1. Chemistry XXI Through this course, we have learned that chemical processes involve energy transformations. Can we apply what we have learned to design chemical systems to harness energy? Unit 8 How do we use chemical systems to harness energy? Chemical systems can thus be used to store energy or to transform it into usable forms.

2. Chemistry XXI M2. Inducing Electron Transitions. M1. Controlling Electron Transfer Analyze electron transfer between coupled systems. Explore the effect of electron transitions in solid systems. The central goal of this unit is to apply and extend central concepts and ideas discussed in this course to design chemical systems to harness energy. Unit 8 How do use chemical systems to harness energy?

3. Chemistry XXI Context To illustrate the power of the concepts, ideas, and ways of thinking discussed in the course, we will focus our attention on understanding how to design cleaner devices for energy transformation. How would your chemistry knowledge be useful in identifying/designing alternative, cleaner sources of energy? Why do we care?

4. Chemistry XXI The Problem There are a wide variety of chemical processes that release energy (exothermic) or transform energy from one form to another. The central question is how to apply chemical concepts, ideas, and ways of thinking to control these energy transformations: Amount, rate, and type of energy production Reversibility Nature of the products

5. Chemistry XXI Module 1: Controlling Electron Transfer Central goal: To analyze electron transfer between coupled chemical systems. Unit 8 How do we use chemical systems to harness energy?

6. Chemistry XXI The Challenge In many chemical processes the redistribution of charge between reacting atoms or ions results in the transformation of potential energy of the reactants into other forms of energy (heat, light). Transformation How do I change it? How can we control the amount, rate, and type of new forms of energy produced?

7. Chemistry XXI The energy released in exothermic chemical reactions essentially results from charge redistribution among reacting atoms. Energy Changes In these processes, electrons move from states of higher to lower potential energy. EpEp 0 If we want to control the amount and rate of energy released, we need to control charge redistribution. How can we do it?

8. Chemistry XXI Basic Components Many systems, natural and artificial, used to control the rate of production of chemical energy include three basic parts: I.A chemical system that undergoes an oxidation: A – ne -   A n+ II. A chemical system that undergoes a reduction: B m+ + me -  B III. A mechanism to a) allow, b) control e - transfer: Wire Molecule (e.g. protein) Reaction Sequence

9. Chemistry XXI Let’s Think Analyze this galvanic cell:  Identify the three basic parts for energy production and harnessing;  Describe in detail what is happening in this system. I. Oxidation: Zn(s) – 2e -  Zn 2+ (aq) II. Reduction: Cu 2+ (aq) + 2e -  Cu(s) III. Electrodes, wire, switch, electrolyte.

10. Chemistry XXI Galvanic Cell Batteries are a combination of one or more electrochemical cells using liquid (wet) or solid (dry) electrolytes.

11. Chemistry XXI Let’s Think Analyze this nano device:  Identify the three basic parts for energy production and harnessing;  Describe in detail what is happening in this system. I. Oxidation: Asc – 2e -  DhAsc II. Reduction: 2H + (aq) + 2e -  H 2 (g) III. Transport proteins, light. ON(+2)ON(+1)

12. Chemistry XXI For these devices to be useful, the  G t for the overall process should be negative (thermodynamically favored): and the activation energy E a for each half-reaction, as well as for the e - transfer process, should be low (~RT) (kinetically favored). Requirements mA + nB m+  mA n+ + nB  G t = m  G 1 + n  G 2 < 0 A – ne -  A n+  G 1 B m+ + me -  B  G 2 x m x n Same # of e - lost and gained. [mA – mxn e - ] + [nB m+ + mxn e - ]  mA n+ + nB

13. Chemistry XXI Given the interest in processes for which  G t < 0, chemists have devised a procedure to determine  G i for many half REDOX reactions using a common standard system (H + /H 2 ) as a reference: Reference System Hydrogen Reference Electrode Electric Potential Difference (E o ) = 0.34 V To understand how this works, let’s analyze this case.

14. Chemistry XXI Cu 2+ /Cu Cu 2+ (aq) + H 2 (g)  Cu(s) + 2H + (aq) The measured electric potential difference across this standard cell is E o reduction (Cu 2+ /Cu) = +0.34 V. H 2 – 2e -  2H + Cu 2+ + 2e -  Cu }  G o (J) = - 2 x 9.6485x10 4 x 0.34 = -6.6x10 kJ (Favored; Cu 2+ is a stronger oxidizer than H + )  G o for the process is a measure of the energy needed to transport electrons across this electric potential:  G o (J) = - n F E o cell n- # of e - transferred in the unit reaction. F- Electric charge of 1 mol of e - (N A q e ) = 9.6485x10 4 C/mol

15. Chemistry XXI Zn 2+ /Zn Now, let’s imagine we couple these processes: Zn 2+ (aq) + H 2 (g)  Zn(s) + 2H + (aq) H 2 – 2e -  2H + Zn 2+ + 2e -  Zn } In this case we find E o reduction (Zn 2+ /Zn) = -0.76 V  G o (J) = -2 x 9.6485x10 4 x (-0.76) = +1.5x10 2 kJ (Not Favored; H + is a stronger oxidizer than Zn 2+ ) Based on these results, would you expect this reaction to be favored or not favored: Cu 2+ (aq) + Zn(s)  Cu(s) + Zn 2+ (aq) Let′s think!

16. Chemistry XXI Cu 2+ (aq) + H 2 (g)   Cu(s) + 2H + (aq)  G o 1 = -6.6x10 kJ Zn(s) + 2H + (aq)   Zn 2+ (aq) + H 2 (g)  G o 2 = -1.5x10 2 kJ Cu 2+ (aq) + Zn(s)   Cu(s) + Zn 2+ (aq)  G o t = -2.2x10 2 kJ Certainly favored!! Cell Potentials We can simplify this analysis by simply looking at the E o reduction values: Zn 2+ + 2e -   Zn E o 2 = -0.76 Implies H + is a stronger oxidizer than Zn 2+ Cu 2+ + 2e -   Cu E o 1 = +0.34 Implies Cu 2+ is a stronger oxidizer than H + Thus, Cu 2+ should be able to oxidize Zn.

17. Chemistry XXI Zn - 2e -  Zn 2+ E o ox = -E o red = +0.76 Cu 2+ + 2e -  Cu E o red = +0.34 Cell Potentials Cu Zn If E o cell > 0 then  G o = -nFE o cell < 0 FAVORED!!!! Cu 2+ (aq) + Zn(s)  Cu(s) + Zn 2+ (aq) E o cell = E o red + E o ox E o cell = +1.10 V

18. Chemistry XXI Activity Series E o red Na Na + -2.71 Zn Zn 2+ -0.76 Fe Fe 2+ -0.40 H2H2 H+H+ 0 Cu Cu 2+ +0.34 Cl - Cl 2 +1.36 Oxidizers Reducers The farther away in the activity series, the more thermodynamically favored the (counterclockwise) reaction between the REDOX pairs.

19. Chemistry XXI Let’s Think Given the available information:  Build the voltaic cell that will generate the maximum E o cell ;  Predict the direction of e - transfer.  Write the chemical equation for the overall process;  Calculate its  G o ; E o cell (V) Cu 2+ + 2e-  Cu0.34Zn 2+ + 2e-  Zn-0.76 2H + + 2e-  H 2 0.0Ag + + e-  Ag0.80

20. Chemistry XXI Let’s Think 2Ag + (aq) + Zn(g)  2Ag(s) + Zn 2+ (aq) Zn 2+ + 2e -  Zn 2Ag + 2e -  2Ag + } E o cell = 0.76 + 0.80 = 1.56 V  G o = -nFE o cell = 3.0x10 2 kJ

21. Chemistry XXI Alternative Energy Sources? Batteries and related electrochemical devices are seen as a very promising route for cleaner portable energy sources. Make a list of actual and potential advantages and disadvantages of these types of devices. Let′s think! Limited power supply Limited durability Need to be disposed or recharged Heavy Expensive Smaller C footprint Portable, Noiseless Rechargable Small, diverse sizes Sealed products DisadvantagesAdvantages

22. Chemistry XXI Two of the most critical issues in battery technology are discharge time and rechargeability. Critical Issues Why does E cell decreases with time of use? Why is it not rechargable? Consider a prototypical alkaline battery:

23. Chemistry XXI As the reaction proceeds, electrodes dissolve and cannot be regenerated by reversing the reaction. Alkaline Battery E initial = 1.54 V With use, the concentration of reactants decreases and the reactions moves towards chemical equilibrium (E cell  0).

24. Chemistry XXI Nernst Equation For a REDOX process of the form: mA + nB m+  mA n+ + nB The value of E cell is determined, in a first approximation, by Nernst equation: As the reaction proceeds, E cell decreases. At equilibrium:

25. Chemistry XXI Depending on their size (amount of reactants) and composition, batteries discharge at different rates. Discharge Rates

26. Chemistry XXI Rechargable Batteries In common rechargable batteries, the products of the reaction tend to get attached to the electrode, so the overall process may be reversed by providing energy. Lead-Acid Battery

27. Chemistry XXI Let’s Think Given the half-reactions that occur during discharge of a lead-acid battery: PbO 2 (s) + 2 H + (aq) + H 2 SO 4 (aq) + 2 e-  PbSO 4 (s) + 2 H 2 O(l) E o red = +1.685 V Pb(s) + H 2 SO 4 (aq) - 2 e-  PbSO 4 (s) + 2 H + (l) E o ox = +0.356 V Write the overall reaction for this process and calculate the cell potential E o cell. PbO 2 (s) + Pb(s) + 2 H 2 SO 4 (aq)  2 PbSO 4 (s) + 2 H 2 O(l) E o cell = E o red + E o ox = 2.041 V

28. Chemistry XXI Let’s Think PbO 2 (s) + Pb(s) + 2 H 2 SO 4 (aq)  2 PbSO 4 (s) + 2 H 2 O(l) E o cell = 2.041 V Write Nernst equation for this system and calculate E cell at 25 o C for [H 2 SO 4 ] = 4.5 M (Typical concentration in a fully charged car battery). E cell = 2.08 V

29. Chemistry XXI Other Options In general, the issue of rechargeability has been addressed by looking for systems in which: a)The electrocehmical process is easily reverted (e.g. Li-ion batteries in laptops, iPhones, etc.)

30. Chemistry XXI Other Options b)The reactants can be replenished when needed (e.g. H 2 fuel cells for cars) O 2 (g) + 4H + + 4e-  2H 2 O(l) 2H 2 (g) – 4e-  4H + E o cell = +1.23 V

31. Chemistry XXI Assess what you know Let′s apply!

32. Chemistry XXI Let′s apply! Calculate Write the overall favored reaction for this electrochemical cell and calculate E o cell. The battery of modern hybrid cars are nickel metal hydride (NiMH). The other is based on nickel alone: NiO(OH)(s) + H 2 O(l) + e-  Ni(OH) 2 (s) + OH - (aq) E o red = +0.490 V M(s) + H 2 O(l) + e-  MH(s) + OH - (aq) E o red = -0.828 V One electrode uses alloys (M) that can soak up hydrogen atoms:

33. Chemistry XXI Let′s apply! Estimate What is the total voltage generate by this type of car battery? Toyota Prius: 28 nickel metal hydride modules—each containing six cells— connected in series. 1.32 x 6 x 28 = 221.8 V NiO(OH)(s) + H 2 O(l) + e-  Ni(OH) 2 (s) + OH - (aq) MH(s) + OH - (aq) – e-  M(s) + H 2 O(l) NiO(OH)(s) + MH(s)  Ni(OH) 2 (s) + M(s) E o cell = E o red + E o ox = +1.318 V

34. Chemistry XXI Write the Nernst Equation for this cell. Analyze this equation and discuss its implications for the properties of this battery. Let′s apply! Analyze NiO(OH)(s) + MH(s)  Ni(OH) 2 (s) + M(s) E o cell = +1.318 V NiO(OH)(s) + H 2 O(l) + e-  Ni(OH) 2 (s) + OH - (aq) MH(s) + OH - (aq) – e-  M(s) + H 2 O(l) E cell = E o cell The cell potential does not depend on the concentration of the electrolyte, which may extend the life f the battery and its rechargeability.

35. Chemistry XXI Working in pairs, identify one central idea introduced in this module.

36. Chemistry XXI Controlling Electron Transfer Summary System used to control the rate of production of chemical energy include three basic parts: I.A chemical system that undergoes an oxidation: A – ne -   A n+ II. A chemical system that undergoes a reduction: B m+ + me -  B III. A mechanism to a) allow, b) control e - transfer: Wire Molecule (e.g. protein) Reaction Sequence

37. Chemistry XXI Standard half-cell potentials E o red, measured in regerence with a standard H + /H 2 electrode, can be used to make predictions about the directionality of a REDOX reaction. Standard Potentials E o red A A n+ -2.71 H+H+ 0 B-B- B m+ +1.36 E o cell = E o red + E o ox mA + nB m+  mA n+ + nB H2H2  G o (J) = - n F E o cell

38. Chemistry XXI For next class, Investigate what are the basic properties of semiconducting materials. What chemical systems tend to be semiconducting?