Essay Cluster 2 by NUR BASYIRA https://padlet.com/m3920823/essaycluster2 Form 5 Chapter 1 - Redox en-us 2021-08-26 00:11:23 UTC 2025-10-29 16:03:22 UTC hello@padlet.com https://padlet-uploads.storage.googleapis.com/1307538379/6f282e17245c61d46faace3d5e66ad8b/Loading_Chemistry___Quiz_Results_Animation.gif m3920823 https://padlet.com/m3920823/essaycluster2/wish/1697075753 2021-08-26 00:27:44 UTC https://padlet.com/m3920823/essaycluster2/wish/1697075753 i) Based on the results, arrange the three metals in order of electropositivity. Explain your answer. m3920823 https://padlet.com/m3920823/essaycluster2/wish/1697092889
  • 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
    • ]]>     2021-08-26 00:34:26 UTC https://padlet.com/m3920823/essaycluster2/wish/1697092889     ii) If Y is copper, name the product formed in Set II. Write the ionic equation for the reaction that take place. m3920823 https://padlet.com/m3920823/essaycluster2/wish/1697114022 2021-08-26 00:41:47 UTC https://padlet.com/m3920823/essaycluster2/wish/1697114022 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. m3920823 https://padlet.com/m3920823/essaycluster2/wish/1697123895
  • 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
    • ]]>     2021-08-26 00:44:48 UTC https://padlet.com/m3920823/essaycluster2/wish/1697123895     c) m3920823 https://padlet.com/m3920823/essaycluster2/wish/1697137019
  • 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-
    • ]]>     2021-08-26 00:49:34 UTC https://padlet.com/m3920823/essaycluster2/wish/1697137019     m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698278614 Aluminium = +3
    Iron = +3]]>     2021-08-26 11:30:31 UTC https://padlet.com/m3920823/essaycluster2/wish/1698278614     i) Name both the compounds based on the IUPAC nomenclature system. m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698280501 Aluminium oxide
    Iron(III) oxide]]>     2021-08-26 11:32:40 UTC https://padlet.com/m3920823/essaycluster2/wish/1698280501     ii) Explain the difference between the names of the two compounds based on the IUPAC nomenclature system. m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698282608
  • 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
    • ]]>     2021-08-26 11:35:03 UTC https://padlet.com/m3920823/essaycluster2/wish/1698282608     m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698283461 2021-08-26 11:36:00 UTC https://padlet.com/m3920823/essaycluster2/wish/1698283461 i) State the role of ferum(II) sulphate solution in Experiment I and Experiment II m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698285475 Experiment I - reducing agent
    Experiment II - oxidizing agent]]>     2021-08-26 11:38:06 UTC https://padlet.com/m3920823/essaycluster2/wish/1698285475     ii) Write the half equations of oxidation and reduction in Experiment II or Experiment III. m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698307698 Experiment I
    oxidation: Mg -> Mg2+ + 2e-
    reduction: Fe2+ + 2e- -> Fe
    Experiment II
    oxidation: Fe -> Fe2+ + 2e-
    reduction: 2H2O + O2 + 4e- -> 4OH-]]>     2021-08-26 12:01:45 UTC https://padlet.com/m3920823/essaycluster2/wish/1698307698     c) Explain the differences in the observation for Experiment I, II andIII based on Redox reaction. m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698318031 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
    ]]>     2021-08-26 12:11:25 UTC https://padlet.com/m3920823/essaycluster2/wish/1698318031     m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698327688 2021-08-26 12:19:08 UTC https://padlet.com/m3920823/essaycluster2/wish/1698327688 role of each substance in term of change of the oxidation number m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698334457
  • 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
    • ]]>     2021-08-26 12:23:10 UTC https://padlet.com/m3920823/essaycluster2/wish/1698334457     half equation at each electrode m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698336869 negative terminal:
    Fe2+ -> Fe3+ + e-
    positive terminal:
    Cl2 + 2e- -> 2Cl-]]>     2021-08-26 12:24:52 UTC https://padlet.com/m3920823/essaycluster2/wish/1698336869     overall ionic equation for the reaction m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698340084 2Fe2+ + Cl2 + 2e- -> 2Fe3+ + 2e- + 2Cl
    2Fe2+ + Cl2  -> 2Fe3+ + 2Cl]]>     2021-08-26 12:27:13 UTC https://padlet.com/m3920823/essaycluster2/wish/1698340084     b) Describe the standard electrode potential and determine oxidising and reducing agent based on their value of standard electrode potentials. m3920823 https://padlet.com/m3920823/essaycluster2/wish/1698510121
  • 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.
    ]]>     2021-08-26 13:59:21 UTC https://padlet.com/m3920823/essaycluster2/wish/1698510121     m3920823 https://padlet.com/m3920823/essaycluster2/wish/1699475881 Set I
    oxidation= copper
    reduction= silver ion, Ag+
    half equation oxidation:  Cu -> Cu2+ + 2e-
    half equation reduction: Ag+ + e- -> Ag
    observation:
    • The solution become blue
    • grey solid deposited
    Set II
    oxidation= zinc
    reduction= iron(III) ion, Fe3+
    half equation oxidation:  Zn -> Zn2+ + 2e-
    half equation reduction: Fe3+ + e- -> Fe2+
    observation:
    • zinc plate become thinner
    • solution turns brown to green
    ]]>     2021-08-27 00:17:19 UTC https://padlet.com/m3920823/essaycluster2/wish/1699475881     m3920823 https://padlet.com/m3920823/essaycluster2/wish/1699585942 2021-08-27 00:59:11 UTC https://padlet.com/m3920823/essaycluster2/wish/1699585942 m3920823 https://padlet.com/m3920823/essaycluster2/wish/1700183317 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
    ]]>     2021-08-27 05:07:13 UTC https://padlet.com/m3920823/essaycluster2/wish/1700183317     m3920823 https://padlet.com/m3920823/essaycluster2/wish/1700186365
  • 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
    • ]]>     2021-08-27 05:08:53 UTC https://padlet.com/m3920823/essaycluster2/wish/1700186365     b) describe the method to extract metal from its ores m3920823 https://padlet.com/m3920823/essaycluster2/wish/1700252451
  • 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
    • ]]>     2021-08-27 05:48:44 UTC https://padlet.com/m3920823/essaycluster2/wish/1700252451     m3920823 https://padlet.com/m3920823/essaycluster2/wish/1700253985
  • 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
    • ]]>     2021-08-27 05:49:47 UTC https://padlet.com/m3920823/essaycluster2/wish/1700253985     m3920823 https://padlet.com/m3920823/essaycluster2/wish/1700318447 2021-08-27 06:27:27 UTC https://padlet.com/m3920823/essaycluster2/wish/1700318447 i) Determine the oxidation number of iron in both compounds and name the compounds based on IUPAC nomenclature. m3920823 https://padlet.com/m3920823/essaycluster2/wish/1700320728 R:
    • oxidation number = +2
    • iron(ii) chloride
    S
    • oxidation number = +3
    • iron(iii) chloride
    ]]>     2021-08-27 06:29:15 UTC https://padlet.com/m3920823/essaycluster2/wish/1700320728     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. m3920823 https://padlet.com/m3920823/essaycluster2/wish/1700323177 oxidising agent - bromine water
    observation: 
    • iron(ii) chloride change from green colour to brown
    • brown precipitate is formed
    ]]>     2021-08-27 06:31:00 UTC https://padlet.com/m3920823/essaycluster2/wish/1700323177     m3920823 https://padlet.com/m3920823/essaycluster2/wish/1700356379 2021-08-27 06:54:05 UTC https://padlet.com/m3920823/essaycluster2/wish/1700356379 m3920823 https://padlet.com/m3920823/essaycluster2/wish/1700357617 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.

    ]]>     2021-08-27 06:55:00 UTC https://padlet.com/m3920823/essaycluster2/wish/1700357617     m3920823 https://padlet.com/m3920823/essaycluster2/wish/1701101417 2021-08-27 16:21:09 UTC https://padlet.com/m3920823/essaycluster2/wish/1701101417 write the half equation at terminals in cell P and cell Q m3920823 https://padlet.com/m3920823/essaycluster2/wish/1701102889 Cell P:
    anode: 4OH- -> O2 + 2H2O + 4e- 
    cathode: Cu2+ + 2e- -> Cu
    Cell Q:
    anode: Mg -> Mg2+ + 2e-
    cathode: Cu2+ + 2e- -> Cu]]>     2021-08-27 16:22:14 UTC https://padlet.com/m3920823/essaycluster2/wish/1701102889     name the reaction occur m3920823 https://padlet.com/m3920823/essaycluster2/wish/1701103165 Cell P :
    Anode: oxiation reaction
    Cathode : reduction reaction
    Cell Q :
    Anode: oxidation reaction 
    Cathode : reduction reaction]]>     2021-08-27 16:22:26 UTC https://padlet.com/m3920823/essaycluster2/wish/1701103165     state the observation m3920823 https://padlet.com/m3920823/essaycluster2/wish/1701103415 Cell P:
    Anode : colourless bubble gass released
    Cathode : brown solid deposited
    Cell Q:
    Anode : magnesium electrode become thinner
    Cathode : copper electrode become thicker]]>     2021-08-27 16:22:37 UTC https://padlet.com/m3920823/essaycluster2/wish/1701103415     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. m3920823 https://padlet.com/m3920823/essaycluster2/wish/1701104537 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

    ]]>     2021-08-27 16:23:30 UTC https://padlet.com/m3920823/essaycluster2/wish/1701104537