1. Week 25 © Pearson Education Ltd 2009 This document may have been altered from the original Explain the terms: redox, oxidation number, half-reaction, oxidising agent and reducing agent for simple redox reactions. Construct redox equations using relevant half equations or oxidation numbers. Interpret and make predictions for reactions involving electron transfer.

2. Redox Remember OIL of electrons RIG There are 3 definitions of oxidation (and so 3 for reduction) Oxidation is Loss of electrons Gain in oxidation number Loss of hydrogen What are the 3 definitions of reduction? Redox reactions are best thought of as electron transfer reactions.

3. Oxidation numbers Using the rules from the table on page 182 assign oxidation numbers in the following compounds: CO 2 Mg(NO 3 ) 2 Σ(C) + 2(O) = 0 Since Mg 2+ has +2 ox no  C + (2x-2) =0 Each NO 3 = -1 C + (-4) = 0 Σ (N) + (3xO) = -1 C = +4 N + (-6) = -1  N -6 = -1  N = +5

4. For some transition metals Work out the oxidation number of the metal in each of the following: MnO 4 - CrO 3 Cr 2 O 7 2- MnO 2 FeCl 4 - V 2 O 5 +7 +6 +4 +3 +5

5. Redox Reactions Any competition or displacement reaction is a redox reaction. E.g. Mg + CuO → MgO + Cu In this reaction the O 2- remains unchanged and is simply passed from the Mg to the Cu. It is a spectator ion. Spectator ions are left out and IONIC EQUATIONS are written. i.e. Mg + Cu 2+ → Mg 2+ + Cu Write HALF EQUATIONS Mg → Mg 2+ + 2e - Loss of e - = oxidation Cu 2+ + 2e - → Cu gain of e - = reduction

6. Oxidising and reducing agents In a competition reaction between metals the MORE REACTIVE metal ends up as the compound. (See previous slide) The more reactive metal is oxidised. The less reactive metal ion is reduced. The more reactive metal gives away its electrons to the less reactive metal ion and thus is the reducing agent. The less reactive metal ion takes the electrons and so is the oxidising agent in the reaction. So oxidising agents are always reduced and lose oxidation number. Reducing agents are always oxidised and increase their oxidation number.

7. General Application These principles can be applied to any reaction as long as oxidation numbers can be assigned.

8. Examples of Redox PCl 3 reacts with chlorine to form PCl 5. PCl 3 + Cl 2 → PCl 5 i) What are the changes in oxidation number for each element? ii) Which substance is the oxidising agent?

9. Examples of Redox PCl 3 reacts with chlorine to form PCl 5. PCl 3 + Cl 2 → PCl 5 i) What are the changes in oxidation number for each element? PCl 3 + Cl 2 → PCl 5 +3 -1 0 +5 -1 ii) Which substance is the oxidising agent? Cl 2 is the oxidising agent

10. For each of the following reactions use the change in oxidation number to deduce the oxidising and reducing agents. i) 2Cu 2+ + 4I - → 2CuI + I 2 ii) MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O iii) 2Cu + 4HCl + O 2 → CuCl 2 + 2H 2 O iv) 2NaOH + Cl 2 → NaOCl + NaCl

11. i) oxidising agent Cu 2+ ; reducing agent I - ii)MnO 2 oxidising agent; HCl red agent iii)O 2 oxidising agent; Cu reducing agent iv)Cl 2 oxidising agent;Cl 2 reducing agent. Above iv) is an example of disproportionation- simultaneous redox. The oxidation number goes up AND down.

12. Constructing Equations From Half Equations See lilac boxes on p. 182 and 183. Do the questions on p. 183

13. Simple Cells Simple Cells. e.g. Mg + 2HCl → MgCl 2 + H 2 This is a redox reaction. The reactions are: Mg → Mg 2+ + 2e - OXIDATION 2H + + 2e - → H 2 REDUCTION. These are half equations. If the 2 half equations take place in separate containers electrons will flow in a wire connecting the 2 half reactions. This arrangement is called a SIMPLE CELL

14. In the example above electrons flow FROM the Mg to the H + in the wire. (The salt bridge makes the electrical connection between the 2 half cells, allowing ions to transfer between them. It is usually a piece of filter paper soaked in potassium nitrate solution. (All nitrates are soluble so no precipitation reactions can occur by accident.).) The current which flows is the cell potential or EMF of the cell. This is measured in volts.

15. What is a half cell? It is made up of an element in 2 oxidation states – generally a metal in contact with an aqueous solution of its salt. An equilibrium is set up on the surface of the metal E.g. Cu 2+ (aq) + 2e -  Cu (s) The electrode potential is the tendency of this half cell to lose or gain electrons and gives an idea of the position of the equilibrium.

16. Week 25 © Pearson Education Ltd 2009 This document may have been altered from the original Standard hydrogen half cell

17. Standard Electrode Potentials Half cell potentials cannot be measured in isolation. Standard electrode potentials are defined by definition as the VOLTAGE (or EMF) MEASURED UNDER STANDARD CONDITIONS WHEN THE HALF CELL IS PART OF AN ELECTROCHEMICAL CELL WITH THE OTHER HALF CELL BEING A STANDARD HYDROGEN ELECTRODE. Or as written on p. 184

18. Standard conditions : 1 atmosphere pressure 1.0M solution 25 o C (298K) A standard hydrogen electrode is a reference electrode. By definition its potential is taken as 0 volts. When a standard half cell is connected to a hydrogen half cell through a high resistance voltmeter (to prevent current flow) the measured potential is taken to be that of the half cell in question.

19. Standard Hydrogen Half Cell See diagram The platinum wire is merely to allow for electrons to pass from one half cell to the other in the simple cell. The half cell equilibrium is: 2H + + 2e -  H 2(g) This sets up on the platinum foil which is covered in porous platinum black in which the electron transfer takes place.

20. Week 25 © Pearson Education Ltd 2009 This document may have been altered from the original Measuring the standard electrode potential of a Zn/Zn 2+ half cell

21. Standard Zinc Half Cell See previous slide. The measured potential for a standard zinc half cell (against the hydrogen half cell) is-0.76v. What does this mean? For the equation: Zn 2+ (aq) + 2e - D Zn (s) E Ө = - 0.76v The negative sign indicates that this equilibrium lies LEFT. The zinc terminal is the negative terminal i.e. it loses electrons which are used to oxidise the H + ions in the hydrogen half cell. Zinc is reduced in this standard cell. Since zinc is a reactive metal which reacts with acid to form hydrogen gas and a zinc salt this is as expected.

22. Standard electrode potentials are always written as REDUCTION POTENTIALS. i.e. oxidised species + electrons → reduced species. It is a measure of the tendency for reduction to occur. Values range from about -3 to +3 volts. So: Cu 2+ + 2e - → Cu E = +0.34v Zn 2+ + 2e - → Zn E = -0.76v. The more positive the potential the more likely it is that reduction will occur i.e. go right. Reactions which go RIGHT more readily than the reduction of hydrogen ions are always given the POSITIVE electrode potential.

23. Calculating Cell EMF Whenever 2 standard half cells are connected together an overall potential is set up and electrons will flow in the wire to the CATHODE (the positive terminal) where reduction takes place. We can write a short hand for the cell in question. e.g. a zinc/copper cell The two half equations are Zn 2+ + 2e - → Zn E = -0.76v Cu 2+ + 2e - → Cu E = +0.34v

24. Zn (s) │ Zn 2+ (aq) ││ Cu 2+ (aq) │ Cu (s) Apply these rules. The half cell with the more POSITIVE potential is placed on the RIGHT. This is where reduction occurs. EMF of the cell is given by: E cell = E R –E L Where E R is the potential of the right hand cell And E L is the potential of the left hand cell. When 2 electrodes combine to form a cell the value for E cell must be at least +0.3v if the reaction is to occur spontaneously So E cell = +0.34- -0.76 = +1.1v This value tells us that the above redox reaction will take place as the cell is written.

25. Zn → Zn 2+ + 2e - Cu 2+ + 2e - → Cu i.e. Zn + Cu 2+ → Zn 2+ + Cu Converting the above into another rule. E cell = the more positive potential – the less positive potential = +0.34- -0.76 = +1.1v as before The more positive potential tells us that that reaction proceeds left to right but that the less positive reaction goes the other way i.e. right to left.

26. Using Standard Electrode Equations to Predict Whether a Reaction Will Occur Is aqueous iodine able to oxidise bromide ions to aqueous bromine? 1. Write down the reaction you are testing. I 2(aq) + 2Br - → 2I - (aq) + Br 2(aq) 2. Split it into 2 half equations (taking account of the DIRECTION of each reaction) and show the electrode potentials. I 2(aq) + 2e - → 2I - (aq) E = + 0.54v 2Br - (aq) → Br 2(aq) + 2e - E = -1.07v* *This is an oxidation potential. The reaction is going in the opposite direction from the standard form and so the sign on the potential is changed. 3. Add the half reactions and the electrode potentials together I 2(aq) + 2Br - (aq) → 2I - (aq) + Br 2(aq) E = 0.54 + (-1.07) = -0.53v The negative value for the cell emf indicates that the reaction will NOT go spontaneously as written and will go back the other way with a positive emf +0.53v.

27. Predict whether the following cell should proceed spontaneously. Cr 3+ + Fe 2+ → Cr 2+ + Fe 3+ Cr 3+ + e - → Cr 2+ E = -0.41v Fe 2+ → Fe 3+ + e - E =-0.77v Cell potential = -0.41+(-0.77) = -1.18v. The reaction will NOT go.

28. The Effect of Changing Conditions Each redox potential is actually an equilibrium constant with the sign and size of the potential indicating the position of the equilibrium under standard conditions. When conditions are not standard the position of the equilibrium and therefore the size of the potential can also change. This is important in reactions where the overall cell potential is low and close to +0.3v. It is possible to make reactions proceed by changing conditions which would NOT proceed under standard conditions.

29. Limitations Consider Le Chatelier when deciding the effect of non standard conditions on the size of a cell potential. E.g. from p.189 Cu 2+ (aq) + 2e -  Cu (s) Increasing the concentration of the aqueous ions causes the equilibrium to shift RIGHT to remove them. This in turn causes the electrode potential to become more positive and this may cause predicted reactions to be wrong.

30. Reaction Rate Redox potentials are about equilibrium NOT rate. Some reactions with a positive E cell greater than 0.3v may go so slowly as to be useless. The larger the difference between the E Ө values the more likely a reaction will take place.