1. Oxidation and Reduction Definitions of oxidation and reduction Oxidation numbers Redox equations

2. Oxidation - reduction  Oxidation is loss of electrons  Reduction is gain of electrons  Oxidation is always accompanied by reduction The total number of electrons is kept constantThe total number of electrons is kept constant  Oxidizing agents oxidize and are themselves reduced  Reducing agents reduce and are themselves oxidized

3. Follow the electrons

4. Oxidation numbers  Metals are typically considered more 'cation-like' and would possess positive oxidation numbers, while nonmetals are considered more 'anion-like' and would possess negative oxidation numbers.  Oxidation number is the number of electrons gained or lost by the element in making a compound

5. Predicting oxidation numbers  Oxidation number of atoms in element is zero in all cases  Oxidation number of element in monatomic ion is equal to the charge  sum of the oxidation numbers in a compound is zero  sum of oxidation numbers in polyatomic ion is equal to the charge  F has oxidation number –1  H has oxidn no. +1; except in metal hydrides where it is – 1  Oxygen is usually –2. Except:  O is –1 in hydrogen peroxide, and other peroxides  O is –1/2 in superoxides KO 2  In OF 2 O is +2

6. Position of element in periodic table determines oxidation number  G1A is +1  G2A is +2  G3A is +3 (some rare exceptions)  G5A are –3 in compounds with metals, H or with NH 4+. Exceptions are in compounds to the right; in which case use rules 3 and 4.  G6A below O are –2 in binary compounds with metals, H or NH 4+. When they are combined with O or with a lighter halogen, use rules 3 and 4.  G7A elements are –1 in binary compounds with metals, H or NH 4+ or with a heavier halogen. When combined with O or a lighter halogen, use rules 3 and 4.

7. Identifying reagents  Those elements that tend to give up electrons (metals) are typically categorized as reducing agents and those that tend to accept electrons (nonmetals) are referred to as oxidizing agents.

8. Iron can reduce Cu 2+ to Cu  The iron nail reduces the Cu 2+ ions and becomes coated with metallic Cu. At the same time, the intensity of the blue color diminishes due to loss of Cu 2+ ions from solution.

9. Any element can be both an oxidizer and reducer depending on relative positions in the activity series  Fe reduced Cu 2+, but Cu can reduce Ag + (lower activity  Fe 2+ is reduced by Zn

10. More active metals are strongly reducing

11. Predicting results of displacement reactions  In this reaction the element metal A displaces the ion metal B from its compound  This will only occur if A lies above B in the activity series  Displacement reaction exercises

12. Nuggets of redox processes  Where there is oxidation there is always reduction Oxidizing agent Reducing agent Is itself reduced Is itself oxidized Gains electrons Loses electrons Causes oxidation Causes reduction

13. Identify redox by change in oxidation numbers: follow the flow of electrons  Reducing agent increases its oxidation number  Oxidizing agent decreases its oxidation number

14. In reaction with metals, nonmetals are always oxidizers  Reactions of elements are always redox  The nonmetal gains electrons, becomes a negative ion  The metal loses electrons, becomes a positive ion  Identification is harder when there are no elements involved: oxidation numbers must be used

15. Balancing redox equations: systematic methods  Oxidation number method – tracking changes in the oxidation numbers  Half-reaction method – tracking changes in the flow of electrons  Same principles, different emphasis  Use is a matter of choice, but familiarity with both is important

16. Oxidation number method  What goes up must come down…  Sum of the changes in oxidation numbers in any process is zero

17. Six habits of the redox equation balancer

18. Walking through the steps  STEP 1: write the unbalanced net ionic equation  In an acid solution, permanganate is reduced by bromide ion to give Mn 2+ ion and bromine

19. STEP 2: Balance the equation for elements other than O and H *  We need to double the bromide ions on the L.H.S. to balance the equation * O and H can remain unbalanced because we will top up with water and hydronium ions later

20. STEP 3: Assign oxidation numbers  Use the rules of oxidation numbers  Element is zero  Monatomic ion: oxidation number is same as charge  Oxide is -2 0 +2+7

21. STEP 4: Identify oxidized and reduced  Mn is reduced from +7 to +2  Net gain of 5 electrons  Br is oxidized from -1 to 0  Net loss of 1 electron 0 +2+7

22. STEP 5: Balance the oxidized and reduced species  For every Mn reduced (decrease in oxidation number of 5), need five Br - oxidized (increase in oxidation number of 1)  Equation becomes  Redox is now complete but material balance is not

23. STEP 6: Material balance with H 2 O and H +  Strategy: add H 2 O to the side that lacks for O and add H + (the reaction is in acid solution) to the other side  In basic solution we add OH - and H 2 O instead of H 2 O and H + respectively  Test equation for both atoms and charges

24. The Half-Reaction method  Any redox process can be written as the sum of two half reactions: one for the oxidation and one for the reduction

25. Six habits of the redox equation balancer

26. STEP 1: the unbalanced equation  Dichromate ion reacts with chloride ion to produce chlorine and chromium (III)

27. STEP 2: identify the oxidized and reduced and write the half reactions  Oxidation half-reaction  Reduction half-reaction

28. STEP 3: Balance the half reactions  Oxidation  Reduction

29. STEP 4: Material balance  As with the oxidation number method, add H 2 O to the side lacking O and add H + to the other side (for reactions in acid solution)  Oxidation reaction – unchanged  Reduction reaction

30. STEP 5: Balance half-reactions for charge by addition of electrons  No explicit calculation of oxidation numbers is required; we balance the charges on both sides of each half-reaction

31. STEP 5 cont: Multiply by factors to balance total electrons  Overall change in electrons must be zero  Multiply the oxidation half reaction by 3

32. STEP 6: Add half reactions and eliminate common items += Atoms and charges balance

33. Redox Titrations  Acid-base titration is used to determine an unknown concentration (either acid or base)  The endpoint is manifested in a color change (of an indicator) or by measuring pH  In redox titrations, the concentration of one of the reagents can be measured, provided there is a sharp distinction between the oxidized and reduced states

34. Using the roadmap  Oxalic acid is oxidized by MnO 4 -  A known quantity of oxalic acid is used to determine the concentration of the MnO 4 -  MnO 4 - has an intense purple colour, whereas Mn 2+ is almost colourless

35. Strategy  A known amount of H 2 C 2 O 4 is used  A solution of KMnO 4 is titrated till the first purple colour – the endpoint. All the H 2 C 2 O 4 is oxidized.  The equation gives the number of moles of MnO 4 -  The volume of solution yields the concentration