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PRE UNIVERSITISEMESTER 2chapter 2electrochemistry2.1Oxidation numberOxidation numbers are a convenient way of determining if a substance has been oxidised or reduced. These numbers are assigned arbitrarily to atoms and are equal to the charge the atom would have if its bonds were purely ionic.All free atoms in element have an oxidation number of zero

For simple ions (and ionic compounds), the oxidation number is the same as the charge of ion

Na = Mg = H2 = Cl2 = P4 =K+ =Ca2+ =B3+ =P3- = O2- =F- =0 0 0 0 0+1+2+3 3 2 1For covalent compounds, the covalent bonds are changed into ionic bonds by assuming that the bonded electrons are on the more electronegative atom. Table below shows some elements oxidation number

ElementOxidation numberNotesGroup I+1--Group 2+2--Group 171 True only to halogen without O in itOxygen2Exception : 1 for peroxide and +2 for F2OHydrogen +1Except : metal hydride MH (H = -1)Na2OB2O3CO2SO3Cl2O7HFH2SNH3CH4H2CO3H2SO4HBrO3In a neutral molecule, the sum of the oxidation numbers of all atom are equals2 Na + O = 02(Na) + (-2) = 0Na = +12 B + 3 O = 02(B) + 3(-2) = 0B = +31 C + 2 O = 01(C) + 2(-2) = 0C = +41S + 3O = 01(S) + 3(-2) = 0S = +62Cl + 7O = 02(Cl) + 7(-2) = 0Cl = +71H + 1F = 01(F) + 1(+1) = 0F = -11S + 2H = 01(S) + 2(+1) = 0S = -21N + 3H = 01(N) + 3(+1) = 0N = -31C + 4H = 01(C) + 4(+1) = 0N = -4 2H + 1C + 3O = 02(+1) +1C + 3(-2) = 0 C = +4 2H + 1S + 4O = 02(+1) +1S + 4(-2) = 0 S = +6 1H + 1Br + 3O = 01(+1) +1Br + 3(-2) = 0 Br = +54.In a molecular ion, the sum of the oxidation numbers of all atoms in the formula unit equals to the charge on the ion.CrO4-Cr2O72-MnO4-C2O42-ClO2-HSO4- 1 Cr + 4 O = -11 Cr + 4(-2) = -1Cr = +7 2 Cr + 7 O = -22 Cr + 7 (-2) = -2Cr = +6 1 Mn + 4 O = -11 Mn + 4(-2) = -1Mn = +7 2 C + 4 O = -22 C + 4(-2) = -2C = +3 1 Cl + 2 O = -11 Cl + 2 (-2) = -1Cl = +31 H + 1 S + 4 O = -11(+1) + 1S + 4(-2) = -1S = +62.2Half equation and redox reaction.Half equation ~ equation which shows how electrons are accept / donate in a chemical reactionWhen a substance is oxidise, electron is . ; e- is written at the .. side equation.When a substance is reduce, electron is . ; e- is written as the side equation.Simple half equation : State the changes of oxidation number and write the half equation.ReactionOxidation no changeReaction Half equationNa Na+Mg Mg2+Al Al3+Cu2+ Cudonatedrightreceivedleft0 +1oxidationNa Na+ + e-0 +2oxidationMg Mg2+ + 2 e-0 +3oxidationAl Al3+ + 3 e-+2 0reductionCu2+ + 2 e- CuReactionOxi. no changeReaction Half equationH2 H+ Cl2 Cl-I- I2O2 O2-Fe2+ Fe3+Pb4+ Pb2+0 +1oxidationH2 2 H+ + 2 e- 0 -1reductionCl2 + 2 e- 2 Cl -1 0oxidation 2 I- I2 + 2 e- 0 -2reductionO2 + 4 e- 2 O2 +2 +3oxidationFe2+ Fe3+ + e-+4 +2reductionPb4+ + 2 e- Pb2+When it comes to the reaction involving molecular ion, the overall charge has to be balanced in such order.Write a skeleton half equation. Determine the reaction (oxidation or reduction) using oxidation numberBalance the charge by adding electrons at the appropriate sideBalance the number of atoms other than oxygen.Based on the changes in number of oxygen, write the number of water molecule formed/used.From the number of water molecule formed/used, write the number of hydrogen ion (H+) required.a)ClO3- Cl-

Half equation : b) CrO42- Cr3+Half equation : c)Cr2O72- Cr3+Half equation : Changes in OS : +5 -1 ; reduction Different in OS = 6, so 6 e- at the LHS of equationClO3- + 6 e- Cl-ClO3- + 6 e- Cl- + 3 H2O6 H+ + ClO3- + 6 e- Cl- + 3 H2O6 H+ + ClO3- + 6 e- Cl- + 3 H2OChanges in OS : +6 +3 ; reduction Different in OS = 3, so 3 e- at the LHS of equationCrO42- + 3 e- Cr3+CrO42- + 3 e- Cr3+ + 4 H2O8 H+ + CrO42- + 3 e- Cr3+ + 4 H2O8 H+ + CrO42- + 3 e- Cr3+ + 4 H2OChanges in OS : +6 +3 ; reduction Different in OS = 3, so 3 e- at the LHS of equationSince there are 2 Cr , so total e- = 6 ; Cr2O72- + 6 e- 2 Cr3+Cr2O72- + 6 e- 2 Cr3+ + 7 H2O14 H+ + Cr2O72- + 6 e- 2 Cr3+ + 7 H2O14 H+ + Cr2O72- + 6 e- 2 Cr3+ + 7 H2Od)MnO4- Mn2+Half equation : ..e)NO2- NO3-Half equation : ..f) CrO2- CrO42-Half equation : ..Changes in OS : +7 +2 ; reduction Different in OS = 5, so 5 e- at the LHS of equationMnO4- + 5 e- Mn2+MnO4- + 5 e- Mn2+ + 4 H2O8 H+ + MnO4- + 5 e- Mn2+ + 4 H2O8 H+ + MnO4- + 5 e- Mn2+ + 4 H2OChanges in OS : +3 +5 ; oxidation Different in OS = 2, so 2 e- at the RHS of equationNO2- NO3- + 2 e- NO2- + H2O NO3- + 2 e-NO2- + H2O 2 H+ + NO3- + 2 e-NO2- + H2O 2 H+ + NO3- + 2 e-Changes in OS : +3 +6 ; oxidation Different in OS = 3, so 3 e- at the RHS of equationCrO2- CrO42- + 3 e- 2 H2O + CrO2- CrO42- + 3 e- 2 H2O + CrO2- 4 H+ + CrO42- + 3 e-2 H2O + CrO2- 4 H+ + CrO42- + 3 e-g) As2O3 As2O5Half equation : When half equation of both oxidation and reduction reaction are written, a redox reaction can be balanced.Example 10 : Cu2+ (aq) + Na (s) Cu (s) + Na+ (aq)Oxidation half equation : .Reduction half equation : Overall equation : ..Example 11 : Fe2+ (aq) + MnO4- (aq) Fe3+ (aq) + Mn2+ (aq) Oxidation half equation : Reduction half equation : Overall equation : ..Changes in OS : +3 +5 ; oxidation Different in OS = 2, so 2 e- at the RHS of equationSince there are 2 As , so total e- = 4 ; As2O3 As2O5 + 4 e-As2O3 + 2 H2O As2O5 + 4 e-As2O3 + 2 H2O As2O5 + 4 e- + 4 H+ As2O3 + 2 H2O As2O5 + 4 e- + 4 H+Na Na+ + e-Cu2+ + 2 e- CuX 2Cu2+ + 2 Na Cu + 2 Na+Fe2+ Fe3+ + e-8 H+ + MnO4- + 5 e- Mn2+ + 4 H2OX 55 Fe2+ 8 H+ + MnO4- Mn2+ + 4 H2O + 5 Fe3+ Example 12 :ClO- (aq) + SO2 (g) Cl- (aq) + SO42-Oxidation half equation : ....Reduction half equation : ..Overall equation :

Example 13 :Cr2O72 + Cl2 ClO3 + Cr3+Oxidation half equation : ..Reduction half equation : .Overall equation : 2 H2O + SO2 4 H+ + SO42- + 2 e-2 H+ + ClO- + 2 e- Cl- + H2OH2O + ClO- + SO2 2 H+ + SO42- + Cl-6 H2O + Cl2 12 H+ + 2 ClO3- + 10 e-14 H+ + Cr2O72- + 6 e- 2 Cr3+ + 7 H2OX 3X 55 Cr2O72- + 34 H+ + 3 Cl2 6 ClO3- + 17 H2O + 10 Cr3+ Other than using half equation, a redox reaction can also be balanced using the change of oxidation number.Supposed we have a reaction : x A + y B productsIf the oxidation of reactant A increased by m while B reduced by n ; Thenx (+ m) + y ( n ) = 0Using a simple reaction : ySn4+ (aq) + xFe2+ (aq) Sn2+ (aq) + Fe3+ (aq)For Sn ; O.N changed from to ; so the difference is ..For Fe ; O.N changed from to ... ; so the difference is ..This will makes the equation become :

Balanced the number of atoms on both side of the equation+4 +2 2 +2 +3+ 1 x (+ 1) + y ( 2) = 0So, x = 2 ; y = 11 Sn4+ (aq) + 2 Fe2+ (aq) Sn2+ (aq) + Fe3+ (aq) Sn4+ (aq) + 2 Fe2+ (aq) Sn2+ (aq) + 2 Fe3+ (aq)Example 14 :Br (aq) + SO42- (aq) SO2 (g) + Br2 (l)For Br ; O.N changed from to . ; so the difference is For S ; O.N changed from to . ; so the difference is This will make the equation become :- 1 0+ 1 +6 +4- 2 x (+ 1) + y ( 2) = 0So, x = 2 ; y = 12 Br (aq) + 1 SO42- (aq) Br2 (l) + SO2 (aq)2 Br (aq) + SO42- (aq) + 4 H+ Br2 (l) + SO2 (aq) + 2 H2OBalanced the number of atoms on both side of the equationExample 15 : CrO42- + Cl- Cr3+ + Cl2 Example 16 :Cr2O72- + NO2- Cr3+ + NO3-

For Cr ; O.N changed from +6 to +3 ; so the difference is 3For Cl ; O.N changed from 1 to 0 ; so the difference is +1x (+ 1) + y ( 3) = 0So, x = 3 ; y = 18 H+ + CrO42- + 3 Cl- Cr3+ + 3/2 Cl2 + 4 H2O16 H+ + 2 CrO42- + 6 Cl- 2 Cr3+ + 3 Cl2 + 8 H2OFor Cr ; O.N changed from +6 to +3 ; so the difference is 3Since there are 2 Cr involved, diff. = 6For N ; O.N changed from +3 to +5 ; so the difference is +2x (+ 2) + y ( 6) = 0So, x = 3 ; y = 1Cr2O72- + 3 NO2- Cr3+ + NO3-8 H+ + Cr2O72- + 3 NO2- 2 Cr3+ + 3 NO3- + 4 H2OIf the redox reaction occur in a basic solution, the number of H+ shall be neutralise by the number of OH-.Example 17 :MnO4- + SO32- MnO2 + SO42- Example 18 : Fe(OH)2 + CrO42 Fe(OH)3 + Cr(OH)3

Oxidation eq : H2O + SO32- 2 H+ + SO42- + 2 e-Reduction eq : 4 H+ + MnO4- + 3 e- MnO2 + 2 H2OOverall : 2 H+ + 2 MnO4- + 3 SO32- 2 MnO2 + 3 SO42- + H2OH2O + 2 MnO4- + 3 SO32- 2 MnO2 + 3 SO42- + 2 OH-X 3X 2 Oxidation eq : H2O + Fe(OH)2 Fe(OH)3 + e + H+Reduction eq : 5 H+ + 3 e + CrO42 Cr(OH)3 + H2OOverall : 2 H2O + 3 Fe(OH)2 + 2 H+ + CrO42 3 Fe(OH)3 + Cr(OH)3 In basic : 4 H2O + 3Fe(OH)2 + CrO42 3Fe(OH)3 + Cr(OH)3 + 2OHX 3Disproportionation reactions~ Substances which are able to undergo self oxidation reduction are called disproportionation~ Examples of disproportionation reaction.18.Cu+ (aq) +Cu+ (aq) Cu2+ (aq) +Cu (s)19.NaOH (aq) +Cl2 (aq) NaCl (aq) + NaOCl (aq) + H2O20.NaOBr (aq)NaBrO3 (aq) +NaBr (aq)+1+200+1-1+1+5-12.3Electrode PotentialWhen a strip of metal, M (s) (known as electrode) is placed in a solution of its aqueous solution, Mn+ (aq), the following equilibrium is established : Mn+ (aq) + n e- M (s)At equilibrium, there is a separation of charge between metal (M) and ions (Mn+) in the solution. as a result, there is a potential difference between the metal and the solution. This potential difference is known as electrode potential and is written as Eo.Electrode potential can be measure under these circumstances whereMetal

Cu2+ + 2e- Cu The positive value of E0 indicates the equilibrium favours to the position. Copper (II) ions (Cu2+), have a greater tendency to ... at copper electrode.Zn2+ + 2e- Zn The negative value of E0 indicates the equilibrium favours to the .. position. Zinc ion (Zn2+) have a greater tendency to .. at zinc electrode.

MM+M+M+M+M+M+rightbe reducedleftbe oxidisedNon Metal

F2 + 2e- 2 F- Cl2 + 2e- 2 Cl- Positive value of ECl2/Cl- and EF2/F- indicates the equilibrium favours to the. position. has a greater tendency to under platinum electrodeThe more positive the value, higher the tendency of non-metal to .In another words, fluorine is a stronger .. agent than chlorine.

Cl2 (g)Cl- (aq) [1.0 M]be reducedFluorine and chlorinerightbe reducedoxidisingMixture of aqueous ion

A potential difference also exists between ions in an aqueous solution. Example : Cr3+ + e- Cr2+ Fe3+ + e- Fe2+ Base on the Eo value, Cr3+ is stable than Cr2+ as equilibrium favour to .. (Eo is negative)Base on the Eo value, Fe3+ is .. stable than Fe2+ as equilibrium favour to .. (Eo is positive)

Ma+ [1.0 M] / Mb+ [1.0 M] morebackwardlessforwardV2.3.1Standard Electrode PotentialDefinition : The standard electrode potential, Eo Mn+ / M of a metal M is the difference between the metal M and the solution of the metal ions of concentration at K and . atm, measured relatively to .Standard Hydrogen Electrode ( S.H.E.)It is impossible to measure the electrode potential for an .. half-cell. It can only be measured for a complete circuit with 2 . , i.e. only differences in electrode potentials are measurable.The standard chosen for electrode potentials is the standard hydrogen electrode (SHE). The standard electrode potentials of other half-cells are measured relative to the SHEs electrode potential.By convention, the standard electrode potential for this reference hydrogen half-cell is taken to be ... 2 H+ (aq) + 2 e- H2 (g) Condition : .... oC ; H2 (g) at atm ; [H+] = 1.00 Mpotentialaqueous1.0 mol dm-32981.0Standard Hydrogen Electrodeincompletehalf cell standard 25 1.0Measuring standard electrode potential of a metal / metal aqueous solutionThe set-up of the apparatus to measure the standard potential electrode, Eo. is described as below :Standard hydrogen electrodeZinc half cellSalt bridge (made of saturated KCl / NaCl)H2 (g)1.0 atmH+ (aq) [1.0 M]25oCPotentiometerZn (s)

Zn2+ (aq) [1.0 M]0.76 VThe chemical cell is set-up by connecting a standard .. half-cell to a standard . electrode.The e.m.f. for the cell is . V. The potentiometer point to the direction of .. electrode in the external circuit, indicating electrons flow from .. to . half-cell.Eq. Zn half-cell :Eq. H half-cell :Overall reaction :The cell notation can be written as :

At zinc electrode ; electrons are ..... ; ... reaction occurAt platinum electrode ; electrons are .. ; . reaction occurSince zinc is oxidised in a SHE, the standard e.m.f value is H+/H2Zn2+/Zn0.76H2H+ / H2Zn2+ / ZnZn (s) Zn2+ (aq) + 2 e-2 H+ (aq) + 2 e- H2 (g)Zn (s) + 2 H+ (aq) H2 (g) + Zn2+ (aq) Zn (s) I Zn2+ (aq) II H+ (aq) , H2 (g) I Pt (s)donatedoxidationreceivedreduction 0.76 VAnother example : silver / silver aqueous solution (Ag / Ag+)The set-up of the apparatus to measure the standard potential electrode, Eo. is described as below :Standard hydrogen electrodeSilver half cellSalt bridge (made of saturated KCl / NaCl)H2 (g)1.0 atmH+ (aq) [1.0 M]25oCPotentiometerAg (s)

Ag+ (aq) [1.0 M]0.80 VThe chemical cell set-up by connecting a standard half-cell to a standard .. electrode.The e.m.f. for the cell is . V. The galvanometer point to the direction of .. electrode in the external circuit, indicating electrons flow from .. to half-cell.Ag half-cell:H2 half-cell:Overall :The cell notation can be written as :

At silver electrode ; electrons are . ; reaction occurAt platinum electrode ; electrons are ; reaction occurSince silver is reduced in a SHE, the standard value is donatedoxidationreceivedreduction+ 0.80 VH+ / H2Ag+ / Ag0.80H+ / H2silverAg+ / AgAg+ (aq) + e- Ag (s)H2 (g) 2 H+ (aq) + 2 e-H2 (g) + 2 Ag+ (aq) 2 Ag (s) + 2 H+ (aq)Pt (s) I H2 (g) , H+ (aq) II Ag+ (aq) I Ag (s)Measuring a standard electrode potential of a gaseous substance The chemical cell set-up by connecting a standard half-cell to a standard electrode. Note that the set-up of the half-cells are the same for gaseous substancesCl2 / ClH+ / H2Standard hydrogen electrodeChlorine half cellSalt bridge (made of saturated KCl / NaCl)H2 (g)1.0 atmH+ (aq) [1.0 M]25oCPotentiometerCl2 (g)1 atmCl- (aq) [1.0 M]1.36 VThe e.m.f. for the cell is .V. The galvanometer point to the direction of .. electrode in the external circuit, indicating electrons flow from .. to . half-cell.Chlorine half-cell :Hydrogen half-cell :Overall ::The cell notation can be written as :

At platinum electrode in the half-cell of hydrogen ; electrons are ; reaction occurAt platinum electrode in the half-cell of chlorine ; electrons are .. ; reaction occurSince chlorine is by SHE, the standard value is .1.36Pt (Cl2) H+ / H2 Cl2 / ClCl2 (g) + 2 e- 2 Cl (aq)H2 (g) 2 H+ (aq) + 2 e-H2 (g) + Cl2 (g) 2 Cl (aq) + 2 H+ (aq)Pt (s) I H2 (g) , H+ (aq) II Cl2 (g), Cl (aq) I Pt (s)donatedoxidationreceivedreductionreduced+ 1.36 VMeasuring a standard electrode potential of a mixture of metal ions.The electrode potential of a mixture of ions can be measured in the similar way, using standard hydrogen electrode (SHE) as the other half-cell of the chemical cellFor example, in a mixture of iron (II) and iron (III) ionStandard hydrogen electrodeFe2+ / Fe3+ half cellSalt bridge (made of saturated KCl / NaCl)H2 (g)1.0 atmH+ (aq) [1.0 M]25oCPotentiometerFe2+ (aq) Fe3+ (aq)[1.0 M]0.77 VThe chemical cell set-up by connecting a standard half-cell to a standard electrode. Note that the set-up of the half-cells are a mixture of iron (II) and iron (III) ion under standard condition with as electrode.The e.m.f. for the cell is . V. The galvanometer point to the direction of ... half cell in the external circuit, indicating electrons flow from . to . half-cell.Fe3+ / Fe2+ half-cell:Hydrogen half-cell:Overall reaction :The cell notation can be written as :

At half-cell of hydrogen ; electrons are ; reaction occurAt half-cell of Fe3+/Fe2+ ; electrons are . ; ... reaction occurSince the mixture is ... by SHE, the value of Fe3+ / Fe2+H+ / H2platinum0.77Fe3+ / Fe2+Fe3+ / Fe2+ H+ / H2Fe3+ (aq) + e- Fe2+ (aq)H2 (g) 2 H+ (aq) + 2 e-H2 (g) + 2 Fe3+(aq) 2 Fe2+ (aq) + 2 H+ (aq)Pt (s) I H2 (g) , H+(aq) II Fe3+(aq), Fe2+(aq) I Pt (s)donatedoxidationreceivedreductionreduced+ 0.77 VThe calomel electrodePlatinum electrode is known as the reference electrode. However, it is relatively difficult to set up and operate under standard condition. It is more easier and safer to use a calomel electrode as a electrode. [calomel = ..]. Diagram of a typical calomel electrode

primarysecondaryMercury base alloy2.4Factors Affecting Electrode PotentialBy convention, the half equation is written with . as the forward reaction.The magnitude of the electrode potential depends on the position of the above equilibriumWhen value is positive ; a reaction is favouredWhen value is negative ; a .. reaction is favouredFactors which affect the position of equilibrium would therefore affect the value of electrode potential1. Nature of metalWhen a metal is highly .., the metal atoms have a greater tendency to become positive ions, leaving the behind on the metal electrode. The electrode potential therefore become more . and the position of equilibrium shift more to (. is favoured)reductionforwardbackwardelectropositiveelectronnegativeleftoxidationMetalHalf equationE (V)Silver+ 0.80Lead 0.13Zinc 0.76Magnesium 2.382. Concentration of metalIf the concentration of the hydrated metal ions is increased in the equilibrium, the position of equilibrium will shift to the , favouring . ; electrode potential become more Pb2+ (aq) + 2 e- Pb (s)E = 0.13 V [ Conc = 1.0 M ]If concentration Pb2+ changed to 0.001 M ; equilibrium shift to . ; E 0.13 VIf concentration Pb2+ changed to 10.0 M ; equilibrium shift to . ; E 0.13 V Ag+ + e- AgPb2+ + 2 e- PbZn2+ + 2 e- ZnMg2+ + 2 e- Mgrightforwardpositivebackward

3 TemperatureMost of the reduction processes are exothermic process. Increasing the temperature will cause the equilibrium to shift to the position of .. process ; which is to the Thus, the electrode potential becomes more .4. Pressure for gaseous speciesFrom what weve learned from chemical equilibria, when pressure increased, equilibrium will shift to the position with .. mole of gas ; while decreasing pressure will cause equilibrium to shift to position with .mole of gas.Eg : Cl2 (g) + 2 e- 2 Cl- (aq) E = + 1.36 VIncreasing pressure will cause equilibrium shift to .. ; E + 1.36 VDecreasing pressure will cause equilibrium shift to .. ; E + 1.36 Vendothermicleftnegative / less positivelessmoreright sideleft side>