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 O 2, I 2, Cl 2, Fe 3+, H 2 O 2, MnO 4  (aq)/H +, Cr 2 O 7 2  (aq)/H + Common reductants are limited to metals, C, CO, H 2, Fe 2+, Br , I , SO 2, (HSO 3  ). Properties are limited to appearance (colour and state), oxidation number (for polyatomic ions and single ions)

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.

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

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

However scientists realised that not all redox reactions involved oxygen e.g. the reaction of zinc metal with copper ions. Zn(s) + Cu 2+ (aq)  Zn 2+ (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 ?

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.

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

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.

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 example in H 2 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 O 2 is 0.. etc

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 O 2  the oxidation number is -2. In an ionic compound, the ions have the same oxidation numbers as they would alone e.g. in Na 2 O the oxidation numbers are still +1 for Na + and -2 for O 2 .

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. H 2 O 2 where O has ON of -1)

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 H 2 SO 4. H 2 S O 4 2 x x ? + 4 x -2 = 0 = 2 + ? = 0 Can you do the math to find ON of S?  Oxidation number of S = = +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

Use the rules to find the oxidation numbers of the each atom in each of the following molecules. NO 2, HNO 3, NO, N 2, N 2 O, HNO 2, N 2 O Note that the N atom has a different ON depending on which atoms it is bonded to!

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

Calculate the oxidation number of S in each of the following ions: SO 4 2  SO 3 2  S 2 O 3 2  S 4 O 6 2  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

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

By assigning oxidation numbers show which of the following reactions is not a redox reaction. (i) CuCO 3  CuO + CO 2 (ii) Cu + 2AgNO 3  Cu(NO 3 ) 2 + 2Ag (iii) Cr 2 O 7 2  + 6Fe 2+  6Fe Cr H 2 O no change in ON’s therefore this is not a redox reaction Ag + reduced Cu oxidised Cr in Cr 2 O 7 2  is reduced Fe 2+ oxidised

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

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! *Off course that’s not including the colours But hey you guys know some of these already

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 Fe 2+ and the species reduced is Cr 2 O 7 2 . The reactions are:Fe 2+  Fe 3+ and Cr 2 O 7 2   Cr  oxidation  reduction

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

Step 5 Balance the charge by adding electrons, e -. To the most positive side. This gives 2 balanced half-equations. Fe 2+  Fe 3+ + e  and Cr 2 O 7 2  + 14H + + 6e   2Cr H 2 O

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. Fe 2+  Fe 3+ + e  is now 6Fe 2+  6Fe e  Cr 2 O 7 2  + 14H + + 6e   2Cr H 2 O Now combine both equations to give the final balanced redox equation – cancelling any electrons, H 2 O or H+ Finally check that the equation is balanced, particularly for charge!! (6 x 2) + (-2) + (14) = +24(2 x 3) + (6 x 3) = +24 6Fe 2+ + Cr 2 O 7 2  + 14H +  2Cr H 2 O + 6Fe 3+ (x6)

Question A solution of Fe 2+ is added to a solution of purple potassium permanganate (KMnO 4 ) which went colourless showing that the Mn 2+ 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 _____ in your booklet

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

Oxidant (aka oxidisng agent) An oxidant accepts electrons and is reduced. (ie oxidants oxidise other substances) Oxidants and Reductant’s Reductant (aka reducing agent) A reductant donates electrons and are oxidised (ie reductants reduce other substances) what was oxidant in the Cr 2 O 7 2- / SO 3 2- reaction?

Look at the Demo then Complete the following KMnO 4 is an oxidising agent and reacts with H 2 O 2. Expt: place 3 mls of KMnO 4 in a boiling tube test tube add 2mls of dilute H 2 SO 4 add 6 mls of H 2 O 2 1.What did you observe in the this reaction? 2.What is the colour of KMnO 4 ? 3.What is the reaction for H 2 O 2 ? 4.Write the ½ reaction for the reduction of the KMnO 4 5.Write the ½ reaction for the oxidation of H 2 O 2 6.Write the full redox reaction by combining both half equations

Weekend Homework Complete all of worksheet one and worksheet 2. 1 a - e

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

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 I 2 Br 2 Cl 2

QUESTION SEVEN exam Group 17 elements, the halogens (F, Cl, Br, I) act as oxidants in reactions. Aqueous chlorine, Cl 2 (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 above 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- I 2 + 2e- Cl 2 + 2e- 2Cl- 2I- + Cl 2 I 2 + 2Cl- Pale green Cl 2 solution oxidises colourless I- ions to form an orange/brown solution of iodine (I 2 ) (E)

Look at the Demo then Complete the following K 2 Cr 2 O 7 (Cr 2 O 7 2- )is an oxidising agent and reacts with NaHSO 3 (HSO 3 - ). 1.What is the colour of Cr 2 O 7 2- ion? 2.Write the reaction for the reduction of the Cr 2 O Write the reaction for the oxidation of HSO Write the full redox reaction by combining both half equations 5.What would you observe in the this reaction?

What’s occurring in this Demo Cu + HNO 3 K 2 Cr 2 O 7 /SO 2 SO 2 gas is bubbled through some acidified K 2 Cr 2 O 7 solution