This reaction is not spontaneous at standard conditions because o cell E 0 no

This reaction is not spontaneous at standard

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This reaction is not spontaneous at standard conditions because ocellE< 0; no reaction occurs. b. Possible reaction: Cl2(g) + 2 I(aq) I2(s) + 2 Cl(aq) ocellE= 0.82 V; this reaction is spontaneous at standard conditions because ocellE> 0. The reaction will occur. Cl2(g) + 2 I(aq) I2(s) + 2 Cl(aq) ocellE= 0.82 V = 0.82 J/C ΔG° = ocellnFE= (2 mol e-)(96,485 C/mol e-)(0.82 J/C) = 1.6 × 105J = 160 kJ E° = n0591.0log K, log K = 0591.0)82.0(20591.0nEo= 27.75, K = 1027.75= 5.6 × 1027c. Possible reaction: 2 Ag(s) + Cu2+(aq) Cu(s) + 2 Ag+(aq) ocellE= 0.46 V; no reaction occurs. d. Fe2+can be oxidized or reduced. The other species present are H+, SO42, H2O, and O2from air. Only O2in the presence of H+has a large enough standard reduction potential to oxidize Fe2+to Fe3+(resulting in ocellE> 0). All other combinations, including the possible reduction of Fe2+, give negative cell potentials. The spontaneous reaction is: 4 Fe2+(aq) + 4 H+(aq) + O2(g) 4 Fe3+(aq) + 2 H2O(l) ocellE= 1.23 0.77 = 0.46 V ΔG° = ocellnFE= (4 mol e)(96,485 C/mol e)(0.46 J/C)(1 kJ/1000 J) = 180 kJ log K = 0591.0)46.0(4= 31.13, K = 1.3 × 103186. a. Cu++ eCu E° = 0.52 V Cu+Cu2++ eE° = 0.16 V 2 Cu+(aq) Cu2+(aq) + Cu(s) ocellE= 0.36 V; spontaneous ΔG° = ocellnFE= (1 mol e)(96,485 C/mol e)(0.36 J/C) = 34,700 J = 35 kJ ocellE= n0591.0log K, log K = 0591.0)36.0(10591.0nEo= 6.09, K = 106.09= 1.2 × 106 b. Fe2++ 2 eFe E° = 0.44 V (Fe2+Fe3++ e) × 2 E° = 0.77 V 3 Fe2+(aq) 2 Fe3+(aq) + Fe(s) ocellE= 1.21 V; not spontaneous
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CHAPTER 18 ELECTROCHEMISTRY 713 c. HClO2+ 2 H++ 2 eHClO + H2O E° = 1.65 V HClO2+ H2O ClO3+ 3 H++ 2 eE° = 1.21 V 2 HClO2(aq) ClO3(aq) + H+(aq) + HClO(aq) ocellE= 0.44 V; spontaneous ΔG° = ocellnFE= (2 mol e)(96,485 C/mol e)(0.44 J/C) = 84,900 J = 85 kJ log K = 0591.0)44.0(20591.0nEo= 14.89, K = 7.8 × 1014 87. (Cr2+Cr3++ e) × 2 Co2++ 2 eCo 2 Cr2+(aq) + Co2+(aq) 2 Cr3+(aq) + Co(s) ocellE= n0591.0log K = 20591.0log(2.79 × 107) = 0.220 V E = E° ]Co[]Cr[]Cr[logn0591.022223= 0.220 V )20.0()30.0()0.2(log20591.022= 0.151 V ΔG = nFE = (2 mol e)(96,485 C/mol e)(0.151 J/C) = 2.91 × 104J = 29.1 kJ 88. 2 Ag+(aq) + Cu(s) Cu2+(aq) + 2 Ag(s) ocellE= 0.80 0.34 = 0.46 V and n = 2 Because [Ag+] = 1.0 M, Ecell= 0.46 V 20591.0log [Cu2+]. Use the equilibrium reaction to calculate the Cu2+concentration in the cell. Cu2+(aq) + 4 NH3(aq) Cu(NH3)42+(aq) K = 432243]NH][Cu[])NH(Cu[= 1.0 × 1013From the problem, [NH3] = 5.0 Mand [Cu(NH3)42+] = 0.010 M: 1.0 × 1013= 42)0.5](Cu[010.0, [Cu2+] = 1.6 × 1018MEcell= 0.46 20591.0log (1.6 × 1018) = 0.46 (0.53) = 0.99 V 89. The Ksp reaction is FeS(s) Fe2+(aq) + S2(aq) K = Ksp. Manipulate the given equations so that when added together we get the Kspreaction. Then we can use the value of ocellEfor the reaction to determine Ksp(by using the equation log K = nE/0.0591). FeS + 2 eFe + S2E° = 1.01 V Fe Fe2++ 2 eE° = 0.44 V Fe(s) Fe2+(aq) + S2(aq) ocellE= 0.57 V
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CHAPTER 18 ELECTROCHEMISTRY 714 log Ksp= 0591.0)57.0(20591.0nEo= 19.29, Ksp= 29.1910= 5.1 × 201090. Al3++ 3 eAl E° = 1.66 V Al + 6 FAlF63-+ 3 eE° = 2.07 V Al3+(aq) + 6 F(aq) AlF63(aq) ocellE= 0.41 V K = ? log K = 0591.0)41.0(30591.0nEo= 20.81, K = 1020.81= 6.5 × 1020 91. e+ AgI Ag + IoAgIE= ?
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