Department of Chemistry University of Texas at Austin

Transcription

1 Electrochemistry Uit Activity RAQ KEY PART I. Lithium is used i two types of batteries, disposable lithium batteries ad rechargeable lithium io batteries. Lithium io is foud i the Earth s crust boud up i certai mierals ad dissolved i the ocea ad dissolved i the hot water that ca be tapped ito from geothermal jets. 1. Lookig at the reductio potetial table, state why oe expects to fid lithium i the atural world as a io ad ot as a free metal The battery idustry employs metallic lithium as the aode material for a wide class of disposable batteries called, lithium batteries. These batteries are lightweight ad have a much loger life tha the cheaper zic-copper cells. They are used i such varyig applicatios as watches, pacemakers ad remote lock car keys (fob). Due to the ever-expadig market for these batteries, there must be a well-accepted method of covertig Li io to Li metal. Lithium is foud at the bottom of the reductio table with the most egative value. It has the lowest reductio potetial. This meas that it is the easiest to oxidize. This also meas that lithium is very reactive ad ready to give up its electros to become more stable. Therefore, whe i cotact with ay of the other materials o the table it will always be spotaeously oxidized ito the lithium io. 2. Oe commo method is to isolate the lithium io i the form of a chloride salt from salt deposits ear geothermal vets. Give that you have isolated the salt, fully describe usig a properly labeled sketch, how you would produce lithium metal from the salt Iclude the type of electrochemical cell you would use to produce the lithium, complete with a descriptio of the half reactio occurrig at the aode, the cathode, the overall cell reactio, ad the miimum potetial ecessary to ru the cell. The lithium chloride salt has a formula uit of: LiCl We would eed to set up a electrolytic cell. First, the LiCl would eed to be molte i order for its ios to be mobile. I would melt solid LiCl ad place to iert electrodes (platium for example) ito the molte solutio. I would the apply a voltage across the electrodes. After ruig the cell, I would collect the solid Lithium at the bottom of my cotaier. Aode (Oxidatio): 2Cl - (l) Cl2 (g) + 2e - Cathode (Reductio): Li + (l) + 1e - Li (s) Overall: 2Cl - + 2Li + 2Li (s) + Cl2(g)

2 Miimum Potetial Necessary: The cell would have a egative potetial (electrolytic): E = E cathode - E aode E = (-3.05) (1.36) = V So, we would have to apply a potetial greater tha 4.41V. 3. Calculate the amout of time it would take to produce 100 g of Li operatig your cell at a curret of 1 Amp. 100 g Li 1 mol Li 2 mol e C 1 secod 6.94 g Li 2 mol Li 1 mol e - 1 C =1,390, secods (23,171.2 miutes, hours, 16.1 days, etc.) 4. Please describe a voltaic cell that could be made usig Li as the aode ad MO2 i a acidic paste i the presece of a graphite cathode. Iclude i your descriptio the balaced REDOX reactio. We would eed to set up two cotaiers, coected by a salt bridge ad a exteral wire. The aode cell would cotai a solid Lithium electrode i a solutio with lithium ios. The cathode cell would cotai the graphite cathode coated with the solid MO2 ad acidic paste i a acidic solutio. Aode: Li (s) Li e - Cathode: MO2 +4H + + 2e - M H2O Overall: 2Li +MO2 + 4H + 2Li + + M H2O

3 OR 5. State the stadard cell potetial of this cell. E = E cathode - E aode E = 1.21 (-3.05) = 4.26 V

4 6. Calculate cell potetial for a battery that has the followig cocetratios: Li io is 0.4 M, the acid cocetratio is 0.1M ad the M 2+ cocetratio is 0.01 M. E E O logq E E log(16) 2 E 4.22 log (0.4)2 (0.01) (0.1) 4 7. Calculate the free eergy of the cell uder stadard coditios. G = -FE G = -(2)(96485)(4.26) = -822 kj PART II. A fuel cell is a device for covertig chemical eergy directly to electrical eergy. Normally, whe a fuel, such as hydroge, atural gas (mostly methae), coal or petroleum, is bured heat eergy is liberated. The heat is the used directly, for example for cookig or heatig a home; or ca be coverted to mechaical eergy via the expasio of a gas. The mechaical eergy ca the be coverted to electrical eergy i a electrical geerator. I a fuel cell, the chemical eergy is coverted directly ito electrical eergy by separatig the reactats ad chaelig the electros through a exteral wire. Basically, the fuel is pumped ito oe half of the device, ad the oxidizig aget is added to the other side. The over all reactio for a hydroge fuel cell is: 2H2 + O2 2 H2O 1. What serves as the fuel? H2 2. What serves as the oxidizig aget? O2 (sice it is reduced) This type of fuel cell allows protos to move across the membrae, but electros are chaeled aroud a exteral wire, such that their flow ca be itercepted for doig electrical work. The diagram at the right shows part of a fuel cell. O oe side hydroge is beig pumped i, o the other side oxyge is beig pumped i. Protos cross a semi-permeable membrae, while the electros move out a exteral wire ad aroud to the other reactat.

5 3. Please redraw the diagram showig where the hydroge eters the cell, i what directio the protos flow, i what directio the electros flow ad where the oxyge eters the cell. Also label the aode ad the cathode ad predict whether or ot you thik those electrodes participate i the REDOX reactio. Label which side is udergoig oxidatio ad which side is udergoig reductio, ad write the correspodig half reactios. Electrodes are iert. The aode is the site at which hydroge gas is oxidized ito the hydroge io ad the cathode is the site at which oxyge gas is reduced ito the oxyge i a water molecule. Electrodes must be solid pieces of metal ad hydroge gas ad oxge gas are ot solids! So the iert electrodes simply provided a meas for the trasfer of electros to the reactig species. Oxidatio: H2 2H + + 2e - Reductio O2 + 4H + + 4e - 2H2O Overall: 2H2 + O2 2H2O 4. Assumig 100% efficiecy how much electrical work ca be doe by a fuel cell supplied with 1 Kg of hydroge ad excess oxyge? E = E cathode - E aode E = = 1.23 V -w = G = -FE G = -(4)(96485)(1.23) = -474,706 J/mol rx -w = G w = 474,706 J/mol rx

6 1000g H2 1 mol H2 1 mol Rx 474,706 J 2 g H2 2 mol H2 1 mol Rx =1.18 x 10 8 J for 1 kg H2 =1.18 x 10 5 kj for 1 kg H2 5. Calculate the value of the equilibrium costat for this cell reactio. Does the equilibrium lie to the reactat or product side? Would you expect the voltage to remai costat, to icrease or to decrease as this fuel cell is ruig? Please explai your aswer choice. E E O logq 0 E O logk E O logk logk 4 K Heavily o the products side! A large K meas the products are favored! As H2 ad O2 are cotiuously beig pumped ito the system, so the voltage should remai costat. We are ever chagig the cocetratios of the reactats because every time a little bit of these reactats are used up, we pump more ito the system!

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