Download 1.Using the table above, decide if the element mercury (Hg) should

Chemistry 241-2012
Exam #2
Name: Answer Key
Reduction Half-Reaction
–
–
F2(g) + 2 e → 2 F (aq)
–
+
–
2+
MnO4 (aq) + 8 H (aq) + 5 e → Mn (aq) + 4 H2O(ℓ)
2+
–
Hg (aq) + 2 e → Hg(ℓ)
+
–7
–
O2(g) + 4 H (aq, 10 M) + 4 e → 2 H2O(ℓ)
+
–
Ag (aq) + e → Ag(s)
2+
–
Hg2 (aq) + 2 e → 2 Hg(ℓ)
3+
–
2+
Fe (aq) + e → Fe (aq)
2+
–
Cu (aq) + 2 e → Cu(s)
2+
–
Sn (aq) + 2 e → Sn(s)
2+
–
Ni (aq) + 2 e → Ni(s)
3+
–
2+
Cr (aq) + e → Cr (aq)
2+
–
Fe (aq) + 2 e → Fe(s)
3+
–
Cr (aq) + 3 e → Cr(s)
2+
–
Zn (aq) + 2 e → Zn(s)
3+
–
Al (aq) + 3 e → Al(s)
2+
–
Mg (aq) + 2 e → Mg(s)
+
–
Na (aq) + e → Na(s)
E° (V)
2.87
1.51
0.855
0.82
0.799
0.789
0.771
0.337
–0.14
–0.25
–0.41
–0.44
–0.74
–0.763
–1.66
–2.37
–2.714
1. Using the table above, decide if the element mercury (Hg) should be found mainly in the core of the earth,
or mainly in the mantle and crust. Explain breifly.
If Fe can reduce Hg22+ to Hg, then Hg will be formed and join the other metals in the core of the Earth. If Fe
cannot reduce Hg22+ to Hg, then Hg will remain in the cationic state and exist as a compound in the
crust/mantle.
Fe --> Fe2+ + 2 e-
Eoanode = -0.44 V
Hg22+ + 2 e- --> 2 Hg
Eocathode = +0.789 V
Can also use the Hg2+ reaction, similar result
Eocell = Eocathode - Eoanode = 0.789 - (-0.44) = +1.23 V
This is positive, so the reaction is favored; Hg is mostly in the Earth's core.
2. Which two reactants would you use from the table above to make the battery with the largest voltage?
What is that voltage?
Na(s) + F2(g)
Eocell = 2.87 V - (-2.714 V) = 5.58 V
3. What is the Eo for a cell that uses the reaction: Cu2+ (aq) + Mg(s) --> Cu(s) + Mg2+(aq)
cathode = Cu2+ (aq) + 2 e- --> Cu(s) Eo = +0.337 V
anode = Mg(s) --> Mg2+(aq) + 2 e-
Eo = -2.37 V
Eocell = Eocathode - Eoanode = 0.337 V - (-2.37 V) = +2.71 V
4. When the Earth formed, nonmetals were the "limiting reactants." What would the Earth's structure be like
if, instead, metals had been the limiting reactants and nonmetals were present in excess?
In this case, there would be no native (elemental) metals. The three kinds of substances would be chemical
compounds formed between metals and nonmetals (ex: FeO, CaCO 3) , compounds between different
nonmetals (ex: CO2), and elemental nonmetals (ex: N2, S).
These would either be mixed or differentiated by density. It's likely the metal-nonmetal compounds would
be more dense and found in the center of the Earth with the less dense metal-less compounds and elements
making up the crust and surface.
5. Which of the following would have the higher melting point? NaBr or PBr3.
Why?
NaBr has the higher melting point because it is an ionic compound with strong ionic bonds. PBr 3 is a slightly
polar molecular compound, which is held in the solid state by comparatively weak intermolecular forces.
6. Below is a table showing the various methods used to extract metals from metal ores.
a. What is the main chemical event involved in producing a metal from its ore? That is, what kind of reaction
takes place?
The metal is reduced from a cationic state to the neutral metal.
b. Why is iron produced by reaction with carbon, but sodium (Na) is produced by electrolysis?
Fe is easy enough to reduce that it can be done with C at reasonable temperatures. Iron could also be done
via electrolysis, but that is more expensive. It's used with Na because Na+ is very hard to reduce and can't be
done chemically (i.e., with C).
c. What is the hard part in producing gold from gold ore?
Separating the very diffuse gold from the surrounding rock. (It's accomplished by leaching the gold with a
cyanide solution, which forms the soluble Au(CN)2- ion.
7. Silicate minerals contain Si, O, and usually one or more metals.
a. What is the effect of changing the ratio of O to Si in the compound on its structure?
As O/Si decreases, the fraction of O atoms bridging between Si atoms increases. As that happens, the structures
become more interlinked and complex. Structures progress from momomeric SiO4- anions to dimers, rings, chains,
double chains, sheets, and, finally, three-dimensional cages.
b. Describe the bonding. What's bonded to what, and what kinds of bonds are involved (ionic, covalent, intermolecular
forces, etc.)?
The Si-O framework is an anion that can be a small or an extended structure. The Si-O framework forms an ionic
compound with metal ions (except in the case of SiO2, when no metals are present).
8. Group 3A compounds are often electron deficient. How do the following compounds minimize the instability of their
electron deficiencies? One sentence for each is enough. You can draw structures but you don't need to.
Aluminum chloride, Al2Cl6
This forms an intermolecular Lewis acid-base dimer where one Cl atom on each AlCl3 donates a pair of electrons to
the neighboring Al atom.
Boron trifluoride, BF3
This forms partial pi bonds between the B and F atoms.
Diborane, B2H6
B-H-B bridges are formed that use only two electrons to bond the 3 atoms together. This is a 3 center- 2 electron
bond.
9. The following questions are about bonding and structure in nonmetal compounds.
a. Why is CO2 a gas, whereas SiO2 is a high-melting solid?
C is smaller than is Si. Therefore, C can form pi bonds to O; Si cannot form pi bonds to O.
In both cases, the C or Si will form a total of four bonds. In the case of CO2, the pi bonds are strong enough that the
entropic favorability of forming small molecules drives formation of the monomeric CO2 molecules.
In the case of SiO2, the enthalpy-favored ability to form 4 single bonds (and no double bonds) means that every O
atom must be bonded to 2 Si atoms, essentially forming bridges, resulting in a networked structure.
b. Elemental oxygen exists as a gas with formula O2. Sulfur exists as an 8-atom ring.
Oxygen's structure favored by:
enthalpy
Sulfur's structure is favored by: enthalpy
entropy
both
entropy
both
c. Why does bromine, Br2, exist as a small molecule instead of an extended structure?
Br does not form pi bonds, but it has only one orbital vacancy, so forms only one bond to another Br atom. That is,
the inability to form extended structures comes from the lack of more than one orbital vacancy with which to form
bonds. It can only form one bond.
d. Would you expect GeO2 to be a small molecule or a large molecule?
Oh, a large molecule. Ge is located below Si on the periodic table so it will not form pi bonds to O. GeO 2 will be similar
to SiO2.
10. The formula for lattice energy, Uo, is:
Compare each term for the two salts: MgBr2 and KBr
a) Circle for which each is larger:
Z+Z-
MgBr2
or
KBr
or
same
M
MgBr2
or
KBr
or
same
(1 – 1/n)
MgBr2
or
KBr
or
same
KBr
or
same
b) Which overall do you expect to have a stronger lattice energy?
MgBr2
or
c) Which term in the equation do you think is most important in any difference between them?
Z+Zd. What is the purpose of the Madelung constant?
The Madelung constant accounts for all the extended attractive and repulsive interactions in the extended structure.
11. Consider the extraction of Ba from the compound BaO, both by simple decomposition and by reaction with carbon.
For each, prepare a graph of ΔGo vs. temperature. Decide on the temperature range for which extraction could occur.
For each, decide if this is an industrially feasible process (assume a maximum viable temperature of 1000 oC).
Data from OWL Tables:
So(Ba) = 67.0 J/k mol
So (BaO) = 70.4 J/K mol
So (O2) = 205.1 J/K mol
So (C) = 5.7 J/K mol
So (CO2) = 213.7 J/K mol
Hf(BaO) = -553.5 kJ/mol
Hf(CO2) = -393.5 kJ/mol
Decomposition:
2 BaO(s) --> 2 Ba(s) + O2(g)
H = 2 x +553.5 kJ/mol = +1107 kJ/mol
S = {[2 x 67.0 + (205.1)] - (2 x 70.4)} J/K mol = +198.3 J/K mol
Crossover Temperature = 5582 K (not feasible)
1500
1000
500
0
0
2000
4000
6000
8000
-500
Reduction with C:
2 BaO(s) + C(s) --> 2 Ba(s) + CO2(g)
H = [-393.5 kJ/mol + 2(0)] - [2 (-553.5 kJ/mol)+ 0] = +713.5 kJ/mol
S = +201.2 J/K mol
Crossover Temperature = 3546 K (also not feasible)
1000
500
0
-500
-1000
0
2000
4000
6000
8000