# data[].attributes.headline
 i) Based on the results, arrange the three metals in order of electropositivity. Explain your answer.

# data[].attributes.subject
 i) Based on the results, arrange the three metals in order of electropositivity. Explain your answer.

# data[].attributes.body
Argentum, Y , X

X can displace Ag from AgNO3 in set I. X is more electropositive than Ag

Y can displace AgNO3 in set II. Y is more electropositive than Ag

Y cannot displace X from X nitrate solution. Y is less electropositive than X

Thus, Y is more electropositive than Ag but less electropositive than X

# data[].attributes.headline
ii) If Y is copper, name the product formed in Set II. Write the ionic equation for the reaction that take place.

# data[].attributes.subject
ii) If Y is copper, name the product formed in Set II. Write the ionic equation for the reaction that take place.

# data[].attributes.headline
b) Copper(II) oxide is an oxide of copper. The oxide can be reduced to copper by hydrogen gas.Write a balanced chemical equation for the reaction between copper(II) oxide with hydrogen gas. Explain the redox reaction in terms of change in oxidation number and identify the oxidising as well as reducing agent for the reaction.

# data[].attributes.subject
b) Copper(II) oxide is an oxide of copper. The oxide can be reduced to copper by hydrogen gas.Write a balanced chemical equation for the reaction between copper(II) oxide with hydrogen gas. Explain the redox reaction in terms of change in oxidation number and identify the oxidising as well as reducing agent for the reaction.

# data[].attributes.body
CuO + H2 -> Cu + H₂O

CuO undergoes reduction reaction

Oxidation number of copper decreases from 2+ to 0

H₂ undergoes oxidation reaction

Oxidation number of hydrogen increases from 0 to 1+

Oxidising agent is copper (II) oxide and reducing agent is Hydrogen gas

# data[].attributes.body
Replace copper with magnesium

The standard electrode potential value of magnesium is more negative than
iron

Magnesium is more electropositive than iron

Magnesium has higher tendency to release electrons than iron to form
magnesium ion, Mg²⁺

Water and oxygen molecules receive electrons to form hydroxide ions, OH⁻

No blue spots present

No iron(II) ions, Fe2+ present

Rusting does not occur

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

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

# data[].attributes.body
Aluminium = +3
Iron = +3

# data[].attributes.headline
i) Name both the compounds based on the IUPAC nomenclature system.

# data[].attributes.subject
i) Name both the compounds based on the IUPAC nomenclature system.

# data[].attributes.body
Aluminium oxide
Iron(III) oxide

# data[].attributes.headline
ii) Explain the difference between the names of the two compounds based on the IUPAC nomenclature system.

# data[].attributes.subject
ii) Explain the difference between the names of the two compounds based on the IUPAC nomenclature system.

# data[].attributes.body
Iron has roman numbers while aluminium does not have roman numbers

Iron is transition metals where as aluminium is not transition metals

Iron has various oxidation numbers where as aluminium does not have various oxidation numbers

# data[].attributes.headline
i) State the role of ferum(II) sulphate solution in Experiment I and Experiment II

# data[].attributes.subject
i) State the role of ferum(II) sulphate solution in Experiment I and Experiment II

# data[].attributes.body
Experiment I - reducing agent
Experiment II - oxidizing agent

# data[].attributes.headline
ii) Write the half equations of oxidation and reduction in Experiment II or Experiment III.

# data[].attributes.subject
ii) Write the half equations of oxidation and reduction in Experiment II or Experiment III.

# data[].attributes.body
Experiment I
oxidation: Mg -> Mg2+ + 2e-
reduction: Fe²⁺ + 2e- -> Fe
Experiment II
oxidation: Fe -> Fe²⁺ + 2e-
reduction: 2H₂O + O2 + 4e⁻ -> 4OH⁻

# data[].attributes.headline
c) Explain the differences in the observation for Experiment I, II andIII based on Redox reaction.

# data[].attributes.subject
c) Explain the differences in the observation for Experiment I, II andIII based on Redox reaction.

# data[].attributes.body
Experiment I

Ferum(II) ions, Fe²⁺ release an electron to form Ferum(III) ions, Fe³⁺

Ferum(II) ions, Fe2+ undergo oxidation

Maganate (VII) ions, MnO⁴⁻ receives electron to form maganese(II) ions, Mn²⁻

Maganate (VII) ions, MnO4-  undergo reduction

Needle of galvanometer deflected showing that the movement of electrons from negative terminal to positive terminal

Experiment II

Magnesium, Mg is more electropositive than Ferum, Fe

Ferum(II) ions, Fe²⁺ receive electrons to form Ferum, Fe

Ferum(II) ions, Fe2+ undergo reduction

Magnesium, Mg release electrons to form magnesium ion, Mg²⁺

Magnesium undergo oxidation

Experiment III

Ferum, Fe is more electropositive than copper, Cu

Ferum, Fe release electrons to form Ferum(II) ions, Fe²⁺

Ferum, Fe undergo rusting/oxidation

# data[].attributes.headline
role of each substance in term of change of the oxidation number

# data[].attributes.subject
role of each substance in term of change of the oxidation number

# data[].attributes.body
Iron(II) sulphate, FeSO4 acts a reducing agent

Iron(II) sulphate, FeSO4 helps halogen X to reduce its oxidation number from 0 to -1

Halogen X acts as an oxidising agent

Halogen X helps iron(II) sulphate, FeSO4 to oxidise its oxidation number from +2 to +3

# data[].attributes.body
negative terminal:
Fe2+ -> Fe³⁺ + e-
positive terminal:
Cl2 + 2e- -> 2Cl⁻

# data[].attributes.body
2Fe²⁺ + Cl2 + 2e- -> 2Fe3+ + 2e⁻ + 2Cl
2Fe²⁺ + Cl2  -> 2Fe3+ + 2Cl

# data[].attributes.headline
b) Describe the standard electrode potential and determine oxidising and reducing agent based on their value of standard electrode potentials.

# data[].attributes.subject
b) Describe the standard electrode potential and determine oxidising and reducing agent based on their value of standard electrode potentials.

# data[].attributes.body
Dip X electrode into a 1.0 moldm^-3 of X nitrate solution until half full while connected to a standard hydrogen electrode and a salt bridge.

Since the standard hydrogen potential is 0.00 V, if the voltmeter reading shows positive value, X will act as an oxidising agent.

If the voltmeter reading shows negative value, X will act as a reducing agent.

If E0 value of X is negative

Half equation of oxidation at negative terminal: X2 -> 2X⁺ + 2e⁻

If E0 value of X is positive

Half equation of reduction at negative terminal: 2H⁺ + 2e- -> H₂

Half equation of oxidation at positive terminal: H2 -> 2H⁺ + 2e⁻

Movement of electron in wire:

The electrons flow throughthe external circuit from the negative terminal to the positive terminal of the cells.

Hence:

if the E0 value of X is positive, then X is an oxidising agent.

if the E0 value of X is negative, then X is a reducing agent.

# data[].attributes.body
Set I
oxidation= copper
reduction= silver ion, Ag⁺
half equation oxidation:  Cu -> Cu ²⁺ + 2e⁻
half equation reduction: Ag⁺ + e⁻ -> Ag
observation:

The solution become blue

grey solid deposited

Set II
oxidation= zinc
reduction= iron(III) ion, Fe ³⁺
half equation oxidation:  Zn -> Zn ²⁺ + 2e⁻
half equation reduction: Fe ³⁺ + e⁻ -> Fe ²⁺
observation:

zinc plate become thinner

solution turns brown to green

# data[].attributes.body
metal used: iron nails, magnesium ribbon, tin strip, copper strip
materials: agar solution, phenolphthalein, potassium hexacyanoferrate (III) solution
procedure:

1. label 3 test tube with I, II and III

2. clean all 3 iron nails, magnesium ribbon, copper strips and tin strips with sand paper

3. coil the three iron nails with tin strip, magnesium ribbon and copper strips respectively

4. place iron nail on three separate test tube as shown in diagram

5. pour the same amount of hot agar solution containing phenolphthalein and potassium hexacyanoferrate (III) solution

6. keep the test tubes left aside for a day

Observation:
Set I - low intensity of blue colour
Set II - low intensity of pink colour
Set III - high intensity of blue colour and low intensity of pink colour

Conclusion: Iron rusts it is in contact with it less electropositive metal while iron does not rusts when it is contacts with more electropositive metal

# data[].attributes.body
rusting of iron occur with the presence water and oxygen

rusting is a redox reaction where oxygen acts as an oxidising agent and iron as reducing agent

the surface area in the centre of water droplet with lower concentration of oxygen become anode

oxidation occurs and iron release electron to form iron(II) ion [Fe -> Fe²⁺ + e⁻]

the edge of water droplet with higher concentrations of oxygen become cathode

reduction occurs and oxygen accept electron to form hydroxide ion [O₂ + H₂O + 4e⁻ -> 4OH⁻]

iron(ii) ion produce react with hydroxide ion to form iron(ii) hydroxide [Fe²⁺ + OH⁻ -> Fe(OH)₂]

iron(ii) hydroxide undergoes continuous oxidation with oxygen to form hydrated iron(iii) oxide, Fe₂O3.XH₂O

thus, iron(iii) oxide is a rust as iron(iii) ion is brown colour

# data[].attributes.headline
b) describe the method to extract metal from its ores

# data[].attributes.subject
b) describe the method to extract metal from its ores

# data[].attributes.body
position in reactivity series of metal determine the method use to extract the metal from ore

potassium, sodium, magnesium and aluminium are more reactive than carbon. thus electrolysis method is used to extract metal

electrolysis is a redox reaction where oxide ion release electron to form oxygen gas and metal ion receive electron to form metal atom

zinc, iron, tin and lead are less reactive than carbon. thus, method of carbon reduction is used via blast furnace

carbon monoxide acts as a reducing agent and undergoes oxidation to form carbon dioxide

carbon monoxide reduce metal oxide to form metal atom

copper and mercury are extracted by direct heating only

lastly, silver and gold exists as elements. thus, no extraction method is needed

# data[].attributes.body
reaction X is not a redox reaction because there is no change of oxidation number for all elements in the compounds of reactants and products

reaction Y is redox because the oxidation number if zinc and copper in the substance have changed

the oxidation number zinc change from 0 to +2

the oxidation number of copper change from +2 to 0

# data[].attributes.headline
i) Determine the oxidation number of iron in both compounds and name the compounds based on IUPAC nomenclature.

# data[].attributes.subject
i) Determine the oxidation number of iron in both compounds and name the compounds based on IUPAC nomenclature.

# data[].attributes.body
R:

oxidation number = +2

iron(ii) chloride

S

oxidation number = +3

iron(iii) chloride

# data[].attributes.headline
ii) Compound R can be converted to compound S in the presence of an oxidising agent.Suggest the oxidising agent and state one observation for the reaction.

# data[].attributes.subject
ii) Compound R can be converted to compound S in the presence of an oxidising agent.Suggest the oxidising agent and state one observation for the reaction.

# data[].attributes.body
oxidising agent - bromine water
observation:

iron(ii) chloride change from green colour to brown

brown precipitate is formed

# data[].attributes.body
Materials and apparatus : 1.0 mol dm^-3 of sulphuric acid, H2S04, 1.0 mol dm-3 of bromine water, 1.0 mol dm^-3 of iron (II) sulphate, U-tube, connecting wires with crocodile clips,galvanometer,retort stand,carbon electrode, dropper and test tube.
Procedure:

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

2. Fill one arm of the U-tube with iron (II) sulphate and the other arm of U-tube with Bromine water by using a dropper.

3. Dip the carbon electrode to the solutions and connect to galvanometer using connecting wore

4. Observe the galvanometer pointer and the colour change of the solutions

5. The deflection of the galvanometer needle shows that there is transfer of electrons
Observation

Observation:

The colour of iron (II) sulphate change from green to brown The colour of bromine water change from brown to colourless

The pointer of galvanometer is deflect

Conclusion:
Transfer of electrons occurs from the iron (II) sulphate to bromine water through connecting wire.

# data[].attributes.headline
write the half equation at terminals in cell P and cell Q

# data[].attributes.subject
write the half equation at terminals in cell P and cell Q

# data[].attributes.body
Cell P:
anode: 4OH⁻ -> O₂ + 2H₂O + 4e⁻
cathode: Cu²⁺ + 2e⁻ -> Cu
Cell Q:
anode: Mg -> Mg²⁺ + 2e⁻
cathode: Cu²⁺ + 2e⁻ -> Cu

# data[].attributes.body
Cell P :
Anode: oxiation reaction
Cathode : reduction reaction
Cell Q :
Anode: oxidation reaction
Cathode : reduction reaction

# data[].attributes.body
Cell P:
Anode : colourless bubble gass released
Cathode : brown solid deposited
Cell Q:
Anode : magnesium electrode become thinner
Cathode : copper electrode become thicker

# data[].attributes.headline
b) A student dissolved iron(II) sulphate salt in distilled water and a green solution formed. But in a few days, the colour of solution change to brown.By using a method of transfer of electrons at a distance, describe how the student can change back the brown solution to green.In your description, suggest the suitable chemicals, apparatus, diagram and procedure.

# data[].attributes.subject
b) A student dissolved iron(II) sulphate salt in distilled water and a green solution formed. But in a few days, the colour of solution change to brown.By using a method of transfer of electrons at a distance, describe how the student can change back the brown solution to green.In your description, suggest the suitable chemicals, apparatus, diagram and procedure.

# data[].attributes.body
Materials: 0.5 mol of Iron(iii) sulphate, Fe₂(S0₄)₃ solutions, 0.5 mol of Potassium iodide, KI solutions, 1 mol of sulphuric acid, H₂S04 solution, 0.5 mol of sodium hydroxide solution.
Apparatus: U-tube, connecting wires with crocodile clips, galvanometer, retort stand, carbon electrodes, dropper, and test tube.
Procedure:

1. Measure and pour 1 mol of sulphuric acid into the u-tube until half full and clamp it vertically.

2. Drop carefully potassium iodide, KI solution into arm X of the U-tube using a dropper until reaches height of 3 cm.

3. Drop carefully iron(iii) sulphate solution into arm Y of the u-tube using a dropper until reaches height of 3 cm.

4. connect the carbon electrodes with the galvanometer using the connecting wires.

5. Dip one of the carbon electrodes into the potassium iodide solution, KI while the other carbon electrode into the ironliii) sulphate solution.

6. observe the direction of deflection of the galvanometer needle and determine the negative and positive terminals for each electrode.

7. leave the apparatus set-up for 30 minutes.

8. record the obsevations

9. after 30 minutes, draw out the solution in arm Y of the u-tube using a dropper and pour it into another test tube.

10. add 0.5 mol of sodium hydroxide into the test tube.

11. record the obsevations.

Observations:

deflection of the galvanometer needle is from arm X to arm Y

Brown colour of Iron (iii) sulphate solutions turns to green.

Green solid percipitate formed in the test tube

Half equation at negative terminal: 2I⁻ -> I₂ + 2e⁻

half equation at positive terminal : Fe²⁺ + e⁻ -> Fe²⁺

colourless solution of potassium iodide turns to brown

Conclusions:

Carbon electrode in arm X is the negative terminal and Carbon electrode in arm Y is the positive terminal.

Iron(ii) ion is present