1. 1. PRE UNIVERSITI SEMESTER 2 CHAPTER 2 ELECTROCHEMISTRY
2. 2. 2.1 Oxidation number Oxidation 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. 1. All free atoms in element have an oxidation number of zero 2. 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 1
3. 3. 3. For 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 Element Oxidation number Notes Group I +1 -- Group 2 +2 -- Group 17 1 True only to halogen without O in it Oxygen 2 Exception : 1 for peroxide and +2 for F2O Hydrogen +1 Except : metal hydride MH (H = -1)
4. 4. Na2O B2O3 CO2 SO3 Cl2O7 HF H2S NH3 CH4 H2CO3 H2SO4 HBrO3 3. In a neutral molecule, the sum of the oxidation numbers of all atom are equals 2 Na + O = 0 2(Na) + (-2) = 0 Na = +1 2 B + 3 O = 0 2(B) + 3(-2) = 0 B = +3 1 C + 2 O = 0 1(C) + 2(-2) = 0 C = +4 1S + 3O = 0 1(S) + 3(-2) = 0 S = +6 2Cl + 7O = 0 2(Cl) + 7(-2) = 0 Cl = +7 1H + 1F = 0 1(F) + 1(+1) = 0 F = -1 1S + 2H = 0 1(S) + 2(+1) = 0 S = -2 1N + 3H = 0 1(N) + 3(+1) = 0 N = -3 1C + 4H = 0 1(C) + 4(+1) = 0 N = -4 2H + 1C + 3O = 0 2(+1) +1C + 3(-2) = 0 C = +4 2H + 1S + 4O = 0 2(+1) +1S + 4(-2) = 0 S = +6 1H + 1Br + 3O = 0 1(+1) +1Br + 3(-2) = 0 Br = +5
5. 5. 4. 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 - Cr2O7 2- MnO4 - C2O4 2- ClO2 - HSO4 - 1 Cr + 4 O = -1 1 Cr + 4(-2) = -1 Cr = +7 2 Cr + 7 O = -2 2 Cr + 7 (-2) = -2 Cr = +6 1 Mn + 4 O = -1 1 Mn + 4(-2) = -1 Mn = +7 2 C + 4 O = -2 2 C + 4(-2) = -2 C = +3 1 Cl + 2 O = -1 1 Cl + 2 (-2) = -1 Cl = +3 1 H + 1 S + 4 O = -1 1(+1) + 1S + 4(-2) = -1 S = +6
6. 6. 2.2 Half equation and redox reaction. Half equation ~ equation which shows how electrons are accept / donate in a chemical reaction When 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. Reaction Oxidation no change Reaction Half equation Na Na+ Mg Mg2+ Al Al3+ Cu2+ Cu donated right received left 0 +1 oxidation Na Na+ + e- 0 +2 oxidation Mg Mg2+ + 2 e- 0 +3 oxidation Al Al3+ + 3 e- +2 0 reduction Cu2+ + 2 e- Cu
7. 7. Reaction Oxi. no change Reaction Half equation H2 H+ Cl2 Cl- I- I2 O2 O2- Fe2+ Fe3+ Pb4+ Pb2+ 0 +1 oxidation H2 2 H+ + 2 e- 0 -1 reduction Cl2 + 2 e- 2 Cl -1 0 oxidation 2 I- I2 + 2 e- 0 -2 reduction O2 + 4 e- 2 O2 +2 +3 oxidation Fe2+ Fe3+ + e- +4 +2 reduction Pb4+ + 2 e- Pb2+
8. 8. When it comes to the reaction involving molecular ion, the overall charge has to be balanced in such order. 1. Write a skeleton half equation. Determine the reaction (oxidation or reduction) using oxidation number 2. Balance the charge by adding electrons at the appropriate side 3. Balance the number of atoms other than oxygen. 4. Based on the changes in number of oxygen, write the number of water molecule formed/used. 5. From the number of water molecule formed/used, write the number of hydrogen ion (H+) required.
9. 9. a) ClO3 - Cl- Half equation : b) CrO4 2- Cr3+ Half equation : c) Cr2O7 2- Cr3+ Half equation : Changes in OS : +5 -1 ; reduction Different in OS = 6, so 6 e- at the LHS of equation ClO3 - + 6 e- Cl- ClO3 - + 6 e- Cl- + 3 H2O 6 H+ + ClO3 - + 6 e- Cl- + 3 H2O 6 H+ + ClO3 - + 6 e- Cl- + 3 H2O Changes in OS : +6 +3 ; reduction Different in OS = 3, so 3 e- at the LHS of equation CrO4 2- + 3 e- Cr3+ CrO4 2- + 3 e- Cr3+ + 4 H2O 8 H+ + CrO4 2- + 3 e- Cr3+ + 4 H2O 8 H+ + CrO4 2- + 3 e- Cr3+ + 4 H2O Changes in OS : +6 +3 ; reduction Different in OS = 3, so 3 e- at the LHS of equation Since there are 2 Cr , so total e- = 6 ; Cr2O7 2- + 6 e- 2 Cr3+ Cr2O7 2- + 6 e- 2 Cr3+ + 7 H2O 14 H+ + Cr2O7 2- + 6 e- 2 Cr3+ + 7 H2O 14 H+ + Cr2O7 2- + 6 e- 2 Cr3+ + 7 H2O
10. 10. d) MnO4 - Mn2+ Half equation : .. e) NO2 - NO3 - Half equation : .. f) CrO2 - CrO4 2- Half equation : .. Changes in OS : +7 +2 ; reduction Different in OS = 5, so 5 e- at the LHS of equation MnO4 - + 5 e- Mn2+ MnO4 - + 5 e- Mn2+ + 4 H2O 8 H+ + MnO4 - + 5 e- Mn2+ + 4 H2O 8 H+ + MnO4 - + 5 e- Mn2+ + 4 H2O Changes in OS : +3 +5 ; oxidation Different in OS = 2, so 2 e- at the RHS of equation NO2 - 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 equation CrO2 - CrO4 2- + 3 e- 2 H2O + CrO2 - CrO4 2- + 3 e- 2 H2O + CrO2 - 4 H+ + CrO4 2- + 3 e- 2 H2O + CrO2 - 4 H+ + CrO4 2- + 3 e-
11. 11. g) As2O3 As2O5 Half 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 equation Since 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- Cu X 2 Cu2+ + 2 Na Cu + 2 Na+ Fe2+ Fe3+ + e- 8 H+ + MnO4 - + 5 e- Mn2+ + 4 H2O X 5 5 Fe2+ 8 H+ + MnO4 - Mn2+ + 4 H2O + 5 Fe3+
12. 12. Example 12 : ClO- (aq) + SO2 (g) Cl- (aq) + SO4 2- Oxidation half equation : .... Reduction half equation : .. Overall equation : Example 13 : Cr2O7 2 + Cl2 ClO3 + Cr3+ Oxidation half equation : .. Reduction half equation : . Overall equation : 2 H2O + SO2 4 H+ + SO4 2- + 2 e- 2 H+ + ClO- + 2 e- Cl- + H2O H2O + ClO- + SO2 2 H+ + SO4 2- + Cl- 6 H2O + Cl2 12 H+ + 2 ClO3 - + 10 e- 14 H+ + Cr2O7 2- + 6 e- 2 Cr3+ + 7 H2O X 3 X 5 5 Cr2O7 2- + 34 H+ + 3 Cl2 6 ClO3 - + 17 H2O + 10 Cr3+
13. 13. 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 products If the oxidation of reactant A increased by m while B reduced by n ; Then x (+ m) + y ( n ) = 0 Using 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) = 0 So, x = 2 ; y = 1 1 Sn4+ (aq) + 2 Fe2+ (aq) Sn2+ (aq) + Fe3+ (aq) Sn4+ (aq) + 2 Fe2+ (aq) Sn2+ (aq) + 2 Fe3+ (aq)
14. 14. Example 14 : Br (aq) + SO4 2- (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) = 0 So, x = 2 ; y = 1 2 Br (aq) + 1 SO4 2- (aq) Br2 (l) + SO2 (aq) 2 Br (aq) + SO4 2- (aq) + 4 H+ Br2 (l) + SO2 (aq) + 2 H2O
15. 15. Balanced the number of atoms on both side of the equation Example 15 : CrO4 2- + Cl- Cr3+ + Cl2 Example 16 : Cr2O7 2- + NO2 - Cr3+ + NO3 - For Cr ; O.N changed from +6 to +3 ; so the difference is 3 For Cl ; O.N changed from 1 to 0 ; so the difference is +1 x (+ 1) + y ( 3) = 0 So, x = 3 ; y = 1 8 H+ + CrO4 2- + 3 Cl- Cr3+ + 3/2 Cl2 + 4 H2O 16 H+ + 2 CrO4 2- + 6 Cl- 2 Cr3+ + 3 Cl2 + 8 H2O For Cr ; O.N changed from +6 to +3 ; so the difference is 3 Since there are 2 Cr involved, diff. = 6 For N ; O.N changed from +3 to +5 ; so the difference is +2 x (+ 2) + y ( 6) = 0 So, x = 3 ; y = 1 Cr2O7 2- + 3 NO2 - Cr3+ + NO3 - 8 H+ + Cr2O7 2- + 3 NO2 - 2 Cr3+ + 3 NO3 - + 4 H2O
16. 16. If the redox reaction occur in a basic solution, the number of H+ shall be neutralise by the number of OH-. Example 17 : MnO4 - + SO3 2- MnO2 + SO4 2- Example 18 : Fe(OH)2 + CrO4 2 Fe(OH)3 + Cr(OH)3 Oxidation eq : H2O + SO3 2- 2 H+ + SO4 2- + 2 e- Reduction eq : 4 H+ + MnO4 - + 3 e- MnO2 + 2 H2O Overall : 2 H+ + 2 MnO4 - + 3 SO3 2- 2 MnO2 + 3 SO4 2- + H2O H2O + 2 MnO4 - + 3 SO3 2- 2 MnO2 + 3 SO4 2- + 2 OH- X 3 X 2 Oxidation eq : H2O + Fe(OH)2 Fe(OH)3 + e + H+ Reduction eq : 5 H+ + 3 e + CrO4 2 Cr(OH)3 + H2O Overall : 2 H2O + 3 Fe(OH)2 + 2 H+ + CrO4 2 3 Fe(OH)3 + Cr(OH)3 In basic : 4 H2O + 3Fe(OH)2 + CrO4 2 3Fe(OH)3 + Cr(OH)3 + 2OH X 3
17. 17. Disproportionation 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) + H2O 20. NaOBr (aq) NaBrO3 (aq) + NaBr (aq) +1 +2 0 0 +1-1 +1 +5 -1
18. 18. 2.3 Electrode Potential When 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 where
19. 19. Metal 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. V34.0E Cu/Cu2 V76.0E Zn/Zn2 M M+ M+ M+ M+ M+ M+ right be reduced left be oxidised
20. 20. Non 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 electrode The more positive the value, higher the tendency of non- metal to . In another words, fluorine is a stronger .. agent than chlorine. V87.2E F/F2 V36.1E Cl/Cl2 Cl2 (g) Cl- (aq) [1.0 M] be reduced Fluorine and chlorineright be reduced oxidising
21. 21. Mixture 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) V41.0E 23 Cr/Cr V77.0E 23 Fe/Fe Ma+ [1.0 M] / Mb+ [1.0 M] more backward less forward V
22. 22. 2.3.1 Standard Electrode Potential Definition : The standard electrode potential, Eo M n+ / 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 M potential aqueous 1.0 mol dm-3 298 1.0 Standard Hydrogen Electrode incomplete half cell standard 25 1.0
23. 23. Measuring standard electrode potential of a metal / metal aqueous solution The set-up of the apparatus to measure the standard potential electrode, Eo. is described as below : Standard hydrogen electrode Zinc half cell Salt bridge (made of saturated KCl / NaCl) H2 (g) 1.0 atm H+ (aq) [1.0 M] 25oC Potentiometer Zn (s) Zn2+ (aq) [1.0 M] 0.76 V
24. 24. The 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 occur At platinum electrode ; electrons are .. ; . reaction occur Since zinc is oxidised in a SHE, the standard e.m.f value is H+/H2Zn2+/Zn 0.76 H2 H+ / H2Zn2+ / Zn Zn (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) donated oxidation received reduction 0.76 V
25. 25. Another 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 electrode Silver half cell Salt bridge (made of saturated KCl / NaCl) H2 (g) 1.0 atm H+ (aq) [1.0 M] 25oC Potentiometer Ag (s) Ag+ (aq) [1.0 M] 0.80 V
26. 26. The 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 occur At platinum electrode ; electrons are ; reaction occur Since silver is reduced in a SHE, the standard value is donated oxidation received reduction + 0.80 V H+ / H2Ag+ / Ag 0.80 H+ / H2 silver Ag+ / Ag Ag+ (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)
27. 27. 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 substances Cl2 / Cl H+ / H2 Standard hydrogen electrode Chlorine half cell Salt bridge (made of saturated KCl / NaCl) H2 (g) 1.0 atm H+ (aq) [1.0 M] 25oC Potentiometer Cl2 (g) 1 atm Cl- (aq) [1.0 M] 1.36 V
28. 28. 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. 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 occur At platinum electrode in the half-cell of chlorine ; electrons are .. ; reaction occur Since chlorine is by SHE, the standard value is . 1.36 Pt (Cl2) H+ / H2 Cl2 / Cl Cl2 (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) donated oxidation received reduction reduced + 1.36 V
29. 29. Measuring 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 cell For example, in a mixture of iron (II) and iron (III) ion Standard hydrogen electrode Fe2+ / Fe3+ half cell Salt bridge (made of saturated KCl / NaCl) H2 (g) 1.0 atm H+ (aq) [1.0 M] 25oC Potentiometer Fe2+ (aq) Fe3+ (aq) [1.0 M] 0.77 V
30. 30. 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 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 occur At half-cell of Fe3+/Fe2+ ; electrons are . ; ... reaction occur Since the mixture is ... by SHE, the value of Fe3+ / Fe2+ H+ / H2 platinum 0.77 Fe3+ / Fe2+ Fe3+ / Fe2+ H+ / H2 Fe3+ (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) donated oxidation received reduction reduced + 0.77 V
31. 31. The calomel electrode Platinum 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 primary secondary Mercury base alloy
32. 32. 2.4 Factors Affecting Electrode Potential By convention, the half equation is written with . as the forward reaction. The magnitude of the electrode potential depends on the position of the above equilibrium When value is positive ; a reaction is favoured When value is negative ; a .. reaction is favoured Factors which affect the position of equilibrium would therefore affect the value of electrode potential 1. Nature of metal When 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) reduction forward backward electropositive electron negative left oxidation
33. 33. Metal Half equation E (V) Silver + 0.80 Lead 0.13 Zinc 0.76 Magnesium 2.38 2. Concentration of metal If 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 V If concentration Pb2+ changed to 10.0 M ; equilibrium shift to . ; E 0.13 V Ag+ + e- Ag Pb2+ + 2 e- Pb Zn2+ + 2 e- Zn Mg2+ + 2 e- Mg right forward positive backward < forward >
34. 34. 3 Temperature Most 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 species From 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 = + 1.36V Increasing pressure will cause equilibrium shift to .. ; E + 1.36V Decreasing pressure will cause equilibrium shift to .. ; E + 1.36V endothermic left negative / less positive less more right side left side > <
35. 35. 2.5 The electrochemical series (ECS) When a series of standard reduction potential of different substances are determined and are arranged in order, a electrochemical series is obtained.
36. 36. Below are some important facts about electrochemical series. Half cell of the standard electrode potential is always written as processes. Due to this reason, sometimes it is also known as The positive / negative sign shows how substances favour to each of the reaction. If the EO is positive, substances favour a .. reaction. In another words, it serve well as . agent. The more positive the value ; the stronger the .. agent If the EO is negative, substances favour .. reaction. In another words, it serve well as agent. The more negative the value ; the stronger the agent The number of electron involve does not affect the standard electrode potential value. If Cl2 (g) + 2 e- 2 Cl- (aq) Eo = + 1.36 V; then Cl2 (g) + e- Cl- (aq) Eo = . V reduction Standard reduction potential (SRP) forward oxidising oxidising backward reducing reducing + 1.36
37. 37. Some substances have more than one Eo value. For example Fe2+ ; H2O2 ; NO2 - ; Cu+. These substances can act as an oxidising or reducing agent. Examples In Fe2+ Fe3+ (aq) + e- Fe2+ (aq) (Fe2+ act as reducing agent) Fe2+ (aq) + 2 e- Fe (s) (Fe2+ act as oxidising agent)
38. 38. 2.6 Redox reaction and electromotive forces (e.m.f.) In standard hydrogen electrode, we had seen on how to measure the standard electrode potential of 3 types of half- cell, which are metal / metal ion half-cell ; non-metal / ion half- cell ; ion / ion half-cell Imagine if we replace the hydrogen half-cell with other half- cell, will we still get the same value? The potential difference between 2 half-cells can be measured using the same way. There are various types of the set-up of a complete chemical cell other than the one introduced during measuring the standard electrode potential, such as Daniel cell (diagram below)
39. 39. A Daniel cell is built using a copper and zinc half-cell A porous pot is used to Substance which has a higher position in ECS (more negative the value of Eo) is the of the cell whereas the substance which has a lower position in ECS (more positive the value of Eo) is the of the cell. The half equation occur at Anode : Cathode : Overall : The e.m.f. of cell is . V Cell notation is written as C u Z n V complete the cell And to separate between the 2 electrolytes anode cathode Zn Zn2+ + 2 e- Eo =+0.76 V Cu2+ + 2 e- Cu Eo = +0.34 V Zn + Cu2+ Zn2+ + Cu Ecell =+1.10 V + 1.10 Zn (s) I Zn2+ (aq) II Cu2+ (aq) I Cu (s) Cu2+ Zn2+
40. 40. A U tube cell is built using iron (II) ion, Fe2+ and bromine water, Br2 H2SO4 is used to . ............... The substance which has more negative / less positive the value of Eo) is the of the cell whereas the substance which has a more positive / less negative value of Eo is the ... of the cell. The half equation occur at Anode : Cathode : Overall : The e.m.f. of cell is ..V Cell notation is written as G complete the cell And to separate between the 2 electrolytes anode cathode Fe2+ Fe3+ + e- Eo = - 0.77 V Br2 + 2 e- 2 Br - Eo = +1.07 V Br2 + 2 Fe2+ 2 Fe3+ + 2 Br- + 0.30 Pt (s) I Fe2+ (aq) , Fe3+ (aq) II Br2 (l) , Br- (aq) I Pt (s) Pt Br2/Br- Fe3+/Fe2+
41. 41. Note the following in a chemical cell : The e.m.f. of a cell is always In another words, we must always subtract the .. standard electrode potential with a . standard electrode potential. Electrons flow from to in the external circuit reaction occur at anode while reaction occur at cathode of the cell. POSITIVE more positive less positive anode cathode Oxidation reduction
42. 42. SRP for both cell : Fe2+ + 2e- Fe E0 = - 0.44 V Mg2+ + 2e- Mg E0 = - 2.38 V Since E0 for Mg2+/Mg is more negative than Fe2+/Fe, so Mg2+/Mg will be oxidised, so SPR of Mg2+/Mg is reversed Oxidation eq : Mg Mg2+ + 2 e- E0 = + 2.38 V Reduction eq : Fe2+ + 2e- Fe E0 = 0.44 V Overall eq : Fe2+ + Mg Fe + Mg2+ Ecell = + 1.94 V Cell diagram : Mg (s) I Mg2+ (aq) II Fe2+ (aq) I Fe (s) G Fe / Fe2+ half cell Mg / Mg2+ half cell
43. 43. G MnO4 - / Mn2+ half cell Ti3+ / Ti2+ half cell SRP : MnO4 - + 5 e- + 8 H+ Mn2+ + 4 H2O E0 = + 1.52 V Ti3+ + e- Ti2+ E0 = 0.37 V Since E0 for Ti3+/Ti2+ is more negative than MnO4 -/Mn2+, so Ti3+/Ti2+ will be oxidised, so SPR of Ti3+/Ti2+ is reversed Oxidation eq : Ti2+ Ti3+ + e- E0 = + 0.37 V Red eq : MnO4 - + 5 e- + 8 H+ Mn2+ + 4 H2O E0 = + 1.52 V Overall: MnO4 - + 5 Ti2+ + 8 H+ Mn2+ + 4 H2O + 5 Ti3+ Ecell = + 1.89V Cell diagram: Pt(s) I Ti2+(aq), Ti3+ (aq)II MnO4 - (aq), Mn2+(aq) I Pt(s)
44. 44. 2.7 Feasibility of a redox reaction If a reaction occurs on its own record when the reactants are mixed, the reaction is a ... reaction Compare the following reaction to distinguish between a spontaneous reaction and not spontaneous reaction. Immerse a zinc plate into HCl 1.0 M Immerse a copper plate into HCl 1.0 M Observation Observation spontaneous -Bubbling is observed -Zinc plate is corroded by HCl - No changes occur
45. 45. We can use e.m.f. to predict the feasibility of the reaction. Supposedly, in the reaction above, the 2 half equation for the reactions can be written as It can also be used to deduce the strength as an oxidising agent in halogen. Halogens are strong oxidising agent. This is supported with the value of standard reduction potential where F2 (aq) + e- F (aq) Eo = + 2.87 V Cl2 (aq) + e- Cl (aq) Eo = + 1.36 V Br2 (aq) + e- Br (aq) Eo = + 1.07 V I (aq) + e- I (aq) Eo = + 0.54 V -Since Zn react with H+, so the 2 half equation can be written Zn Zn2+ + 2e- E0 = + 0.76 V 2 H+ + 2e- H2 E0 = + 0.00 V Zn + 2 H+ Zn2+ + H2 E = + 0.76 V Since Ecell is positive, the reaction is spontaneous -Since Cu react with H+, so the 2 half equation can be written Cu Cu2+ + 2e- E0 = - 0.34 V 2 H+ + 2e- H2 E0 = + 0.00 V Cu + 2 H+ Cu2+ + H2 E = - 0.34 V Since Ecell is negative, the reaction is nonspontaneous Stronger oxidisingagent
46. 46. Observation Half equation & overall equation Chlorine in Tetrachloromethane is added to aqueous potassium bromide (KBr). Bromine in tetrachloromethane is added to aqueous potassium iodide (KI) Iodine is tetrachloromethane is added to aqueous potassium chloride (KCl) Pale yellow solution in CCl4 turned brown when shaken with KBr. 2Br- Br2 + 2e- Eo = - 1.07 V Cl2 + 2e- 2Cl- Eo = + 1.36 V Cl2 + 2Br- Br2 + 2Cl Ecell = + 0.29 V Brown solution in CCl4 turned purple when shaken with KI 2I- I2 + 2e- Eo = - 0.54 V Br2 + 2e- 2Br- Eo = + 1.07 V Br2 + 2I- I2 + 2Br Ecell = + 0.53 V No changes occur. Purple solution remain after shaken with KCl 2Cl- CI2 + 2e- Eo = - 1.36 V I2 + 2e- 2I- Eo = + 0.54 V I2 + 2 Cl- Cl2 + 2 I Ecell = - 0.82 V
47. 47. a) Iron nail are placed in zinc sulphate b) Copper is placed in concentrated nitric acid solution (Assume NO2 (g) is produced) c) Chlorine gas is bubbled into acidified potassium dichromate d) potassium iodide is added to acidified potassium manganate (VII) solution Reactant : Fe and Zn2+ Suitable half equation Fe Fe2+ + 2e- Eo = + 0.44 V Zn2+ + 2e- Zn Eo = - 0.76 V Fe + Zn2+ Fe2+ + Zn Ecell = - 0.32V Since Ecell is negative, reaction is not spontaneous (cannot react) Reactant : Cu and NO3 - Suitable half equation Cu Cu2+ + 2e- Eo = - 0.34 V NO3 + 2H+ +e NO2 + H2O E0= +0.81 V Cu + 2 NO3 + 4H+ 2NO2 +2H2O + Cu2+ Ecell = + 0.47 V Since Ecell is positive, reaction is spontaneous (can react) Reactant : Cl2 and Cr2O7 2 Suitable half equation Cl2 + 2H2O 2HOCl + 2H+ + 2e Eo = 1.64 V Cr2O7 2- + 6e- + 14H+ 2Cr3+ + 7H2O Eo = + 1.33 V Cr2O7 2- + 6HOCl + 2H+ 3Cl2 + H2O Ecell = 0.31V Since Ecell is negative, reaction is non spontaneous (cannot react) Reactant : I- and MnO4 - Suitable half equation 2I- I2 + 2 e- Eo = - 0.54 V MnO4 - + 8 H+ + 5e- Mn2+ + 4 H2O Eo = + 1.52 V 10 I- + 2 MnO4 - + 16H+ 2 Mn2+ + 8 H2O + 5 I2 Ecell = + 0.98 V Since Ecell is positive, reaction is spontaneous (can react)
48. 48. e) Calcium metal is added to water f) Acidified potassium dichromate solution is added to a solution of iron (II) sulphate Reactant : Ca and H2O Suitable half equation 2H2O + 2e- H2 + 2OH- Eo = - 0.83 V Ca Ca2+ + 2e- Eo = + 2.87 V Ca + 2H2O H2 + Ca2+ + 2 OH- Ecell = + 2.04 V Since Ecell is positive, reaction is spontaneous (can react) Reactant : Fe2+ and Cr2O7 2- Suitable half equation Cr2O7 2- + 6e- + 14H+ 2Cr3+ + 7H2O Eo = + 1.33 V Fe2+ Fe3+ + e- Eo = - 0.77 V Cr2O7 2- + 6Fe2+ + 14H+ 2Cr3+ + 7H2O + 6Fe3+ Ecell = + 0.56 V Since Ecell is positive, reaction is spontaneous (can react)
49. 49. Among the oxidation states available in d-orbital, +2 and +3 oxidation states are the most common states available in the d- block elements. The stability of the oxidation state can be explained in terms of electrochemistry. The standard reduction potential of a few transition metals is given in the table below. Half equation of reduction Eo (V) Stable ion Cr3+ + e- Cr2+ 0.41 Cr3+ Ti3+ + e- Ti2+ 0.37 Ti3+ V3+ + e- V2+ 0.26 V3+ Fe3+ + e- Fe2+ + 0.77 Fe2+ Mn3+ + e- Mn2+ + 1.51 Mn2+ Co3+ + e- Co2+ + 1.82 Co2+
50. 50. The action of dilute acids on metal are usually carried out in the presence of oxygen. We must therefore determine whether oxygen has any effect on such reactions. For example, in oxidation of iron (II) ion .. and can also react in the same way as iron does. For the case of cobalt and manganese, it does not react in the same way as iron does. Consider the reaction of cobalt (II) ion with acid in the absence / presence of oxygen Action of acids on iron (II) ion in the absence of air (oxygen) Action of acids on iron (II) ion in the presence of air Fe2+ Fe3+ + e- E0 = - 0.77 V 2 H+ + 2 e- H2 E0 = + 0.00 V 2 Fe2+ + 2H+ 2 Fe3+ + H2 Ecell = - 0.77 V Fe2+ Fe3+ + e- E0 = - 0.77 V O2 + 4H+ + 4e 2H2O E0 = +1.23 V 4Fe2+ + O2 + 4H+ 4Fe3+ + 2H2O Ecell = + 0.46 V Ti2+ V2+ Cr2+
51. 51. Action of acids on cobalt in the absence of air Action of acids on cobalt in the presence of air Co2+ Co3+ + e- E0 = - 1.82 V 2 H+ + 2 e- H2 E0 = + 0.00 V 2 Co2+ + 2H+ 2 Co3+ + H2 Ecell = - 1.82 V Co2+ Co3+ + e- E0 = - 1.82 V O2 + 4H+ + 4e 2H2O E0 = +1.23 V 4Co2+ + O2 + 4H+ 4Co3+ + 2H2O Ecell = - 0.59 V Graph below shows the relative stability of ions which exist in different oxidation state
52. 52. Other than the presence of oxygen, the presence of ligands can also affect the stability of ions. Consider the following electrode reactions for cobalt : [Co(NH3)6]3+ + e- [Co(NH3)6]2+ E0 = + 0.10 V O2 + 4 H+ + 4 e- 2 H2O E0 = + 1.23 V [Co(H2O)6]3+ + e- [Co(H2O)6]2+ E0 = + 1.82 V In water, Co2+ is stable toward oxidation, even in the presence of oxygen, since the E for the reaction is -0.59 (based on the calculation above). Therefore, +2 is more stable than +3 oxidation state in aqueous solution (ligand is water) When aqueous NH3 is added to the solution of Co2+, the complex ion [Co(NH3)6]2+ is formed (since NH3 is a strong ligand, water molecule can easily displaced by NH3 ligand). Eq : [Co(H2O)6]2+ + 6 NH3 [Co(NH3)6]2+ + 6 H2O When [Co(NH3)6]2+ is formed, it can react easily with acids with the presence of air
53. 53. When [Co(NH3)6]2+ is formed, it can react easily with acids with the presence of air [Co(NH3)6]2+ [Co(NH3)6]3+ + e- E0 = - 0.10 V O2 + 4 H+ + 4 e- 2 H2O E0 = + 1.23 V 4 [Co(NH3)6]2+ + O2 + 4 H+ 2 H2O + 4 [Co(NH3)6]3+ Ecell = + 1.13 V Therefore, although Co3+ is not stable in the presence of air, but Co3+ is stable in ammonia aqueous solution. Hence, this is one of the ways to prepare an ion solution which is not stable in air.
54. 54. 2.7 Nernst Equation and Its application All the electrochemical cells that discussed so far are standard Eo. [At ..oC ; atm ; . M] If concentration of ions and temperature change, it will affec the value of electrode potential. At this moment, we can use an equation to study the changes of concentration of ions using Nernst Equation. 25 1.00 1.00 y x 0 ]ionsproduct[ ]ionsttanreac[ ln nF RT EE R = 8.31 J mol-1 K-1 T = 250C = 298 K F (Faraday constant) F = 96500 C mol-1 y x 0 ]ionsproduct[ ]ionsttanreac[ lg )96500(n )303.2)(298)(31.8( EE y x 0 ]ionsproduct[ ]ionsttanreac[ lg n 059.0 EE
55. 55. Ag / Ag+ (1.5 mol dm-3) Cu / Cu2+ (2.0 mol dm-3) Cl2 / Cl- (0.50 mol dm-3) Fe3+ (0.800 mol dm-3) / Fe2+ (1.30 mol dm-3) 1 ]Ag[ lg n 059.0 EE 1 0 Ag+ + e- Ag E0 = + 0.80 V 1 )5.1( lg 1 059.0 80.0E 1 E = + 0.81 V 1 ]Cu[ lg n 059.0 EE 12 0 Cu2+ + 2e- Cu E0 = + 0.34 V 1 )0.2( lg 2 059.0 34.0E 1 E = + 0.35 V 2 1 20 ]Cl[ ]Cl[ lg n 059.0 EE Cl2 + 2e- 2 Cl- E0 = + 1.36 V 2 1 )50.0( )1( lg 2 059.0 36.1E E = + 1.38 V 12 13 0 ]Fe[ ]Fe[ lg n 059.0 EE Fe3+ + e- Fe2+ E0 = + 0.77 V )30.1( )800.0( lg 1 059.0 77.0E 1 E = + 0.758 V
56. 56. Ti3+ (1.20 mol dm-3) / Ti2+ (0.700 mol dm-3) MnO4 - (1.10 mol dm-3) ; H+ (0.800 mol dm-3) / Mn2+ (17.0 mol dm-3) 12 13 0 ]Ti[ ]Ti[ lg n 059.0 EE Ti3+ + e- Ti2+ E0 = - 0.37 V )700.0( )20.1( lg 1 059.0 37.0E E = - 0.356 V ]Mn[ ]H][MnO[ lg n 059.0 EE 2 8 40 MnO4 - + 8 H+ + 5e- 4 H2O + Mn2+ E0 = + 1.52 V )0.17( )800.0)(10.1( lg 5 059.0 52.1E 8 E = + 1.50 V
57. 57. 13.7.1 Nernst Equation and e.m.f. of a chemical cell. Consider the following redox reaction in an chemical cell p A + q B r C + s D At 25oC, Nernst Equation : *For pure solids and liquids, it will not appear in the equation y x 0 ]ionsproduct[ ]ionsttanreac[ ln nF RT EE sr qp cell ]D[]C[ ]B[]A[ lg n 059.0 EE
58. 58. a) Cr (s) Cr3+ (0.010 mol dm-3) Ni2+ (0.20 mol dm-3) Ni (s) Oxidation : Cr Cr3+ + 3 e E0 = + 0.74 V Reduction : Ni2+ + 2 e Ni E0 = 0.25 V Overall : 2 Cr + 3 Ni2+ 3 Ni + 2 Cr3+ Ecell = + 0.49 V E = + 0.51 V 23 32 cell ]Cr[ ]Ni[ lg 6 059.0 EE 2 3 )010.0( )20.0( lg 6 059.0 49.0E
59. 59. b) Mg (s) Mg2+ (0.500 mol dm-3) Fe3+ (1.80 mol dm-3) , Fe2+ (0.750 mol dm-3) Pt (s) Oxidation : Mg Mg2+ + 2e E0 = + 2.38 V Reduction : Fe3+ + e- Fe2+ E0 = + 0.77 V Overall : Mg + 2 Fe3+ Mg2+ + 2 Fe2+ Ecell = + 3.15 V E = + 3.18 V 222 23 cell ]Fe][Mg[ ]Fe[ lg 2 059.0 EE 2 2 )750.0)(500.0( )80.1( lg 2 059.0 15.3E
60. 60. c) Pt (s) Sn2+ (0.300 mol dm-3), Sn4+ (0.500 mol dm-3) Mn3+ (1.20 mol dm-3) , Mn2+ (0.250 mol dm-3) Pt (s) Oxidation : Sn2+ Sn4+ + 2e E0 = 0.15 V Reduction : Mn3+ + e- Mn2+ E0 = + 1.49 V Overall : Sn2+ + 2 Mn3+ Sn4+ + 2 Mn2+ Ecell = + 1.34 V E = + 1.37 V 224 232 ]][[ ]][[ lg 2 059.0 MnSn MnSn EE cell 2 2 )250.0)(500.0( )20.1)(300.0( lg 2 059.0 34.1 E
61. 61. 2.7.1 Nernst Equation and Equilibrium Constant, KC Consider the following reaction : Cu2+ (aq) + Zn (s) Zn2+ (aq) + Cu (s) E0 = + 1.10 V The Kc of the reaction can be expressed as As time past, the concentration of [Cu2+] while [Zn2+] .. Using standard reduction potential of copper and zinc Cu2+ (aq) + 2 e- Cu (s) Eo = + 0.34 V ; when [Cu2+] ; equilibrium shift to Eo Zn2+ (aq) + 2 e- Zn (s) Eo = 0.76 V ; when [Zn2+] ; equilibrium shift to Eo ... ]Cu[ ]Zn[ Kc 2 2 decrease increase decrease left decrease increase right increase
62. 62. V + 0.34 0 Time / s 0.76 Half cell of Cu Half cell of Zn
63. 63. Applied to Nernst equation where When system achieved equilibrium at room temperature, Nernst equation is simplified to : Since and at equilibrium, Ecell = 0, so y x 0 ]ionsproduct[ ]ionsttanreac[ ln nF RT EE x y 0 ]ionreactat[ ]ionproduct[ lg n 059.0 EE ]Cu[ ]Zn[ Kc 2 2 C 0 Klg n 059.0 EV0
64. 64. Consider the following reaction : 2 Fe3+ (aq) + Cu 2 Fe2+ (aq) + Cu2+ (aq) a) Calculate the E0 of the cell b) Calculate the Kc for the reaction. Consider the following reaction : Fe3+ (aq) + Ag (s) Fe2+ (aq) + Ag+ (aq) a) Calculate the E0 of the cell b) Calculate the Kc for the reaction. Fe3+ + e- Fe2+ E0 = + 0.77 V Cu 2 e- + Cu2+ E0 = - 0.34 V 2 Fe3+ + Cu Cu2+ + 2 Fe2+ Ecell = + 0.43 V 23 222 C ]Fe[ ]Cu[]Fe[ K C 0 Klg 2 059.0 EV0 CKlg 2 059.0 43.0V0 lg KC = 14.57 KC = 3.77 x 1014 mol dm-3 Fe3+ + e- Fe2+ E0 = + 0.77 V Ag e- + Ag+ E0 = 0.80 V Fe3+ + Ag Fe2+ + Ag+ Ecell = 0.03 V ]Fe[ ]Fe][Ag[ K 3 2 C CKlg 1 059.0 03.0V0 CKlg 1 059.0 03.0V0 lg KC = 0.5085 KC = 0.310 mol dm-3
65. 65. 2.7.2 Nernst Equation and solubility product, Ksp Similar to calculating equilibrium constant, solubility product of a sparingly soluble salt can also be calculated in the same way mentioned above. Consider the following reaction of dissociation of AgCl (s) AgCl (s) Ag+ (aq) + Cl- (aq) Half-cell : AgCl (s) + e- Ag (s) + Cl- (aq) Eo = + 0.22 V Half-cell : Ag (s) Ag+ (aq) + e- Eo = - 0.80 V Overall Eq: AgCl (s) Ag+ (aq) + Cl- (aq) E0 = - 0.58 V Using Nernst equation and given Ksp = [Ag+][Cl] x y 0 ]ionreactat[ ]ionproduct[ lg n 059.0 EE 1 ]Cl][Ag[ lg n 059.0 EE 0 sp 0 Klg n 059.0 EE
66. 66. At equilibrium ; E = 0 ; so replace in the equation lg Ksp = 9.83 mol2 dm-6 Ksp = 1.48 x 1010 mol2 dm-6 The question may be extend to calculate the solubility from the solubility product calculated Note that Eo is always negative for sparingly soluble salt as Ksp is spKlg 1 059.0 V58.00 very small
67. 67. Example : Given HgCl2 + 2 e- Hg + 2 Cl- Eo = + 0.27 V and Hg2+ + 2 e- Hg Eo = + 0.85 V. Calculate the Ksp for HgCl2. Oxidation Hg Hg2+ + 2 e E0 = 0.85 V Reduction HgCl2 + 2 e- Hg + 2 Cl- E0 = + 0.27 V Overall HgCl2 Hg2+ + 2 Cl- Ecell = 0.58 V sp 0 Klg n 059.0 EE spKlg 2 059.0 V58.00 lg Ksp = 19.66 Ksp = 2.18 x 10-20 mol3 dm-9
68. 68. 13.7.3 Nernst Equation and pH of a Solution Under SHE, if the solution of H+ is not 1.00 mol dm-3, the Ecell can be calculated using Nernst equation. Assuming the metal is oxidise by SHE, the cell notation is written as M (s) I M2+ (aq) II H+ (aq) (x mol dm-3) , H2 (g) I Pt (s) Standard hydrogen electrode Zinc half cell Salt bridge (made of saturated KCl / NaCl) H2 (g) 1.0 atm H+ (aq) [x M] 25oC M (s) M2+ (aq) [1.0 M]
69. 69. The overall reaction can be written as M (s) + 2 H+ (aq) M2+ (aq) + H2 (g) E0 = + a V y x 0 ]ionsproduct[ ]ionsttanreac[ lg n 059.0 EE 12 2 0 ]M[ ]H[ lg 2 059.0 EE ]1[ ]H[ lg2 2 059.0 EE 0 pH059.0EE 0
70. 70. Example : The e.m.f. of of the following cell at 25oC is 0.093 V Pb (s) I Pb2+ (1.00 mol dm-3) II H+ (test solution), H2 (g) I Pt (s) Calculate the pH of the solution Overall equation : Pb + 2 H+ Pb2+ + H2 E0 = + 0.13 V E = E0 0.059 pH 0.059 pH = + 0.13 0.093 pH = 0.63 Calculate the e.m.f. of the chemical cell compared relatively to SHE at [H+] = 0.0030 mol dm-3 for a calcium half cell.
71. 71. 2.9 Type of cell. A battery is a galvanic cell, or a series of combined galvanic cells, that can be used as a source of direct electric current at a constant voltage. Although the operation of a battery is similar in principle to that of the galvanic cells. A battery has the advantage of being completely self-contained and requiring no auxiliary components such as salt bridges. Here we will discuss several types of batteries that are in widespread use. Lithium ion battery Fuel Cell
72. 72. 2.9.1 Lithium ion battery Figure below shows a schematic diagram of a lithium-ion battery. The anode is made of a conducting carbonaceous material, usually graphite, which has tiny spaces in its structure that can hold both Li atoms and Li+ ions. During the discharge of the battery, the half-cell reactions are Anode : Li(s) Li+ + e- Cathode : Li+ + CoO2 + e- LiCoO2 (s) Overall : Li (s) + CoO2 LiCoO2 (s) Ecell = + 3.4 V
73. 73. The cathode is made of a transition metal oxide such as CoO2, which can also hold Li+ ions. Because of the high reactivity of the metal, non - aqueous electrolyte (organic solvent plus dissolved salt) must be used. The advantage of the battery is that lithium has the most negative standard reduction potential and hence the greatest reducing strength. Furthermore, lithium is the lightest metal so that only 6.941 g of Li (its molar mass) are needed to produce 1 mole of electrons. A lithium-ion battery can be recharged literally hundreds of times without deterioration. These desirable characteristics make it suitable for use in cellular telephones, digital cameras, and laptop computers.
74. 74. 2.9.2 Fuel Cell Fossil fuels are a major source of energy, but conversion of fossil fuel into electrical energy is a highly inefficient process. Consider the combustion of methane: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) H = - x kJ mol-1 To generate electricity, heat produced by the reaction is first used to convert water to steam, which then drives a turbine that drives a generator. An appreciable fraction of the energy released in the form of heat is lost to the surroundings at each step; even the most efficient power plant converts only about 40 percent of the original chemical energy into electricity. Combustion reactions are redox reactions, it is more desirable to carry them out directly by electrochemical means, thereby greatly increasing the efficiency of power production
75. 75. This objective can be accomplished by a device known as a fuel cell, a galvanic cell that requires a continuous supply of reactants to keep functioning. In its simplest form, a hydrogen-oxygen fuel cell consists of an electrolyte solution, such as potassium hydroxide solution, and two inert electrodes. Hydrogen and oxygen gases are bubbled through the anode and cathode compartments where the following reactions take place.
76. 76. Anode : 2 H2 (g) + 4 OH- (aq) 4 H2O (l) + 4 e- Eo = + 0.83 V Cathode : O2 (g) + 2 H2O (l) + 4e- 4 OH-(aq) Eo = + 0.40 V Overall : 2 H2 (g) + O2 (g) 2 H2O(l) Eo cell = + 1.23 V The electrodes used have a two-fold function. They serve as electrical conductors, and they provide the necessary surfaces for the initial decomposition of the molecules into atomic species, prior to electron transfer. They are also known as electrocatalysts. Metals such as platinum, nickel, and rhodium are good electrocatalysts.
77. 77. 2.9.2.1 Propane-oxygen fuel cell In addition to the H2-O2 system, a number of other fuel cells have been developed. Among these is the propane-oxygen fuel cell. The half-cell reactions are Anode : C3H8 (g) + 6 H2O (l) 3 CO2 (g) + 20 H+ (aq) + 20 e- Cathode : 5 O2 (g) + 20 H+ (aq) + 20 e- 10 H2O (l) Overall : C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) Unlike batteries, fuel cells do not store chemical energy. Reactants must be constantly resupplied, and products must be constantly removed from a fuel cell. In this respect, a fuel cell resembles an engine more than it does a battery. However, the fuel cell does not operate like a heat engine and therefore is not subject to the same kind of thermodynamic limitations in energy conversion
78. 78. Properly designed fuel cells may be as much as 70 percent efficient, about twice as efficient as an internal combustion engine. In addition, fuel-cell generators are free of the noise, vibration, heat transfer, thermal pollution, and other problems normally associated with conventional power plants.
79. 79. 2.10 Electrolysis ~ decomposition of a substance by direct current electricity. Electrolyte substance that can conduct electricity when in aqueous solution or in molten state Electrolytic cell cell consisting of 2 electrodes immersed in an electrolyte for carrying out electrolysis In an electrolytic cell, the following apply Positive terminal is called as whereas the negative terminal is called as At anode, . process occur where as at cathode, ..process occur Cations are attracted to .. while anions are attracted to Electrons flow from the to .... in the external circuit. A anode cathode oxidation reduction cathode anode anode cathode
80. 80. 2.11 Faradays Law Faradays Law stated that 1 Faraday is the quantity of electricity (9.65 x 104 C) that must be supplied to an electrolytic cell in order to produce one mole of electrons for reactions in the cell. The extension of Faraday Law is stated in Faradays First Law, where it stated that the mass of a substance produced at an electrode during electrolysis is . to the quantity of electricity (in Coulumb) passed. From the statement above, the factors that influence the mass of a substance liberated during electrolysis are The greater the number of electrons transferred, the greater the mass of the product. The longer the time taken, the greater the electrical current produced, the more the mass produced Based on the statement above, the quantity of electrical current can be calculate according to time where Q = electric current I = current t = time in second proportional Q = I t
81. 81. Example 1 : Calculate the mass of silver formed when a current of 0.200 A is applied to a electrolytic cell filled with aqueous silver nitrate for 2 hour. Example 2 : An aqueous solution of copper (II) sulphate is electrolysed using a current of 0.50 A for 4 hours. Calculate the mass of Copper, Cu deposited at cathode. Example 3 : Calculate the mass of chromium formed when 1.20 A of current is directed into molten chromium (III) chloride for 3 hours. Example 4 : Calculate the time taken to produce 18.0 g of silver from silver nitrate by a current of 0.900A Q = It @ Q = (0.200)(2 x 60 x 60) = 1440 C Eq : Ag+ + e- Ag mol of e- = Q / F @ mol = 1440 / 96500 mol of e- = 0.0149 mol Since 1 e- = 1 Ag ; mol Ag = 0.0149 mol Mass Ag = 0.0149 x 108 = 1.61 g Q = It @ Q = (0.50)(4 x 60 x 60) = 7200 C Eq : Cu2+ + 2 e- Cu mol of e- = Q / F @ mol = 7200 / 96500 mol of e- = 0.0746 mol Since 2 e- = 1 Cu ; mol Cu = 0.0373 mol Mass Cu= 0.0373 x 63.5 = 2.4 g Q = It @ Q = (1.20)(3 x 60 x 60) = 12960 C Eq : Cr3+ + 3 e- Cr mol e- = Q / F @ mol = 12960 / 96500 mol of e- = 0.134 mol Since 3 e- = 1 Cr ; mol Cr = 0.0448 mol Mass Cr = 0.0448 x 52.0 = 2.33 g Mol of Ag = 18.0 / 108 = 0.167 mol Eq : Ag+ + e- Ag 1 Ag = 1 e- ; so mol e- = 0.167 mol Q = mol e- x F @ Q = 0.167 x 96500 Q = 16083 C Q = It @ t =16083 / 0.900 t = 1.78 x 104 s
82. 82. Example 5 : Calculate the time required to form 200g of lead from molten lead (II) bromide by a current of 1.50 A. Example 6 : Calculate the time required to form 10 g of aluminium from molten aluminium oxide by a current of 10 A Mol of Pb = 200 / 207 = 0.966 mol Eq : Pb2+ + 2 e- Pb 1 Pb = 2 e- ; so mol e- = 1.93 mol Q = mol e- x F @ Q = 1.93 x 96500 Q = 186473 C Q = It @ 186473 / 1.50 t = 1.24 x 105 s Mol of Al = 10 / 27 = 0.370 mol Eq : Al3+ + 3 e- Al 1 Al = 3 e- ; so mol e- = 1.11 mol Q = mol e- x F @ Q = 1.11 x 96500 Q = 107222 C Q = It @ t = 107222 / 10 t = 1.1 x 104 s
83. 83. 14.2.1 Faradays Second Law ~ stated that if the same quantity of electricity is passed through different electrolytes, the mass of substance liberated at electrode is inversely proportional to the charge of ions. 1 Faraday Silver (I) nitrate Copper (II) sulphate Chromium (III) chloride Sulphuric acid
84. 84. Half equation occur at cathode for From the diagram above, 1 F will discharge . mol of Ag+ ions ; mol of Cu2+ ion ; mol of Cr3+ ion ; . mol of O2. Type of electrode Half equation Mol of metal deposited Silver Copper Carbon in CrCl3 Carbon in H2SO4 (occur at anode) Mol of non-metal Ag+ (aq) + e- Ag (s) 1 mol Cu2+ (aq) + 2 e- Cu (s) 1/2 mol Cr3+ (aq) + 3 e- Cr (s) 1/3 mol 2 H2O O2 + 4 H+ + 4 e 1/4 mol 1 1/2 1/3 1/4
85. 85. Example 7 : Calculate the mass of copper deposited under the same cell if the amount of silver formed under the same amount of quantity charge is 1.8 g. An electric current produced 0.56 g of aluminium from molten aluminium oxide. If the same current was used to electrolysed molten lead (II) bromide, calculate the mass of lead deposited. Mol of Ag = 1.8 / 108 = 0.01667 mol Eq : Ag+ + e- Ag Since 1 Ag = 1 e- ; mol e- = 0.01667 mol For Cu ; Cu2+ + 2 e Cu Since 2 mol e- = 1 mol Cu So mol of Cu = 0.008333 mol Mass of Cu = 0.008333 x 63.5 = 0.53 g Mol of A1 = 0.56 / 27 = 0.02074 mol Eq : Al3+ + 3 e- Al Since 1 Al = 3 e- ; mol e- = 0.06222 mol For Pb ; Pb2+ + 2 e Pb Since 2 mol of e = 1 mol of Pb So mol of Pb = 0.03111 mol Mass of Pb = 0.03111 x 207 = 6.44 g
86. 86. Predicting the product for electrolysis In electrolysis, there may be more than one type of cation / anion inside the electrolytes. Under such circumstance, since an electrode can only discharge one cation / anion, the ion must be choose under certain guidelines. The selectivity of ions are based on electrochemical series
87. 87. No matter it is an electrolytic cell or chemical cell, At anode, . reaction occur ; electrons are At cathode, . reaction occur ; electrons are Electrolytes can only discharge under 2 conditions : ..... state or solution. When in molten state, the electrolytes contain only the cation and anion of the substance involve Molten lead (II) bromide : PbBr2 (l) Pb2+ + 2 Br Molten aluminium oxide : Al2O3 (l) 2 Al3+ + 3 O2 Molten barium chloride : BaCl2 (l) Ba2+ + 2 Cl Molten silver (I) iodide : AgI (l) Ag+ + I oxidation donated reduction received molten aqueous
88. 88. Electrolyte Subst. present Half equation at anode Substance at anode Half equation at cathode Substance at cathode PbBr2 (l) Al2O3 (l) NaCl (aq) Pb2+ Br Br2 + 2e- 2Br- E0 = + 1.07 V Rev: 2Br- 2e- + Br2 bromine Pb2+ + 2e- Pb E0 = 0.13 V lead Al3+ O2 O2 + 4e- 2O2- E0 = ? V Rev: 2O2- O2 + 4e- oxygen Al3+ + 3e- Al E0 = 1.67 V Aluminu m Na+ Cl Cl2 + 2e- 2Cl- E0 = +1.36 V Rev: 2Cl- Cl2 + 2e- chlorine Na+ + e- Na E0 = 2.71 V Sodium
89. 89. No matter it is an electrolytic cell or chemical cell, At anode, . reaction occur ; electrons are At cathode, . reaction occur ; electrons are When in aqueous solution, not only it contains the cation and anion of substance involve, but it also involves water. thus there is a selectivity of ion occur In the terms of E0, a more .. value will be selected for discharge at cathode, while a more . value will be selected for discharge at anode. At anode : E0 = . V At cathode : E0 = . V However, water is a weak electrolytes. At 250C and 1 atm, the E0 value varies with the solution used. So in deciding which E0 value we should used, we need to consider if the solution is different or not. Under neutral condition, where [H+] = [OH-], using Nernst Equation, Eo values are (At [H+] = [OH-]1.0 x 10-7 mol dm-3) At anode : E0 = .. V At cathode : E0 = .. V oxidation donated reduction received positive negative 4 H+ + O2 + 4 e 2 H2O + 1.23 2 H2O + 2 e 2 OH + H2 0.83 4 H+ + O2 + 4 e 2 H2O + 0.81 2 H2O + 2 e 2 OH + H2 0.41
90. 90. Electrolyte Subst. present Half equation at anode Substance at anode Half equation at cathode Substance at cathode NaCl (aq) Na+ Cl H2O Cl2 + 2e- 2Cl- E0 = +1.36 V 4 H+ + O2 + 4 e 2 H2O E0 = +1.23 V Rev: 2 H2O 4 H+ + O2 + 4 e oxygen Na+ + e- Na E0 = 2.71 V 2 H2O + 2 e 2 OH + H2 E0 = 0.83 V Hydroge n
91. 91. Electrolyte Ions present Half equation at anode Substance Half equation at cathode Substance CuSO4 (aq) PbI2 (aq) KOH (aq) Cu2+ SO4 2 H2O S2O8 2- +2e- 2SO4 2- E0 = +2.01 V 4 H+ + O2 + 4 e 2 H2O E0 = +1.23 V Rev: 2 H2O 4 H+ + O2 + 4 e oxygen Cu2+ + 2e- Cu E0 = + 0.34 V 2 H2O + 2 e 2 OH + H2 E0 = 0.83 V Copper Pb2+ I H2O I2 + 2 e- 2 I- E0 = +0.54 V 4 H+ + O2 + 4 e 2 H2O E0 = +1.23 V Rev: 2 I I2 + 2 e iodine Pb2+ + 2e- Pb E0 = 0.13 V 2 H2O + 2 e 2 OH + H2 E0 = 0.83 V Lead K+ OH H2O O2 + 2 H2O + 4 e- 4 OH E0 = + 0.40 V 4 H+ + O2 + 4 e 2 H2O E0 = +1.23 V Rev: 4 OH 2 H2O + O2 + 4 oxygen K+ + e- K E0 = 2.92 V 2 H2O + 2 e 2 OH + H2 E0 = 0.83 V hydrogen
92. 92. Effect on concentration towards the selectivity of the ions to discharge In general, an ion with a very high concentration is preferentially discharged. For example if Pb2+ ion and Cu2+ ion are mixed under the same concentration, ion is preferred to be discharge at cathode as it has a lower position in electrochemical series. However, if the concentration of Pb2+ ion concentration is raised much higher than Cu2+ ion, . ion is more readily to be discharged. Another example is potassium chloride in aqueous solution. Under dilute solution of KCl, .. will be selected at anode and ... gas is given out, as it has a lower position in electrochemical series. Cu2+ Pb2+ H2O oxygen
93. 93. However, if concentrated KCl is used as electrolyte, the concentrated of Cl- increase, and is selected to be discharge and ... gas is given out. Still, if the position in electrochemical series differ too much, like K+ and water in the example above, K+ ion be discharge as the position is much too high. At the end, . gas is given out at cathode. If chloride ion is replaced with fluoride ion, F ion, and concentration of F ion is increased, ....... is still be preferred as F ion has a . position in electrochemical series In the table below, predict the element that is expected to form when electrolyse. Cl chlorine will not hydrogen water very high
94. 94. Electrolyte Ions present Half equation at anode Substance Half equation at cathode Substance Concentrate NaCl Concentrated PbBr2 Concentrated LiF Na+ Cl H2O Cl2 + 2e- 2Cl- E0 = +1.36 V 4 H+ + O2 + 4 e 2 H2O E0 = +1.23 V Rev: 2 Cl- Cl2 + 2 e chlorine Na+ + e- Na E0 = 2.71 V 2 H2O + 2 e 2 OH + H2 E0 = 0.83 V Hydroge n Pb2+ Br H2O 4 H+ + O2 + 4 e 2 H2O E0 = +1.23 V Br2 + 2e- 2Br- E0 = + 1.07 V Rev: 2Br- 2e- + Br2 bromine Pb2+ + 2e- Pb E0 = 0.13 V 2 H2O + 2 e 2 OH + H2 E0 = 0.83 V lead Li+ F H2O F2 + 2e- 2F- E0 = +2.87 V 4 H+ + O2 + 4 e 2 H2O E0 = +1.23 V Rev: 2 H2O 4 H+ + O2 + 4 e oxygen Li+ + e- Li E0 = 3.04 V 2 H2O + 2 e 2 OH + H2 E0 = 0.83 V Hydrogen
95. 95. 2.12.3 Overvoltage Overvoltage ~ the difference between electrode potential and discharge potential. In another words, Overvoltage is the voltage that must be applied to an electrolytic cell in addition to the theoretical voltage to cause an electrode reaction to occur. Example of over voltage phenomenon is the electrolysis of aqueous sodium chloride. Consider the electrolysis of sodium chloride in aqueous solution where the substance presence are Na+, Cl-, H2O Half equation for substance attracted to anode Half equation for substance attracted to cathode Cl2 (g) + 2 e- 2 Cl- (aq) E0 = + 1.36 V 4 H+ + O2 (g) + 4 e- 2 H2O (l) E0 = + 0.81 V (conc. H+ = 1 x 10-7 mol dm-3) Na+ (aq) + e- Na (s) E0 = - 2.71 V 2 H2O (l) + 2 e- 2 OH- + H2 (g) E0 = - 0.41 V (conc. OH- = 1 x 10-7 mol dm-3)
96. 96. From the E0 values at anode, it is suggested that H2O should be preferentially oxidized at the anode. However, by experiment we find that the gas liberated at the anode is Cl2, not O2. In studying electrolytic processes, we sometimes find that the voltage required for a reaction is considerably higher than the electrode potential indicates. The overvoltage is the difference between the electrode potential and the actual voltage required to cause electrolysis. In this case, overvoltage for O2 formation is quite high. Therefore, under normal operating conditions Cl2 gas is actually formed at the anode instead of O2. As for the selectivity at cathode, H2O is selected since the E0 value for H2O is less negative than Na+.
97. 97. Thus the half equation occur at both anode and cathode are Anode Cathode Overall As the overall reaction shows, the concentration of the Cl2 ions decreases during electrolysis and that of the OH- ions increases. Therefore, in addition to H2 and Cl2, the useful by-product NaOH can be obtained by evaporating the aqueous solution at the end of the electrolysis.
98. 98. 2.12.4 Electrorefining and electroplating The purification of a metal by means of electrolysis is called electrorefining. For example, impure copper obtained from ores is converted to pure copper in an electrolytic cell that has impure copper as the anode and pure copper as the cathode. The electrolyte is an aqueous solution of copper sulphate
99. 99. At the impure Cu anode, copper is oxidized along with more easily oxidized metallic impurities such as zinc and iron. Less easily oxidized impurities such as silver, gold, and platinum fall to the bottom of the cell as anode mud, which is reprocessed to recover the precious metals. At the pure Cu cathode, ions are reduced to pure copper metal, but the less easily reduced metal ions (and so forth) remain in the solution Half equations occur for electrorefining process of copper above are At anode Cu (s) Cu2+ (aq) + 2 e- At cathode Cu2+ (aq) + 2 e- Cu (s) Thus, the net cell reaction simply involves transfer of copper metal from the impure anode to the pure cathode, hence purified the copper.
100. 100. Closely related to electrorefining is electroplating, the coating of one metal on the surface of another using electrolysis. For example, steel automobile bumpers are plated with chromium to protect them from corrosion, and silver-plating is commonly used to make items of fine table service. The object to be plated is carefully cleaned and then set up as the cathode of an electrolytic cell that contains a solution of ions of the metal to be deposited (as shown in diagram above) Half equations occur for electroplating process of silver above are At anode Ag (s) Ag+ (aq) + e- At cathode Ag+ (aq) + e- Ag (s)
101. 101. 2.13 Industrial Electrolysis : In this Chapter, we shall discussed the manufacturing of aluminium and chlorine gas using the principle of electrolysis Part 1 : Getting pure aluminium oxide (alumina) from bauxite. 1st step: Removal of impurities from the ore by dissolving powdered bauxite in hot concentrated sodium hydroxide solution. 2nd step: Insoluble impurities are filtered off. Filtrate contain aluminium and silicon ions. Aluminium ion is precipitated as aluminium hydroxide which is filtered out later as white gelatinous precipitate. 3rd step: Aluminium hydroxide is filtered, washed, dried and finally heated out to 12000C to produce pure aluminium oxide (alumina), Al2O3 Al2O3 + 2 NaOH + 3 H2O 2 NaAl(OH)4 SiO2 + 2 NaOH Na2SiO3 + H2O Use acid : 2 [Al(OH)4]- + 2 H+ 2 Al(OH)3 + 2 H2O Or use CO2 : 2 [Al(OH)4]- + CO2 2 Al(OH)3 + H2O + CO3 2- 2 Al(OH)3 Al2O3 + 3 H2O
102. 102. Part 2 : Extracting aluminium out from aluminium oxide Hall-Heroult process:- A process of electrolysing aluminium oxide (alumina) to extract out aluminium. Aluminium metal is extracted by the cell electrolytic reduction of alumina. Melting point of alumina is 20300C. To lower the temperature of the electrolyte, alumina is dissolved in molten cryolite (Na3AlF6), to maintain a temperature at about 9600C. When alumina dissolve in molten cryolite : Al2O3 (s) 2 Al3+ (l) + 3 O2 (l)
103. 103. Electrolyte mixture is then placed in carbon-lined iron vat (cathode). The heating effect of the electric current melts the electrolyte mixture, producing Na+, Al3+, O2- and F- ions. Half equation occur at cathode Half equation occur at anode Na+ + e- Na E0 = 2.71 V Al3+ + 3e- Al E0 = 1.66 V F2 + 2 e- 2 F- E0 = + 2.87 V O2 + 4e- 2O2- E0 = + 1.++V Al3+ + 3e- Al E0 = 1.66 V 2O2- O2 + 4e- E0 = + 1.++V
104. 104. Aluminium alloy parts are anodized to greatly increase the thickness of this layer for corrosion resistance. The corrosion resistance of aluminium alloys is significantly decreased by certain alloying elements or impurities : copper, iron, and silicon, tend to be most susceptible. By making an aluminium the anode of cell in which dilute sulphuric acid is the electrolytes, it is possible to produce a thicker and harder film of aluminium oxide on the surface of metal. Al C
105. 105. 2.13.1.2 Recycling aluminium Environmental pollution arises as cans are littered everywhere. The best solution of environmental pollution is recycling. The benefits of recycling can be seen by comparing the energy consumed in the extraction of aluminium from the bauxite or using Hall process with that consumed when aluminium is recycle.
106. 106. Pure aluminium has a rather low melting print of 6600C, thus requiring only 26.1 kJ mol-1 of energy. On comparison between the Hall process and recycling, Energy used in recycling = 26.1 297 100% = 8.8% This means that about 91% of the energy is saved for every 1 mole of the aluminium produced through recycling
107. 107. In industrial process, chlorine gas, together with sodium metal, is prepared using molten sodium chloride (brine) using mercury-cathode cell. Mercury is specially used to attract the sodium formed in cathode and form an alloy named amalgam This method is not environment friendly as the mercury used is poisonous. Half equation occur at cathode Half equation occur at anode Na+ + e- Na E0 = 2.71 V 2Cl- Cl2 + 2e- E0 = 1.36 V
108. 108. Similar to the mercury-cathode cell, the electrolytes used in diaphragm cell is also .. The process inside the diaphragm cell is known as the chlor- alkali process. When sodium chloride dissociates under the effect of an electric current, the chloride ions are discharged. Half equation : brine (sodium chloride) 2 Cl- Cl2 + 2 e-
109. 109. Titanium is chosen as the anode because it resists corrosion by the very reactive chlorine At cathode, since the sodium ion (Na+) is attracted to cathode through the diaphragm, the selectivity to discharge is between sodium ion and water molecule. Standard reduction potential of sodium : Standard reduction potential of water : Since has a higher E0 value, is discharge and is produced. The level of brine (left or anode position) is always placed higher than the water (right or cathode position) to ........ Na+ + e- Na E0 = 2.71 V 2 H2O + 2 e 2 OH + H2 E0 = 0.83 V water water hydrogen gas prevent water from crossing to brine portion. This will dilute the solution and chlorine will not be discharged.
110. 110. 13.8 Corrosion of metal Corrosion is the oxidative deterioration of a metal, such as the conversion of ... to .. 2 main important components for rusting are .. and A possible mechanism for rusting, consistent with the known facts, is illustrated in Figure below metal metal oxide oxygen water
111. 111. At anode : Fe (s) Fe2+ (aq) + 2e- Eo = +0.44 V At cathode : O2(aq) + 2H2O(l) + 4e 4OH(aq) Eo = + 0.40 V 2 Fe (s) + O2 (aq) + 2 H2O 2 Fe2+ + 4 OH [or 2 Fe(OH)2] Ecell = + 0.84 V Fe(OH)2 (aq) + OH Fe(OH)3 + e Eo = +0.56 V O2(aq) + 2H2O(l) + 4e 4OH(aq) Eo = + 0.40 V 4 Fe(OH)2 (aq) + O2(aq) + 2H2O(l) 4 Fe(OH)3 (aq) Ecell = + 0.96 V Forming rust : 2 Fe(OH)3 (s) Fe2O3.x H2O + (3 x) H2O In alkaline / neutral condition
112. 112. In acidic condition At anode : Fe (s) Fe2+ (aq) + 2e- Eo = +0.44 V At cathode : O2(aq) + 4 H+(aq) + 4e 2H2O(l) Eo = + 1.23 V 2 Fe (s) + O2 (aq) + 4 H+ 2 Fe2+ + 2 H2O Ecell = + 1.67 V At anode : Fe2+ (aq) Fe3+ (aq) + e- Eo = -0.77 V At cathode : O2(aq) + 4 H+(aq) + 4e 2H2O(l) Eo = + 1.23 V 4 Fe2+ (aq) + O2 (aq) + 4 H+(aq) 4 Fe3+ (aq) + 2H2O(l) Ecell = + 0.46 V Forming rust : 2 Fe3+ (aq) + 4 H2O (l) Fe2O3.H2O (s) + 6 H+ (aq)
113. 113. 13.8.2 Prevention of rusting Various methods are used to prevent / slowing down rusting. Methods Explanation Alloying Iron is alloyed with nickel and chromium to form . The chromium forms an impervious oxide layer on the surface of iron increasing its resistance to corrosion. Chromium at the same time .. the steel Barrier Painting the iron object Use grease / oil to coat the moving parts of machine Coating ironwith chromium (plating) or zinc (galvanising) Sacrificial Also known as protection Metal with a . position in electrochemical series is connected to iron. Under such way, ... Will be oxidised first before iron. Stainless steel decorated / coated anodic higher reactive metal
114. 114. Additional Slide (from old syllabus)
115. 115. 13.7.4 Effect of pH on Electrode Potential of a Half cell Some reaction involve H+ ions. Examples of the are 14 H+ + Cr2O7 2- + 6 e- 2 Cr3+ + 7 H2O 8 H+ + MnO4 - + 5 e- Mn2+ + 4 H2O 6 H+ + ClO3 - + 6 e- Cl- + 3 H2O Under standard conditions, the [H+] is 1.00 mol dm-3. varying the concentration of [H+] and hence its pH, would change the electrode potential of the half cell. Consider the following half cell reaction : MnO4 - (aq) + 8 H+ (aq) + 5 e- Mn2+ (aq) + 4 H2O (l) E = + 1.52 V Using Nernst Equation, Ecell can be expressed as When [MnO4 -] = [Mn2+] = 1.00 mol dm-3- ; Ecell = ]Mn[ ]H][MnO[ lg 5 059.0 52.1E 2 8 4 ]H[lg8 5 059.0 52.1E pH0944.052.1E
116. 116. [H+] / mol dm-3 1.0 0.1 0.01 0.00001 pH 0 1 2 5 Ecell (V) + 1.52 + 1.43 + 1.33 +1.05 It is then compare to the standard electrode potential of chlorine, bromine and iodine Cl2 + 2 e- 2 Cl- Eo = + 1.36 V @ 2 Cl- Cl2 + 2 e- Eo = - 1.36 V Br2 + 2 e- 2 Br- Eo = + 1.07 V @ 2 Br- Br2 + 2 e- Eo = - 1.07 V I2 + 2 e- 2 I- Eo = + 0.54 V @ 2 I- I2 + 2 e- Eo = - 0.54 V Under pH = 1 , Ecell of manganate (VII) ion is ................ V are able to by manganate (VII) ion as it is still a strong agent. Under pH = 2, Ecell of manganate (VII) ion is .............. V. Only are able to by manganate (VII) ion. cannot oxidise as the reaction is not (Ecell = ) Under pH = 5, Ecell of manganate (VII) ion is .............. V. Only are able to by manganate (VII) ion. . cannot oxidise as the reaction is not (Ecell = ) + 1.43 Cl- , Br- , I- oxidise oxidising + 1.33 Br- , I- oxidise Cl- spontaneous - 0.03 V + 1.05 I- oxidise Cl- , Br- spontaneous -0.03 V for Br- -0.31 V for Cl-
117. 117. Primary cell Dry cell Alkaline cell Diagram Anode Equation occur at anode Cathode Equation occur at cathode Electrolytes zinc Zn Zn2+ + 2 e Manganese (IV) oxide 2 MnO2 + 2 NH4 + 2 e- Mn2O3 + 2 NH3 + H2O Ammonium chloride and zinc chloride paste zinc Zn + 2OH H2O + ZnO + 2 e Manganese (IV) oxide 2 MnO2 + 2 H2O + 2 e- Mn2O3 + 2 OH Potassium hydroxide
118. 118. Primary cell Mercury cell Lead acid accumulator Diagram Anode Equation occur at anode Cathode Equation occur at cathode Electrolytes Zinc Zn + 2OH H2O + ZnO + 2 e Mercury (II) oxide HgO + H2O + 2 e- Hg + 2 OH Potassium hydroxide Lead Pb + HSO4 PbSO4 + H+ + 2 e Lead (IV) oxide PbO2 + 3 H+ + HSO4 + 2 e PbSO4 + 2 H2O Sulphuric acid
119. 119. Fuel Cell Lithium ion cell Diagram Anode Lithium metal Equation occur at anode Li (s) Li+ (aq) + e- Cathode Manganese (IV) oxide Equation occur at cathode MnO2 (s) + Li+ + e- LiMnO2 Electrolytes Lithium chlorite (VII), LiClO4 Hydrogen 2 H2 + 4 OH 2 H2O + 4 e Oxygen O2+ 2 H2O + 4 e 4 OH Hot potassium hydroxide (aq)
120. 120. 13.9 Dental Filling The material commonly used to fill decaying teeth is an . (a base alloy). The component in dental filling of amalgam are . , and .. The standard electrode potential of these electrode system are : Hg2 2+ (aq) / Ag2Hg3 (s) Eo = + 0.85 V Sn2+ (aq) / Ag3Sn (s) Eo = 0.55 V Sn2+ (aq) / Sn8Hg (s) Eo = 0.13 V The diagram shows the reaction take place when gold is contact with dental amalgam, which result a electrochemical cell. act as the anode of the cell, while act as the cathode and .. act as the electrolyte. Since tin is more electro than gold, hence tin will corrode to form Sn2+ and mixed with saliva. This will result an unpleasant taste in the mouth. If the dental amalgam is in contact with an aluminium foil, an electrochemical cell will also produced. Unlike gold, aluminium is more .. than any of the electrode above, which makes aluminium serves as an of the cell, while . as the cathode of cell. This will result a weak current flow between the electrode and cause an unpleasant sensation in the tooth. amalgam mercury mercury silver tin dental amalgam gold saliva positive electropositive anode amalgam