My distinguished padlet by Khalida Azman https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9 Made with the strength to succeed en-us 2021-08-25 07:26:39 UTC 2023-03-24 19:07:55 UTC hello@padlet.com Q7 qgyscvjw6w https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695600502 (a) SIMILARITIES
  • Electrodes are dipped into electrolyte
  • Oxidation reaction at the anodes
  • Reduction reaction at the cathodes
  • Electron flow from the anodes to the cathodes through the connecting wires
      DIFFERENCES
1. Cell P
  •   Half equation
    • Copper A = Cu -> Cu2+ + 2e-
    • Copper B = Cu2+ + 2e- -> Cu
  • Reaction occur
    • Copper A = Oxidation
    • Copper B = Reduction
  • Observation
    • Copper A becomes thinner
    • Coppers B thickens
    • Blue solution remain unchanged
 1. Cell Q
  •   Half equation
    • Copper = Cu2+ + 2e- -> Cu
    • Magnesium = Mg -> Mg2+ + 2e-
  • Reaction occur
    • Copper = Reduction
    • Magnesium = Oxidation
  • Observation
    • 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

Diagram : (refer above)

Procedure :
  1. Measure and pour 1 mol dm-3 of sulphuric acid, H2SO4 into the U-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 Fe2+ :
  • 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
]]> 2021-08-25 07:31:40 UTC https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695600502 Q6 qgyscvjw6w https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695600625 (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

(c) Reducing agent: Potassium iodide 
      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
]]> 2021-08-25 07:31:45 UTC https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695600625 Q5 qgyscvjw6w https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695600745
  • The surface of iron with lower concentration of oxygen becomes anode in which oxidation occurs.
  • Iron atom release electrons to form Fe2+.
    • 4Fe → Fe2+ + 4e- (half equation at anode)

    1. Electrons flow through iron to the end which has more concentration of oxygen.
    2. The end becomes a cathode which reduction occurs.
    3. Oxygen molecules gain electrons and reduce to form OH-. 
    4e- +O2 + H20 → 4OH- (half equation at cathode) 

    1. Fe2+ then combines with OH- to form iron(ii)hydroxide.
    Fe2+ + 4OH- → Fe(OH)2

    1. Iron(ii)hydroxide then undergoes further oxidation by oxygen to form Fe2O3.xH20. 
    Fe(OH)2 → Fe2O3.xH2O
    Thus rust happens.



    b) explain the extraction of metal from its ore 
     (draw electrolysis)
    1. extraction of metals using electrolysis for metals more reactive than carbon. Exp aluminium ore
    2. At anode oxide ions donate electrons and undergoes oxidation.
    half eqn : 2O2- -> 02 + 4e-
    3. at cathode Al3+ is reduced by receiving electrons to form molten aluminium, Al. Reduction of half eqn at cathode : Al3+ + 3e- -> Al
    4. Ionic eqn : 6O2- + 4Al3+ -> 3O2 + 4Al


    ]]>     2021-08-25 07:31:50 UTC https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695600745     Q4 qgyscvjw6w https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695601089 (a)
    Set 1
    •  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 2
    • 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), K3Fe(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(III) 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 II - Magnesium undergoes oxidation or corrosion instead of iron. 
      • Set III - Iron undergoes oxidation or corrosion instead of copper 
    ]]>     2021-08-25 07:32:03 UTC https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695601089     Q3 qgyscvjw6w https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695601191  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+ -> 2Fe 3+ + 2e
     At electrode M :
     ⁃ Br2 + 2e -> 2Br-

    Overall ionic equation:
     ⁃ 2Fe 2+ + Br2 -> 2Fe 3+ + 2Br-

    b) 

    -*lukis diagram standard hydrogen potential + label ( boleh rujuk buku teks / tele)*
    - Half equation of hydrogen 
    2H+ + 2e- <=> H2
    - E* value of standard hydrogen electrode potential 
    E* = 0.00V

    -*lukis diagram standrad electride potential + label ( rujuk diagram kat atas)*
    - 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 


    ]]>     2021-08-25 07:32:07 UTC https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695601191     Q2 qgyscvjw6w https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695601349 a)i) oxidation number if aluminium: +3
           oxidation number of iron: +3
       ii) Al2O3: Aluminium oxide
           Fe2O3: Iron (III) oxide
      iii) - Aluminium has only one oxidation number whereas iron has 2 oxidation number.
    - The roman numeral III indicated 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 whereas aluminium is not a transition element that as only one oxidation number.

    b)i) Experiment I: reducing agent
           Experiment II: 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 oxidizing agent
    • Iron (II) ions, Fe2+ undergoes oxidation process while manganate (VII) ions, Mn04- undergoes a reduction process
    • Half equation oxidation: Fe2+ → Fe3+ + 2e- 
    • Half  equation reduction: Mn04- + 8H+ + 5e → Mn 2+ + 4H2O
    • Iron (II) ion, Fe2+ enode value is lower than manganate ion (VII), Mn04- ion
    • Iron (II), Fe2+ releases 2 electrons and is oxidized to iron (III) ion, Fe3+ Manganate (VII) ion, MnO4- receive 7 electrons and  reduced to manganese (II) ions, Mn2+.
    Experiment II 
    • Magnesium, Mg is a reducing agent while iron (II) ions, Fe2+ is an oxidizing agent
    • Magnesium, Mg undergoes a temporary oxidation process while iron (II) ions,  Fe2+ undergoes a reduction process
    • Half equation oxidation: Mg → Mg2+ + 2e
    • Half equation reduction: Fe 2+ + 2e → Fe
    • Enode value of Magnesium ion, Mg2+ s is lower than iron(II) ion, Fe2+
    • Magnesium releases 2 electrons to produce magnesium ion, Mg2+.  Iron (II) ions, Fe2+ receive 2 electrons to produce iron atom, Fe.
    Experiment III
    • Iron atom is a reducing agent
    • Iron atom undergoes oxidation process
    • Half equation oxidation: Fe → Fe2+ + 2e- 
    • Enode value of iron metal, Fe is lower than copper metal, Cu
    • Iron atom releases 2 electrons to form iron(II) ions, Fe2+
    ]]>     2021-08-25 07:32:13 UTC https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695601349     Q1 qgyscvjw6w https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695601449 i) Ag, Y , X
       - X can displace silver nitrate 
       - Because X is more electropositive than silver
       - Y can displace silver nitrate
       - Because Y is more electropositive than silver
       - Y cannot displace X nitrate solution
       - Because Y is less electropositive than X

    ii) Copper (II) nitrate, 
        Half equation of oxidation : Cu→ Cu2+ + 2e-
        Half equation of reduction : 2Ag+ + 2e- → 2Ag
        Overall ionic equation: Cu + 2Ag+ → Cu2+ + 2Ag

    b) 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 H2 increase from 0 to +1
        - Cu2+ ion is oxidising agent 
        - H2 is reducing agent

    c) Magnesium 
        - When iron is in contact with more electropositive metal like magnesium, rusting of iron is slower
        - Magnesium atom release electron to form magnesium ion 
       - Magnesium corrodes or undergoes oxidation instead of iron 
       - The observation is high intensity of pink spot and no blue  spots is present in the test tube
       - Half equation of oxidation : Mg → Mg2+ + 2e-
       - Half equation of reduction : 2H2O + O2 + 4e- → 4OH-
    ]]>     2021-08-25 07:32:18 UTC https://padlet.com/qgyscvjw6w/6zdukjry63i31ze9/wish/1695601449