Department of Chemistry University of Texas at Austin

Transcription

1 Electrochemistry Uit Activity Cocetratio of Electrochemical Cells KEY MINI LECTURE CONCENTRATION ELECTROCHEMICAL CELLS Cells are remarkably complex systems. However, we ca uderstad some key characteristics ad fuctios of cells from simple models. Oe of the most remarkable aspects of may cells is that they utilize eergy to maitai differet cocetratios of ios iside the cell compared to the solutio outside. This differece i ios the creates a electrical potetial that the cells ca use for sigalig (ad which ca be cotrolled via io cocetratios). For a typical muscle cell i your heart there are three key ios that have cocetratios differeces iside ad outside of the cell: Na +, K +, ad Ca 2+. Let s look at the simplest situatio of just the K + ios. The cocetratio of potassium io iside the cell is higher tha outside: [K + ]i > [K + ]out. Therefore K + should be costatly diffusig out of the cell. However, the egative couter ios (chloride, proteis,.) caot pass through the membrae. This meas that as the K + diffuses out a egative charge builds up i the cell. At some poit, equilibrium is reached where the drivig force for the outward diffusio (the cocetratio differece) is balaced by the electric force from the potetial differece. This is the rest potetial of the cell. We ca determie the rest potetial for the K + from the Nerst equatio. The reactio is simply the diffusio the K + K + i K + out 1. For this reactio, what is the reactio quotiet Q? Q K out K i 2. For this reactio what is the std potetial? Sice the two half reactios are the same AND the cocetratios of the ios uder stadard coditios are both 1M (aka the same cocetratios) the, the stadard potetial is zero. Nobody wis the free eergy game. That is, there is NO potetial differece. 3. Usig the Nerst equatio, write a expressio for the potetial based o the cocetratios iside ad outside the cell. (ote: for physiological coditios the costat i the Nerst equatio is slightly larger as the temperature is higher). We ca use the room temperature Nerst Equatio:

2 E E O RT F l Q E RT F l [K out] E log [K out] Alteratively, we ca adjust the Nerst Equatio to accout for body temperature which is 37 C or 310K: E E O RT F l Q E RT F OR l [K out] J E E O molk 310K C mole E E O E ad log) log [K out] log [K out] [K out] 2.3 log (2.3 is the coversio betwee l 4. How large a potetial ca you build up (polarize)? It depeds o the cocetratio differece. Give the typical cocetratios for a muscle cell i your heart are [K + ]i = 150 mm ; [K + ]out = 4 mm, what is this potetial? E logq E log E V E 97mV (Covertig betwee mm ad M does ot matter because the uits cacel out iside the logarithm expressio ayway.) MINI LECTURE ELECTROLYTIC CELLS

3 Alumium metal is produced primarily from reductio of alumium oxide via electrolysis. The actual process ivolves a umber of compouds icludig Na3AlF6 ad Al2O3 but the geeral process ca be simplified to be: Al e - Al C + 2O 2- CO2 + 4e - 1. What is the overall balaced equatio for this reactio? What reactio would occur at the aode? The cathode? Our goal is to produce alumium metal. So alumium metal must be oe of our products. The other reactio must iclude electros o the opposite side of the equatio from the alumium reductio so the two half reactios would be: Reductio ½ Reactio: Al e - Al Oxidatio ½ Reactio: C + 2O 2- CO2 + 4e - The two half-reactios are balaced i terms of elemets ad charge idividually. I order to combie them though, we must elimiate the electros by multiplyig the reactios with coefficiets. The lowest commo multiple of 3 ad 4 is 12. Reductio ½ Reactio: 4x(Al e - Al) Oxidatio ½ Reactio: 3x(C + 2O 2- CO2 + 4e - ) Combie: Reductio ½ Reactio: 4Al e - 4Al Oxidatio ½ Reactio: 3C + 6O 2-3CO2 + 12e - Balaced Overall Reactio: 4Al C + 6O 2-4Al + 3CO2 Oxidatio ALWAYS occurs at the aode. The oxidatio of carbo would occur: At Aode: C + 2O 2- CO2 + 4e - Reductio ALWAYS occurs at the cathode. The reductio of alumium would occur. At Cathode: Al e - Al 2. What total charge, i Coulombs, is eeded to produce 1 kg of Al (27 g mol -1 )? This is just a dimesioal aalysis problem: 1 kg Al 1000 g Al 1 mol Al 3 mol e - 96,485 C 1 kg Al 27 g Al 1 mol Al 1 mol e - = C = MC (mega Coulombs)

4 We could have used the balaced equatio to compare moles of Al to moles of electros i fourth colum of this dimesioal aalysis. I the balaced equatio, 4 moles of Alumium are reduced whe 12 moles of electros are beig used. However, the ratio of 4 moles Al to 12 moles electros is the SAME ratio as 1 mole Al to 3 moles electros. 4:12 = 1:3. 3. Assumig you wated to produce 1 kg of Al i oe hour, what curret would be required? C Curret, I, is measured i Amps which has uits (Coulombs per secod). So the s total charge would be the curret multipled by the time i secods. Oe hour has 3,600 secods (60 secodsx60 miutes = 1 hour). Q = (I)(t) I = Q/t I = ( C) (3600 secods) = Amps 4. The electrical power eeded is the curret times the voltage of the electrochemical cell (Amps x Volts = Watts). This reactio eeds to be ru with a applied potetial of approximately 5 V to efficietly drive the electrolysis. For the give curret, how much power would be eeded to produce the 1 kg of Al? Power = (I)(V) = ( Amps)(5V) = Watts 5. Electrical eergy is sold i uits of kw-hr (power x time). Give that you used the above power for 1 hr, how may kw-hr of eergy did you eed? Power = Watts = kw Eergy = (P)(t) = ( kw)(1hr) = kw-hr 6. The cost of 1 kw-hr varies greatly, but assumig it is approximately $0.10/kWhr, how much would electricity cost aloe be for the productio of your 1 kg of Al? Cost = (Eergy)(Price) = ( kw-hr)(0.10$) = $ The commodities cost of Al is curretly about $1.50/kg. What fractio of this cost is the electricity? Fractio of Electricity per Total Cost = ($1.49)/($1.50) = or about 99.26% 8. I the productio of your 1 kg of Al, how may kg of carbo are cosumed? 1 kg Al 1000 g Al 1 mol Al 3 mol C 12 g C 1 kg C

5 1 kg Al 27 g Al 4 mol Al 1 mol C 1000 g C = 0.33 kg C