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Redox Difficult but necessary Obviously: Oxidation is adding oxygen Reduction is removing oxygen 2H2 + O2 2H2O 2FeO + C 2Fe + CO2 But also oxidation is removal of hydrogen And reduction is adding hydrogen And Oilrig Oxidation is loss of electrons: Cu – 2eˉ Cu2+ Reduction is gain: Cu2+ + 2eˉ Cu Notice the charge increases Notice the charge has reduced Both often happen in one reaction, so it is a redox reaction 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 e.g. Mg + CuO MgO + Cu As ions: Note the oxygen ions are spectator ions, they aren’t actually involved, so we get: Mg + Cu2+ + O2ˉ Mg2+ + Cu + O2ˉ Mg + Cu2+ Mg2+ + Cu So the magnesium has reduced the copper And the copper has oxidised the magnesium 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 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. 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. 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 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) 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 Practice: What is the oxidation state of: Chromium in CrO42Hydrogen and magnesium in MgH2 Both elements in water H2O Chlorine in HClO Sodium and chlorine in Na2ClO3 Carbon in carbonate CO32Iron in Fe3O4 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) 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 Balancing Just when you thought you had got it.... Consider this redox change: MnO4-(aq) Mn2+(aq) Continued 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... Try: VO43-(aq) V2+(aq) VO43-(aq) + 8H+(aq) + 3e- V2+(aq) + 4H2O MnO4-(aq) MnO2(s) MnO4-(aq) + 4H+(aq) + 3e- MnO2(s) + 2H2O CrO42-(aq) Cr2+(aq) CrO42-(aq) + 8H+(aq) + 4e- Cr2+(aq) + 4H2O SO42-(aq) S8(s) 8SO42-(aq) + 64H+(aq) + 48e- S8(s) + 32H2O SO42-(aq) + 8H+(aq) + 6e- S(s) + 4H2O Some specific half-equations of oxidising agents: Oxygen plus metal: O2 + 4e- 2O2chlorine plus metal: Cl2 + 2e- 2ClSulphur plus metal: S + 2e- S2In hydrogen peroxide, oxygen is in a -1 state. Is this likely to be a stable compound? H2O2 + 2H+ +2e- 2H2O 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 Some specific half-equations of reducing agents: Oxygen plus metal: O2 + 4e- 2O2chlorine plus metal: Cl2 + 2e- 2ClSulphur plus metal: S + 2e- S2In hydrogen peroxide, oxygen is in a -1 state. Is this likely to be a stable compound? H2O2 + 2H+ +2e- 2H2O