1. Redox Difficult but necessary

2. Obviously: Oxidation is adding oxygen 2H 2 + O 2  2H 2 O Reduction is removing oxygen 2FeO + C  2Fe + CO 2 But also oxidation is removal of hydrogen And reduction is adding hydrogen

3. And Oilrig Oxidation is loss of electrons: Cu – 2eˉ  Cu 2+ Notice the charge increases Reduction is gain: Cu 2+ + 2eˉ  Cu Notice the charge has reduced Both often happen in one reaction, so it is a redox reaction

4. More… A redox reaction: 2Al 2 O 3  4Al + 3O 2 aluminium reduced +3 to 0, oxygen oxidised, -2 to 0 Not a redox reaction: PbCl 2 + 2NaI  PbI 2 + 2NaCl Pb stays at +2, Cl stays at -1, Na stays at +1, I stays at -1

5. e.g. Mg + CuO  MgO + Cu As ions: Mg + Cu 2+ + O 2 ˉ  Mg 2+ + Cu + O 2 ˉ Note the oxygen ions are spectator ions, they aren’t actually involved, so we get: Mg + Cu 2+  Mg 2+ + Cu So the magnesium has reduced the copper And the copper has oxidised the magnesium

6. Oxidation numbers / states Represent charges where there aren’t any They are an “accounting trick” to keep track of how atoms have control over electrons Apply to ions and covalently bonded atoms The oxidation numbers of elements are zero e.g.. Fe(s), and even O 2

7. Working them out Rules for assigning: (these rarely change) F is always -1 O is -2, except in OF 2 Group 7 are -1, except with O or F Group 1 metals are +1 Group 2 metals are +2 H is +1, except in hydrides, e.g. NaH Al is +3 The total for an ion is its charge (e.g. -1 for CN - ) More electronegative atoms get negative numbers The total for a compound is 0, even in O 2, Cl 2 etc.

8. Note: the allocation of a high oxidation number does not necessarily mean that electrons have been lent and borrowed. E.g. in CrO 4 2 ˉ the oxidation number of chromium is +6, yet it is covalently bonded to the oxygens and the energy required to remove 6 electrons would be prohibitive. All the +6 tells us is that the electrons probably spend more time near the oxygens.

9. Working out Overall charge on a compound ion is the sum of the oxidation states: E.g. for MnO 4 ˉ in KMnO 4 Oxidation state of Mn is +7 because overall charge is -1 and oxygens are -8 (-2 x 4) So -1 = +7 – 8 Write MnO 4 ˉ as manganese(VII) oxide or manganate(VII) Manganate(VII) compounds are common oxidising agents

10. e.g. oxidation states in CaSO 4 Ca is +2 O is -2 X 4 =-8 Uncharged compound so total oxidation number is 0 So sulphur is +6 (0 = +2 -8 +6) Ca O 4 S Call it calcium sulphate(VI)

11. e.g. the thiosulphate anion S 2 O 3 2 ˉ This is a common reducing agent, it donates electrons to reduce other chemicals Overall = oxygens + sulphurs -2 = (3 x -2) + (2 x sulphur)  -2 = (-6) + 4  2 x sulphur = +4  sulphur = +2 This is the sulphur(II) oxide (or thiosulphate) anion

12. Practice: What is the oxidation state of: Chromium in CrO 4 2- Hydrogen and magnesium in MgH 2 Both elements in water H 2 O Chlorine in HClO Sodium and chlorine in Na 2 ClO 3 Carbon in carbonate CO 3 2- Iron in Fe 3 O 4

13. Naming: If there is any doubt about the oxidation state, usually transition metals, it must be given: CuCl 2 Copper(II) chloride CuCl 3 Copper(III) chloride NaNO 3 Sodium nitrate(V) in NO 3 ˉ we count nitrogen using -6 for 3 oxygens, to make -1 for the negative charge, so N is +5 From -1=+5-6 (remember, overall charge is the total of oxidation states)

14. Redox or not? Cl 2 + 2KBr  2KCl + Br 2 Cl: 0 to -1, Br: -1 to 0 Cl reduced, Br oxidised MnO 2 + 4HCl  MnCl 2 + Cl 2 + 2H 2 O Mn from +4 to +2, some Cl from -1 to 0 Mn reduced, Cl oxidised 2CrO 4 2 ˉ +2H +  Cr 2 O 7 2 ˉ + H 2 O Cr is +6 before and after, nothing else changes either – not redox

15. Balancing Just when you thought you had got it.... Consider this redox change: MnO 4 - (aq)  Mn 2+ (aq) Continued

16. Continued.... MnO 4 - (aq)  Mn 2+ (aq) In water Add oxygen in H 2 O to balance.... Giving MnO 4 - (aq)  Mn 2+ (aq) + 4H 2 O (l) Assume an acidic solution to balance H.... Giving MnO 4 - (aq) + 8H +  Mn 2+ (aq) + 4H 2 O (l) Sort-out electrons for charge and redox.... MnO 4 - (aq) + 8H + + 5e -  Mn 2+ (aq) + 4H 2 O (l) +7+2 In fact we’ve always done this, but it was easy examples...

17. Try: VO 4 3- (aq)  V 2+ (aq) MnO 4 - (aq)  MnO 2 (s) CrO 4 2- (aq)  Cr 2+ (aq) SO 4 2- (aq)  S 8 (s) VO 4 3- (aq) + 8H + (aq) + 3e -  V 2+ (aq) + 4H 2 O MnO 4 - (aq) + 4H + (aq) + 3e -  MnO 2 (s) + 2H 2 O CrO 4 2- (aq) + 8H + (aq) + 4e -  Cr 2+ (aq) + 4H 2 O 8SO 4 2- (aq) + 64H + (aq) + 48e -  S 8 (s) + 32H 2 O SO 4 2- (aq) + 8H + (aq) + 6e -  S (s) + 4H 2 O

18. Some specific half-equations of oxidising agents: Oxygen plus metal: O 2 + 4e -  2O 2- chlorine plus metal: Cl 2 + 2e -  2Cl - Sulphur plus metal: S + 2e-  S 2- In hydrogen peroxide, oxygen is in a -1 state. Is this likely to be a stable compound? H 2 O 2 + 2H + +2e -  2H 2 O

19. More.... Concentrated sulphuric acid: 2H 2 SO 4 + Cu  CuSO 4 + 2H 2 O + SO 2 ½ equation: SO 4 2- + 2e - +4H +  SO 2 +2H 2 O Conc. nitric acid: Cu + 4HNO 3  Cu(NO 3 ) 2 + 2H 2 O + 2NO 2

20. Some specific half-equations of reducing agents: Oxygen plus metal: O 2 + 4e -  2O 2- chlorine plus metal: Cl 2 + 2e -  2Cl - Sulphur plus metal: S + 2e-  S 2- In hydrogen peroxide, oxygen is in a -1 state. Is this likely to be a stable compound? H 2 O 2 + 2H + +2e -  2H 2 O