1.5a Learning Outcomes define oxidation number, oxidation state define oxidation and reduction in terms of change of oxidation numbers define oxidising agent and reducing agent state the rules for oxidation numbers (exclude peroxides, except for hydrogen peroxide)   calculate oxidation numbers of transition metals in their compounds and of other elements use oxidation numbers in nomenclature of transition metal compounds

Learning Outcomes arrange the electrochemical series of metals in order of their ease of oxidation (reactions, other than displacement reactions, not required)

1.5 Oxidation & Reduction Also known as Redox Means something in terms of: Addition or Removal Oxygen / Hydrogen Electrons Oxidation Numbers Introduction to topic via three simple examples

Redox in Terms of O2 / H2 Oxidation = addition of oxygen to a substance (or removal of hydrogen) C + O2 CO2 Reduction is loss of oxygen (or addition of hydrogen) CuO + H2  Cu + H2O

Summary of Addition /Removal of Oxygen & Hydrogen C2H5OH Oxidation = -H2 CH3CHO Ox. Agent CH3CHO Reduction = +H2 C2H5OH Red. Agent Oxidation = +O CH3CHO CH3COOH Ox. Agent Reduction = -O CH3COOH CH3CHO Red. Agent Ox. Agent = Na2Cr2O7 / H+ Red. Agent = H2 / Ni

Oxidation in Terms of Electrons OILRIG Oxidation involves Loss of electrons Reduction Involves Gain of Electrons Species Oxidised is Reducing Agent Species Reduced is the Oxidising Agent

Example 1 of 3 Sodium + chlorine  sodium chloride Na + Cl  Na+ + Cl- OILRIG Oxidn involves loss / Redn involves gain Na loses an electron [oxidised] Cl gains an electron [reduced]

Example 2 of 3 Magnesium + Oxygen  magnesium oxide Mg + O  MgO => Oxidn / Redn by inspection? Mg  Mg+2 loses 2 electrons [oxidation] O  O-2 gains 2 electrons [reduction]

Example 3 of 3 Zinc loses electrons (oxidised) Zinc +copper sulphate  Zinc sulphate+ Copper Zn + CuSO4  ZnSO4 + Cu Zn + Cu+2  Zn+2 + Cu Zinc loses electrons (oxidised) Copper gains electrons (reduced)

Oxidising Agent A substance that causes oxidation in another substance – the oxidising agent is itself reduced

Reducing agent A substance that causes reduction in another substance.

Summary of Redox in terms of Electrons Oxidation is loss of electrons; Reduction is gain of electrons CuO + H2  Cu + H2O CuO is made up of Cu+2 and O-2 Overall: Cu+2  Cu [gains 2 electrons] reduced Overall: H2  H2+2[loses 2 electrons] oxidised Overall: O-2  O-2 [ no change]

Summary of Redox in terms Oxidation numbers Oxidation Number: The charge that an atom has or appears to have assuming that the compound is ionic. Electrons always go to the most electronegative element Oxidation is an increase on oxidation number Reduction is a decrease in oxidation number.

Oxidation Number Rule 1 Elements on their own = 0 H2 = 0 Zn = 0 Cl2 = 0

Oxidation Number Rule 2 Ions = same as charge Cu +2 = +2 O-2 = -2 Cl-1 = -1

Oxidation Number Rule 3 Charges of all elements in a compound = 0 CuSO4 Cu = +2 S = +6 O4 = -8 [O = -2] Total = +2 +6 –8 = 0

Oxidation Number Rule 4 Oxygen = -2 Exceptions are peroxides O = -1 [H2O2, Na2O2 ] OF2 O = +2, F = -1 (Oxygen diflouride)

Oxidation Number Rule 5 Hydrogen = +1 Exceptions are the metal hydrides NaH Na = +1, H = -1 (Sodium hydride) MgH2 Mg = +2 each H = -1 Magnesium Hydride

Oxidation Number Rule 6 Halogens [ F, Cl, Br, I] are always –1 except when joined to a more electronegative element Cl2O Cl = +1, O = -2

Oxidation Number Rule 7 In a complex ion the sum of all the charges = the charge on the ion. SO4-2 (Sulfate anion) S = +6, O4 = -8 [O = -2] +6 –8 = -2 What about Sulfite anion = SO3-2?

Redox Equations MnO4- + H2S + H+  Mn+2 + S + H2O 2 -8 +2 MnO4- + H2S + H+  Mn+2 + S + H2O + 7 +1 -2 +2 -2 Ox => Loses 2e- L.C.M. = 10 Red => Gains 5e- 2 MnO4- + H2S + H+  Mn+2 + S + H2O 5 2 MnO4- + H2S + H+  Mn+2 + S + H2O 5 6 2 5 8 -2 +6 +4

Redox Equations Zn + NO3- + H+  Zn+2 + N2H4 + H2O 7 +5 -6 +4 Zn + NO3- + H+  Zn+2 + N2H4 + H2O +5 -2 +2 -2 +1 Ox. => Loses 2e- Red. => Gains 7e- L.C.M. = 14 7 Zn + NO3- + H+  Zn+2 + N2H4 + H2O 2 7Zn + 2NO3- + H+  Zn+2 + N2H4 + H2O 16 7 6 +16 +14 -2

Redox Equations MnO4- + H+ + Cl-  Mn+2 + Cl2 + H2O -8 MnO4- + H+ + Cl-  Mn+2 + Cl2 + H2O -1 +2 +7 -2 Red => Gains 5e- Ox => Loses 1e- L.C.M. = 5 MnO4- + H+ + Cl-  Mn+2 + Cl2 + H2O 5 MnO4- + H+ + 5Cl-  Mn+2 + Cl2 + H2O 8 5/2 4 -1 +8 -5 +2

Redox Equations ClO3- + I- + H+  I2 + Cl- + H2O -6 ClO3- + I- + H+  I2 + Cl- + H2O +5 -2 -1 -1 Red => Gains 6e- Ox => Loses 1e- L.C.M. = 6 ClO3- + I- + H+  I2 + Cl- + H2O 6 ClO3- + 6I- + H+  I2 + Cl- + H2O 6 3 3 -1 -1 -6 +6

Redox Equations Cr2O72- + Cl- + H+  Cr+3 + H2O + Cl2 -12 -14 Cr2O72- + Cl- + H+  Cr+3 + H2O + Cl2 +6 -2 -1 +3 Red => EACH Gains 3e- => 2 x 3e- = 6 Ox => Loses 1e- L.C.M. = 6 Cr2O72- + Cl- + H+  Cr+3 + H2O + Cl2 6 Cr2O72- + 6Cl- + H+  Cr+3 + H2O + Cl2 14 2 7 3 -2 -6 +14 +6

Redox Equations Cr2O72- + H+ + S  SO2 + OH- + Cr2O3 2 -12 -14 -4 -6 Cr2O72- + H+ + S  SO2 + OH- + Cr2O3 -2 +6 -2 +4 -2 +3 Red => EACH Gains 3e- => 2 x 3e- = 6 Ox => Loses 4e- L.C.M. = 12 2 Cr2O72- + H+ + S  SO2 + OH- + Cr2O3 3 2Cr2O72- + H+ + 3S  SO2 + OH- + Cr2O3 2 3 2 2 -2 -4 +2

Redox Equations MnO4- + H2O2 + H+  Mn+2 + O2 + H2O 2 -8 +2 -2 MnO4- + H2O2 + H+  Mn+2 + O2 + H2O +7 -2 +1 -1 +2 Red => Gains 5e- Ox => EACH loses 1e- => 2 x 1e- = 2 L.C.M. = 10 2 MnO4- + H2O2 + H+  Mn+2 + O2 + H2O 5 2MnO4- + 5H2O2 + H+  Mn+2 + O2 + H2O 6 2 5 8 +4 -2 +6

Exam Q’s Q.4 ’02 (e) What is the oxidation number of sulfur in Na2S2O3? (e) + 2 / 2 / 0 and 4 (but only if both are given – no marks for 0 or 4 on its own) (6) 10. (a) Define oxidation number. (4) (i) Using oxidation numbers, identify which species is being oxidised and which species is being reduced in the following reaction. (12) MnO4− + Cl− + H+ → Mn2+ + Cl2 + H2O oxidised: Cl¯ / Cl (–1) (6) [OR Cl (–1 to 0) (3) reduced: MnO4¯/ Mn(+7) / Mn(7) / Mn(VII) (6) [OR Mn (7 to 2) (3) reduced (3)]

(ii) Hence, or otherwise, balance the equation. (9) MnO4¯ + 5Cl¯ + 8H+ Mn 2+ + 2Cl2 + 4H2O (9)

2004 Q.4 (g) What is the oxidation number (i) of oxygen in H2O2 and (ii) of bromine in KBrO3? (g) (i) – 1 (3) (ii) + 5 / 5 / V (3) 2005 Q.11 (a) (i) Define oxidation in terms of change in oxidation number. (a) (i) DEFINE: increase in oxidation number (4) (ii) What is observed when chlorine gas is bubbled into an aqueous solution of sodium bromide? Explain your answer in terms of oxidation and reduction. (9) (ii) WHAT: solution turns red-brown (red, orange, yellow) (3 EXPLAIN: bromide ions oxidised to bromine / Br¯ to Br2 / Br(-1) to Br(0) (3) chlorine reduced to chloride ions / Cl2 to Cl¯ / Cl(0) to Cl(-1) (3)

2006 q.10 (b) Define oxidation in terms of change in oxidation number. (4) What is the oxidation number of (i) chlorine in NaClO and (ii) nitrogen in NO3¯? (6) DEFINE: increase (4) WHAT: (i) +1 [Accept 1] (3) (ii) +5 [Accept 5] (3) State and explain the oxidation number of oxygen in the compound OF2. (6) EXPLAIN: oxygen is more electropositive / less electronegative / fluorine is more electronegative / fluorine is less electropositive [Allow even if ox. no. incorrect.] (3)

Using oxidation numbers or otherwise, identify the reducing agent in the reaction between acidified potassium manganate(VII) and potassium iodide solutions represented by the balanced equation below. Use your knowledge of the colours of the reactants and products to predict the colour change you would expect to see if you carried out this reaction. (9) 2MnO4¯ + 10I¯ + 16H+ 2Mn2+ + 5I2 + 8H2O IDENTIFY: potassium iodide (KI) solution / potassium iodide (KI) / iodide (I¯) / I( –1 to 0) (3) COLOURS: purple / violet / maroon (3) to brown / red / orange/ yellow (3)

2007 Q.10 (c) The halogens are good oxidising agents. (i) How does the oxidation number of the oxidising agent change during a redox reaction? (4) (c) (i) HOW: it decreases (4) (ii) Assign oxidation numbers in each of the following equations to show clearly that the halogen is the oxidising agent in each case. (12) Br2 + 2Fe2+ → 2Br– + 2Fe 3+ oxidation number of Br in Br2 = 0 (3) oxidation number of Br in Br¯ = – 1 (3) …reduced => ox. agent Cl2 + SO32– + H2O → Cl– + SO4 2– + H+ oxidation number of Cl in Cl2 = 0 (3) oxidation number of Cl in Cl¯ = – 1 (3) …reduced => ox. agent

Q.10 2008(b) Define oxidation in terms of (i) electron transfer, (ii) change in oxidation number. (7) DEFINE: (i) loss (decrease) of electrons (ii) increase (rise, gain) in oxidation number (4 + 3) (iii) For the redox reactions shown below, use oxidation numbers to identify the species oxidised in the first reaction and the oxidising reagent in the second reaction. (6) ClO¯ + I¯ + H+  Cl¯ + I2 + H2O I2 + S2O32–  I¯ + S4O62– 1st I¯ (iodide ion) / I (–1) (3) ….oxidised itself 2nd I2 (iodine molecule) / I (0) / I2 /(3).. reduced itself

(iv) Using oxidation numbers or otherwise balance both equations. (12) ClO¯ + I¯ + H+  Cl¯ + I2 + H2O Balanced: ClO¯ + 2I¯ + 2H+ → Cl¯ + I2 + H2O I2 + S2O32–  I¯ + S4O62– Balanced: I2 + 2S2O32– → 2I¯ + S4O62–

Electrochemical Series Electrochemical Series – Metals listed in order of Decreasing ability as Reducing Agents (themselves oxidised) Ca  Ca+2 + 2e- (Oxidation)

Metals Electrochemical Series [K] => Best Reducing Agent – becomes easily oxidized [Na] [Ca] [Mg] [Al] [Zn] [Fe] [Sn] [Pb] [H] [Cu] [Ag] => Worst Reducing Agent – most difficult in list to oxidize