ELECTROCHEMISTRY INVOLVES TWO MAIN TYPES OF PROCESSES

1. ELECTROCHEMISTRY INVOLVES TWO MAIN TYPES OF PROCESSES • A. Voltaic(galvanic) cells – which are spontaneous chemical reactions (battery) • B. Electrolytic cells – which are non-spontaneous and require external e− source (DC power source) • C. BOTH of these fit into the category entitled Electrochemical cells

2. Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

3. Oxidation and Reduction • A species is oxidized when it loses electrons. • Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.

4. Oxidation and Reduction • A species is reduced when it gains electrons. • Here, each of the H+ gains an electron and they combine to form H2.

5. Oxidation and Reduction • What is reduced is the oxidizing agent. • H+ oxidizes Zn by taking electrons from it. • What is oxidized is the reducing agent. • Zn reduces H+ by giving it electrons.

6. Assigning Oxidation Numbers • Elements in their elemental form have an oxidation number of 0. • The oxidation number of a monatomic ion is the same as its charge.

7. Voltaic Cells In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

8. Voltaic Cells • We can use that energy to do work if we make the electrons flow through an external device. • We call such a setup a voltaic cell.

9. Voltaic Cells • A typical cell looks like this. • The oxidation occurs at the anode. • The reduction occurs at the cathode.

10. Voltaic Cells The flow of electrons is always from the anode to the cathode through the wire

11. Voltaic Cells Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

12. Voltaic Cells • Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. • Cations move toward the cathode. • Anions move toward the anode.

13. Voltaic Cells • In the cell, then, electrons leave the anode and flow through the wire to the cathode. • As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.

14. Voltaic Cells • As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. • The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

15. Voltaic Cells Animation

16. Electromotive Force (emf) • Water only spontaneously flows one way in a waterfall. • Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

17. Cell Potential (Electromotive Force (emf)) • The potential difference between the anode and cathode in a cell is called the electromotive force (emf). • It is also called the cell potential, and is designated Ecell.

18. Cell Potential (Electromotive Force (emf)) • Each potential is measured against a standard , which is the standard hydrogen electrode [consists of a piece • of inert Platinum that is bathed by hydrogen gas at 1 atm]. The hydrogen electrode is assigned a value • of ZERO volts.

19. Cell Potential (Electromotive Force (emf)) • standard conditions—1 atm for gases, 1.0M for solutions and 25°C for all (298 K) • • naught (°)--we use the naught to symbolize standard conditions [Experiencing a thermo flashback?] • That means • Ecell, Emf, or εcell become • Ecello , Emfo , or εcello when measurements are taken at standard conditions. You’ll soon learn how these change when the conditions are nonstandard!

20. Reduction Potential Chart • elements that have the most positive reduction potentials are easily reduced (in general, non-metals) • elements that have the least positive reduction potentials are easily oxidized (in general, metals) • The table can also be used to tell the strength of various oxidizing and reducing agents. • Can also be used as an activity series. Metals having less positive reduction potentials are more active and will replace metals with more positive potentials.

21. Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated.

22. Writing a Cell Diagram-Standard Cell notation“ion sandwich in alphabetical order” • The reduced species is written on the right hand side—this electrode is the cathode • The oxidized species is placed on the left hand side and this is the anode. • For example the cell diagram of a zinc and copper galvanic cell: • A vertical line (I ) represents the different phases present in each half cell-phase boundary. (if in same phase use comma instead) • A double vertical line II represents the salt bridge connecting the two half cells • Anode I solution II cathode solution I cathode

23. Zn(s) I Zn +2 II Cu+2 I Cu (s) •  Zinc and Lead(s) (plumbous) • Copper and iron+3 and +2 (use copper II)

24. What do you do when one of your electrodes lacks a SOLID METAL??You will need an inert conductor (usually platinum ) put in parentheses (Pt)Different species of the same phase are separated by a comma • Fe+3 (aq) + Cu(s)  Cu+2 + Fe2+

25. Calculating Standard Cell Potential (Ecello, Emfo , or εcello ) • Decide which element is oxidized or reduced using the standard reduction potential chart. Element with more positive reduction potential gets reduced! • Write both equations as is from the chart with their voltage • Reverse the equation that will be oxidized and change the sign of the voltage • Balance the two half reactions BUT DO NOT MULTIPLY VOLTAGE VALUES • Add the two half reactions and their voltages together • Ecello = Eooxidation +Eoreduction where o means standard conditions 1atm , 1M and 25oC

26. Calculate Cell Potential E cello •   Zinc and lead (s) Plumbous • Copper and iron+3 and +2 (use copper II) 1. H2 (g) + I2 (g)  2H+ (aq) + 2 I – (aq)

27. Calculate Cell Potential E celloDraw a diagram of the galvanic cell for the reaction and label completely • Fe+3 (aq) + Cu(s)  Cu+2 + Fe2+

28. Work(J) Charge ( C) 1 V = 1 Cell Potential , Electrical Work, And Free energy Cell potential is measured in volts (V). It is the work that can be accomplished when electrons are transferred through a wire.

29. Work(J) Charge ( C) 1 V = 1 Cell Potential , Electrical Work, And Free energy 1 joule of work is required or produced when one coloumb of charge is transferred between two points

30. Cell Potential , Electrical Work, And Free energy • Using the table of standard reduction potentials, calculateDG o for the reaction: • Cu+2(aq) + Fe(S)  Cu (s) + Fe+2 (aq) • IS it spontaneous?

31. Cell Potential , Electrical Work, And Free energy DG o = - nFEo • G= Gibb’s Free energy • n= numbers of moles of electrons • F= Faraday constant 96500 J/V . Mol • The redox reaction will be spontaneous when DG o = _________ • The redox reaction will be spontaneous when Eo = __________

32. Cell Potential , Electrical Work, And Free energy DG o = - nFEo + DG o = - RT ln K = • You can derive the equation: for Ecell at standard conditions at equilibrium Eo = RT ln K nF

33. Eo = RT ln K nF

34. RT nF ln Q E = E − Nernst Equation • Why? It is used to calculate the voltage generated by the combination of two half-cells when the conditions are not standard!!!!!!! • n= numbers of electrons transferred between species • F= Faraday constant 96500 J/V . mol • R=ideal gas constant 8.314 /K mol • T= Kelvin Temperature • Eo= the voltage generated if the conditions • were standard • Q= reactions quotient [products]x • [reactants]y

35. If the reaction below is carried out using solutions that are 5.0 M Zn +2 and 0.3 M Cu+2 at 298 K, what is the actual cell voltage? Zn (s) + Cu+2 Cu(s) + Zn+2 • Step 1: Work out Eo cell assuming standard conditions: • Step 2: Calculate Q (don’t forget _____ and _______ are omitted from Q) • Step 3: Find n • Step 4: Plug everything in! (Answer 1.06 V)

36. Assigning Oxidation Numbers • Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. • Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1. • Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.

37. Assigning Oxidation Numbers • Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. • Fluorine always has an oxidation number of −1. • The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.

38. Assigning Oxidation Numbers • The sum of the oxidation numbers in a neutral compound is 0. • The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

39. Balancing Oxidation-Reduction Equations Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.

40. Balancing Oxidation-Reduction Equations This involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction.

41. Half-Reaction Method • Assign oxidation numbers to determine what is oxidized and what is reduced. • Write the oxidation and reduction half-reactions.

42. Half-Reaction Method • Balance each half-reaction. • Balance elements other than H and O. • Balance O by adding H2O. • Balance H by adding H+. • Balance charge by adding electrons. • Multiply the half-reactions by integers so that the electrons gained and lost are the same.

43. Half-Reaction Method • Add the half-reactions, subtracting things that appear on both sides. • Make sure the equation is balanced according to mass. • Make sure the equation is balanced according to charge.

44. Half-Reaction Method Consider the reaction between MnO4− and C2O42− : MnO4−(aq) + C2O42−(aq) Mn2+(aq) + CO2(aq)

45. +7 +3 +2 +4 MnO4− + C2O42- Mn2+ + CO2 Half-Reaction Method First, we assign oxidation numbers. Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized.

46. Oxidation Half-Reaction C2O42− CO2 To balance the carbon, we add a coefficient of 2: C2O42− 2 CO2

47. Oxidation Half-Reaction C2O42− 2 CO2 The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side. C2O42− 2 CO2+ 2 e−

48. Reduction Half-Reaction MnO4− Mn2+ The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side. MnO4− Mn2++ 4 H2O

49. Reduction Half-Reaction MnO4− Mn2+ + 4 H2O To balance the hydrogen, we add 8 H+ to the left side. 8 H+ + MnO4− Mn2+ + 4 H2O

50. Reduction Half-Reaction 8 H+ + MnO4− Mn2+ + 4 H2O To balance the charge, we add 5 e− to the left side. 5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O