Batteries Self contained electrochemical cell •! Primary batteries (not rechargeable) •! Secondary batteries (rechargeable) •! Research Needed to Improve Batteries: CHEM112 LRSVDS Batteries and Corrosion 1 Dry Cell (Flashlight Battery) Anode: Zn(s) Zn(s) ! Zn2+(aq) + 2e- Cathode: NH4Cl + MnO2 + graphite paste 2NH4+(aq) + 2MnO2(s) + 2e- ! Mn2O3(s) + 2NH3(aq) + H2O CHEM112 LRSVDS Batteries and Corrosion 2 Battery Connection in Series Total emf is the sum of the individual emf’s CHEM112 LRSVDS Batteries and Corrosion 3 Dry Cell Battery: Alkaline Version ALKALINE CELL BATTERY: NH4Cl is replaced by KOH. Provides up to 50% more energy. Zn is used as a powder mixed with the electrolyte. Anode: Zn(s) + 2OH- ! Zn(OH)2(aq) + 2eCathode: 2H2O(s) + 2MnO2(s) + 2e- !2MnO(OH)(s) + 2OH-(aq) CHEM112 LRSVDS Batteries and Corrosion 4 Lead/Acid Batteries DURING DISCHARGE: Anode: Pb(s) + SO42-(aq) ! PbSO4(s) + 2e- E° Cathode: PbO2(s) + SO42-(aq) + 4H+ + 2e- ! PbSO4(s) + 2H2O Overall: Pb(s) + PbO2(s) + 2H2SO4 ! 2PbSO4(s) + 2H2O CHEM112 LRSVDS Batteries and Corrosion 5 12 V Automotive Lead-Acid Battery CHEM112 LRSVDS Batteries and Corrosion 6 Rechargeable Nickel-Cadmium Batteries Anode: Cd metal Cd(s) + 2OH-(aq) ! Cd(OH)2(s) + 2eCathode: NiO2(s) NiO2(s) + 2H2O + 2e- ! Ni(OH)2(s) + 2OH-(aq) Overall: E° = Cd(s) + NiO2(s) + 2H2O ! Cd(OH)2(s) + Ni(OH)2(s) PROBLEMS: CHEM112 LRSVDS Batteries and Corrosion 7 Batteries: Rechargeable NiMH: Metal Hydride During Discharge: Cathode: NiO(OH)(s) NiO(OH)(s) + 2H2O + 2e! ! Ni(OH)2(s) + 2OH! Anode: NiMH MH + OH! !H2O + M + e! NiMH= Nickel + metal alloy with dissolved H atoms M = ZrNi2 or LaNi5 (intermetallic compounds) E0cell = 1.2 V Advantages: Light weight, Last longer Uses: Batteries in hybrid cars CHEM112 LRSVDS Batteries and Corrosion 8 Batteries: Rechargeable Lithium Ion Battery During Discharge: Anode: solid (in graphite) Li(s) ! Li+ + eCathode : Li+ + CoO2 + e- ! LiCoO2 Overall: Li(s) + CoO2 ! LiCoO2 E0cell = 3.7 V Advantages: Light weight, high energy density Uses: Cell phones, laptops CHEM112 LRSVDS Batteries and Corrosion 9 Hydrogen Oxygen FUEL CELL Assets: !! !! !! !! Drawbacks: !! !! !! Anode: 2H2(g) + 4OH-(aq) ! 4H2O(l) + 4e- Cathode: O2(g) + 2H2O(l) + 4e- ! 4OH-(aq) Overall: 2H2(g) + O2(g) ! 2H2O(l) E0cell = 1.23 V CHEM112 LRSVDS Batteries and Corrosion 10 Sample Problem: What is Ecell for a fuel cell running in air (PO2 = 0.2 atm), at pH = 2, with PH2 = 1 atm? O2(g) + 4 H+(aq) + 4e! ! 2 H2O(l) 2 H+(aq) + 2e! ! H2(g) Eored = +1.229 V Eored = 0 CHEM112 LRSVDS Batteries and Corrosion 11 Methanol Fuel Cells Methanol can be made from CO2 + H2. Use of CO2 sequestered from power plants “Carbon neutral fuel” Anode: CH3OH (g) + H2O(g) ! CO2(g)+ 6H+ + 6e- Cathode: 3/2 O2(g) + 6H+ + 6e- ! 3H2O(g) Overall: CH3OH (g) + 3/2 O2(g) ! CO2(g)+ 2H2O(l) Assets: Drawbacks: CHEM112 LRSVDS Batteries and Corrosion 12 Corrosion: Spontaneous electrochemical process •! What is needed for corrosion to occur? (besides Fe) !! !! !! !! •! How do we prevent corrosion? !! !! CHEM112 LRSVDS Batteries and Corrosion 13 Corrosion E° E° Overall reaction in acid: CHEM112 LRSVDS Batteries and Corrosion 14 How Can Corrosion Be Prevented? Cathodic Protection CHEM112 LRSVDS Batteries and Corrosion 15 Corrosion CATHODIC PROTECTION OF IRON Galvanized Steel: Tin cans: CHEM112 LRSVDS Batteries and Corrosion 16 Corrosion CATHODIC PROTECTION OF IRON PIPES CHEM112 LRSVDS Batteries and Corrosion 17 ELECTROLYSIS Definition: An electrolytic cell consists of two electrodes in: Electrolysis forces the reaction to run in the reverse: The anode: The cathode: CHEM112 LRSVDS Batteries and Corrosion 18 COMMERCIAL APPLICATIONS OF ELECTROLYSIS: Production of metals Na Al Purification of Metals Cu Electroplating Ag Au CHEM112 LRSVDS Batteries and Corrosion 19 ELECTROLYSIS OF MOLTEN NaCl What are the reactions at the electrodes? E° Cathode: Anode: ___________________________ Overall: CHEM112 LRSVDS Batteries and Corrosion 20 ELECTROLYSIS OF AQUEOUS NaCl ********What species are present in the system? Which will be the CATHODE? E° Which will be the ANODE? What are the reactions at the electrodes? Cathode: Anode: ______________________________ Overall: What is left behind? THIS IS THE CHLOR-ALKALAI PROCESS; 3-5% OF WORLD’S ELECTRICITY USED TO DRIVE THIS REACTION CHEM112 LRSVDS Batteries and Corrosion 21 ELECTROLYSIS OF AQUEOUS Na2SO4 DOES THIS PROCESS PRODUCE Na(s)? ********What species are present in the system? Which will be the CATHODE? E° Which will be the ANODE? What are the reactions at the electrodes? Cathode: Anode: ______________________________ Overall: What is left behind? CHEM112 LRSVDS Batteries and Corrosion 22 Electrometallurgy of Aluminum Hall Process Electrolysis Cell is used to produce aluminum. Problem: Al2O3 melts at 2000°C. Why is this a problem? Solution: molten cryolite, Na3AlF6 What happens to a melting point when impurities are added? Anode: C(s) + 2O2-(l) ! CO2(g) + 4eCathode: 3e- + Al3+(l) ! Al(l) What happens to the graphite rods during the reaction? To produce 1000 kg of Al, we need: •! kg of bauxite, •! kg of cryolite, •! kg of C anodes •! of energy. CHEM112 LRSVDS Batteries and Corrosion 23 Electrorefining of Copper: What are the reactions at the electrodes? Anode: impure Cu ore; mixture of metals (Cu, Ni, Fe, Zn, Ag, Au, Pb...) Cathode: thin sheet of pure Cu As the electrolysis reaction proceeds, what happens to Cu? Other metals? CHEM112 LRSVDS Batteries and Corrosion 24 ELECTROLYSIS CALCULATIONS 1 mole of e- = charge of 1 Faraday = 96,485 Coulombs = charge on 1 mole of e1 Ampere = 1 coulomb/second 1 coulomb = 1 Amp-sec Electromotive Force (EMF) = Cell potential force that causes electrons to flow (voltage) 1 Watt = 1 Amp-Volt 1 Joule = 1 coul-Volt = 1 Amp-sec-Volt = 1 Watt-sec 1 kW-hour = (1000 Watt)(3600 sec) = 3.6 x 106 Watt-sec = 3.6 x 106 Joules! CHEM112 LRSVDS Batteries and Corrosion 25 ELECTROLYSIS CALCULATIONS If Electrolysis gives 1.00 g of Cu (63.54 g/mol) from CuSO4 according to the following reaction, How many Faradays (F) of charge are required? Cu2+ + 2e- ! Cu How many Coulombs of charge is this? CHEM112 LRSVDS Batteries and Corrosion 26 CALCULATIONS CONTINUED If 1.00 g of Cu is obtained in 1 hour using 3030.1 C of charge, how many amps of current are required? If 2 amps were used, how long would it take to produce 1 g of Cu? How can we calculate the maximum amount of work required to run an electrolytic cell?! CHEM112 LRSVDS Batteries and Corrosion 27