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)! Zn 2+ (aq) + 2e - Cathode: NH 4 Cl + MnO 2 + graphite paste 2NH 4+ (aq) + 2MnO 2 (s) + 2e -! Mn 2 O 3 (s) + 2NH 3 (aq) + H 2 O 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: NH 4 Cl 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) + 2e - Cathode: 2H 2 O(s) + 2MnO 2 (s) + 2e -!2MnO(OH)(s) + 2OH - (aq) CHEM112 LRSVDS Batteries and Corrosion 4
DURING DISCHARGE: Lead/Acid Batteries Anode: E Pb(s) + SO 4 2- (aq)! PbSO 4 (s) + 2e - Cathode: PbO 2 (s) + SO 4 2- (aq) + 4H + + 2e -! PbSO 4 (s) + 2H 2 O Overall: Pb(s) + PbO 2 (s) + 2H 2 SO 4! 2PbSO 4 (s) + 2H 2 O 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) + 2e - Cathode: NiO 2 (s) NiO 2 (s) + 2H 2 O + 2e -! Ni(OH) 2 (s) + 2OH - (aq) Overall: E = Cd(s) + NiO 2 (s) + 2H 2 O! Cd(OH) 2 (s) + Ni(OH) 2 (s) PROBLEMS: CHEM112 LRSVDS Batteries and Corrosion 7 NiMH: Batteries: Rechargeable Metal Hydride During Discharge: Cathode: NiO(OH)(s) NiO(OH)(s) + 2H 2 O + 2e!! Ni(OH) 2 (s) + 2OH! Anode: NiMH MH + OH!!H 2 O + M + e! NiMH= Nickel + metal alloy with dissolved H atoms M = ZrNi 2 or LaNi 5 (intermetallic compounds) E 0 cell = 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 + + e - Cathode : Li + + CoO 2 + e -! LiCoO 2 Overall: Li(s) + CoO 2! LiCoO 2 E 0 cell = 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: 2H 2 (g) + 4OH - (aq)! 4H 2 O(l) + 4e - Cathode: O 2 (g) + 2H 2 O(l) + 4e -! 4OH - (aq) Overall: 2H 2 (g) + O 2 (g)! 2H 2 O(l) E 0 cell = 1.23 V CHEM112 LRSVDS Batteries and Corrosion 10
Sample Problem: What is E cell for a fuel cell running in air (P O2 = 0.2 atm), at ph = 2, with P H2 = 1 atm? O 2 (g) + 4 H + (aq) + 4e!! 2 H 2 O(l) E o red = +1.229 V 2 H + (aq) + 2e!! H 2 (g) E o red = 0 CHEM112 LRSVDS Batteries and Corrosion 11 Methanol Fuel Cells Methanol can be made from CO 2 + H 2. Use of CO 2 sequestered from power plants Carbon neutral fuel Anode: CH 3 OH (g) + H 2 O(g)! CO 2 (g)+ 6H + + 6e - Cathode: 3/2 O 2 (g) + 6H + + 6e -! 3H 2 O(g) Overall: CH 3 OH (g) + 3/2 O 2 (g)! CO 2 (g)+ 2H 2 O(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 Definition: ELECTROLYSIS 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? E Which will be the CATHODE? 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 Na 2 SO 4 DOES THIS PROCESS PRODUCE Na(s)? ********What species are present in the system? E Which will be the CATHODE? 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: Al 2 O 3 melts at 2000 C. Why is this a problem? Solution: molten cryolite, Na 3 AlF 6 What happens to a melting point when impurities are added? Anode: C(s) + 2O 2- (l)! CO 2 (g) + 4e - Cathode: 3e - + Al 3+ (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 e - 1 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 10 6 Watt-sec = 3.6 x 10 6 Joules! CHEM112 LRSVDS Batteries and Corrosion 25 ELECTROLYSIS CALCULATIONS If Electrolysis gives 1.00 g of Cu (63.54 g/mol) from CuSO 4 according to the following reaction, How many Faradays (F) of charge are required? Cu 2+ + 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