1. (a)	Because Al3+ and V (s) overlap on the predominance diagram, it is theoretically possible to completely reduce VO2+ using Al as the reducing agent. 5 Al (s)  +  3 VO2+ (aq)  +  12 H+ (aq)  --->  5 Al3+ (aq)  +  3 V (s)  +  6 H2O (aq) For the 1/2 reaction VO2+ (aq)  +  4 H+ (aq)  +  5 e-  --->  V (s)  +  2 H2O (aq)      Eo = [(1 V)(1) + (0.33 V)(1) + (-0.25 V)(1) + (-1.13 V)(2)] / 5 = -0.236 V Therefore, ΔEorxn = 1.65 V - 0.236 V = 1.41 V
1. (b)	1.25 g of Al will release (1.25 g Al)*(1 mol Al / 26.96 g Al)*(3 mol e- / 1 mol Al) = 0.139 mol e- 4.00 g of VO2Cl is (4.00 g) / (118.39 g/mol) = 0.0338 mol VO2+ Reduction to VO2+ would take 0.0338 mol of electrons, leaving 0.139 - 0.0338 = 0.105 mol Reduction to V3+ would take another 0.0338 mol of electrons leaving 0.105 - 0.0338 = 0.071 mol Reduction to V2+ would take another 0.0338 mol of electrons leaving 0.071 - 0.0338 = 0.038 mol Reduction to V (s) would take 2*0.0338 = 0.0676 mol of electrons, which is more than we have left! Therefore, the products of the reaction would be a mixture of V2+ and V (and Al3+).
1. (c)	For the reduction of Sb2O5 to elemental Sb, 3 Sb2O5 (s)  +  10 Al (s)  +  30 H+ (aq)  --->  6 Sb (s)  +  10 Al3+ (aq)  +  15 H2O (l) (3.00 g Sb2O5) * (1 mol Sb2O5/323.50 g Sb2O5) * (10 mol Al / 3 mol Sb2O5) * (26.96 g Al / 1 mol Al)      = 0.833 g of Al required For the 1/2 reaction, Sb2O5 (s)  +  10 H+ (aq)  +  10 e-  --->  2 Sb (s)  +  5 H2O (l)      Eo = [(0.70 V)(2) + (0.15 V)(3)] / 5 = 0.37 V Therefore, ΔEorxn = 1.65 + 0.37 = 2.02 V
2. (a)	For the disproportionation reaction, 4 RuO4- (aq)  +  4 H+ (aq)  --->  3 RuO4 (s)  +  RuO2 (s)  +  2 H2O (l) ΔEorxn = 1.53 V - 0.99 V = 0.54 V Because this is positive, the reaction as written is spontaneous, and RuO4- is unstable relative to its disproportionation products.
2. (b)	Converting to pH 7, For the 1/2 reaction, RuO4- (aq)  --->  3 RuO4 (s)  +  e- Eo remains the same (-0.99 V) For the 1/2 reaction, RuO4- (aq)  +  4 H+ (aq)  +  3 e-  --->  RuO2 (s)  +  2 H2O (l) Eo = 1.53 - (4/3)(0.0592)(7) = 0.977 V Therefore, ΔEorxn = 0.977 - 0.99 = -0.013 V Because this is negative, the disproportionation reaction is no longer spontaneous.  Thus, the anion RuO4- is stable under these conditions.
3.	First reduction of Co releases 2 e- for each Co: Co (s)  +  2 OH- (aq)  --->  Co(OH) (s)  +  2 e- Therefore, each Ag must be reduced by 2 electrons, which causes AgO to be reduced fully to Ag.  The reaction stops because Co(OH)2 and Ag overlap (and because Ag cannot be further reduced). For the 1/2 reaction, AgO (s)  +  H2O (l)  +  2 e-  --->  Ag (s)  +  2 OH- (aq)      Eo = [(0.602)(1) + (0.343)(1)] / 2 = 0.473 V Therefore, the overall voltage generated will be ΔEorxn = 0.473 + 0.733 = 1.206 V
4.	The following thermochemical reactions (with the correct signs) are known: CaF2 (s)  --->  Ca (s)  +  F2 (g) -Hof = 1215 kJ/mol Ca (s)  --->  Ca (g) Hovap = 178 kJ/mol F2 (g)  --->  2 F (g) HoF-F = 154 kJ/mol Ca (g)  --->  Ca+ (g) HoIE1 = 590 kJ/mol Ca+ (g)  --->  Ca2+ (g) HoIE2 = 1145 kJ/mol 2 F (g)  --->  2 F- (g) 2(-HoEA) = -(2*328) = -656 kJ/mol Therefore, ΔHoL = -[1215+178+154+590+1145-656] = -2626 kJ/mol (Note - the number for Hof in Zumdahl is slightly different, which gives ΔHoL = -2630 kJ/mol)

 

Last updated, January 31, 2005
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