1. 14 Redox Equilibria 14.1 Redox Equations (Review) 14.2 Electrode Potentials and the Electrochemical Series 14.3 Predicting the Direction of Redox Reactions 14.4 Uses for Electrochemical Cells

2. 14.1 Redox Equations Learning Objectives: 1.Calculate the oxidation states of different elements in a chemical reaction. 2.Identify which species have been oxidised and which have been reduced in a redox reaction. 3.Write balanced redox equations.

3. Rules for Finding Oxidation Numbers 1.Uncombined elements = 0 2.Hydrogen = +1 (except metal hydrides = -1) 3.Group 1 = +1 4.Group 2 = +2 5.Aluminium = +3 6.Oxygen = -2 (except peroxides = -1, compounds with F = +2) 7.Fluorine = -1 8.Chlorine = -1 (except compounds with F and O = +)

4. Oxidation States Balance Out to Equal Charge of the Molecule

5. Balancing Redox Equations 1.Write balanced half equations. Remember H + or H 2 O molecules may need to be added to balance. 2.Balance the electrons. 3.Add the half equations and cancel anything appearing on both sides.

6. Ex: Chlorine will oxidise iron(II) to iron(III) and is itself reduced to chloride ions. Step 1: Write half equations Cl 2 (g) + 2e -  2Cl - (aq) reduction (0  -1) Fe 2+ (aq)  Fe 3+ (aq) + e - oxidation (+2  +3) Step 2: Balance for electrons Cl 2 (g) + 2e -  2Cl - (aq) 2Fe 2+ (aq)  2Fe 3+ (aq) + 2e - Step 3: Add half equations together and cancel out electrons Cl 2 (g) + 2Fe 2+ (aq)  2Cl - (aq) + 2Fe 3+ (aq)

7. In acid solution, dichromate(VI) ions (Cr 2 O 7 2- ) will oxidise sulphate(IV) ions (SO 3 2- ) to sulphate(VI) (SO 4 2- ) ions and are themselves reduced to Cr 3+. Step 1: Write half equations Reduction Cr 2 O 7 2- (aq)  Cr 3+ (aq) Balance Cr 2 O 7 2- (aq)  2Cr 3+ (aq) Cr 2 O 7 2- (aq) + 14H + (aq)  2Cr 3+ (aq) + 7H 2 O (l) Electrons+6  +3 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O (l)

8. In acid solution, dichromate(VI) ions (Cr 2 O 7 2- ) will oxidise sulphate(IV) ions (SO 3 2- ) to sulphate(VI) (SO 4 2- ) ions and are themselves reduced to Cr 3+. Step 1: Write half equations Oxidation SO 3 2- (aq)  SO 4 2- (aq) Balance SO 3 2- (aq) + H 2 O (l)  SO 4 2- (aq) + H + (aq) Electrons+4  +6 SO 3 2- (aq) + H 2 O (l)  SO 4 2- (aq) + 2H + (aq) + 2 e -

9. In acid solution, dichromate(VI) ions (Cr 2 O 7 2- ) will oxidise sulphate(IV) ions (SO 3 2- ) to sulphate(VI) (SO 4 2- ) ions and are themselves reduced to Cr 3+. Step 2: Balance electrons Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O (l) 3SO 3 2- (aq) + 3H 2 O (l)  3SO 4 2- (aq) + 6H + (aq) + 6 e - Step 3: Add the half equations and cancel. Cr 2 O 7 2- (aq) + 14H + (aq) + 6e - + 3SO 3 2- (aq) + 3H 2 O (l)  2Cr 3+ (aq) + 7H 2 O (l) + 3SO 4 2- (aq) + 6H + (aq) + 6 e - Cr 2 O 7 2- (aq) + 8H + (aq) + 3SO 3 2- (aq)  2Cr 3+ (aq) + 4H 2 O (l) + 3SO 4 2- (aq)

10. Let’s try it! Acidified manganite(VII) ions (MnO 4 - ) can be reduced to Mn 2+ by Fe 2+ ions. The Fe 2+ ions are oxidised to Fe 3+. Write a balanced equation for this reaction. MnO 4 - (aq) + 8H + (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) + 4H 2 O (l) + 5Fe 3+ (aq)

11. 14.2 Electrode Potentials and the Electrochemical Series Learning Objectives: 1.Write half equations for reactions at electrodes. 2.Describe what is meant by “standard electrode potential”. 3.Describe how standard electrode potentials are measured. 4.Calculate electrode potentials for an electrochemical cell. 5.Draw a representation of an electrochemical cell.

12. Electrochemical Cells If two different metals are dipped into salt solutions containing their own ions and connected by a wire an electric current will flow. The electrons will flow from the more reactive metal (gives up electrons more easily to form + ions) to the less reactive metal. Reduction happens at one electrode and oxidation happens at the other. ZnCu Zn 2+ Cu 2+ e-e-

13. Electrochemical Cells Zn is more reactive than Cu. Zn is oxidised (loses electrons so is a reducing agent). Zn  Zn 2+ + 2e - Cu 2+ ions in solution are reduced (gains electrons so is an oxidising agent) Cu 2+ + 2e -  Cu ZnCu Zn 2+ Cu 2+ e-e-

14. Potential Difference (Voltage) Zn acquires a positive charge. Zn is said to gain a negative electrical potential. This represents how easily electrons are lost. However, this electrical potential cannot be measured. But the potential difference between two electrodes can be measured using a voltmeter.

15. The Standard Hydrogen Electrode The standard hydrogen electrode is used as a “control” to compare metals. Bubble H 2 gas into a solution of H + ions. A platinum electrode is used to conduct electricity. Platinum is used as it is a unreactive metal. Standard conditions: H + 1.00 mol/dm -3, 100 kPa, 298 K The electrode potential is defined as zero, so the potential difference of any half cell attached to is the electrode potential of that metal.

16. The Electrochemical Series Electrode potentials can be put into a table called the electrochemical series. Usually listed with most negative at the top (best reducing agents) To least negative (positive) at the bottom (best oxidising agents). All reactions are shown as reduction reactions. If oxidation takes place, switch the sign of the electrode potential.

17. Electrode Potentials Can Be Used to Calculate the Cell Potential Cell potential (E), also called electromotive force (e.m.f.), is the potential difference (voltage) between two half cells. This can be calculated using the electrode potentials.

18. Calculating Cell Potentials 1.Look up the electrode potentials for the two half cells. 2.Write the reaction for the more negative electrode potential as an oxidation reaction and switch the sign of the electrode potential. 3.Add the two electrode potentials together.

19. Example: Zinc/Copper Electrochemical Cell Connecting two half cells creates an electrochemical cell. What reactions are taking place? Zn has a more negative electrode potential, so is the stronger reducing agent, meaning it is oxidised. Zn  Zn 2+ + 2e - E = +0.76 V Copper is then reduced. Cu 2+ + 2e -  CuE = +0.34 V Half ReactionE θ (V) Zn 2+ + 2e - Zn-0.76 Cu 2+ + 2e - Cu+0.34

20. Example: Zinc/Copper Electrochemical Cell Zn  Zn 2+ + 2e - E θ = +0.76 V Cu 2+ + 2e -  Cu E θ = +0.34 V To find the cell potential, add the two electrode potentials together (remember to be careful of the signs!). E θ cell = (+0.76 V) + (+0.34 V) = +1.10 V Half ReactionE θ (V) Zn 2+ + 2e - Zn-0.76 Cu 2+ + 2e - Cu+0.34

21. Drawing Electrochemical Cells To simplify, there is a standard shorthand way to represent electrochemical cells.

22. Example: Lithium Cell The two half-equations below show the reactions at the positive and negative electrodes of a lithium cell. Li + (aq) + e -  Li (s) E θ = -3.04 V Li + (aq) + MnO 2 (s) + e -  LiMnO 2 (s) E θ = -0.13 V A platinum rod is used as the positive electrode. Draw a representation of the cell.

23. Li + (aq) + e -  Li (s) E θ = -3.04 V Li + (aq) + MnO 2 (s) + e -  LiMnO 2 (s) E θ = -0.13 V The top equation has the more negative electrode potential, so will go on the left side of the cell representation (oxidation will happen, so we flip it). Li (s)  Li + (aq) + e - Li (s) I Li + (aq) This is the first half cell. The second equation will go on the right side and remain as it is (reduction). Li + (aq) + MnO 2 (s) + e -  LiMnO 2 (s) Li (s) l Li + (aq) II Li + (aq) I MnO 2 (s), LiMnO 2 (s) Now add in the Pt electrode at the end. Li (s) l Li + (aq) II Li + (aq) I MnO 2 (s), LiMnO 2 (s) I Pt

24. 14.3 Predicting Redox Reactions Learning Objectives: 1. Use electrode potentials to predict if a redox reaction will occur.

25. Predicting if a redox reaction will occur 1.Write the two half equations for the redox reaction but write them both as reduction reactions. 2.Use the electrochemical series to find which has the more negative electrode potential. 3.Write the equation with the more negative electrode potential as an oxidation reaction. 4.Add the half equations together. 5.The reaction is feasible in this direction, but not the other.

26. Example: Will zinc react with aqueous copper ions? 1.Half equations as reductions Zn 2+ + 2e -  Zn Cu 2+ + 2e -  Cu 2.Find the more negative electrode potential Zinc E = -0.76VCopper = +0.34 V 3.Write this as oxidation Zn  Zn 2+ + 2e - 4.Add half equations together Zn + Cu 2+  Zn 2+ + Cu 5.Is it feasible? Yes, the reaction is in the correct direction.