1. OXIDATION & REDUCTION "NEVER SAY, 'I TRIED IT ONCE AND IT DID NOT WORK.'" -- ERNEST RUTHERFORD Ch.10 J.C. Rowe Windsor University School of Medicine

2. Electron Transfer  Ionic Bond Formation

3. Antoine Lavoisier  A French Chemist  Father of Modern Chemistry  ID’ed role of O in combustion  Named O & H  Assisted in formation of metric system  Predicted Si & proved S as an element  OXIDATION ! 26 Aug 1743 – 8 May 1794

4. OXIDATION  Oxidation is a chemical combination with oxygen  Oxidation is a loss of combined Hydrogen  Oxidation is a loss of electron  Oxidation is an increase in oxidation number Combining with OxygenRemoval of Electron

5. REDUCTION  Reduction is a chemical combination with Hydrogen  Reduction is a loss of combined Oxygen  Reduction is a gain of electron  Reduction is a decrease in oxidation number Removal of OxygenGain of Electron

6. Oxidation and reduction in terms of oxygen transfer Definitions Oxidation is gain of oxygen. Reduction is loss of oxygen.

7. Oxidation and reduction in terms of hydrogen transfer Definitions Oxidation is loss of hydrogen. Reduction is gain of hydrogen

8. Oxidation and reduction in terms of electron transfer Definitions Oxidation is loss of electrons. Reduction is gain of electrons.

9. Reducing Agent vs. Oxidising Agent  A reducing agent reduces something else  Reducing agents remove oxygen from another substance or give hydrogen or electron to it.  An oxidising agent is substance which oxidises something else  Oxidising agents give oxygen to another substance or remove hydrogen or electron from it. Reducing AgentOxidising Agent

10. Oxidising and Reducing agents  If you look at the equation above, the magnesium is reducing the copper(II) ions by giving them electrons to neutralise the charge. Magnesium is a reducing agent.  Looking at it the other way round, the copper(II) ions are removing electrons from the magnesium to create the magnesium ions. The copper(II) ions are acting as an oxidising agent.

11. A. Oxidation & redox changes B. Writing oxidation numbers C. Oxidation numbers in chemical names Oxidation Number

12. OXIDATION STATES (OXIDATION NUMBERS)  Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state.  Oxidation involves an increase in oxidation state  Reduction involves a decrease in oxidation state

13. How to work out oxidation states ?  learn some simple rules, and do some very simple sums! The oxidation state of an uncombined element is zero. This applies whatever the structure of the element - whether it is, for example, Xe or Cl 2 or S 8, or whether it has a giant structure like carbon or silicon.

14. work out oxidation states Cont’d. The sum of the oxidation states of all the atoms or ions in a neutral compound is zero. The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion

15. work out oxidation states Cont’d. The more electronegative element in a substance is given a negative oxidation state. The less electronegative one is given a positive oxidation state. Remember that fluorine is the most electronegative element with oxygen second. Some elements almost always have the same oxidation states in their compounds :

16. element usual oxidation state exceptions Group 1 metalsalways +1 Group 2 metalsalways +2 Oxygenusually -2 except in peroxides and F 2 O Hydrogen usually +1 except in metal hydrides where it is -1 Fluorinealways -1 Chlorineusually -1 except in compounds with O or F

17. Examples of working out oxidation states  What is the oxidation state of chromium in Cr 2+ ? For a simple ion like this, the oxidation state is the charge on the ion - in other words: +2 (Don't forget the + sign.)  What is the oxidation state of chromium in CrCl 3 ? This is a neutral compound so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1. If the oxidation state of chromium is n: n + 3(-1) = 0 thus n = +3 (Again, don't forget the + sign!)

18. Examples of working out oxidation states Cont’d.  What is the oxidation state of chromium in Cr(H 2 O) 6 3+ ? This is an ion and so the sum of the oxidation states is equal to the charge on the ion. The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr 3+. The oxidation state is +3.

19. Examples of working out oxidation states Cont’d.  What is the oxidation state of chromium in the dichromate ion, Cr 2 O 7 2- ? The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present. 2n + 7(-2) = -2 thus n = +6

20. Examples of working out oxidation states Cont’d.  What is the oxidation state of copper in CuSO 4 ? You might recognise this as an ionic compound containing copper ions and sulphate ions, SO 4 2-. To make an electrically neutral compound, the copper must be present as a 2+ ion. The oxidation state is therefore +2.  You might recognise the formula as being copper(II) sulphate. The "(II)" in the name tells you that the oxidation state is 2.  You will know that it is +2 because you know that metals form positive ions, and the oxidation state will simply be the charge on the ion.

21. Using oxidation states In naming compounds  You will have come across names like iron(II) sulphate and iron(III) chloride.  The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. That tells you that they contain Fe 2+ and Fe 3+ ions.

22. This can also be extended to the negative ion. Iron(II) sulphate is FeSO 4.  The sulphate ion is SO 4 2-. The oxidation state of the sulphur is +6 (work it out!). The ion is more properly called the sulphate(VI) ion.  So FeSO 4 is properly called iron(II) sulphate(VI),  In fact, because of the easy confusion between these names, the old names sulphate and sulphite are normally still used in introductory chemistry courses.

23. There is also a compound FeSO 3 with the old name of iron(II) sulphite  The sulphite ion is SO 3 2-. The oxidation state of the sulphur is +4 (work that out as well!). This ion is more properly called the sulphate(IV) ion. The ate ending simply shows that the sulphur is in a negative ion.  So FeSO 3 is properly called iron(II) sulphate(IV).

24. Using oxidation states to identify what's been oxidised and what's been reduced  This is easily the most common use of oxidation states. Remember: Oxidation involves an increase in oxidation state Reduction involves a decrease in oxidation state  In each of the following examples, we have to decide whether the reaction involves redox, and if so what has been oxidised and what reduced.

25. Example 1:  This is the reaction between magnesium and hydrochloric acid or hydrogen chloride gas:  Have the oxidation states of anything changed? Yes they have - you have two elements which are in compounds on one side of the equation and as uncombined elements on the other. Check all the oxidation states to be sure:.

26. Example 1 Cont’d. The magnesium's oxidation state has increased - it has been oxidised. The hydrogen's oxidation state has fallen - it has been reduced. The chlorine is in the same oxidation state on both sides of the equation - it hasn't been oxidised or reduced.

27. Example 2:  The reaction between sodium hydroxide and hydrochloric acid is:

28. Example 2 Cont’d.  Checking all the oxidation states: Nothing has changed. This isn't a redox reaction.

29. Redox Applied

30. A. Metals B. Non-metals C. Making predictions Electrochemical Series

31.  a serial arrangement of metallic elements or ions according to their electrode potentials determined under specified conditions; the order shows the tendency of one metal to reduce the ions of any other metal below it in the series

32.  Metals. When a metal reacts, it becomes a positive ion and so gives up electrons. It acts as reducing agent.  Nonmetals accept electrons to form negative ions. They act as oxidising agents

33. "I do not know what I may appear to the world, but to myself I seem to have been only like a boy playing on the sea-shore, and diverting myself in now and then finding a smoother pebble or a prettier shell than ordinary, whilst the great ocean of truth lay all undiscovered before me." --- Sir Issac Newton