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Chemistry 2.7 (AS 90306) Describe oxidation-reduction reactions

Chemistry 2.7 (AS 90306) Describe oxidation-reduction reactions Questions may involve any of the following: the properties of common oxidants and reductants, and the products of their reaction.

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Chemistry 2.7 (AS 90306) Describe oxidation-reduction reactions

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  1. Chemistry 2.7 (AS 90306) • Describe oxidation-reduction reactions • Questions may involve any of the following: the properties of common oxidants and reductants, and the products of their reaction. • Common oxidants are limited to O2, I2, Cl2, Fe3+, H2O2, MnO4(aq)/H+, Cr2O72(aq)/H+ • Common reductants are limited to metals, C, CO, H2, Fe2+, Br, I, SO2, (HSO3). • Properties are limited to appearance (colour and state), oxidation number (for polyatomic ions and single ions)

  2. Chemistry 2.7 (AS 90306) • Describe oxidation-reduction reactions • Questions may involve any of the following: writing balanced oxidation–reduction equations classifying balanced half-equations as oxidation or reduction identifying the oxidant and/or reductant from a given reaction describing the ability of halogens to act as oxidants in reactions with other elements, water or halide ions principles of simple electrolytic cells.

  3. Homework for the Holidays Read Unit 18 pages 71 – 74 in your year 12 Pathfinder Text Complete at least Q’s 1, 2 and 3 page 73 from your text book, do more if you can.

  4. OXIDATION - REDUCTION • Oxidation was originally defined as a gain of oxygen eg Mg reacts with O2 to form magnesium oxide, MgO. • 2Mg(s) + O2(g)  2MgO(s) Simlarly reduction was the removal of oxygen e.g. CO reduces Fe2O3 and produces Fe and CO2. Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)

  5. However scientists realised that not all redox reactions involved oxygen e.g. the reaction of zinc metal with copper ions. Zn(s) + Cu2+(aq)  Zn2+ (aq) + Cu(s) Do you remember doing this? What metal is giving up its electrons and being oxidised ? What metal ion is accepting these electrons and being reduced ? What did you observe ?

  6. So the Definition of oxidation/reduction now is: An oxidation-reduction reaction (or redox reaction) is one that involves the transfer of electrons from one species to another.

  7. In the reaction below the Zn metal has been oxidised as it has lost electrons, and the Cu2+ has been reduced as it has gained electrons. 1. Zn  Zn2+ + 2e 2. Cu2+ + 2e Cu Remember mnemonic - OIL RIG oxidation is loss reduction is gain Oxidation (loss of e’s) Reduction (gain of e’s)

  8. Oxidation Numbers A useful tool in recognising redox reactions, and for determining what is being oxidised and what is reduced in a reaction, involves the use of oxidation numbers The oxidation number (symbol ON) describes the “degree” to which an element has been oxidised or reduced. Chemists have developed a number of rules you must learn for assigning oxidation numbers Note: Oxidation numbers are always quoted per atom.

  9. Oxidation number rules • The oxidation number (or state) is a number that can be assigned to each atom in an element, compound or ion, using a set of 6 rules. • These rules are as follows: Rule 1 The oxidation number of an atom in any element is zero. For examplein H2 the oxidation number of H is 0. The oxidation number of C in carbon is 0 The oxidation of Mg in a piece of Mg is 0 The oxidation number of O in O2 is 0 .. etc

  10. Rule 2 The oxidation number of an atom in a monatomic ion is the same as the charge on the ion e.g. in Na+ the oxidation number is +1, in O2 the oxidation number is -2. In an ionic compound, the ions have the same oxidation numbers as they would alone e.g. in Na2O the oxidation numbers are still +1 for Na+ and -2 for O2.

  11. Rule 3 In compounds each hydrogen atom usually has an oxidation number +1 (the exception is in the metal hydrides e.g.NaH where oxidation number of H= -1). Rule 4 In compounds each oxygen atom has an oxidation number -2 (except in peroxides e.g. H2O2 where O has ON of -1)

  12. Rule 5 In a molecule the sum of the oxidation numbers of all the atoms is zero. Using rules 3, 4 and 5 it is possible to calculate the oxidation numbers of all atoms. Eg Find the oxidation number of S in H2SO4. H2 S O4 2 x +1 + 1x ? + 4 x -2 = 0 = 2 + ? + - 8 = 0 Can you do the math to find ON of S?  Oxidation number of S = +8 - 2 = +6 Hint ask : What’s the ON of H in compounds? (R3) What’s the ON of O in compounds? (R4) Then use these with R5 to find the ON of S

  13. Use the rules to find the oxidation numbers of the each atom in each of the following molecules. NO2, HNO3, NO, N2, N2O, HNO2, N2O4 +1 +1 +5 -2 +2 -2 -2 0 +4 -2 +1 +3 -2 +4 -2 Note that the N atom has a different ON depending on which atoms it is bonded to!

  14. Rule 6 In polyatomic ions the sum of the oxidation numbers of all the atoms is equal to the charge on the ion. In the ion Cr2O72 the oxidation number of Cr is calculated as follows: • 2 x Cr + 7 x O • 2 x ? + 7 x -2 = -2 • 2 x ? + -14 = -2 • 2 x Cr = 12 • Oxidation number of Cr = = +6 Charge on the polyatomic ion (+14 to BS) Note: Oxidation numbers are always quoted per atom.

  15. Calculate the oxidation number of S in each of the following ions: SO42 SO32 S2O32 S4O62 S + (4 x -2) = -2 ON of S = +6 S + (3 x -2) = -2  ON of S = +4 (2 x S) + (3 x -2) = -2 ON of S = +2 4 x S + (6 x -2) = -2 ON of S = +2.5

  16. We now can identify which species is oxidised and which is reduced in a reaction by applying this fact: An increase in oxidation number corresponds to oxidation A decrease in oxidation number corresponds to reduction. This is an exceptionally important thing to remember!

  17. By assigning oxidation numbers show which of the following reactions is not a redox reaction. (i) CuCO3  CuO + CO2 (ii) Cu + 2AgNO3 Cu(NO3)2 + 2Ag (iii) Cr2O72+ 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O Not a redox reaction no change in ON’s -2 -2 +4 +2 +2 -2 +4 Ag reduced Cu oxidised +2 0 -2 0 +1 +5 -2 +5 +6 +3 +2 +3 Cr2O72reduced Fe 2+ oxidised

  18. Looks like you’ve mastered the whole assigning ON thing! Now it’s time to apply them in balancing more complex redox reactions

  19. Balancing Redox Equations. The following method for balancing more complex redox equations is commonly called the ion-electron half-equation method. *These are 6 more steps we must memorise This is all you need to memorize – honest!

  20. Step 1 Identify the species undergoing oxidation and the species undergoing reduction Remember to use ON for this: The species that decreases in ON is reduced The species that increases in ON is oxidised e.g. when a solution of potassium dichromate reacts with iron II nitrate the species oxidised is Fe2+ and the species reduced is Cr2O72. The reactions are:Fe2+Fe3+ and Cr2O72Cr3+ +3 oxidation +2 +6 +3 reduction

  21. Step 2 Balance all atoms undergoing a change in oxidation number in each half equation. Fe2+ Fe3+ and Cr2O722Cr3+ Step 3 Balance the number of O atoms by adding the appropriate number of water molecules. Fe2+ Fe3+ and Cr2O722Cr3+ + 7H2O Step 4 Balance the H atoms by adding H+ ions. Fe2+ Fe3+ and Cr2O72 + 14H+ 2Cr3+ + 7H2O

  22. Step 5 Balance the charge by adding electrons, e-. This gives 2 balanced half-equations. Fe2+ Fe3+ + e and Cr2O72 + 14H++ 6e2Cr3+ + 7H2O

  23. Step 6 To obtain an overall balanced equation the 2 half equations must be added together. Before doing this the equations may have to be multiplied so that the number of electrons in each half-equation is the same. In this way, the electrons will be eliminated in the final equation. Fe2+ Fe3+ + e is now 6Fe2+ 6Fe3+ + 6e Cr2O72 + 14H+ + 6e 2Cr3+ + 7H2O Now combine both equations to give the final balanced redox equation – cancelling any electrons, H2O or H+ Finally check that the equation is balanced, particularly for charge!! (x6) 6Fe2+ + Cr2O72 + 14H+ 2Cr3+ + 7H2O + 6Fe3+ (6 x 2) + (-2) + (14) = +24 (2 x 3) + (6 x 3) = +24

  24. Question A solution of Fe2+ is added to a solution of purple potassium permanganate (KMnO4) which went colourless showing that the Mn2+ ion had formed. Using the steps for balancing redox equations write the balanced redox equation. Hint – The Fe 2+ is oxidised to Fe 3+ Now turn to page 230 in your lab book

  25. Common oxidants and reductants Then colours of species – issue sheet to be coloured Turn to page 236 in your lab books “more oxidants” Read the experiment carefully

  26. Oxidants and Reductants revisited Oxidant (aka oxidisng agent) An oxidant is the substance that a______ electrons and is _________ ccepts reduced Reductant (aka reducing agent) A reductant is the substance that d______ electrons and is _______ onates oxidised

  27. Look at the Demo then Complete the following KMnO4 is an oxidising agent and reacts with H2O2 . What is the colour of KMnO4? Write the reaction for the reduction of the KMnO4 Write the reaction for the oxidation of H2O2 Write the full redox reaction by combining both half equations What would you observe in the this reaction?

  28. Reactions of Halogens (has appeared in exams) Halogens e.g. chlorine, Cl2 (a yellow-green gas), bromine, Br2 (an orange/brown liquid), and iodine, I2 (a shiny black solid) are all reduced to their respective colourless halide ions, Cl, Br, I. The order of oxidising strength is Cl2 > Br2 > I2 In other words Cl2 will oxidise Br- ions to form Br2 the reaction is: Cl2 + 2Br- 2Cl- + Br2 (brown bromine appears) But !! 2Cl- + Br2 no reaction Why? Any halogen is able to oxidise the halide ion from a weaker halogen e.g. Cl2 can successfully oxidise I to I2. But I2 cannot oxidise Cl- because it is weaker oxidising agent

  29. How could you introduce Cl- ions into a solution? How could you introduce Br- ions into a solution? What are the colours of I- Br- Cl- What are the colours of I2 Br2 Cl2

  30. QUESTION SEVEN 2.7 2005 exam Group 17 elements, the halogens (F, Cl, Br, I) act as oxidants in reactions. Aqueous chlorine, Cl2(aq), can react with a solution containing iodide ions, I–(aq). Write balanced half-equations for the oxidation and reduction reactions that occur below. Then use these to write a balanced equation for the overall oxidation-reduction reaction that occurs. oxidation: reduction: overall equation: Use the balanced equation to predict expected observations for this reaction, and justify these observations by referring to the species involved. 2I- I2 + 2e- Cl2 + 2e- 2Cl- 2I- + Cl2I2 + 2Cl- Colourless Cl2 solution oxidises colourless I- ions to form an orange/brown solution of iodine (I2) (E)

  31. Look at the Demo then Complete the following K2Cr2O7 (Cr2O72-)is an oxidising agent and reacts with NaHSO3 (HSO3-). What is the colour of Cr2O72- ion? Write the reaction for the reduction of the Cr2O72- Write the reaction for the oxidation of HSO3- Write the full redox reaction by combining both half equations What would you observe in the this reaction?

  32. What’s occurring in this Demo Cu + HNO3 K2Cr2O7/SO2 SO2 gas is bubbled through some acidified K2Cr2O7 solution

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