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Lecture 284/11/05. Primary Batteries (non-rechargeable) Oxyride battery Not responsible for this one E° = 1.7 V Cathode: NiOOH, MnO 2 Anode:Zinc New battery.

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Presentation on theme: "Lecture 284/11/05. Primary Batteries (non-rechargeable) Oxyride battery Not responsible for this one E° = 1.7 V Cathode: NiOOH, MnO 2 Anode:Zinc New battery."— Presentation transcript:

1 Lecture 284/11/05

2 Primary Batteries (non-rechargeable) Oxyride battery Not responsible for this one E° = 1.7 V Cathode: NiOOH, MnO 2 Anode:Zinc New battery Vacuum pouring technology Should last 2X longer than alkaline battery

3 Not responsible for this chemistry E° = 3.4 V Cathode: Li + (in CoO 2 ) + e - + CoO 2  LiCoO 2 Anode:Li(s) in polymer  Li + + e - Secondary Batteries (rechargeable) Lithium Ion battery

4 Secondary Batteries (rechargeable) Lead Acid battery E° = 2.04 V Anode:Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + + 2e - Cathode:PbO 2 (s) + HSO 4 - (aq) + 3H + + 2e -  PbSO 4 (s) + 2H 2 O Combine 6 cells to get 12 V

5 Fuel Cell (Proton Exchange Membrane – PEM) E° = 0.7 V Anode:H 2 (g)  2H + + 2e - Cathode:O 2 (g) + 2H 2 O + 4e -  4OH - Net: O 2 (g) + 2H 2 (g)  2H 2 O Big difference between fuel cell and battery is that the reactants can be replenished with fuel cell

6 Electrolysis Need to put in energy 2 kinds Molten Aqueous Overpotential (or overvoltage) Extra voltage need to overcome electron transfer limitations between electrode and solution

7 Electrolysis of molten salts Cathode:2Na + (l) + 2e -  2Na(l)E° = -2.71 V Anode:2Cl - (l)  Cl 2 (g) + 2e - E° = 1.36 V Net:2NaCl (l)  Cl 2 (g) + 2Na(l)E° = -4.07 V

8 Electrolysis of aqueous solutions What would happen if you put electricity into a solution of KI? Possible oxidation reactions: 2I - (aq)  I 2 (s) + 2e - E° = 0.535 V 6H 2 O  O 2 (g) + 4H 3 O + + 4e - E° = 1.229 V Possible reduction reactions: K + (aq) + e -  K(s) E° = -2.925 V 6H 2 O + 2e -  H 2 (g) + 2OH - E° = -0.8277 V Actual Reaction 2I - (aq) + 6H 2 O  I 2 (s) + H 2 (g) + 2OH - E° = -1.363 V

9 General rules in aqueous solutions Reduction 6H 2 O + 2e -  H 2 (g) + 2OH - E° = -0.8277 V A metal or other species has to have a reduction potential greater than -0.8277V, otherwise water will be the species reduced Oxidation 6H 2 O  O 2 (g) + 4H 3 O + + 4e - E° = 1.229 V A metal or other species has to have a reduction potential less than 1.229 V, otherwise water will be the species oxidized

10 Counting electrons


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