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Through this course, we have learned that chemical processes involve energy transformations.

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

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Through this course, we have learned that chemical processes involve energy transformations.

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

  2. Unit 8How do use chemical systems to harness energy? 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. Analyze electron transfer between coupled systems. M1. Controlling Electron Transfer . Explore the effect of electron transitions in solid systems. M2. Inducing Electron Transitions

  3. How would your chemistry knowledge be useful in identifying/designing alternative, cleaner sources of energy? Why do we care? 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.

  4. 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 The Problem There are a wide variety of chemical processes that release energy (exothermic) or transform energy from one form to another.

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

  6. TransformationHow do I change it? How can we control the amount, rate, and type of new forms of energy produced? 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).

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

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

  9. 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- Zn2+(aq) II. Reduction: Cu2+(aq) + 2e- Cu(s) III. Electrodes, wire, switch, electrolyte.

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

  11. ON(+2) ON(+1) 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- H2(g) III. Transport proteins, light.

  12. Same # of e-lost and gained. x m x n [mA – mxn e-] + [nBm+ +mxn e-] mAn+ + nB Requirements For these devices to be useful, the DGt for the overall process should be negative (thermodynamically favored): A – ne-  An+DG1 Bm+ + me- B DG2 mA + nBm+ mAn+ + nBDGt = mDG1+ nDG2 < 0 and the activation energy Ea for each half-reaction, as well as for the e-transfer process, should be low (~RT) (kinetically favored).

  13. Hydrogen Reference Electrode Electric Potential Difference (Eo) = 0.34 V To understand how this works, let’s analyze this case. Reference System Given the interest in processes for which DGt < 0, chemists have devised a procedure to determine DGifor many half REDOX reactions using a common standard system (H+/H2) as a reference:

  14. DGo for the process is a measure of the energy needed to transport electrons across this electric potential: DGo (J) = - nFEocell n- # of e- transferred in the unit reaction.F- Electric charge of 1 mol of e- (NAqe) = 9.6485x104 C/mol Cu2+/Cu H2 – 2e-  2H+ Cu2+ + 2e- Cu } Cu2+(aq) + H2(g)  Cu(s) + 2H+(aq) The measured electric potential difference across this standard cell is Eoreduction(Cu2+/Cu) = +0.34 V. DGo (J) = - 2 x 9.6485x104 x 0.34 = -6.6x10 kJ (Favored; Cu2+ is a stronger oxidizer than H+)

  15. Based on these results, would you expect this reaction to be favored or not favored: Cu2+(aq) +Zn(s) Cu(s) +Zn2+(aq) Let′s think! Zn2+/Zn Now, let’s imagine we couple these processes: H2 – 2e-  2H+ Zn2+ + 2e- Zn } Zn2+(aq) + H2(g)  Zn(s) + 2H+(aq) In this case we find Eoreduction (Zn2+/Zn) = -0.76 V DGo (J) = -2 x 9.6485x104 x (-0.76) = +1.5x102 kJ (Not Favored; H+ is a stronger oxidizer than Zn2+)

  16. Cu2+(aq)+Zn(s) Cu(s)+Zn2+(aq) DGot = -2.2x102 kJ Certainly favored!! Cell Potentials Cu2+(aq) + H2(g) Cu(s) + 2H+(aq) DGo1 = -6.6x10 kJ Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) DGo2 = -1.5x102 kJ We can simplify this analysis by simply looking at the Eoreduction values: Cu2+ + 2e-   Cu Eo1 = +0.34 Implies Cu2+ is a stronger oxidizer than H+ Zn2+ + 2e-  Zn Eo2 = -0.76Implies H+is a stronger oxidizer than Zn2+ Thus, Cu2+ should be able to oxidize Zn.

  17. Cu2+(aq)+Zn(s)  Cu(s)+Zn2+(aq) Eocell = Eored + Eoox Eocell = +1.10 V Zn Cu If Eocell > 0 then DGo= -nFEocell < 0 FAVORED!!!! Cell Potentials Cu2+ + 2e- Cu Eored =+0.34 Zn - 2e-Zn2+Eoox= -Eored=+0.76

  18. Reducers Oxidizers Activity Series Na Zn Fe H2 Cu Cl- -2.71 -0.76 -0.40 0 +0.34 +1.36 Eored Na+ Zn2+ Fe2+ H+ Cu2+ Cl2 The farther away in the activity series, the more thermodynamically favored the (counterclockwise) reaction between the REDOX pairs.

  19. Let’s Think • Given the available information: • Build the voltaic cell that will generate the maximum Eocell; • Predict the direction of e- transfer. • Write the chemical equation for the overall process; • Calculate its DGo;

  20. Zn2+ + 2e-  Zn 2Ag + 2e- 2Ag+ } 2Ag+(aq) + Zn(g)  2Ag(s) + Zn2+(aq) Let’s Think DGo = -nFEocell= 3.0x102kJ Eocell= 0.76 + 0.80 = 1.56 V

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

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

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

  24. At equilibrium: Nernst Equation For a REDOX process of the form: mA + nBm+ mAn+ + nB The value of Ecellis determined, in a first approximation, by Nernst equation: As the reaction proceeds, Ecell decreases.

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

  26. 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. Let’s Think Given the half-reactions that occur during discharge of a lead-acid battery: PbO2(s) + 2H+(aq) + H2SO4(aq) + 2e- PbSO4(s) + 2H2O(l) Eored = +1.685 V Pb(s) + H2SO4(aq) - 2e- PbSO4(s) + 2H+(l)Eoox = +0.356 V Write the overall reaction for this process and calculate the cell potential Eocell. PbO2(s) + Pb(s) + 2H2SO4(aq)  2PbSO4(s) + 2H2O(l) Eocell = Eored+ Eoox = 2.041 V

  28. Let’s Think PbO2(s) + Pb(s) + 2H2SO4(aq)  2PbSO4(s) + 2H2O(l) Eocell = 2.041 V Write Nernst equation for this system and calculate Ecell at 25oC for [H2SO4] = 4.5M(Typical concentration in a fully charged car battery). Ecell= 2.08 V

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

  30. Other Options • The reactants can be replenished when needed (e.g. H2fuel cells for cars) O2(g) + 4H+ + 4e-  2H2O(l) 2H2(g) – 4e-  4H+ Eocell= +1.23 V

  31. Let′s apply! Assess what you know

  32. Let′s apply! One electrode uses alloys (M) that can soak up hydrogen atoms: M(s) + H2O(l) + e- MH(s) + OH-(aq) Eored= -0.828 V Calculate The battery of modern hybrid cars are nickel metal hydride (NiMH). The other is based on nickel alone: NiO(OH)(s) + H2O(l) + e- Ni(OH)2(s) + OH-(aq)Eored = +0.490 V Write the overall favored reaction for this electrochemical cell and calculate Eocell.

  33. Let′s apply! NiO(OH)(s) + MH(s)  Ni(OH)2(s) + M(s)Eocell = Eored + Eoox = +1.318 V Toyota Prius:28 nickel metal hydride modules—each containing six cells—connected in series. Estimate MH(s) + OH-(aq) – e- M(s) + H2O(l) NiO(OH)(s) + H2O(l) + e- Ni(OH)2(s) + OH-(aq) What is the total voltage generate by this type of car battery? 1.32 x 6 x 28 = 221.8 V

  34. Let′s apply! Analyze MH(s) + OH-(aq) – e- M(s) + H2O(l) NiO(OH)(s) + H2O(l) + e- Ni(OH)2(s) + OH-(aq) NiO(OH)(s) + MH(s)  Ni(OH)2(s) + M(s) Eocell = +1.318 V Write the Nernst Equation for this cell. Analyze this equation and discuss its implications for the properties of this battery. Ecell= EocellThe cell potential does not depend on the concentration of the electrolyte, which may extend the life f the battery and its rechargeability.

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

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

  37. mA + nBm+ mAn++ nB Eocell= Eored+ Eoox Standard Potentials Standard half-cell potentials Eored, measured in regerence with a standard H+/H2 electrode, can be used to make predictions about the directionality of a REDOX reaction. A H2 B- Eored -2.71 +1.36 0 An+ H+ Bm+ DGo (J) = - nFEocell

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

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