1. Oxidation Number In standard elemental form = 0 MgBr 2 Ne 0 0 0 Ion = charge Na + Ba 2+ Cl - +1 +2 -1 Fluoride = -1 Oxygen (in compounds) = -2 (except peroxide = -1; superoxide = - 1 / 2 ; with F = +2) Hydrogen (in compoundds) = +1 (except hydride = -1) Group 1 compound = +1; Group 2 compound = +2 Aluminium compound = +3

2. Oxidation Number Sum of oxidation numbers = charge on species Neutral compounds NaCl BaO SO 2 +1; -1 = 0 +2; -2 = 0 +4; 2x(-2) = 0 Compound ions NH 4 + SO 4 2- NO 2 - -3; 4x(+1) = +1 +6; 4x(-2) = -2 +3; 2x(-2) = -1 sulphate (VI) nitrate (III)/nitrite Try: Cu in Cu 2 O H in LiH C in CO Cr in Cr 2 O 7 2- Mn in KMnO 4

3. Redox Electron transfer OIL RIG Oxidation – increase in oxidation number Mg ⇋ Mg 2+ + 2e - 0  +2 Reduction – decrease in oxidation number ClO - + 2 H + + 2e - ⇋ Cl - + H 2 O +1-2 2x(+1)-1 2x(+1) -2 Change in oxidation state = number of electrons ‘transferred’

4. Reducing Agents Oxidises other species Is itself reduced Stronger oxidising agents oxidise weaker oxidising agents Cl 2 + 2e - ⇋ 2 Cl - Reduction Cl 2 + 2 Br -  2 Cl - + Br 2 Br 2 + 2 I -  2 Br - + I 2 Oxidising Agents Reduces other species Is itself oxidised Stronger reducing agents reduce weaker reducing agents Cl 2 + 2 Br -  2 Cl - + Br 2 Br 2 + 2 I -  2 Br - + I 2

5. Redox half equations Conventionally written to show reduction Mg 2+ + 2 e - ⇋ Mg + 2 0 ‘Extra’ oxygen removed by H + (‘acidic conditions’) ClO - + 2 H + + 2e - ⇋ Cl - + H 2 O +1-2 2x(+1)-1 2x(+1) -2 ‘Extra’ oxygen added by water Mn 2+ + 4H 2 O ⇋ MnO 4 - + 8 H + + 5 e -

6. A redox half equation Cr 2 O 7 2- ⇋ 2 Cr 3+ -2 = 7x(-2) + 2x(+?) Cr oxidation state = (-2+14)/2 = +6; Chromium (VI) Change in oxidation state: +6  +3 3 e - per Cr  6 e - taken in Cr 2 O 7 2- + 6 e - ⇋ 2 Cr 3+ 7 oxygens to be removed Each O removed by 2 H + Require 14 H + Making 7 H 2 O Cr 2 O 7 2- + 14 H + + 6 e - ⇋ 2 Cr 3+ + 7 H 2 O

7. Combining half equations Requires an oxidation and a reduction Electrons cancel out Magnesium produces hydrogen when it reacts with dilute acid Mg 2+ + 2 e - ⇋ Mg 2 H + + 2 e - ⇋ H 2 (half equations given as reductions) –Convert Mg 2+ reduction to oxidation Mg ⇋ Mg 2+ + 2 e - –Combine equations Mg + 2 H + + 2 e -  Mg 2+ + H 2 + 2 e - –Cancel out electrons Mg + 2 H +  Mg 2+ + H 2

8. Combining half equations Oxidation of iron (II) to iron (III) by acidified manganate (VII) MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O Fe 3+ + e - ⇋ Fe 2+ Reverse the oxidation reaction Fe 2+ ⇋ Fe 3+ + e - Balance the electrons 5 Fe 2+ ⇋ 5 Fe 3+ + 5 e - Combine MnO 4 - + 8H + + 5e - + 5 Fe 2+  Mn 2+ + 4H 2 O + 5 Fe 3+ + 5 e - Cancel electrons MnO 4 - + 8H + + 5 Fe 2+  Mn 2+ + 4H 2 O + 5 Fe 3+

9. Disproportionation The same species is simultaneously oxidised and reduced Cl 2 + 2 OH -  ClO - + Cl - + H 2 O 0 +1 -1

10. Electrochemistry To observe and measure electrode potentials To understand why electrode potentials occur To recognise that differences in electrode potential are related to the chemicals involved and the conditions under which the measurements are taken

11. On whiteboards Complete these half equations and combine to give full equation MnO 4 - ⇋ Mn 2+ Sn 2+ ⇋ Sn 4+ What would you see during the reaction? If you were using this reaction to find the concentration of tin (II) what would you put into the burette – and why?

12. On whiteboards Complete these half equations and combine to give full equation MnO 4 - + 8 H + + 5 e - ⇋ Mn 2+ + 4 H 2 O Sn 2+ ⇋ Sn 4+ + 2 e - 2 MnO 4 - + 16 H + + 5 Sn 2+  2 Mn 2+ + 8 H 2 O + 5 Sn 4+ What would you see during the reaction? If you were using this reaction to find the concentration of tin (II) what would you put into the burette – and why?

13. Apparatus set up Prevented by high resistance voltmeter Zn(s)|Zn 2+ (aq) Cu 2+ (aq)|Cu (s) Electron flow

14. What is happening? 1 moldm -3 Zn 2+ (aq) Zn (s) 2 e - Oxidation is an anodic process i.e. the electrode at which oxidation occurs is the anode Zn(s)|Zn 2+ (aq) Zn (s) ⇋ Zn 2+ (aq) + 2 e - E Θ = +0.76 V Data books publish reduction potentials: Zn 2+ (aq) + 2 e - ⇋ Zn (s) E Θ = -0.76 V Change of state

15. The standard electrode i.e. metals that react with dilute acid i.e. metals that don’t react with dilute acid – can be reduced by hydrogen gas

16. The standard electrode in use E Θ cell = E Θ rhs - E Θ lhs E Θ H 2 half cell defined as 0  E Θ cell = 0 - E Θ Zn half cell  E Θ Zn half cell = - 0.76 V It is simpler to have the H 2 electrode on the lhs Completes circuit…without reacting

17. A way to visualise things 0 V 2 H + + 2 e - H 2 Reduction potentials + ve stronger oxidising agents - ve weaker oxidising agents + 0.34 V Cu 2+ + 2 e - Cu + 0.80 V Ag + + e - Ag -0.76 V Zn 2+ + 2 e - Zn E Θ cell = difference between points on line The right hand half equation goes forward (reduction – oxidising agent) The left hand half equation reverses (oxidation – reducing agent)

18. Zinc-copper cell Zn(s)|Zn 2+ (aq) Cu 2+ (aq)|Cu (s) E Θ = +1.10 V Standard conditions: 298 K, 1 moldm -3 (& 1 atm) Zn(s)|Zn 2+ (aq) Zn (s) ⇋ Zn 2+ (aq) + 2 e - E Θ = +0.76 V Data books publish reduction potentials: Zn 2+ (aq) + 2 e - ⇋ Zn (s) E Θ = -0.76 V Cu 2+ (aq)|Cu(s) Cu 2+ (aq) + 2 e- ⇋ Cu (s) E Θ = + 0.34 V E Θ cell = E Θ rhs – E Θ lhs = 0.34 –(-0.76) = + 1.10 V

19. Zinc-iron cell Zn(s)|Zn 2+ (aq) Fe 2+ (aq)|Fe (s) E Θ = +0.32 V Standard conditions: 298 K, 1 moldm -3 (& 1 atm) Zn(s)|Zn 2+ (aq) Zn (s) ⇋ Zn 2+ (aq) + 2 e - E Θ = +0.76 V Data books publish reduction potentials: Zn 2+ (aq) + 2 e - ⇋ Zn (s) E Θ = -0.76 V Fe 2+ (aq)|Fe(s) Fe 2+ (aq) + 2 e- ⇋ Fe (s) E Θ = - 0.44 V E Θ cell = E Θ rhs – E Θ lhs = -0.44 –(-0.76) = + 0.32 V

20. Iron-copper cell Fe(s)|Fe 2+ (aq) Cu 2+ (aq)|Cu (s) E Θ = +0.78 V Standard conditions: 298 K, 1 moldm -3 (& 1 atm) Fe(s)|Fe 2+ (aq) Fe (s) ⇋ Fe 2+ (aq) + 2 e - E Θ = +0.44 V Data books publish reduction potentials: Fe 2+ (aq) + 2 e - ⇋ Fe (s) E Θ = -0.44 V Cu 2+ (aq)|Cu(s) Cu 2+ (aq) + 2 e- ⇋ Cu (s) E Θ = + 0.34 V E Θ cell = E Θ rhs – E Θ lhs = 0.34 –(-0.44) = +0.78 V

21. Adding Electrode Potentials Zn(s)|Zn 2+ (aq) Fe 2+ (aq)|Fe (s) E Θ = +0.32 V Fe(s)|Fe 2+ (aq) Cu 2+ (aq)|Cu (s) E Θ = +0.78 V Zn(s)|Zn 2+ (aq) Cu 2+ (aq)|Cu (s) E Θ = +1.10 V

22. Two solutions E Θ for a redox between two aqueous substances, e.g. Fe 2+ /Fe 3+ is performed using a mixture of both solutions (both at 1 moldm -3 ), and an inert electrode, e.g. Pt. Pt|Fe 2+ (aq),Fe 3+ (aq) MnO 4 -,Mn 2+ |Pt electron flow E Θ cell = 1.52 – (+ 0.77) = + 0.75

23. Try these Ni (s)|Ni 2+ (aq) Sn 4+ (aq), Sn 2+ (aq)|Pt (s) Pt (s)|I - (aq), ½I 2 (aq) Ag + (aq)|Ag (s)

24. Effect of concentration Zn 2+ (aq) + 2 e - Zn (s) – 0.76 V dilution shifts equilibrium left less reduction E more –ve/less +ve increased concentration shifts equilibrium right more reduction E less –ve/more +ve Summary Lower [ ] Higher [ ]

25. A way to visualise things 0 V 2 H + + 2 e - H 2 Reduction potentials + ve stronger oxidising agents - ve weaker oxidising agents + 0.34 V Cu 2+ + 2 e - Cu + 0.80 V Ag + + e - Ag -0.76 V Zn 2+ + 2 e - Zn Dilution shifts reduction equilibrium left (less positive/more negative)

26. A way to visualise things 0 V 2 H + + 2 e - H 2 Reduction potentials + ve stronger oxidising agents - ve weaker oxidising agents + 0.34 V Cu 2+ + 2 e - Cu + 0.80 V Ag + + e - Ag -0.76 V Zn 2+ + 2 e - Zn Increased concentration shifts reduction equilibrium right (more positive/less negative)

27. Reduction or oxidation? E Θ measures reduction potential The more +ve E Θ the more likely to be reduced (gain electrons) Cu 2+ + 2 e - ⇋ Cu +0.34 V The more -ve E Θ the more likely the right hand species is to be oxidised (lose electrons) Zn 2+ + 2 e - ⇋ Zn -0.76 V Zn ⇋ Zn 2+ + 2 e - +0.76 V Zn + Cu 2+  Zn 2+ + Cu E Θ cell = 0.34 – (-0.76) = + 1.10 V i.e. zinc displaces copper from solution

28. Reduction or oxidation? E Θ measures reduction potential The more +ve/less -ve E Θ the more likely to be reduced (gain electrons) Fe 2+ + 2 e - ⇋ Fe -0.44 V The more –ve/less +ve E Θ the more likely to be oxidised (lose electrons) Zn 2+ + 2 e - ⇋ Zn -0.76 V Zn ⇋ Zn 2+ + 2 e - +0.76 V Zn + Fe 2+  Zn 2+ + Fe E Θ cell = -0.44 – (-0.76) = + 0.32 V i.e. zinc displaces iron from solution

29. Reduction or oxidation? E Θ measures reduction potential The more +ve/less -ve E Θ the more likely to be reduced (gain electrons) MnO 4 - + 8 H + + 5 e - ⇋ Mn 2+ + 4 H 2 O +1.52 V The more –ve/less +ve E Θ the more likely to be oxidised (lose electrons) Cl 2 + 2 e - ⇋ 2 Cl - + 1.36 V 2 Cl - ⇋ Cl 2 + 2 e - - 1.36 V 2 MnO 4 - + 16 H + + 10 Cl -  2 Mn 2+ + 8 H 2 O + 5 Cl 2 E Θ cell = 1.52 – 1.36 = + 0.16 V Note the E Θ cell calculation ignores stochiometry

30. Will it go? A +ve E Θ cell is equivalent to a –ve ΔH Θ A redox reaction with an E Θ > 0.3 V will probably go to completion High activation energy (kinetic stability) might prevent anything from happening E Θ is under standard conditions – change concentration or temperature changes the value Gases can escape! MnO 2 + 4 H + + 2 e - ⇋ Mn 2+ + 2 H 2 O +1.23 V Cl 2 + 2 e - ⇋ 2 Cl - + 1.36 V 2 Cl - ⇋ Cl 2 + 2 e - - 1.36 V MnO 2 + 4 H + + 2 Cl -  Cl 2 + Mn 2+ + 2 H 2 O E Θ cell = 1.23 + (– 1.36) = - 0.13 V Chlorine gas can be made by adding hot concentrated hydrochloric acid to manganese (IV) oxide

31. Will it go? Sn 4+ + 2 e - ⇋ Sn 2+ +0.15 V PbO 2 + 4 H + + 2 e - ⇋ Pb 2+ + 2 H 2 O+ 1.47 V Will tin (IV) oxidise lead (II)? Will lead (IV) oxidise tin (II)? Sn 4+ + 2 e - ⇋ Sn 2+ ; Pb 2+ + 2 H 2 O ⇋ PbO 2 + 4 H + + 2 e - = + 0.15 - (+1.47) = - 1.32 V NO PbO 2 + 4 H + + 2 e - ⇋ Pb 2+ + 2 H 2 O; Sn 2+ ⇋ Sn 4+ + 2 e - + 1.47 - (+ 0.15) = + 1.32 V YES

32. Relationship of thermodynamic values o E cell x zF/T = ΔS total = RlnK o E cell α ΔS total α lnK o ΔS total +ve reaction is regarded as spontaneous o If Ea high the reactants can still be kinetically stable o ΔS total > + 200 JK -1 mol -1 o K > 10 10 o E > +0.6 V goes to completion

33. Using the information in the Fuel Cells document and your data book discuss critically the use of fuel cell powered vehicles. You need consider utility, environment and cost. 2 H 2 (g) + O 2 (g)  2 H 2 O (l) ΔH = -572 kJmol -1 In a hydrogen fuel cell, the half- equations that occur at the electrodes are: 2H 2 (g) ⇋ 4H + (aq) + 4e - (oxidation half-equation) O 2 (g) + 2 H 2 O(l) + 4e- ⇋ 4OH - (aq) (reduction half-equation) 2H 2 (g) + O 2 (g) + 2 H 2 O(l)  4H + (aq) + 4OH - (aq)  4 H 2 O(l) 2H 2 (g) + O 2 (g)  2 H 2 O(l) Catalysts & membrane

34. Breathalyser Potassium dichromate –Tube containing orange potassium dichromate and sulfuric acid absorbed onto silica matrix –Alcohol reduces Cr 2 O 7 2- to Cr 3+ –The more dichromate that changed colour the higher the alcohol level Ethanol/oxygen fuel cell –Electrodes: Ag or Pt –Permeable glass membrane –Electrolyte: NaOH

35. Breathalyser –Negative terminal: CH 3 CH 2 OH + H 2 O ⇋ CH 3 COOH + 4H + + 4e - –Positive terminal: O 2 + 4H + + 4e - ⇋ 2H 2 O –Current α alcohol in breath –Calibrated using air with known amounts of ethanol present Infrared spectroscopy –Taken to police station –Length 2950 peak α alcohol