The document discusses redox reactions, including definitions of oxidation, reduction, oxidizing agents and reducing agents. It provides examples of redox reactions like magnesium burning in oxygen, bromine water oxidizing iron(II) ions, and the displacement of metals in salt solutions. It also describes how electrons are transferred during redox reactions and how redox reactions can occur at a distance using a salt bridge.

1. Mg 2+ O 2- KMnO 4

2. Redox reactions are a chemical reactions involving simultaneously (serentak) oxidation and reduction processes Oxidation & reduction : a addition (gain) or loss (elimination) of oxygen or hydrogen accepting(receives) or donating of electrons change in oxidation number A Redox Reactions : Oxidation is the process of gaining oxygen & elimination of hydrogen reduction is the process of losing oxygen & addition of hydrogen oxidising agent is the substance which experiences reduction & receives electrons . reducing agent is the substance which experiences oxidation & donates electrons .

3. PbO – oxidising agent (experiences reduction Mg – reducing agent ( experiences oxidation) Mg + PbO MgO + Pb Gain of oxygen ---- oxidation Loss of oxygen ---- reduction

4. Cl 2 – oxidising agent ( undergoes reduction ) --- chlorine oxidises hydrogen sulphide to sulphur H 2 S – reducing agent ( undergoes oxidation) --- hydrogen sulphide reduces chlorine to hydrogen chloride Redox reaction H 2 S + Cl 2 2HCl + S Loss of hydrogen --- oxidation Gain of hydrogen --- reduction

5. B Transfer of Electrons oxidation involves the loss (releases) of electrons reduction involves a gain ( receives) in electrons Chlorine – oxidising agent Sodium – reducing agent metals are oxidised & its loss their electrons to form cations non- metal are reduced & its receive electrons to form anions . 2Na(s) + Cl 2 (g) 2NaCl(s) Na Na + + e ( loss of electron) Cl 2 + 2e - 2Cl - ( gain of electron) Oxidation Process Reduction Process

7. Solution : Determine the oxidation and reduction process , oxidising and reducing agent that occurs in the reactions below . Mg + PbO MgO + Pb Anode : Cu Cu 2+ + 2e - ; Cu 2+ + 2e - Cu 2CuO + C 2Cu + CO 2 Fe 2 O 3 + 3C 2Fe + 3CO Mg + CuO MgO + Cu

8. oxidation is the increase in oxidation number reduction is the decrease in the oxidation number O.N of ions is same value to the charge of the ion. Ex : Na + , K + , H + is +1 Mg 2+ , Ca +2 is +2 O 2- , S 2- is -2 O.N for atom or molecule in a neutral elements are zero ( 0 ) example : O 2 , N 2 , Na , Mg, Br 2 is 0 . O.N Reducing agent --- iron (II) chloride Oxidising agent --- chlorine gas C : Change in Oxidation Number ( O.N ) 2FeCl 2 + Cl 2 2FeCl 3 Oxidation number decreases (0 -> -1) Oxidation number increases ( +2 -> +3) reduction oxidation +2 +3 -1 0 -1

9. (i) The total oxidation number of all the atoms is equal to the charge on the ion . (ii) the total oxidation numbers for all atoms in neutral compound is zero . Example : (i) ClO 3 - , oxidation number of chlorine is X X + 3(-2) = -1 X - 6 = -1 X = +5 (ii) The oxidation number of S in MgSO 4 +2 + X + 4 (-2) = 0 +2 + X - 8 = 0 X = + 6 the total oxidation number for dichromate (VI) ion, Cr 2 O 7 2- is -2 , manganate (VII) , MnO 4 -1 is -1 The charge of chlorate Oxidation number of S

10. Test Yourself : Calculate the oxidation number of the following elements : Manganese , Mn in potassium manganate (VII) , KMnO 4 Manganese, Mn in manganate(VII) ion, MnO 4 - Chromium, Cr, in potassium dichromate(VI), K 2 Cr 2 O 7 Cromium, Cr,in chromate(VI) ion, CrO 4 2- Iron in iron(II) chloride , FeCl 2 Iron in iron(III) chloride , FeCl 3 Carbon , C in sodium carbonate, Na 2 CO 3 In each of the cases above, the oxidation number of each element is represented by the value of X .

11. The oxidising agent is the substance that receives electrons experiences a reduction(pengurangan) in the oxidation number . The reducing agent is the substance that : loses electrons experiences an increase (penambahan) in the oxidation number . Example : Fe 2+ Fe 3+ + e - (O.N ) +2 +3 Br 2 + 2e - 2Br - (O.N) 0 -1 Make sure that you add the electron on the side of the half equation that has the bigger oxidation number

12. Redox reactions need to shown in the form of : half - equations ionic equations Fe 2+ Fe 3+ Cl - Cl 2 Br - I - Br 2 Green solution Brown solution Yellow solution Colourless solution MnO 4 - Purple solution Cr 2 O 7 2- Cu 2+ Orange solution Blue solution Cr 3+ I 2 Example : Half equation : Fe 2+ Fe 3+ + e - -------- (1) X 2 Br 2 + 2e - 2Br - -------- (2) Ionic equation : 2Fe 2+ + Br 2 2Fe 3+ + 2Br - Combined

13. A . Redox Reaction ( The combustion of Magnesium in oxygen) Oxygen oxidises magnesium to magnesium ion . Magnesium releases electrons to form Magnesium ion . Half –equation : Mg Mg 2+ + 2e - ------------(1) O.Number 0 +2 (oxidation) Magnesium reduces oxygen to oxide ion . Oxygen atom receives electrons to form oxide ion . Half- equation : O 2 + 2e - O 2- ------------(2) O.Number 0 -2 (reduction ) The overall equation : (1) + (2) 2Mg + O 2 2MgO Observation :- Combusts with a white shiny flame A white solid is formed Oxidising agent : Oxygen Reducing agent : Magnesium combined Reactants

14. Bromine water oxidises iron(II) ion , Fe 2+ to iron(III) ion , Fe 3+ Iron(II) ion releases electron to form iron(III) ion. Half-equation : Fe 2+ Fe 3+ + e - ------------(1)X 2 O. Number +2 +3 ( oxidation) Iron(II) ion reduces bromine ,Br 2 to bromide ion , Br - Bromine receives electron to form bromide ion , Br -1 Half-equation : Br 2 + 2e - 2Br - ------------(2) O.Number 0 -1 ( reduction ) Observation : The green iron(II) sulphate solution changes to brown . The brown coloured bromine water is decolourised . Oxidising agent : Bromine water , Br 2 Reducing agent : Iron(II) ion, Fe 2+ Ionic Equation : 2 Fe 2+ + Br 2 2Fe 3+ + 2Br - other oxidising agent : Cl 2 ,KMnO 4 ,K 2 Cr 2 O 7 ,HNO 3 concentrated,H 2 O 2 Combined B. The change of iron(II) ions to iron(III) ions ( Fe 2+ Fe 3+ )

15. C : The change of iron(III) ion, Fe 3+ to iron(II) ion , Fe 2+ Iron (III) ion oxidises Zn atom to zinc ion , Zn 2+ Iron(III) ion receives electron to form iron(II) ion The brown iron(III) chloride solution changes to green . Half-equation : Fe 3+ + e - Fe 2+ --------(1) X 2 O.Number +3 +2 ( Reduction) Iron(III) ion -------- oxidising agent ( oxidation number decrease ) . Zn reduces iron(III) ion to iron(II) ion . Zinc atom releases electrons to form zinc ion , Zn Zinc powder dissolves . Zinc metal -------- reducing agent ( oxidation number increase ) Half-equation : Zn Zn 2+ + 2e - -------(2) O. Number 0 +2 ( oxidation ) The ionic equation : 2Fe 3+ + Zn 2Fe 2+ + Zn 2+ Other reducing agent : metals that are more electropositive than iron // SO 2 , H 2 S gas // Na 2 SO 3 , SnCl 2 solution combined

16. D : The Displacement (penyesaran) of Metal from its Salt Solution The element is more electropositive in the E.S, the higher the tendency (kecenderungan) to release electrons to form positive ions . More electropositive , oxidised more easily & act as a reducing agent The higher the position in the E.S. can displace other elements that are lower in the E.S . The displacement reaction between Zn & CuSO 4 solution . Zn more electropositive than copper . Zn releases two electron to form zinc ion , Zn 2+ Zn reduces copper(II) ion ,Cu 2+ to copper , Cu Copper(II) ion oxidises Zn to zinc ion ,Zn 2+ K, Na , Ca , Mg , Al , Zn , Fe , Sn , Pb , H , Cu , Hg , Ag , Au Most electropositive Least electropositive

17. Observation : The blue CuSO 4 solution fades or becomes colourless . A brown solid is formed . The Zn piece is corroded or dissolves . Copper is displaced by zinc from the copper(II) sulphate solution . Half-equations : Zn Zn 2+ + 2e - ----------(1) O . Number 0 +2 ( oxidation ) Cu 2+ + 2e Cu ----------(2) O. Number +2 0 ( reduction ) Ionic equation : Zn + Cu 2+ Zn 2+ + Cu Zn ------ reducing agent Copper(II) ion ------- oxidising agent Zn loses electrons & is oxidised to Zn 2+ Cu 2+ receives electrons & is reduced to Cu

18. E : Displacement of Halogens from Halide Solutions Halogen ----- Group 17 examples : Cl 2 ( chlorine) ------- yellow Br 2 ( bromine) ------- brown I 2 (iodine) ------- yellow or brown can be differentiated by shaking the solution with a little CCl 4 Halogens are reduced to halide ions Halogen ----- oxidising agent The more reactive halogen can displace less reactive halogens from its halide solutions. Group 17 : Flourine Chlorine Bromine Iodine Solution Reactivity decreases, higher act as a oxidising agent

19. Example Chlorine water react with sodium bromide solution Chlorine water , Cl 2 oxidises bromide ion, Br - to bromine ,Br 2 Bromide ion , reduces chlorine , Cl 2 to chloride ion , Cl - Bromide ion, Br - releases electrons to form bromine ,Br 2 Half-equation : 2Br - Br 2 + 2e - ------ (1) -1 0 oxidation colourless brown Chlorine , Cl 2 receives electrons to form chloride ion , Cl - Half-equation : Cl 2 + 2e - 2Cl - ------- (2) 0 -1 reduction yellow decolourised(colourless) Ionic equation : (1) + (2) Cl 2 + 2Br - Br 2 + 2Cl - Chlorine ----- oxidising agent Bromide ion ----- reducing agent Chlorine displaces bromine from the sodium bromide solution.

20. Confirmatory Test for the Bromine, Chlorine and Iodine By adding and shaking the halogen solution in tetrachloromethane (CCl 4 ) liquid Solution Colour in water Colour in CCl 4 Concentrated Dilute Iodine Brown Yellow Purple Bromine Brown Yellow Brown Chlorine Light greenish yellow Colourless Colourless

21. F : Transfer of Electrons at a Distance If two chemicals are separated at a distance by an electrolyte solution in a U-tube acts as a salt bridge . used to separate two solutions but allows ions to pass (flow) through to complete the circuit . examples : H 2 SO 4 , KNO 3 , Na 2 SO 4 solution The electrons that are released from reducing agent (negative electrode) will flow out through outer circuit to the oxidising agent ( positive electrode) reduction oxd

22. Example : The Reaction Between Bromine Water and Iron(II) Sulphate solution Iron(II) ion, Fe 2+ releases electron & is oxidised to iron(III) ion , Fe 3+ Fe 2+ Fe 3+ + e -1 ---------- (1) O. Number + 2 +3 ( oxidation) The green solution ,(Fe 2+ ) changes to brown , Fe 3+ The electrons that are released collect at the carbon electrode that is immersed in FeSO 4 It act as the negative terminal . Bromine ,Br 2 receives electron & is reduced to bromide ion, Br -1 Half equation : Br 2 + 2e - 2Br -1 --------------(2) brown colour decolourised O.Number 0 -1 (reduction) the carbon electrode in bromine water act as the positive terminal the ionic equation :- 2Fe 2+ + Br 2 2Fe 3+ + 2Br -1 O. Number +2 0 +3 -1

23. oxidising agent ----- Bromine water , Br reducing agent ----- Iron (II) ion, Fe The galvanometer needle is deflected because the movement of electrons from the negative electrode to the positive electrode produces an electric current . Reduction Oxidation (Positive terminal) (Negative terminal)

24. Test Yourself The figure shows a U-tube redox cell . Write a summary of the redox reaction for the reaction between Iron(II) sulphate, FeSO 4 solution and the acidified potassium manganate (VII) , KMnO 4 solution. Can dilute sulphuric acid be replaced with dilute hydrochloric acid ? Give the reason for your answer .

26. Observation : Electrode (-) : The green coloured iron(II) nitrate solution changes to brown Electrode (+) : The purple coloured acidified potassium manganate(VII) solution is decolourised . Half equation : Electrode (-) : Fe 2+ Fe 3+ + e - ----------- ( oxidation) Electrode (+) : MnO 4 - + 8H + + 5e - Mn 2+ + 4H 2 O (reduction) Ionic Equation : 5Fe 2 + MnO 4 - + 8H + Fe 3+ + Mn 2+ + 4H 2 O Oxidising agent : manganate(VII) ion Reducing agent : Iron (II) ion . Confirmatory test for the product( Fe ) that is formed. Add sodium hydroxide solution, a brown precipitate is formed . (b) Can . Hydrochloric acid also allows the transfer of ions to occur . Solution :

27. Redox Reaction in a simple voltaic cells The porous pot ( pasu berliang) --- to separate the two solutions but allows the ions to flow through it to complete circuit . the transfer of electrons occur from reducing agent to the oxidising agent through an outer circuit . The negative electrode ( anode ) ----- metal which is more electropositive in the E.S. The positive electrode ( cathode ) ----- metal which is less electropositive in the E. S. electron flows from the negative electrode to the positive electrode . two types of Daniell cell that uses a porous pot :

28. At the negative electrode( anode) : Zn is more electropositive than copper Zn has more tendency to releases two electrons to form zinc ion,Zn 2+ Zn rod acts as the negative electrode . Zn Zn 2+ + 2e ------ oxidation process occurs The electrons will flows from the zinc rod to the copper rod through the outer circuit an electric current is produced . At the cathode : copper ion, Cu 2+ receives two electrons to form copper atom, Cu & undergoes reduction process . Cu 2+ + 2e Cu ------ reduction process Copper(II) ion oxidises (mengoksidakan) zinc, Zn to zinc ion, Zn 2+ Cu 2+ ----------- oxidising agent Zinc reduces( menurunkan) copper(II) ion,Cu 2+ to copper atom, Cu Zn ------------- reducing agent

29. Overall Ionic Equation : Zn + Cu 2+ Zn 2+ + Cu Observation : cathode – the blue copper(II) sulphate solution becomes fade/ colourless --- a brown solid forms at the copper rod // the copper rod thickens // the mass of the copper will increases. anode ---- the zinc rod dissolves / corrodes/ becomes thinner(menipis) Cell symbol : Zn(s) / Zn 2+ (aq) // Cu 2+ (aq) / Cu(s) 0 +2 +2 0 Oxidation Reduction

31. G . Corrosion of Metals occur when a metal loses electrons & is oxidised to form the metal ion . the metal is corroded example : Iron loses electrons to form iron(II) ion , Fe 2+ Fe Fe 2+ + 2e -1 ------- oxidation O. Number 0 +2 Iron is corroded . If magnesium loses electrons to form magnesium ion Mg 2+ , magnesium is corroded. Mg Mg 2+ + 2e - ------- oxidation The metals is more electropositive in E.S. , corrode much easier . because the metals more tendency to release electrons to form metal ions Example : Al corrodes more easily compared to copper . because Al is more electropositive than copper . the rusting requires water and oxygen Metal corrosion

32. RUSTING OF IRON At the end of the water droplet ( Anode / negative terminal ) the iron , Fe loses electrons and is oxidised to iron(II) ion, Fe Iron(II) ion dissolves in water Iron is corroded . Fe Fe 2+ + 2e ------- oxidation The electrons flows to the edge(pinggir) of the water droplet through the iron Stage 1 corrosion

33. Iron(II) hydroxide , Fe(OH) 2 is then oxidised by oxygen to form hydrated Iron(III) oxide, (brown solid ) or rust . equation : Fe(OH) 2 Fe 2 O 3 .3H 2 O (rust) O 2 in the air Stage 4 Iron(II) ion , Fe 2+ & hydroxide ion , OH - combine to form iron(II) hydroxide ( green solid ) Fe 2+ (aq) + 2OH -1 (aq) Fe(OH) 2 (s) Stage 3 At the edge of the water droplet ( cathode / positive terminal ) Electrons are received by oxygen & water to form OH ions through reduction O 2 + 2H 2 O + 4e 4OH - ------ Reduction Stage 2 Iron rusting

34. is a process that occurs when two metals come into contact(bersentuhan) with an electrolyte . the more electropositive metal will donate (release) electrons & is corroded If the iron comes into contact with metal that is more electropositive ,it will not corrode . the corrosion of iron can be accelerated by the presence of electrolytes such as acid & salt solution. K Na Ca Mg Al Zn Fe Sn Pb H Cu Hg Ag Au Example : Electrochemical Corrosion of Metals More easily corroded Difficult to be corroded Tendency for corrosion increases

35. Example : The effect of rusting when iron comes into contact with other metals ( Mg, Cu , Zn , Sn) Hypothesis : Iron is protected from rusting when it comes in contact with more electropositive metals, but rusts when it contact with less electropositive metals . Potassium hexacyanoferrate(III) solution ---- to detect the presence of iron(II) ions . Iron nail is corroded , the dark blue colour will be seen in the solid agar . Phenolphthalein will turn pink in colour if OH - ions are present. The gas bubbles formed are hydrogen gas . The iron nail in test tube B is not corroded , while zinc is corroded because Zn is more electropositive than iron . In test tube A ,the iron nail is corroded because Iron more electropositive than copper . B A rusting

36. PREVENTION THE RUSTING OF IRON Coating a layer of metal such as Al or Sn on food tins Applying paint,oil or grease on surface such as engine Wrapping the iron with a layer of plastic . Ex: hangers Applying a coat of Al such as car bumpers or water pipes Iron sheet used as house roofs Are galvanised with a layer of zinc Iron is alloyed with other metals such as chromium or nickel to produce stainless steel Huge iron construction structure such as bridges protected from corrosion by using sacrificial metals(logam korban) such as Mg & Zn

37. Reactivity Series (R.S) of Metals A : Metals with Oxygen Metal is heated in oxygen to produce metal oxide .A more reactive metal will displace a less reactive metal from its oxide. Observation : Mg ----- burn very rapidly & vigorously with a very bright flame ------ metal oxide colour : white powder ( Hot & cold ) Zinc ---- burns rapidly , glows brightly ----- metal oxide colour : yellow when hot & white when it is cold . Iron ----- burns rapidly, glows less brightly than Zn ----- metal oxide colour brown when hot & cold Cu ----- very slow reaction ----- metal oxide colour black ( hot & cold ) Pb ----- burns slowly ----- metal oxide colour : brown when hot & yellow when colour

38. K Na Ca Mg Al C Zn Fe Sn Pb Cu Ag Au Reactivity decreases Example : The reaction between Lead(II) oxide with Carbon Observation : burn brightly : produces a grey solids Inference : Carbon is more reactive than Lead Equation : PbO + C Pb + CO 2 If carbon is more reactive than metal X , a flame or glows(baraan) can be seen . If carbon is less reactive than metal Y ,the flame or glows will not be seen when carbon react with metal oxide Y is heated . The position of Carbon in the R. S.

40. PREPARED BY PN ZAINAB BINTI AYUB 22 JULY 2008