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Redox

Redox. Difficult but necessary. 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. And Oilrig. O xidation i s l oss of electrons:

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Redox

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  1. Redox Difficult but necessary

  2. Obviously: • Oxidation is adding oxygen • 2H2 + O2 2H2O • Reduction is removing oxygen • 2FeO + C  2Fe + CO2 • But also • oxidation is removal of hydrogen • And reduction is adding hydrogen

  3. And Oilrig • Oxidation is loss of electrons: • Cu – 2eˉ  Cu2+ • Notice the charge increases • Reduction is gain: • Cu2+ + 2eˉ  Cu • Notice the charge has reduced • Both often happen in one reaction, so it is a redox reaction

  4. More… • A redox reaction: • 2Al2O3 4Al + 3O2 • aluminium reduced +3 to 0, oxygen oxidised, -2 to 0 • Not a redox reaction: • PbCl2 + 2NaI PbI2 + 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 + Cu2+ + O2ˉ  Mg2+ + Cu + O2ˉ • Note the oxygen ions are spectator ions, they aren’t actually involved, so we get: • Mg + Cu2+ Mg2+ + 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 O2

  7. Working them out • Rules for assigning: (these rarely change) • F is always -1 • O is -2, except in OF2 • 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 O2, Cl2 etc.

  8. Note: the allocation of a high oxidation number does not necessarily mean that electrons have been lent and borrowed. • E.g. in CrO42ˉ 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 MnO4ˉ in KMnO4 • Oxidation state of Mn is +7 because overall charge is -1 and oxygens are -8 (-2 x 4) • So -1 = +7 – 8 • Write MnO4ˉ as manganese(VII) oxide or manganate(VII) • Manganate(VII) compounds are common oxidising agents

  10. e.g. oxidation states in CaSO4 • 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 O4 S • Call it calcium sulphate(VI)

  11. e.g. the thiosulphate anion S2O32ˉ • 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 CrO42- • Hydrogen and magnesium in MgH2 • Both elements in water H2O • Chlorine in HClO • Sodium and chlorine in Na2ClO3 • Carbon in carbonate CO32- • Iron in Fe3O4

  13. Naming : • If there is any doubt about the oxidation state, usually transition metals, it must be given: • CuCl2 Copper(II) chloride • CuCl3 Copper(III) chloride • NaNO3 Sodium nitrate(V) • in NO3ˉ 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? • Cl2 + 2KBr  2KCl + Br2 • Cl: 0 to -1, Br: -1 to 0 • Cl reduced, Br oxidised • MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O • Mn from +4 to +2, some Cl from -1 to 0 • Mn reduced, Cl oxidised • 2CrO42ˉ +2H+  Cr2O72ˉ + H2O • 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: • MnO4-(aq) Mn2+(aq) • Continued

  16. Continued.... • MnO4-(aq) Mn2+(aq) • In water • Add oxygen in H2O to balance.... • Giving MnO4-(aq) Mn2+(aq) + 4H2O(l) • Assume an acidic solution to balance H.... • Giving MnO4-(aq) + 8H+ Mn2+(aq) + 4H2O(l) • Sort-out electrons for charge and redox.... • MnO4-(aq) + 8H+ + 5e- Mn2+(aq) + 4H2O(l) • +7 +2 • In fact we’ve always done this, but it was easy examples...

  17. Try: • VO43-(aq) V2+(aq) • MnO4-(aq) MnO2(s) • CrO42-(aq) Cr2+(aq) • SO42-(aq) S8(s) • VO43-(aq) + 8H+(aq) + 3e-  V2+(aq) + 4H2O • MnO4-(aq) + 4H+(aq) + 3e- MnO2(s) + 2H2O • CrO42-(aq) + 8H+(aq) + 4e- Cr2+(aq) + 4H2O • 8SO42-(aq) + 64H+(aq) + 48e- S8(s) + 32H2O • SO42-(aq) + 8H+(aq) + 6e- S(s) + 4H2O

  18. Some specific half-equations of oxidising agents: • Oxygen plus metal: • O2 + 4e- 2O2- • chlorine plus metal: • Cl2 + 2e- 2Cl- • Sulphur plus metal: • S + 2e-  S2- • In hydrogen peroxide, oxygen is in a -1 state. Is this likely to be a stable compound? • H2O2 + 2H+ +2e- 2H2O

  19. More.... • Concentrated sulphuric acid: • 2H2SO4 + Cu  CuSO4 + 2H2O + SO2 • ½ equation: SO42- + 2e-+4H+ SO2 +2H2O • Conc. nitric acid: • Cu + 4HNO3 Cu(NO3)2 + 2H2O + 2NO2

  20. Some specific half-equations of reducing agents: • Oxygen plus metal: • O2 + 4e- 2O2- • chlorine plus metal: • Cl2 + 2e- 2Cl- • Sulphur plus metal: • S + 2e-  S2- • In hydrogen peroxide, oxygen is in a -1 state. Is this likely to be a stable compound? • H2O2 + 2H+ +2e- 2H2O

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