Download CHEM 116-Dr. Babb`s Sections Answer Key to Lecture Problem

CHEM 116-Dr. Babb’s Sections
Answer Key to Lecture Problem Sheet Questions for Chapters 20, 21, and 23.
199.
First complex: Co(NH3)6Cl3 + 3 AgNO3 –> Co(NH3)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(NH3)6]Cl3.
Second complex: Co(NH3)5Cl3 + 2 AgNO3 –> Co(NH3)6Cl+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(NH3)5Cl]Cl2.
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Third complex: Co(NH3)4Cl3 + AgNO3 –> Co(NH3)4Cl2+ + 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(NH3)4Cl2]Cl.
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.
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).
Incoming ligands destabilize the metal cation d-orbitals causing the d-orbitals to increase in
energy. The dz2 and dx2-y2 (eg) levels are destabilized more than the dxy, dxz and dyz (t2g) levels.
The dz2 and dx2-y2 orbitals are destabilized the most since the incoming ligands are coming in
pointing directly toward the lobes of these two orbitals.
Crystal field splitting in octahedral field ()O) - difference in energy between eg and t2g levels.
Weak field ligand - ligand that causes splitting between t2g and eg levels to be small.
Strong field ligand - ligand that causes splitting between t2g and eg levels to be large.
High spin complex - results when metal d-electrons singly occupy both t2g and eg levels
before pairing; occurs when weak field ligands are present.
Low spin complex - results when metal d-electrons singly occupy t2g and then pair up in t2g
before singly occupying eg levels; occurs when strong field ligands are present.
CN- > NO2- > en > NH3 > OH- > H2O > F- > Cl- > Br- > IStrong field ligands
Weak field ligands
result in low spin
result in high spin
Complex
Metal cation
Ligand Type
# unpaired eweak field
three
A.
[Cr(H2O)6]+3 Cr+3 (d3)
-3
+3
6
B.
[CoF6]
Co (d )
weak field
four
C.
[Co(CN)6]-3 Co+3 (d6)
strong field
zero
+3
+3
1
D.
[Ti(H2O)6]
Ti (d )
weak field
one
E.
[V(H2O)6]+3 V+3 (d2)
weak field
two
F.
[MnCl6]-3
Mn+3 (d4)
weak field
four
-3
G.
[FeF6]
+3
5
weak
field
five
Fe (d )
H.
[Fe(CN)6]-3 Fe+3 (d5)
strong field
one
+2
+2
8
I.
[Ni(NH3)6]
Ni (d )
strong field
two
J.
[Cr(CN)6]-3 Cr+3 (d3)
strong field
three
222.
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K.
[Mn(CN)6]-3 Mn+3 (d4)
strong field
two
+2
L.
[Ni(H2O)6]
weak field
two
Ni+2 (d8)
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
CuNO3 colorless
Bi(NO3)3 colorless
Cu(NO
)
colored
AsCl3 colorless
3 2
Fe(C2H3O2)3 colored
HgCl2 colorless
KNO3 colorless
Pb(ClO4)2 colorless
Zn(NO3)2 colorless
[Ni(H2O)6]Cl2 colored
Mn(NO3)3 colored
SbCl3 colorless
Formation of a metal complex usually increases the solubility of a salt because Kf for
formation of most metal complexes is >>1.
Ag+(aq) + 2 NH3(aq) <–> [Ag(NH3)2]+(aq) K =1.7×107
f
Cu+2(aq) + 4 NH3(aq) <–> [Cu(NH3)4]+2(aq) K =1.1×1013
f
Ag+(aq) + 2 CN-(aq) <–> [Ag(CN)2]-(aq)
21
K
f=1×10
+3
Al (aq) + 4 OH (aq) <–> [Al(OH)4] (aq) K =2.1×1034
f
Cr+3(aq) + 4 OH-(aq) <–> [Cr(OH)4]-(aq) K =8×1029
f
Zn+2(aq) + 4 OH-(aq) <–> [Zn(OH)4]-2(aq) K =2.8×1015
f
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(NH3)2]+.
AgCl(s) <–> Ag+(aq) + Cl-(aq)
Ksp=1.8×10-10
Ag+(aq) + 2 NH3(aq) <–> [Ag(NH3)2]+(aq) K =1.7×107
f
Net Eq:
AgCl(s) + 2 NH3(aq) <–> [Ag(NH3)2]+(aq) + Cl-(aq) K=Ksp×Kf=3.1×10-3
Limited NH3: Cd+2(aq) + 2 NH3(aq) + 2 H2O <–> Cd(OH)2(s) + 2 NH4+(aq)
Excess NH3: Cd(OH)2(s) + 4 NH3(aq) <–> [Cd(NH3)4]+2(aq) + 2 OH-(aq)
Limited NH3: Zn+2(aq) + 2 NH3(aq) + 2 H2O <–> Zn(OH)2(s) + 2 NH4+(aq)
Excess NH3: Zn(OH)2(s) + 4 NH3(aq) <–> [Zn(NH3)4]+2(aq) + 2 OH-(aq)
Limited NH3: Cu+2(aq) + 2 NH3(aq) + 2 H2O <–> Cu(OH)2(s) + 2 NH4+(aq)
Excess NH3: Cu(OH)2(s) + 4 NH3(aq) <–> [Cu(NH3)4]+2(aq) + 2 OH-(aq)
Limited NH3: Ag+(aq) + NH3(aq) + H2O <–> AgOH(s) + NH4+(aq)
Excess NH3: AgOH(s) + 2 NH3(aq) <–> [Ag(NH3)2]+(aq) + OH-(aq)
227.
Limited NH3: Ni+2(aq) + 2 NH3(aq) + 2 H2O <–> Ni(OH)2(s) + 2 NH4+(aq)
Excess NH3: Ni(OH)2(s) + 6 NH3(aq) <–> [Ni(NH3)6]+2(aq) + 2 OH-(aq)
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)
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Limited NaOH: Sn+2(aq) + 2 OH-(aq) <–> Sn(OH)2(s)
Excess NaOH: Sn(OH)2(s) + OH-(aq) <–> [Sn(OH)3]-(aq)
Solubility of AgBr in 0.75 M NH3 = 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 NH3 than in pure water due to formation of the very stable complex ion,
[Ag(NH3)2]+.
[Ni+2]=7.6×10-10 M; [NH3] = 0.70 M
Yes, AgCl will precipitate.
pH=10.1
1.2 M NH3
)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.
-904.6 kJ
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.
)S= +; entropy is increasing and this favors spontaneous process in forward direction.
)S= -; entropy is decreasing and this favors nonspontaneous process in forward direction.
Sgas >> Sliquid > Ssolid
A.
)S = +
E.
)S = B.
)S = +
F.
)S = +
)S = C.
D.
)S = +
4 NH3(g) + 5 O2(g) –> 4 NO(g) + 6 H2O(g); )S = + because )ngas=+1
2 H2O(g) + N2(g) –> N2H4(l) + O2(g);
)S = - because )ngas= -2
+180.8 J/K
)G = Gibbs Free Energy; )G = )H - T)S;
)G=-, spontaneous
)G=+, nonspontaneous
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.
)G/(25 C)= -1325.2 kJ; This reaction is spontaneous at 25 C. This reaction can be made
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nonspontaneous at any temperature above the crossover temperature of 9825 K.
A.
-391.08 kJ; spontaneous at 25 C
B.
+231.10 kJ; nonspontaneous at 25 C
)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
At 25 C:
)G = +86.6 kJ; nonspontaneous
)G = -29.68 kJ; spontaneous
At 800 C:
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/.
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)
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=2.0×108 and compare it to K.
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/.
At 25 C, )G = -1388 kJ; reaction is spontaneous
At 750 C, )G = -1374 kJ; reaction is spontaneous
A.
At 25 C, )G/= !407 kJ and K = 2.2×1071; reaction is spontaneous under standard
conditions.
B.
A temperature of 741 K (468 C) or above is needed to regenerate the P4O10. This
temperature is the crossover temperature. At the crossover temperature the numerical
value of K is 1.
C.
Yes, at 25 C the P4O10 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/(pwater)6
where the pwater must be in units of atmospheres.
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
to electrical energy (an electrical current is generated).
electrolytic cell:
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.
The solution turns blue indicating that Cu is oxidized to blue Cu(II) and at the same time
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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)
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.
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.
Cu(s)|Cu+2(1 M)||Ag+(1 M)|Ag(s)
Pb(s)|Pb+2(1 M)||F2(1 atm)|F-(1 M)|Pt(s)
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.
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.
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- —> H2(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.
Best oxidizing agent= F2. 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)
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
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
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in forward direction; Anode = Pt(s) on left side;
Cathode = Na on right side.
Best oxidizing agent = Ag+ because has highest positive standard reduction potential (E/red)
Best reducing agent = Na because has most negative standard reduction potential (E/red) or
conversely has most positive standard oxidation potential (E/oxid).
Ecell(non-std cond) = E/cell + (-RT/nF)lnQ
Ecell= +5.66 V; Cell is spontaneous under this set of non-standard conditions.
Ecell= +0.98 V; Cell is spontaneous under this set of non-standard conditions.
E/cell = +(RT/nF)lnK or K = exp(E/cellnF/RT)
K = 7.3×1035
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.
A.
0.22 g Ga
B.
52 mL O2
3.22 hr