CHEM 116-Dr. Babb s Sections Answer Key to Lecture Problem Sheet Questions for Chapters 20, 21, and 23. 199. First complex: Co(NH 3 ) 6 Cl 3 + 3 AgNO 3 > Co(NH 3 ) 6 +3 + 3 AgCl(s); the three Cl - are outside the coordination sphere since they act as any other ion in a precipitation reaction. The best way to designate the chemical formula of the first complex is as [Co(NH 3 ) 6 ]Cl 3. Second complex: Co(NH 3 ) 5 Cl 3 + 2 AgNO 3 > Co(NH 3 ) 6 Cl +2 + 2 AgCl(s); two of the three Cl - are outside the coordination sphere since they act as any other ions in a precipitation reaction. The best way to designate the chemical formula of the second complex is as [Co(NH 3 ) 5 Cl]Cl 2. Third complex: Co(NH 3 ) 4 Cl 3 + AgNO 3 > Co(NH 3 ) 4 Cl + 2 + AgCl(s); one of the three Cl - is outside the coordination sphere since it acts as any other ion in a precipitation reaction. The best way to designate the chemical formula of the third complex is as [Co(NH 3 ) 4 Cl 2 ]Cl. 200. Complex IV would yield two moles of water/mol of complex when heated. Only in complex IV are two moles/molecules of water outside the coordination sphere and therefore weakly bound and easily lost during heating. 214. Bonding between ligand and metal cation is considered to be electrostatic, either ionic (if ligand is an anion) or ion-dipole (if ligand is neutral molecule). 215. Incoming ligands destabilize the metal cation d-orbitals causing the d-orbitals to increase in energy. The d z2 and d x2-y2 (e g ) levels are destabilized more than the d xy, d xz and d yz (t 2g ) levels. The d z2 and d x2-y2 orbitals are destabilized the most since the incoming ligands are coming in pointing directly toward the lobes of these two orbitals. 218. Crystal field splitting in octahedral field () O ) - difference in energy between e g and t 2g levels. Weak field ligand - ligand that causes splitting between t 2g and e g levels to be small. Strong field ligand - ligand that causes splitting between t 2g and e g levels to be large. High spin complex - results when metal d-electrons singly occupy both t 2g and e g levels before pairing; occurs when weak field ligands are present. Low spin complex - results when metal d-electrons singly occupy t 2g and then pair up in t 2g before singly occupying e g levels; occurs when strong field ligands are present. 219. CN - - > NO 2 > en > NH 3 > OH - > H 2 O > F - > Cl - > Br - > I - Strong field ligands Weak field ligands result in low spin result in high spin 220. Complex Metal cation Ligand Type # unpaired e- A. [Cr(H 2 O) 6 ] +3 Cr +3 (d 3 ) weak field three B. [CoF 6 ] -3 Co +3 (d 6 ) weak field four C. [Co(CN) 6 ] -3 Co +3 (d 6 ) strong field zero D. [Ti(H 2 O) 6 ] +3 Ti +3 (d 1 ) weak field one E. [V(H 2 O) 6 ] +3 V +3 (d 2 ) weak field two F. [MnCl 6 ] -3 Mn +3 (d 4 ) weak field four G. [FeF 6 ] -3 Fe +3 (d 5 ) weak field five H. [Fe(CN) 6 ] -3 Fe +3 (d 5 strong field one I. [Ni(NH 3 ) 6 ] +2 Ni +2 (d 8 ) strong field two J. [Cr(CN) 6 ] -3 Cr +3 (d 3 ) strong field three
K. [Mn(CN) 6 ] -3 Mn +3 (d 4 ) strong field two L. [Ni(H 2 O) 6 ] +2 Ni +2 (d 8 ) weak field two 222. Colored compounds result when metal d-orbitals are partially filled with electrons. Colorless compounds result when metal d-orbitals are empty or totally filled with electrons. NaBr colorless CuNO 3 colorless Bi(NO 3 ) 3 colorless AsCl 3 colorless Cu(NO 3 ) 2 colored Fe(C 2 H 3 O 2 ) 3 colored HgCl 2 colorless KNO 3 colorless Pb(ClO 4 ) 2 colorless Zn(NO 3 ) 2 colorless [Ni(H 2 O) 6 ]Cl 2 colored SbCl 3 colorless Mn(NO 3 ) 3 colored 223. Formation of a metal complex usually increases the solubility of a salt because K f for formation of most metal complexes is >>1. 224. Ag + (aq) + 2 NH 3 (aq) < > [Ag(NH 3 ) 2 ] + (aq) K f =1.7 10 7 Cu +2 (aq) + 4 NH 3 (aq) < > [Cu(NH 3 ) 4 ] +2 (aq) Kf =1.1 10 13 Ag + (aq) + 2 CN - (aq) < > [Ag(CN) 2 ] - (aq) K f =1 10 21 Al +3 (aq) + 4 OH - (aq) < > [Al(OH) 4 ] - (aq) K f =2.1 10 34 Cr +3 (aq) + 4 OH - (aq) < > [Cr(OH) 4 ] - (aq) K f =8 10 29 Zn +2 (aq) + 4 OH - (aq) < > [Zn(OH) 4 ] -2 (aq) K f =2.8 10 15 225. AgCl is more soluble in a solution of aqueous ammonia than in pure water due to formation of the very stable metal complex ion, [Ag(NH 3 ) 2 ] +. AgCl(s) < > Ag + (aq) + Cl - (aq) Ksp=1.8 10-10 Ag + (aq) + 2 NH 3 (aq) < > [Ag(NH 3 ) 2 ] + (aq) K f =1.7 10 7 Net Eq: AgCl(s) + 2 NH 3 (aq) < > [Ag(NH 3 ) 2 ] + (aq) + Cl - (aq) K=K sp K f =3.1 10-3 226. Limited NH 3 : Cd +2 (aq) + 2 NH 3 (aq) + 2 H 2 O < > Cd(OH) 2 (s) + 2 NH + 4 (aq) Excess NH 3 : Cd(OH) 2 (s) + 4 NH 3 (aq) < > [Cd(NH 3 ) 4 ] +2 (aq) + 2 OH - (aq) Limited NH 3 : Zn +2 (aq) + 2 NH 3 (aq) + 2 H 2 O < > Zn(OH) 2 (s) + 2 NH + 4 (aq) Excess NH 3 : Zn(OH) 2 (s) + 4 NH 3 (aq) < > [Zn(NH 3 ) 4 ] +2 (aq) + 2 OH - (aq) Limited NH 3 : Cu +2 (aq) + 2 NH 3 (aq) + 2 H 2 O < > Cu(OH) 2 (s) + 2 NH + 4 (aq) Excess NH 3 : Cu(OH) 2 (s) + 4 NH 3 (aq) < > [Cu(NH 3 ) 4 ] +2 (aq) + 2 OH - (aq) Limited NH 3 : Ag + (aq) + NH 3 (aq) + H 2 O < > AgOH(s) + NH 4+ (aq) Excess NH 3 : AgOH(s) + 2 NH 3 (aq) < > [Ag(NH 3 ) 2 ] + (aq) + OH - (aq) Limited NH 3 : Ni +2 (aq) + 2 NH 3 (aq) + 2 H 2 O < > Ni(OH) 2 (s) + 2 NH + 4 (aq) Excess NH 3 : Ni(OH) 2 (s) + 6 NH 3 (aq) < > [Ni(NH 3 ) 6 ] +2 (aq) + 2 OH - (aq) 227. Limited NaOH: Al +3 (aq) + 3 OH - (aq) < > Al(OH) 3 (s) Excess NaOH: Al(OH) 3 (s) + OHG(aq) < > [Al(OH) 4 ] - (aq) Limited NaOH: Zn +2 (aq) + 2 OH - (aq) < > Zn(OH) 2 (s) Excess NaOH: Zn(OH) 2 (s) + 2 OH - (aq) < > [Zn(OH) 4 ] -2 (aq) Limited NaOH: Pb +2 (aq) + 2 OH - (aq) < > Pb(OH) 2 (s) Excess NaOH: Pb(OH) 2 (s) + 2 OHG(aq) < > [Pb(OH) 4 ] -2 (aq)
Limited NaOH: Cr +3 (aq) + 3 OH - (aq) < > Cr(OH) 3 (s) Excess NaOH: Cr(OH) 3 (s) + OH - (aq) < > [Cr(OH) 4 ] - (aq) Limited NaOH: Sn +2 (aq) + 2 OH - (aq) < > Sn(OH) 2 (s) Excess NaOH: Sn(OH) 2 (s) + OH - (aq) < > [Sn(OH) 3 ] - (aq) 228. Solubility of AgBr in 0.75 M NH 3 = 2.3 10-3 M. Solubility of AgBr in pure water = 7.3 10-7 M. Note that AgBr in 3200 times more soluble in 0.75 M NH 3 than in pure water due to formation of the very stable complex ion, [Ag(NH 3 ) 2 ] +. 229. [Ni +2 ]=7.6 10-10 M; [NH 3 ] = 0.70 M 230. Yes, AgCl will precipitate. 231. ph=10.1 232. 234. 1.2 M NH 3 )H= - (exothermic); this favors spontaneous process in forward direction but does not mean that overall the reaction is spontaneous. )H= +(endothermic); this favors nonspontaneous process in forward direction but does not mean that overall the reaction is nonspontaneous. 235. -904.6 kj 236. Entropy = S = amount of randomness or disorder in a system. As amount of randomness/disorder increases, S increases. S does not have negative values but )S, which is the change in entropy during a process, can have negative values. 237. )S= +; entropy is increasing and this favors spontaneous process in forward direction. )S= -; entropy is decreasing and this favors nonspontaneous process in forward direction. 238. S gas >> S liquid > S solid A. )S = + E. )S = - B. )S = + F. )S = + C. )S = - D. )S = + 239. 4 NH 3 (g) + 5 O 2 (g) > 4 NO(g) + 6 H 2 O(g); )S = + because )n gas =+1 2 H 2 O(g) + N 2 (g) > N 2 H 4 (l) + O 2 (g); )S = - because )n gas = -2 241. +180.8 J/K 243. )G = Gibbs Free Energy; )G = )H - T)S; )G=-, spontaneous )G=+, nonspontaneous 244. CASE#1: )H= +; )S= +; )G = ±, dep. on temp.; non-spont at low T; spont at high T CASE#2: )H= +; )S= - )G = +, independent of temp; non-spont at all T CASE#3: )H= -; )S= + )G = -, independent of temp; spont at all T CASE#4: )H= -; )S= - )G = ±, dep. on temp.; spont at low T; non-spont at high T. Crossover temperature: temperature at which reaction can be made to change spontaneity (i.e. spontaneous >nonspontaneous or nonspontaneous >spontaneous). To calculate crossover temperature: T = )H/)S. Crossover temperatures cannot be calculated for Cases #2 and #3 because these reactions are always either spontaneous or nonspontaneous at all temperatures. 245. )G/(25 C)= -1325.2 kj; This reaction is spontaneous at 25 C. This reaction can be made
nonspontaneous at any temperature above the crossover temperature of 9825 K. 246. A. -391.08 kj; spontaneous at 25 C B. +231.10 kj; nonspontaneous at 25 C 247. )G/= Gibbs free energy under standard conditions and at specified temperature. Standard Conditions: all gases having partial pressure of 1 atm; all solutes having concentrations of 1 M; solids, liquids and gases in pure form. )G = Gibbs free energy under nonstandard conditions and at specified temperature. Equation relating )G/ to )G: )G = )G/ + RTlnQ 248. At 25 C: )G = +86.6 kj; nonspontaneous At 800 C: )G = -29.68 kj; spontaneous NOTE: To calculate )G at 25 C, )G/ (at 25 C) could be calculated from standard Gibbs free energies of formation )G/ f since these are tabulated specifically at a temperature of 25 C. However to calculate )G at 800 C, )G/(at 800 C) had to be calculated from the equation )G/=)H/ - T )S/. 249. Derivation of equation relating )G/ to K: )G = )G/ + RTlnQ; However at equilibrium the following conditions apply: 1. )G=0; 2. Q=K; Therefore, the original equation becomes: 0 = )G/ + RTlnK; Rearranging for )G/: )G/ = -RTlnK; Rearranging for K: K = exp(-)g//rt) 250. A. )G/(at 25 C) = -42 kj/mol B. Under the conditions given in Part B, the reaction is nonspontaneous in the forward direction. To determine this fact, either 1.) calculate )G=+5kJ under this set of nonstandard conditions OR 2.) calculate Q=2.0 10 8 and compare it to K. 251. K(at 25 C)=3.01 10-7 ; K(at 200 C)=0.15 NOTE: To calculate K at 200 C, )G/ at 200 C must first be calculated from the equation )G/=)H/! T )S/. 252. At 25 C, )G = -1388 kj; reaction is spontaneous At 750 C, )G = -1374 kj; reaction is spontaneous 253. A. At 25 C, )G/=!407 kj and K = 2.2 10 71 ; reaction is spontaneous under standard conditions. B. A temperature of 741 K (468 C) or above is needed to regenerate the P 4 O 10. This temperature is the crossover temperature. At the crossover temperature the numerical value of K is 1. C. Yes, at 25 C the P 4 O 10 will spontaneously adsorb the water vapor. To determine this calculate )G= -351 kj under this set of nonstandard conditions. NOTE: To calculate )G, Q must first be calculated. For this reaction, Q = 1/(p water ) 6 where the p water must be in units of atmospheres. 254. electrochemistry: study of interconversion between chemical and electrical energy electrochemical cell: cell that interconverts chemical and electrical energy galvanic cell: cell that converts chemical energy of a spontaneous chemical reaction electrolytic cell: to electrical energy (an electrical current is generated). cell that converts electrical energy to chemical energy (an electrical current is used to drive a non-spontaneous chemical reaction). Batteries are considered to be galvanic cells. 255. The solution turns blue indicating that Cu is oxidized to blue Cu(II) and at the same time
Ag(I) is reduced to Ag(s). Net Ionic eqn: Cu(s) + 2 Ag + > Cu +2 + 2 Ag(s). Oxidation Half-Rxn: Cu(s) > Cu +2 Reduction Half-Rxn: Ag + > Ag(s) 256. No, a reaction between Ag(s) and Cu +2 would not occur. If the forward reaction between Cu(s) and Ag + is spontaneous; then, the reverse reaction between Ag(s) and Cu +2 is nonspontaneous. 257. A. electrode - conductor of electricity (usually made of metal) B. Anode - electrode at which oxidation occurs; Cathode - electrode at which reduction occurs. C. Electrons always flow from the anode to the cathode. By convention, the anode is given a (-) sign; while, the cathode is given a (+) sign. D. Salt bridge - used to keep the solutions in each half-cell electrically neutral. In anode compartment, as oxidation occurs positive ions are generated so negative ions flow out of salt bridge. In cathode compartment, as reduction occurs positive ions are lost so negative ions flow into the salt bridge. 258. Cu(s) Cu +2 (1 M) Ag + (1 M) Ag(s) 259. Pb(s) Pb +2 (1 M) F 2 (1 atm) F - (1 M) Pt(s) 260. Net Cell Reaction: Sn +2 + 2 Fe +3 > Sn +4 + 2 Fe +2 Pt on tin side is anode. Pt on iron side is cathode. 261. Net cell potential (E/ cell ) is measured by placing a voltmeter in line and between the two half cells. )G/ = -nf E/ cell where n = # mole of electrons transferred and F = 96500 C/mol e - (Faraday constant) Note: If E/ cell >0, reaction is spontaneous under standard conditions. If E/ cell < 0, reaction is non-spontaneous under standard conditions. If E/ cell =0, reaction is at equilibrium under standard conditions. 262. 263. Since E/ cell >0, this reaction is spontaneous under standard conditions. )G/= -89 kj. E/ cell = E/ (anode, oxid) + E/ (cathode, red) S.H.E. = standard hydrogen electrode. Half Rxn: 2H + + 2 e - > H 2 (g), E/ SHE =0 Volts. The S.H.E. is used as the zero point for standard cell potentials. All half-cell potentials are measured relative to the S.H.E. which is assigned a standard reduction potential of zero. 264. Best oxidizing agent= F 2. Best oxidizing agent wants to be reduced the most so has highest positive standard reduction potential (E/ red ). Worst oxidizing agent = Na +. Worst oxidizing agent wants to be reduced the least so has most negative standard reduction potential (E/ red ). Worst reducing agent = F - ; Best reducing agent = Na(s) 265. Cu(s) + 2 Ag + (aq) > Cu +2 (aq) + 2 Ag(s); E/ cell = -(0.34 V) + 0.80 V = +0.46 V; spontaneous in forward direction Cu +2 (aq) + 2 Ag(s) > Cu(s) + 2 Ag + (aq) E/ cell = -(0.80) V + 0.34 V = -0.46 V; non-spontaneous in forward direction 266. A. Fe(s) + 2 Fe +3 > 3 Fe +2 E/ cell = -(-0.45 V) + 0.77 V = +1.22 V; spontaneous in forward direction; Anode = Fe(s) on left side; Cathode = Pt on right side. B. Ni(s) + 2 Ag + > Ni +2 + 2 Ag(s) E/ cell = -(-0.26 V) + 0.80 V = +1.06 V; spontaneous in forward direction; Anode = Ni(s) on left side; Cathode = Ag(s) on right side. C. Fe +2 + Na + > Fe +3 + Na(s) E/ cell = -(0.77 V) + -2.71V = -3.48 V; non-spontaneous
in forward direction; Anode = Pt(s) on left side; Cathode = Na on right side. 267. Best oxidizing agent = Ag + because has highest positive standard reduction potential (E/ red ) 268. Best reducing agent = Na because has most negative standard reduction potential (E/ red ) or conversely has most positive standard oxidation potential (E/ oxid ). 269. E cell (non-std cond) = E/ cell + (-RT/nF)lnQ 270. E cell = +5.66 V; Cell is spontaneous under this set of non-standard conditions. 271. 272. E cell = +0.98 V; Cell is spontaneous under this set of non-standard conditions. E/ cell = +(RT/nF)lnK or K = exp(e/ cell nf/rt) 273. K = 7.3 10 35 274. E/ cell = -4.07 V; because reaction is non-spontaneous in forward direction, the only way to make this reaction proceed is to use an electrical current to drive the reaction. An electrolytic cell would then result. To deposit 5.0 g liquid sodium in 1 hr, an electrical current of 5.8 A was used. 275. A. 0.22 g Ga B. 52 ml O 2 276. 3.22 hr