(Solution Manual) Electrochemical Engineering

Fuller, 1

Engineering

Solution Manual

1st Edition

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Solution Manual for Electrochemical Engineering, The downloadable file consists of 532 pages, and the total size of the file is 3.77 MB. Purchase now only for highly discounted price.

Author(s): Fuller

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Solutions Manual for Electrochemical Engineering, 1e Thomas Fuller (All Chapters ) Chapter 1 Problem 1 1/1 Electrochemical Engineering , Thomas F. Fuller and John N. Harb 1.1 The original "International Ampere" was defined electrochemically as the current required to deposit 1.118 mg of silver per second from a solution of silver nitrate. Using this definition, how does the international ampere compare to the SI version? (Note that the SI version is based on the Ampere force law). 1.118 × 10−3g Ag s�mol Ag 107 .8682 g Ag�9648 5.33 C equivalent�1 equivalent mol Ag= 1. 00022 A The international Ampere is slightly larger than the SI one, which is the current required to create a force of 2 ×10 -7 N for two parallel wires separated by a distance of 1 m. J=1A I=1AI1m -*F=2xKT7NFile:problem 1-2.EES 2/9/2015 7:50:28 AM Page 1 EES Ver. 9.724: #2355: For use only by students and faculty in Chemical and Biological Engineering Georgia Tech. Chapter 1, problem 2 deposition of Mo from molten salt MW = 95.94 [g/mol] F = 96485 [coulomb/mol] m = 12.85 [g] I = 7 [coulomb/s] t = 3600 [s] m = I · t · MW F · z z=1.95 SOLUTION Unit Settings: SI C kPa kJ mass deg F = 96485 [coulomb/mol] I = 7 [coulomb/s] m = 12.85 [g] MW = 95.94 [g/mol] t = 3600 [s] z = 1.95 No unit problems were detected.File:problem 1-3.EES 2/9/2015 7:51:17 AM Page 1 EES Ver. 9.724: #2355: For use only by students and faculty in Chemical and Biological Engineering Georgia Tech. Chapter 1, problem 3 amount of hydrogen needed to run fuel cell F = 96485 [coulomb/mol] Power = 50000 [W] V = 0.7 [V] cell voltage Power = I · V t = 3600 · 3 · 1 [s] MW = 0.002 [kg/mol] molecular weight of hydrogen n = 2 m = I · t · MW F · n SOLUTION Unit Settings: SI C kPa kJ mass deg F = 96485 [coulomb/mol] I = 71429 [coulomb/s] m = 7.995 [kg] MW = 0.002 [kg/mol] n = 2 Power = 50000 [W] t = 10800 [s] V = 0.7 [V] No unit problems were detected.File:problem 1-4.EES 2/9/2015 7:52:00 AM Page 1 EES Ver. 9.724: #2355: For use only by students and faculty in Chemical and Biological Engineering Georgia Tech. Chapter 1, problem 4 Al production from Electrolysis MW = 26.982 [g/mol] F = 96485 [coulomb/mol] I = 200000 [coulomb/s] t = 24 · 3600 [s]  = 0.95 m =  · I · t · MW F · z z = 3 cf1 = 1000 [g/kg] prod rate = m cf1 SOLUTION Unit Settings: SI C kPa kJ mass deg cf1 = 1000 [g/kg]  = 0.95 F = 96485 [coulomb/mol] I = 200000 [coulomb/s] m = 1.530E+06 [g] MW = 26.98 [g/mol] prod rate = 1530 [kg] t = 86400 [s] z = 3 No unit problems were detected.File:problem 1-5.EES 2/9/2015 7:52:36 AM Page 1 EES Ver. 9.724: #2355: For use only by students and faculty in Chemical and Biological Engineering Georgia Tech. "!Chapter 1, problem 5" "Power for Chlor-alkali process"MW=70.9052[g/mol]; "molecular weight of Cl2" F=96485 [coulomb/mol] m=I*MW/(F*z) z=2cf1=24*365*3600 [s/year]prod_rate=45e12 [g/year]m=prod_rate/cf1 V=3.4 [V] P=I*V SOLUTION Unit Settings: SI C kPa kJ mass deg cf1 = 3.154E+07 [s/year] F = 96485 [coulomb/mol] I = 3.883E+09 [coulomb/s] m = 1.427E+06 [g/s] MW = 70.91 [g/mol] P = 1.320E+10 [W] prod rate = 4.500E+13 [g/year] V = 3.4 [V] z = 2 No unit problems were detected.File:problem 1-6.EES 2/9/2015 7:53:15 AM Page 1 EES Ver. 9.724: #2355: For use only by students and faculty in Chemical and Biological Engineering Georgia Tech. "!Chapter 1, problem 6" "deposition of Fe from molten salt"MW=55.845[g/mol]F=96485 [coulomb/mol]m=50 [g]I=25 [coulomb/s]m=I*t*MW/(F*z)z=3"calculate the volume of chlorine gas" MWg=70.9052 [g/mol]R=8.314 [J/mol-K]Tk=273 [K]p=1e5 [N/m^2] "from stoichiometry, get moles of Cl2"n=(m/MW)*(3/2) p*V=n*R*Tk SOLUTION Unit Settings: SI C kPa kJ mass deg F = 96485 [coulomb/mol] I = 25 [coulomb/s] m = 50 [g] MW = 55.85 [g/mol] MWg = 70.91 [g/mol] n = 1.343 [mol] p = 100000 [N/m2] R = 8.314 [J/mol-K] t = 10366 [s] Tk = 273 [K] V = 0.03048 [m3] z = 3 No unit problems were detected.File:problem 1-7.EES 2/9/2015 7:53:47 AM Page 1 EES Ver. 9.724: #2355: For use only by students and faculty in Chemical and Biological Engineering Georgia Tech. "!Chapter 1, problem 7" "Al production from Electrolysis"MW=26.982[g/mol]F=96485 [coulomb/mol] I=150000 [coulomb/s] t=24*(3600 [s]) eta=0.89m=eta*I*t*MW/(F*z)z=3cf1=1000 [g/kg]prod_rate=m/cf1 SOLUTION Unit Settings: SI C kPa kJ mass deg cf1 = 1000 [g/kg]  = 0.89 F = 96485 [coulomb/mol] I = 150000 [coulomb/s] m = 1.075E+06 [g] MW = 26.98 [g/mol] prod rate = 1075 [kg] t = 86400 [s] z = 3 No unit problems were detected.File:problem 1-8.EES 2/9/2015 7:54:52 AM Page 1 EES Ver. 9.724: #2355: For use only by students and faculty in Chemical and Biological Engineering Georgia Tech. "!Chapter 1, problem 8" "Electrodeposition of copper" "use a basis of 1 m2"A=1 [m^2] MW=0.063546[kg/mol] rho=8930 [kg/m^3] F=96485 [coulomb/mol] cd=1750 [A/m^2] I=cd*Aloading=1.22 [kg/m^2]m=loading*Aangle=165/360 d=loading/rho; "desired thickness of deposit" eta=0.95m=eta*I*t*MW/(F*z)z=2 cf=3600 [s/h] t=angle*cf/rotationrate SOLUTION Unit Settings: SI C kPa kJ mass deg A = 1 [m2] angle = 0.4583 cd = 1750 [A/m2] cf = 3600 [s/h] d = 0.0001366 [m]  = 0.95 F = 96485 [coulomb/mol] I = 1750 [coulomb/s] loading = 1.22 [kg/m2] m = 1.22 [kg] MW = 0.06355 [kg/mol]  = 8930 [kg/m3] rotationrate = 0.7404 [1/h] t = 2228 [s] z = 2 No unit problems were detected.File:problem 1-9.EES 2/9/2015 7:55:39 AM Page 1 EES Ver. 9.724: #2355: For use only by students and faculty in Chemical and Biological Engineering Georgia Tech. Chapter 1, problem 9 amount of lithium in battery MW = 6.941 [g/mol] molecular weight of Li Cap = 1.32 [A-h] cf1 = 3600 [coulomb/A-h] F = 96485 [coulomb/mol] m = Cap · cf1 · MW F · z z = 1 SOLUTION Unit Settings: SI C kPa kJ mass deg Cap = 1.32 [A-h] cf1 = 3600 [coulomb/A-h] F = 96485 [coulomb/mol] m = 0.3419 [g] MW = 6.941 [g/mol] z = 1 No unit problems were detected.File:problem 1-10.EES 8/6/2015 12:23:33 PM Page 1 EES Ver. 9.724: #2355: For use only by students and faculty in Chemical and Biological Engineering Georgia Tech. problem 1-10 corrosion of steel plate m = 50 [g] n = 2 F = 96485.33 [Coulomb/mol] m = I · t · MW n · F t = 3600 · 24 · 365 [s] MW = 55.845 [g/mol] SOLUTION Unit Settings: SI C kPa kJ mass deg F = 96485 [Coulomb/mol] I = 0.005479 [A] m = 50 [g] MW = 55.85 [g/mol] n = 2 t = 3.154E+07 [s] No unit problems were detected.File:problem 1-11.EES 8/6/2015 12:26:38 PM Page 1 EES Ver. 9.724: #2355: For use only by students and faculty in Chemical and Biological Engineering Georgia Tech. PROBLEM 1-11 accelerated corrosion i = 0.14 [A/m2] n = 2 F = 96485.33 [Coulomb/mol] m = i · t · A · MW n · F t = 360000 [s] A = 0.01 [m2] MW = 55.845 [g/mol] molecular weight of iron mr = 0.11 [g]  = mr mfaradaic efficiency cur = i · A oxygen evolved n1 = 4 mo = i · t · A · 1 –  n1 · F SOLUTION Unit Settings: SI C kPa kJ mass deg A = 0.01 [m2] cur = 0.0014 [A]  = 0.7542 F = 96485 [Coulomb/mol] i = 0.14 [A/m2] m = 0.1459 [g] mo = 0.000321 [mol] mr = 0.11 [g] MW = 55.85 [g/mol] n = 2 n1 = 4 t = 360000 [s] No unit problems were detected.File:problem 1-12.EES 8/6/2015 12:28:28 PM Page 1 EES Ver. 9.724: #2355: For use only by students and faculty in Chemical and Biological Engineering Georgia Tech. "!PROBLEM 1-12" r= 1e-9 [m]eo=8.854e-12 [farad/m]q=1.602e-19 [coulomb]W=q*q/(4*pi*eo*r) phi=W/q SOLUTION Unit Settings: SI C kPa kJ mass deg eo = 8.854E-12 [farad/m]  = 1.44 [V] q = 1.602E-19 [coulomb] r = 1.000E-09 [m] W = 2.307E-19 [J] No unit problems were detected.Chapter 2 Problem 2.1 1/1 Electrochemical Engineering, Thomas F. Fuller and John N. Harb 2.1 Write the associated electrochemical reactions and calculate the standard potential , Uθ from ∆Go for the following cells a. Chlor -alkali process to produce hydrogen and chlorine from a brine of NaCl (aqueous salt solution). Use the hydrogen reaction for an alkaline solution. b. Acetic acid/oxygen fuel cell with acidic electrolyte, where the acetic acid reacts to form liquid water and carbon dioxide. The reaction at the negative electrode is 2H2O+CH3COOH→2CO2+8H++8e− a) Cl 2+2e−↔2Cl− H2+2OH−↔2H2O+2e− _________________________________ overall H2+Cl2+2NaOH↔2H2O+2NaCl ∆𝐺𝑅𝑅=∆𝐺𝑓products−∆𝐺𝑓reactants ∆𝐺𝑅𝑅𝑜=2∆𝐺𝑓,NaCl𝑜+2∆𝐺𝑓,H2O𝑜−2∆𝐺𝑓,NaOH𝑜 ∆𝐺𝑅𝑅𝑜=2{−393. 1−237 .19−(−419. 2)}=−422 .34 kJ mol−1 𝑈𝜃=−∆𝐺𝑅𝑅𝑜 𝑛𝑛=4223440 (2)96485=2.188 V b) multiply second reaction by 2 to balance electrons and add together 2H 2O+CH3COOH→2CO2+8H++8e− 4H++4e−+O2↔2H2O _________________________________ overall CH3COOH +2O2↔2CO2+2H2O ∆𝐺𝑅𝑅𝑜=2∆𝐺𝑓,CO2𝑜+2∆𝐺𝑓,H2O𝑜−∆𝐺𝑓,CH3COOH𝑜 ∆𝐺𝑅𝑅𝑜={−(2)394. 39−(2)237. 19−(−389 )}=−874 .18 kJ mol−1 𝑈𝜃=−∆𝐺𝑅𝑅𝑜 𝑛𝑛=874180 (8)96485=1.113 V Chapter 2 Problem 2.2 1/1 Electrochemical Engineering, Thomas F. Fuller and John N. Harb Does the redox reaction as written below proceed spontaneously at 25 °C and standard conditions? 2Ag++H2→2Ag +2H+ Ag ++e−↔Ag (Uθ=0.7991 V) 1 2H2↔H++e− (Uθ=0 V) _________________________________ overall Ag++12H2→Ag+H+ (U=0.7991 V) 𝑈𝜃=−∆𝐺𝑅𝑅𝑜 𝑛𝑛=0.7991 V Thus, ∆G is less than zero, and the reaction would proceed spontaneously as written at standard conditions. Chapter 2 Problem 2.3 1/1 Electrochemical Engineering, Thomas F. Fuller and John N. Harb What is the standard half -cell potential for the oxidation of methane under acidic conditions? The reaction for methane is as follows: CH4(g)+2H2O(ℓ)→CO2+8H++8e− Which element is oxidized and how does its oxidation state change? CH 4(g)+2H2O(ℓ)→CO2+8H++8e− (Uθ=? V) 8H++8e−↔4H2 (Uθ=0 V) _________________________________ overall CH4(g)+2H2O(ℓ)→CO2+4H2 Find the change in standard Gibbs energy of formation for the overall reaction ∆𝐺𝑅𝑅𝑜=∆𝐺𝑓,CO2𝑜−2∆𝐺𝑓,H2O𝑜−∆𝐺𝑓,CH4𝑜 ∆𝐺𝑅𝑅𝑜={−394 .359−2(−237. 19)−(−50.5)}=−130. 52 kJ mol−1 𝑈𝜃=−∆𝐺𝑅𝑅𝑜 𝑛𝑛=130520 8𝑛=0.1691 V The oxidation state of carbon changes. Chapter 2 Problem 2.4 1/1 Electrochemical Engineering, Thomas F. Fuller and John N. Harb What is the standard cell potential for a methane /oxygen fuel cell? The oxidation of methane produces CO 2 as shown in problem 2.3, but here assume the product water is a gas, rather than a liquid. CH 4(g)+2H2O(g)→CO2+8H++8e− (Uθ=? V) 8H++8e−+2O2↔4H2O(g) (Uθ=0 V) _________________________________ overall CH4(g)+2O2→CO2+2H2O(g Find the change in standard Gibbs energy of formation for the overall reaction ∆𝐺𝑅𝑅𝑜=∆𝐺𝑓,CO2𝑜+2∆𝐺𝑓,H2O𝑜−∆𝐺𝑓,CH4𝑜 ∆𝐺𝑅𝑅𝑜={−394 .359 +2(−228. 572 )−(−50.5)}=−801 kJ mol−1 𝑈𝜃=−∆𝐺𝑅𝑅𝑜 𝑛𝑛=801000 8𝑛=1.038 V Chapter 2 Problem 2.5 1/1 Electrochemical Engineering, Thomas F. Fuller and John N. Harb Let’s consider the oxidation of methane in a fuel cell that utilizes an oxygen conductor (O2-) rather than a proton conductor as the electrolyte. a. At which electrode (oxygen or methane) is O2- produced and at which is it consumed? b. In which direction does O2- move through the electrolyte? Why? c. Propose two electrochemical half -cell reactions. d. Does θU change for this fuel cell relative to a fuel cell that utilizes a proton conductor? Why or why not? a. O 2- is produced at the cathode; and O2- is consumed at the anode O2+4 e−→2O2− b. oxygen ions move from the cathode to the anode, since they are negatively charged this still represents a positive current from the anode to the cathode. c. CH 4(g)+4O2−→CO2+2H2O+8e− 2O2+8 e−→4O2− _________________________________ overall CH4(g)+2O2→CO2+2H2O(g) d. No, the standard potential depends on the overall reaction and the reference pressure, but since the overall reaction is the same regardless of the ion conductor. Chapter 2 Problem 2.6 1/1 Electrochemical Engineering, Thomas F. Fuller and John N. Harb Determine the equilibrium potential of the cell shown below. The reaction at positive electrode (right) is AgCl+e−↔Ag+Cl− Uθ=0.222 V and for the negative (left) H2↔2H++2e− Uθ=0 _________________________________ overall H2+2AgCl↔2Ag+2HCl Uθ=0.222 V where the overall potential for the cell is the positive (right) – negative (left). 𝑈=𝑈𝜃 cell−𝑅𝑅 𝑛𝑛ln∏𝑎𝑖𝑠𝑖, (2-11) 𝑈=𝑈𝜃 cell−𝑅𝑅𝑛𝑛ln𝑎Ag2𝑎HCl2 𝑎H2𝑎AgCl2, The activity of solids are assumed to be one, and n =2 𝑈=𝑈𝜃 cell+𝑅𝑅 2𝑛ln𝑎H2−𝑅𝑅 𝑛ln𝑎HCl, Hydrogen is a gas, assume ideal 𝑎H2=𝑓H2 𝑓𝑜=𝑝H2 𝑝𝑜 For the aqueous HCl 𝑎HCl=𝑎+𝜈+𝑎−𝜈−=𝑚+𝛾+𝑚−𝛾−=𝑚2𝛾±2 substituting gives 𝑈=𝑈𝜃 cell+𝑅𝑅 2𝑛ln�𝑝H2 𝑝𝑜�−2𝑅𝑅 𝑛ln�𝑚𝛾±� αβ Pt(s), H2(g) HCl(aq) Ag(s)α AgCl (s)γChapter 2 Problem 2.7 1/2 Electrochemical Engineering, Thomas F. Fuller and John N. Harb Consider the electrochemical reactions shown below. Mercury(I) chloride, also known as calomel , is a solid used in reference electrodes. The two reactions are − ++↔ 2e Zn Zn2 2Hg 2Cl 2e ClHg2 2 +↔+− − a. What is the overall chemical reaction? b. Develop an expression for U , the equilibrium potential of the cell. c. Write down an expression for the standard potential of the cell in terms of the standard Gibbs energies of formation. d. Use standard half -cell potentials from the table to determine the sta ndard Gibbs energy of formation for aqueous ZnCl 2. Why is this value different than the value for solid ZnCl 2? e. What is the standard Gibbs energy of formation for Hg 2Cl2? a) The reacti on at positive electrode is Hg2Cl2+2e−↔2Hg +2Cl− Uθ=0.2676 V and for the negative Zn↔Zn2++2e− Uθ=-0.7618 V _________________________________ overall Hg2Cl2+Zn↔2Hg +Zn2++2Cl− Uθ=1.029 V b) 𝑈=𝑈𝜃 cell−𝑅𝑅 𝑛𝑛ln∏𝑎𝑖𝑠𝑖, (2-11) 𝑈=𝑈𝜃 cell−𝑅𝑅𝑛𝑛ln𝑎Hg2𝑎ZnCl2 𝑎Zn𝑎Hg2Cl2, The activity of solid and liquid Hg are assumed to be one, and n =2 𝑈=𝑈𝜃 cell−𝑅𝑅 2𝑛ln𝑎ZnCl2 , ν+=1, ν-=2, ν =3 For the aqueous ZnCl 2 𝑎ZnCl2=𝑎+𝜈+𝑎−𝜈−=𝑚+𝛾+𝑚−2𝛾−2=4𝑚3𝛾±3 substituting gives 𝑈=𝑈𝜃 cell−𝑅𝑅 𝑛ln2−3𝑅𝑅 2𝑛ln�𝑚𝛾±� Chapter 2 Problem 2.7 2/2 Electrochemical Engineering, Thomas F. Fuller and John N. Harb c,d) From the table of standard potentials Cl2+2e−↔2Cl− Uθ=1.3595V Zn↔Zn2++2e− Uθ=-0.7618 V Given that the solids and liquids are in their standard state, we can right the change in Gibbs energy as ∆𝐺𝑅𝑅𝑜=∆𝐺𝑓,ZnCl2𝑜=−𝑛𝑛𝑈𝜃=−2(96485 )(2.121 𝑉)=−409 .3 kJ e) Hg 2Cl2+2e−↔2Hg +2Cl− Uθ=0.2676 V Zn↔Zn2++2e− Uθ=-0.7618 V _________________________________ Zn+Hg2Cl2↔ZnC l2(aq)+2Hg Uθ=1.029V ∆𝐺𝑅𝑅𝑜=−𝑛𝑛𝑈𝜃=−198, 740 J mol−1 ∆𝐺𝑅𝑅𝑜=∆𝐺𝑓,ZnCl2𝑜−∆𝐺𝑓,Hg2Cl2𝑜=−210 .6 kJ mol−1 The values are different because the standard states are different for a solid versus solution. There is some change in energy and entropy involved in the dissolution of the solid. Chapter 2 Problem 2.8 1/1 Electrochemical Engineering, Thomas F. Fuller and John N. Harb The lithium air cell offers the possibility of a very high energy battery. At the negative electrode, -e Li Li +↔+ . At the positive electrodes the following reactions are postulated 2Li++O2+2e−↔Li2O2 . 2Li++1 2O2+2e−↔Li2O . Estimate the standard potential for each of the two possible reactions at the positive electrode paired with a lithium anode. The reacti on at positive electrode is 2Li++O2+2e−↔2Li2O2 and for the negative 2Li↔2Li++2e− _________________________________ overall 2Li+O2↔2Li2O2 ∆𝐺𝑅𝑅𝑜=∆𝐺𝑓,Li2O2𝑜=−571 .116 kJ=−𝑛𝑛𝑈𝜃 𝑈𝜃=−∆𝐺𝑅𝑅𝑜 𝑛𝑛=−571116 2𝑛=2.96 V For the second reaction, at the positive electrode is 2Li++1 2O2+2e−↔Li2O and for the negative 2Li↔2Li++2e− _________________________________ overall 2Li+12O2↔Li2O ∆𝐺𝑅𝑅𝑜=∆𝐺𝑓,Li2O𝑜=−561 .2 kJ=−𝑛𝑛𝑈𝜃 𝑈𝜃=−∆𝐺𝑅𝑅𝑜 𝑛𝑛=−571200 2𝑛=2.91 V Chapter 2 Problem 2.9 1/1 Electrochemical Engineering, Thomas F. Fuller and John N. Harb Develop an expression for the equilibrium potential for the cell below. The first reaction is the negative electrode of the Edison cell (battery). Fe+2OH−→Fe(OH)2+2𝑒− − −→++ 4OH OH2 e4 O2 2 The Gibbs energy of formation for Fe(OH) 2 is -486.6 kJ/mol. The reacti on at positive electrode is O2+4e−+2H2O↔4OH− Uθ=0.401 V and for the negative Fe+2OH−→Fe(OH)2+2e− Uθ=? _________________________________ overall 2Fe +O2+2H2O↔Fe(OH)2 𝑈=𝑈𝜃 cell−𝑅𝑅 𝑛𝑛ln∏𝑎𝑖𝑠𝑖, (2-11) 𝑈=𝑈𝜃 cell−𝑅𝑅𝑛𝑛ln𝑎Fe(OH)22 𝑎Fe2𝑎O2𝑎H2O2, 𝑈=𝑈𝜃 cell+𝑅𝑅 2𝑛ln𝑎H2O+𝑅𝑅 4𝑛ln�𝑝𝑂2 𝑝𝑜� 𝑈𝜃=−∆𝐺𝑅𝑅𝑜 𝑛𝑛 ∆𝐺𝑅𝑅𝑜=2∆𝐺𝑓,Fe(OH)2𝑜−2∆𝐺H2O𝑜=2{−486 ,600−(−237, 2=129 )}=−499 .5 kJ mol−1 𝑈𝜃=−∆𝐺𝑅𝑅𝑜 𝑛𝑛=499500 (4)96485=1.29 V 𝑈=1.29+𝑅𝑅 2𝑛ln𝑎H2O+𝑅𝑅 4𝑛ln�𝑝𝑂2 𝑝𝑜� Chapter 2 Problem 2.10 1/1 Electrochemical Engineering, Thomas F. Fuller and John N. Harb Develop an expression for the equilibrium potential of a hydrogen- oxygen fuel cell operating under acidic conditions. The two electrochemical reactions are H2↔2H++2e− and O2+4H++4e−↔2H2O Use the data in the Appendix C for standard Gibbs energy of formation. Compare with the value calculated from standard electrode potentials to identify whether the standard state for water in the table of Appendix A is liquid or gas. The reacti on at positive electrode is O 2+4H++4e−↔2H2O Uθ=1.229 V and for the neg ative H2↔2H++2e− Uθ=0 V _________________________________ overall H2+1 2O2↔H2O 𝑈=𝑈𝜃 cell−𝑅𝑅𝑛𝑛ln∏𝑎𝑖𝑠𝑖, (2-11) 𝑈=𝑈𝜃 cell+𝑅𝑅𝑛𝑛ln𝑎O21/2𝑎H2 𝑎H2O, 𝑈=𝑈𝜃 cell+𝑅𝑅 2𝑛ln�𝑝𝐻2 𝑝𝑜��𝑝𝑂2 𝑝𝑜�0.5 𝑎H2O 𝑈𝜃=−∆𝐺𝑅𝑅𝑜 𝑛𝑛=1.229 V, where n =2 Therefore, ∆𝐺𝑅𝑅𝑜=−𝑛𝑛𝑈𝜃=−2(96485 )1.229 =−237 kJ mol−1 From the Appendix C liquid ∆𝐺𝑓,H2O𝑜=−237 kJ mol−1 gas ∆𝐺𝑓,H2O𝑜=−229 kJ mol−1 Thus, the 1.229 V corresponds to liquid water Chapter 2 Problem 2.11 1/1 Electrochemical Engineering, Thomas F. Fuller and John N. Harb Write the associated electrochemical reactions, and calculate the standard potential , Uθ, from ∆Go for the following cells a. Propane fuel cell with solid oxygen conductor electrolyte b. Electrolysis of aluminum, where aluminum is produced from Al 2O3 and carbon. Note that carbon is oxidized at the anode. a) The reacti on at positive electrode is 5O2+20e−↔10O2− and for the negative C3H8+10O2−↔3CO2+4H2O+20e− _________________________________ overall C3H8+5O2↔3CO2+4H2O Since the oxygen ion does not appear in the overall reaction, the standard potential does not depend on its activity. ∆𝐺 𝑅𝑅𝑜=−𝑛𝑛𝑛𝜃=3∆𝐺𝑓,CO2𝑜+4∆𝐺𝑓,H2O𝑜−∆𝐺𝑓,C3H8𝑜 ∆𝐺𝑅𝑅𝑜=−𝑛𝑛𝑛𝜃=3(−394. 3)+4(−228. 6)−(−24.3)=−2070 kJ mol−1 𝑛𝜃=−2,070 ,000 20𝐹=1.074 V b) The overall reaction is 2Al 2O3+3C↔4Al +3CO2 . ∆𝐺𝑅𝑅𝑜=−𝑛𝑛𝑛𝜃=3∆𝐺𝑓,CO2𝑜−2∆𝐺𝑓,Al2O3𝑜 ∆𝐺𝑅𝑅𝑜=−𝑛𝑛𝑛𝜃=3(−394. 3)−2(−1582 .3) 𝑛𝜃=−1,981,500 12𝐹=1.711 V The actual reactions are complex, here is one set of reactions 3C+6O 2−↔3CO2+12e− 2Al2O3+12e−↔4Al +6O2−