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Redox lecture slides, conducted in Yishun Junior College (Singapore) in 2011.

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- 1. REDOX REACTIONS A.k.a. A study on why things explode
- 2. But I Know Every Rock and Tree and Creature… <ul><li>Has a life, has a spirit, has a name. </li></ul>
- 3. Oxidation of Food: What a Waste! <ul><li>Fruits and Vegetables oxidised when left in open air </li></ul><ul><ul><li>Solution: Seal in plastic wrap </li></ul></ul><ul><ul><li>More radical: Add lemon juice to the cut fruit </li></ul></ul>
- 4. Oxidation of… <ul><li>Oxidation of nutrients causes increased activity of cells, leading to aging skin </li></ul><ul><ul><li>Solution: Beauty products? </li></ul></ul>People!
- 5. What will I learn? <ul><li>What is a redox reaction? </li></ul><ul><li>What is oxidation and reduction? (and how to identify them) </li></ul><ul><li>What are oxidising and reducing agents? </li></ul><ul><li>What is oxidation state? </li></ul><ul><li>What is oxidation and reduction in terms of oxidation state? </li></ul><ul><li>How to assign the oxidation state of an atom? </li></ul><ul><li>How to use oxidation state to determine which species is oxidised / reduced </li></ul>
- 6. What is a redox reaction? <ul><li>Redox – red uction + ox idation </li></ul><ul><li>Both processes occur simultaneously </li></ul><ul><li>Hence, one species is oxidised, another is reduced </li></ul><ul><li>So, what is oxidation, and what is reduction? </li></ul><ul><li>3 different versions of the definition: </li></ul>
- 7. Redox gain of electrons loss of electrons gain in hydrogen loss of hydrogen loss of oxygen gain in oxygen Reduction Oxidation
- 8. Oxidation and Reduction <ul><li>In terms of Oxygen: </li></ul><ul><ul><li>Oxidation: Gain of oxygen in a species </li></ul></ul><ul><ul><ul><li>E.g. Mg is oxidized to MgO </li></ul></ul></ul><ul><ul><li>Reduction: Loss of oxygen in a species </li></ul></ul><ul><ul><ul><li>E.g. H 2 O is reduced to H 2 </li></ul></ul></ul><ul><ul><li>Note: It’s the gain or loss of O, not O 2- </li></ul></ul>
- 9. Oxidation and Reduction <ul><li>In terms of Hydrogen: </li></ul><ul><ul><li>Oxidation: Loss of hydrogen in a species </li></ul></ul><ul><ul><ul><li>E.g. H 2 O is oxidised to O 2 </li></ul></ul></ul><ul><ul><li>Reduction: Gain of hydrogen in a species </li></ul></ul><ul><ul><ul><li>E.g. O 2 is reduced to H 2 O 2 </li></ul></ul></ul><ul><ul><li>Note: It’s the gain or loss of H, not H + </li></ul></ul>
- 10. Oxidation and Reduction <ul><li>In terms of Electrons (OIL RIG: Oxidation Is Loss, Reduction Is Gain): </li></ul><ul><ul><li>Oxidation: Loss of electrons in a species </li></ul></ul><ul><ul><ul><li>E.g. Mg is oxidized to MgO (Mg from 12 electrons to 10 electrons in Mg 2+ ) </li></ul></ul></ul><ul><ul><li>Reduction: Gain of electrons in a species </li></ul></ul><ul><ul><ul><li>E.g. O 2 is reduced to H 2 O 2 (O from 8 electrons to 9 electrons per O in O 2 2- ) </li></ul></ul></ul>
- 11. Oxidising and Reducing agent <ul><li>An oxidising agent is a chemical species that causes the other reactant in a redox reaction to be oxidised, and it is always reduced in the process. </li></ul><ul><li>A reducing agent is a chemical species that causes the other reactant in a redox reaction to be reduced, and it is always oxidised in the process. </li></ul>
- 12. Oxidising and Reducing Agents <ul><li>Remember: </li></ul><ul><ul><li>An oxidising agent is itself REDUCED when it oxidises something </li></ul></ul><ul><ul><li>A reducing agent is itself OXIDISED when it reduces something </li></ul></ul><ul><ul><li>2Mg + O 2 -> 2MgO </li></ul></ul><ul><ul><li>Mg is oxidised, and thus is the reducing agent </li></ul></ul><ul><ul><li>O 2 is reduced, and thus is the oxidising agent </li></ul></ul>
- 13. List of common Oxidising and Reducing Agents <ul><li>Realise something? </li></ul><ul><ul><li>H 2 O 2 is both an oxidising and a reducing agent! </li></ul></ul><ul><ul><li>If a stronger oxidising agent is present, H 2 O 2 is reducing </li></ul></ul>
- 14. Oxidation and Reduction in ‘A’ Level <ul><li>In terms of Oxidation States: </li></ul><ul><ul><li>Oxidation: Gain in oxidation state in a species </li></ul></ul><ul><ul><ul><li>E.g. Mg is oxidized to MgO (Mg from 0 to +2 in Mg 2+ ) </li></ul></ul></ul><ul><ul><li>Reduction: Gain of electrons in a species </li></ul></ul><ul><ul><ul><li>E.g. O 2 is reduced to H 2 O 2 (O from 0 to -1 in O 2 2- ) </li></ul></ul></ul><ul><li>Note: Oxidation states are always written in + x or – x , never just x or x - (e.g. Oxidation State of Mg in MgO is +2 , not 2 or -2 ) </li></ul>
- 15. Common Oxidation States Chemical species Oxidation state and remarks Any element eg Fe, O 2 , S 8 zero Oxygen in any compound -2 except in peroxides example H 2 O 2 or Na 2 O 2 then oxygen atom has oxidation state of -1 or in F 2 O , then oxygen atom has oxidation state of +2 Fluorine in any compound -1 being most electronegative Hydrogen in any compound +1 except in metal hydrides example NaH then hydrogen atom has oxidation state of -1 as metals have a greater tendency to lose electrons Chlorine, bromine, iodine -ve oxidation state if bonded to less electronegative element eg NaCl; then Cl = -1. +ve oxidation state if bonded to more electronegative element eg ClO - , then Cl = +1; ClO 3 - , then Cl = +5
- 16. Determine the oxidation state of… <ul><li>H in H 2 O </li></ul><ul><li>N in NH 4 + </li></ul><ul><li>S in S 2 O 3 2- </li></ul><ul><li>Cr in Cr 2 O 7 2- </li></ul>
- 17. Determine the oxidation state of… <ul><li>* Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom </li></ul><ul><li>1) H in H 2 O </li></ul><ul><li>Let the oxidation state of H be x . </li></ul><ul><li>Thus, in H 2 O, 2 x + (-2) = 0 </li></ul><ul><li>x = 1 </li></ul>
- 18. Determine the oxidation state of… <ul><li>* Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom </li></ul><ul><li>2) N in NH 4 + </li></ul><ul><li>Let the oxidation state of N be x . </li></ul><ul><li>Thus, in NH 4 + , x + 4(+1) = +1 </li></ul><ul><li>x = -3 </li></ul>
- 19. Determine the oxidation state of… <ul><li>* Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom </li></ul><ul><li>3) S in S 2 O 3 2- </li></ul><ul><li>Let the oxidation state of S be x . </li></ul><ul><li>Thus, in S 2 O 3 2- , 2 x + 3(-2) = -2 </li></ul><ul><li>x = +2 </li></ul>
- 20. Determine the oxidation state of… <ul><li>* Sum of all oxidation states of all atoms = Overall Charge of molecule / ion / atom </li></ul><ul><li>4) Cr in Cr 2 O 7 2- </li></ul><ul><li>Let the oxidation state of Cr be x . </li></ul><ul><li>Thus, in Cr 2 O 7 2- , 2 x + 7(-2) = -2 </li></ul><ul><li>x = +6 </li></ul>
- 21. Example 1 <ul><li>Let the oxidation state of Mn be x . </li></ul><ul><li>Thus, in MnO 4 - , x + 4(-2) = -1 </li></ul><ul><li>x = +7 </li></ul><ul><li>Manganese is reduced from oxidation state of +7 in MnO 4 - to +2 in Mn 2+ , while iron is oxidised from oxidation state of +2 in Fe 2+ to +3 in Fe 3+ . </li></ul>+7 +2 +2 +3
- 22. A Special Redox: Disproportionation <ul><li>Definition: A disproportionation reaction is a redox reaction in which one species is simultaneously oxidised and reduced . </li></ul><ul><li>Chlorine is simultaneously reduced from oxidation state of 0 in Cl 2 to -1 in Cl - , and oxidised from oxidation state of 0 in Cl 2 to +1 in ClO - . </li></ul>0 +1 -1
- 23. Disproportionation Reaction <ul><li>Example: Is this a disproportionation reaction? </li></ul><ul><li>This is NOT a disproportionation reaction </li></ul><ul><li>Disproportionation requires that the same atom is both oxidised and reduced simultaneously. </li></ul><ul><li>In this case, different atoms (of nitrogen) are oxidised and reduced. </li></ul>-3 +5 +1
- 24. Non-redox reactions <ul><li>The oxidation states of the elements remained unchanged in the following reactions: </li></ul><ul><li>Neutralisation reactions: </li></ul>
- 25. Non-redox reactions <ul><li>The oxidation states of the elements remained unchanged in the following reactions: </li></ul><ul><li>Precipitation reactions: </li></ul>
- 26. Non-redox reactions <ul><li>The oxidation states of the elements remained unchanged in the following reactions: </li></ul><ul><li>Complex formation: </li></ul>Tetraammine copper(II) complex (deep blue solution) ligand
- 27. Non-redox reactions <ul><li>The oxidation states of the elements remained unchanged in the following reactions: </li></ul><ul><li>Another reaction: </li></ul>
- 28. Interlude: More real-life redox!
- 29. Balancing redox reactions <ul><li>To make calculations in redox titrations, you need a balanced equation </li></ul>
- 30. Balancing redox reactions <ul><li>Example: Try to balance the following reaction by trial and error. </li></ul><ul><li>Possible answer: </li></ul>
- 31. Balancing redox reactions <ul><li>Example: Try to balance the following reaction by trial and error. </li></ul><ul><li>Actual answer: </li></ul><ul><li>Note: You might not even be told at the beginning that H + is reactant, H 2 O is product. </li></ul>
- 32. The half-equation method <ul><li>Write down the given reactants and products of the reaction </li></ul><ul><li>Identify the atoms in the given species that are undergoing oxidation / reduction and construct the unbalanced oxidation / reduction half-equations </li></ul><ul><li>Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance the “odd” atoms (“odd” atoms refer to atoms other than oxygen and hydrogen) </li></ul></ul><ul><ul><li>Balance oxygen atoms by adding H 2 O molecules </li></ul></ul><ul><ul><li>Balance hydrogen atoms by adding H + ions </li></ul></ul><ul><ul><li>Balance charges by adding electrons </li></ul></ul><ul><li>Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal </li></ul><ul><li>Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation. </li></ul>
- 33. Oh yeah, the blue bottle… <ul><li>“ Angry bottle” </li></ul><ul><li>The blue colour is methylene blue in the oxidised form </li></ul><ul><li>Before the bottle was shaken, the methylene blue is in the reduced state (colourless) </li></ul><ul><li>After the bottle was shaken, the oxygen in the air oxidised the methylene blue to the oxidised state (blue) </li></ul><ul><li>After a while, the glucose (reducing sugar) in the solution reduces the methylene blue back to the reduced state (colourless) </li></ul>
- 34. End of Lecture 1 of Redox “ I have never let my schooling interfere with my education.” Mark Twain
- 35. The half-equation method <ul><li>Example: Balance the following reaction: </li></ul>
- 36. The half-equation method <ul><li>Step 1: Write down the given reactants and products of the reaction </li></ul>
- 37. The half-equation method <ul><li>Step 2: Identify the atoms in the given species that are undergoing oxidation / reduction and write the unbalanced oxidation / reduction half-equations </li></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>Reduced Oxidised
- 38. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance the atoms undergoing oxidation / reduction </li></ul></ul><ul><li>Reduction half-equation: </li></ul>
- 39. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance oxygen atoms by adding H 2 O molecules </li></ul></ul><ul><li>Reduction half-equation: </li></ul>
- 40. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance oxygen atoms by adding H 2 O molecules </li></ul></ul><ul><li>Reduction half-equation: </li></ul>
- 41. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance hydrogen atoms by adding H + ions </li></ul></ul><ul><li>Reduction half-equation: </li></ul>
- 42. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance hydrogen atoms by adding H + ions </li></ul></ul><ul><li>Reduction half-equation: </li></ul>
- 43. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance charges by adding electrons </li></ul></ul><ul><li>Reduction half-equation: </li></ul>
- 44. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance charges by adding electrons </li></ul></ul><ul><li>Reduction half-equation: </li></ul>
- 45. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>
- 46. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance the atoms undergoing oxidation / reduction </li></ul></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>
- 47. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance oxygen atoms by adding H 2 O molecules </li></ul></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>
- 48. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance hydrogen atoms by adding H + ions </li></ul></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>
- 49. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance hydrogen atoms by adding H + ions </li></ul></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>
- 50. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance charges by adding electrons </li></ul></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>
- 51. The half-equation method <ul><li>Step 3: Balance both the half-equations using the following steps: </li></ul><ul><ul><li>Balance charges by adding electrons </li></ul></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>
- 52. The half-equation method <ul><li>Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal </li></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>
- 53. The half-equation method <ul><li>Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal </li></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>
- 54. The half-equation method <ul><li>Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal </li></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>x 2 x 5
- 55. The half-equation method <ul><li>Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal </li></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>x 2 x 5
- 56. The half-equation method <ul><li>Step 4: Multiply the balanced half-equations by appropriate integers such that the number of electrons in both half-equations are equal </li></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>x 2 x 5
- 57. The half-equation method <ul><li>Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation. </li></ul><ul><li>Reduction half-equation: </li></ul><ul><li>Oxidation half-equation: </li></ul>x 2 x 5
- 58. The half-equation method <ul><li>Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation. </li></ul>+6H +
- 59. The half-equation method <ul><li>Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation. </li></ul><ul><li>Balanced Equation: </li></ul>+6H +
- 60. The half-equation method <ul><li>Step 5: Add the resulting half-equations together, and eliminate any common species on both sides to obtain the balanced equation. </li></ul><ul><li>Balanced Equation: </li></ul>

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