Download Exam 3 Review Key

Chemistry 204 Exam III Review Key
Spring 2011
1. Bromide ions and water compete to coordinate to copper ions. When Cu2+ is in aqueous
solution, the light blue hexaaquacopper(II) complex is the dominant species. However, in
the presence of a large concentration of bromide ions, violet-colored tetrabromocuprate(II)
forms. Upon dilution with water, the solution becomes light blue again.
a) Write the formulas of the two copper complex ions.
[CuBr4]2-, [Cu(OH2)6]2+
b) Explain the differences in color in the two complex ions.
Br- is a weak-field ligand, while H2O is a stronger-field ligand. Weaker-field complexes
absorb longer wavelengths (less energy) and transmit shorter wavelength color; opposite for
stronger-field complexes.
2. Using some of the key ideas proposed in ligand field theory, explain why:
a) Ethylenediamine is a weaker field ligand than CO
b) Water is a weaker field ligand than ammonia
a) Ethylenediamine forms sigma bonds, but isn’t a pi-acceptor or donor (acid/base). It won’t
change the t2g orbital energy. CO is a pi-acid, so it lowers the t2g energy and increases
splitting.
b) Water is a pi-base, since it uses one lone pair to sigma bond to the metal, and the leftover
lone pair can be donated to the metal in a pi fashion. Ammonia is not a pi-base since it has no
more lone pairs after donating one in a sigma bond. Pi-bases increase the energy of the t2g
orbitals and decrease splitting.
3. Given the coordination compound [CoCl(NCS)(OH2)4]Cl·H2O, draw ALL the possible
isomers. Remember to consider ALL types of isomerization!
9 types total:
2 linkage isomers (NCS or SCN), each of which has cis and trans form (4 total)
1 ionization isomer (Cl- replaces SCN- which have same charge), cis or trans (2 total)
3 hydration isomers (water replaces Cl-, SCN-, or NCS-)
4. Calculate the solubility of silver chloride in 0.10 M KCN.
Ksp: AgCl (s) = 1.6 x 10-10
Kf: Ag(CN)2- (aq) = 5.6 x 108
This is like homework 12.77, 12.78 (5th) and 11.75, 11.76 (4th)
Overall reaction: AgCl (s) + 2CN- (aq) à Ag(CN)2- (aq) + Cl- (aq)
K = Ksp x Kf = 0.0896
K = ([Ag(CN)2-][Cl-])/[CN-]2 = 0.0896
Set up ICE table with overall reaction
I
C
E
AgCl + 2CN- à Ag(CN)2- + Cl0.10
0
0
-2s
+s
+s
0.10-2s
s
s
K = s2/(0.10-2s)2
s = 0.0187 M
5. World War I saw the use of lewisite, an arsenic-based poison gas. The organic compound
dimercaprol (BAL) was administered as an antidote. Many decades later in World War II,
navy personnel painting ship hulls fell ill to lead poisoning, no doubt from the lead in the
paint. Ethylenediaminetetraacetic acid (EDTA) was introduced as a treatment. However,
both BAL and EDTA had moderate-to-severe side effects. Today, medicine has improved,
and dimercaptosuccinic acid (DMSA) has replaced the former compounds as the primary
treatment for arsenic, lead, and mercury poisoning – with many fewer side effects.
BAL
DMSA
EDTA
a) Circle all the possible ligand sites.
SH, OH, and N sites (3 on BAL, 4 on DMSA, 6 on EDTA)
NOTE: C=O oxygens are not Lewis bases and will not act as ligand sites
b) Lead’s primary mode of toxicity is its interference with enzyme function – it mimics
other essential metals that take part in enzymatic reactions and displaces them.
Considering the fact that sulfhydryl (-SH) groups are found on many enzymes, how
might EDTA and DMSA work to treat lead poisoning?
*The chelates’ ligand sites replace the enzyme’s –SH bonds to the lead cation, effectively
displacing the lead from the enzyme. The chelate then carries the lead away safely.
c) Mercury, in both its +1 and +2 oxidation states, can form inorganic salts. One such salt,
mercury(II) chloride, is highly toxic. It has a solubility of just above 6% at room
temperature. The body must excrete mercury in the urine to get rid of it, but that’s
difficult to do when it’s an insoluble precipitate. How might DMSA help treat mercury
poisoning in this case?
*When mercury is an insoluble salt, it sticks around in the body and can’t be excreted. When
DMSA chelates to it, it forms a soluble complex ion and can be excreted. In other words…
you pee out the DMSA-mercury chelate complex.
6. Sketch the titration curve for the amino acid glutamate, which has the following structure
at pH=7:
pKa=9
pKa=4
pKa=3
Draw the dominant form of glutamate as the titration progresses and indicate the isoelectric point
on the titration curve. (hint: when pH=pKa, there are equal amounts of the protonated and
unprotonated forms.)
•
•
•
•
•
Below pH=3: R-NH3+, both carboxyl groups R-COOH
Between pH=3 to pH=4: R-NH3+, pKa=3 carboxyl group R-COO-, pKa=4 carboxyl group RCOOH
Between pH=4 to pH=9: R-NH3+, both carboxyl groups R-COOAbove pH=9: R-NH2, both carboxyl groups R-COOIsoelectric point: occurs at first halfway point, pH=3.5
7. Calculate the concentration of all chemical species in a 0.05M solution of H2SeO3 (selenous
acid) and determine the pH. Ka1=2.3x10-3; Ka2=5.3x10-9
[H2SeO3] = 0.0404M (must use quadratic equation), [HSeO3-] = [H3O+] = 9.64x10-3M, [SeO32-] =
5.3x10-9M, [OH-] = 1.04x10-12M
8. Complete the following statements concerning redox reactions:
a) The species that gains electrons is reduced and is called the oxidizing agent.
The species that loses electrons is oxidized and is called the reducing agent.
b) When balancing redox half reactions in aqueous solution, add water to the side deficient in
oxygen, then add H+(aq) to the side deficient in hydrogen.
c) If the reaction arrow in a redox half reaction is reversed, the sign of Eo is also reversed.
If a redox half reaction is multiplied by an integer, Eo stays the same.
9. Balance the following reactions in acidic solution and identify the oxidizing and reducing
agents:
a) Cr2O72-(aq) + CH3OH(aq) à Cr3+(aq) + HCO2H(aq)
16H+ + 2Cr2O72- + 3CH3OH à 4Cr3+ + 3HCO2H + 11H2O
Cr2O72- is oxidizing agent
CH3OH is reducing agent
b) As(s) + ClO3 -(aq) à H3AsO3(aq) + HClO(aq)
6H2O + 3H+ + 4As + 3ClO3 - à 4H3AsO3 + 3HClO
As is reducing agent
ClO3 - is oxidizing agent
10. Balance the following reactions in basic solution and identify the oxidizing and reducing
agents:
a) H2O2(aq) + Cl2O7(aq) à ClO2-(aq) + O2(g)
2OH - + Cl2O7 + 4H2O2 à 2ClO2 - + 5H2O + 4O2
Cl2O7 is oxidizing agent
H2O2 is reducing agent
b) O3(aq) + Br-(aq) à O2(g) + BrO3-(aq)
3O3 + Br - à 3O2 + BrO3 O3 is oxidizing agent
Br - is reducing agent
11. For each of the following redox couples below, the type of cell is specified. Determine the
oxidizing agent, the reducing agent, give the standard cell notation, and specify its standard
cell potential.
a) Oxidizing agent: Co2+; reducing agent: Ti3+;
Pt(s)|Ti3+(aq), Ti2+(aq)||Co2+(aq)|Co(s); +0.09 V
b) Oxidizing agent: U3+; reducing agent: La3+
La(s)|La3+(aq)||U3+(aq)|U(s); +0.73 V
c) Oxidizing agent: H+; reducing agent: Fe3+
Pt(s)|Fe3+(aq), Fe2+(aq)|| H2(g)|H+(aq)|Pt(s) -0.77 V
d) Oxidizing agent: Ag+; reducing agent: O3
Pt(s)|O3(g), O2(g)|OH-(aq)|| Ag+(aq)|Ag(s); -0.44V
(this is for basic conditions, used because this is how the reaction is given in the Standard
Potentials Table in the book’s appendix. In acid, the cell would have the same set-up, only there
would be H+ on the right and no OH- on the left)
12. Calculate the reaction quotient, Q, for the cell reaction, given the measured cell potential.
Balance the equations using the smallest whole-number coefficients.
(Remember the Nernst Equation relates Q, K to electronic potentials. To correctly determine n, the
number of electrons transferred, we need the complete balanced cell reactions. See previous
problems for assistance in this.)
a) Q = 101
b) Q = 104
13. The unknown metal ions are reduced, increasing the weight of the unknown metal electrode.
The cell can be written as:
Fe(s)|Fe2+(aq)||M+(aq)|M(s), meaning that the M+/M electrode is the cathode, because this is where the
reduction is occurring.
Fe2+ + 2e- à Fe(s) Eo = -0.44V
Eocathode = +1.24V + Eoanode
Eocathode = +1.24 V + (-0.44V) = +0.80 V
(Remember there are two ways of determining Eocell: (1) by using the equation Eocell = Eoreduction –
Eooxidation or (2) by reversing the half-reaction for the cell we know to be oxidized, and inverting its
sign, then simply adding the two Eo’s. Don’t be confused; these are just two ways of doing the exact
same thing)
14. Calculate the potential of each of the following cells. Is the cell reaction spontaneous as written
or spontaneous in the opposite direction?
(It is spontaneous (galvanic) as given if the Eocell is positive. If negative, the reaction is nonspontaneous as written (electrolytic))
a) 0.326 V; spontaneous as written
b) -0.685 V; not spontaneous as written
15. Consider an electrochemical cell in which one half contains Cr2O72- and Cr3+ ions, and the
other half contains MnO2 solid in a solution of Mn2+ ions. Assume the cell is galvanic under
standard conditions.
a) Determine the overall balanced equation for the cell reaction.
Cr2O72-(aq) + 2 H+(aq) + 3 Mn2+(aq) à 2 Cr3+(aq) + 3 MnO2(s) + H2O(l)
b) What is Eocell for this reaction?
Eo = +0.10 V
c) At what pH would the cell become electrolytic? Assume all concentrations except H+ are 1 M.
We want Ecell < 0, because 0 is the threshold between electrolytic and galvanic cells. Also, we
know the equation for Q, in terms of concentrations. If all other concentrations are 1, we have:
Q = 1/[H+]2 (use the answer to A and the definition of Q to determine this)
Since we can set Ecell = 0, and substituting Q we can use the Nernst equations to get
Eocell = RT/(nF)*ln(1/[H+]2)
Solving for [H+] with n = 6 and Eocell = +0.10, we get
[H+] = 8.49E-6
And
pH = 5.07