(II) Oxide

iii- Aluminium has only one oxidation number whereas iron has two 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 elements which normally have more than one oxidation number whereas aluminium is not a transition element that has only one oxidation number

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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 where reduction occurs
- Reduction half-equation at cathode: O2 + 2H20 + 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

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.Half Equation: 2H+ +2e- <-> H2, E0 of standard hydrogen electrode potential= 0.00V

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E0 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
E0 value is more negative:
- X atom on the right side is a stronger reducing agent
- It is easier for X atom to release electron and undergo oxidation
- Conversely, X ion on the left side is difficult to accept electron

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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 :

1. Measure and pour 1 mol dm-3 of sulphuric acid, H2SO4 into the 7-tube until half full

and clamp it vertically.

2. Measure and pour 0.5 mol dm-3 of potassium bromide, KBr solution into one arm using dropper.

3. Measure and pour 0.5 mol dm-3 of iron (III) sulphate, Fe2(SO4)3 solution into other arm of U-tube.

4. Dip the carbon electrodes into the solutions and connect to the galvanometer using

connecting wires.

5. Observe the galvanometer pointer and the colour changes of the solutions of potassium

bromide, KBr solution and iron (III) sulphate, Fe2(SO4)3 solution.

Observation :

- Oxidising agent: Iron (III) sulphate solution

- Half equation of 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 reducing agent : 2Br- → Br2 + 2e-

- The colourless potassium bromide solution turns brown

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

- For metals that are less reactive than carbon, carbon reduction process is used. For example, iron from iron ore
- At high temperature, carbon reacts with oxygen to form carbon dioxide. C+O2 → CO2
- Then, CO2 reacts with remaining carbon to form carbon monoxide. CO2 + C → 2CO
- Carbon and carbon monoxide becoming reducing agents and undergo oxidation
- Iron (III) Oxide becomes oxidising agent and undergo reduction
- Chemical equation:
2Fe2O3 + 3C → 4Fe + 3CO2
Fe2O3 + 3CO → 2Fe + 3CO2
- Carbon and carbon monoxide reduce iron(III) oxide to molten iron

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c) Reducing agent: Potassium Iodide
Procedure:
1. Fill the U-tube half-full with dilute sulphuric acid and clamp it vertically
2. Using dropper, fill one arm of U-tube with potassium iodide and the other arm of U-tube with bromine water
3. Dip the carbon electrodes into the solution and connect the galvanometer using connecting wires as shown in the diagram
4. Observe the galvanometer pointer and the colour changes of the solutions
Observation:
- Brown colour of Bromine water turns colourless
Conclusion:
- Br2 molecules reduced to Br- ions

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a) Similarities:
- Electrodes are dipped into electrolyte
- Oxidation reaction at anodes
- Reduction reaction at cathode
- Electron flow from anodes to cathodes through connecting wires
Differences: (diagram)

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a) - Reaction X is not a redox reaction because no change in oxidation number on any ions in X
- Reaction Y is a redox reaction because Zn atom had an increase in oxidation number from 0 to +1 while 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- Bromine water
- Green colour solution turns brown