Worked Out Solutions Exam Three Spring, 2013 1. Ligands are Lewis bases: electron pair donors. CH4 has no lone pairs of

electrons in its Lewis structure—it is not a Lewis base, and it is not a ligand. Choice E

2. In this voltaic cell, Pt is the anode and Ag is the cathode. Ag+ is reduced at the cathode, not the anode. H2 is oxidized at the anode, not the cathode. Electrons flow from the anode to the cathode: from Pt to Ag In the salt bridge, cations flow toward the cathode, or from the H+ side to the Ag+ side. So this statement is correct. Pt is an inert electrode. It is the location where electron transfer takes place, as H2 gives off electrons in the anode of this cell. But the Pt itself is unchanged. Choice D

3. Co/Co2+ is the anode, Ce+4/Ce+3 is the cathode.

Eo Co → Co2+ + 2e- +0.28 (change sign when we reverse the reaction) 2 { Ce4+ + e- → Ce3+ } x Co + 2Ce+4 → Co2+ + 2Ce3+ 1.88 0.28 + x = 1.88 x = 1.60 V Choice B

4. A spontaneous reaction with have a positive Eo for the overall reaction.

From Eo chart, a positive voltage if something on the top left (gets reduced more easily) reacts with something in the bottom right (gets oxidized more easily). From the choices, the only reaction which satisfies this condition and produces a positive cell voltage will be: Fe3+(aq) + 2I-(aq) → I2(s) + Fe2+(aq) Choice C

5. The 2nd Law of Thermodynamics states that the entropy of the universe increases for all spontaneous processes. This is equivalent to: the free energy of the system decreases (ΔG < 0) for all spontaneous processes, which is the way we usually evaluate spontaneity.

Choice E

6. This is a common ion effect problem. The initial concentration of I- ion in the

solution, coming from NaI, is 0.10mol0.250L

= 0.40 M

PbI2(s) Pb2+(aq) + 2I-(aq) Solid x 0.40+2x x(0.40+2x)2 = 1.4 x 10-8

Solving: x = 8.75 x 10-8 M

8.75 x 10-8 molL

x 0.250 L = 2.2 x 10-8 mol dissolve Choice E

7. N2(g) + 3H2(g) → 2NH3(g) ΔGo = 2(-16.6) = -33.2 kJ ΔG = ΔGo + RT ln Q

ΔG = -33.2 + 8.3141000

(298) ln (0.10)2

(2.00)(5.00)3 = -58.3 kJ Choice B

8. The speed of a reaction has nothing to do with spontaneity. Thermodynamic

spontaneity is unrelated to kinetics. It’s possible to have spontaneous reaction which goes very slowly due to high activation energy.

Choice E

9. H2O() → H2O(s) at -10oC This is the freezing of water The reaction is exothermic (ΔH < 0) since bonds of the solid (intermolecular forces

of the lattice) are being formed from the more loosely-bonded liquid The reaction goes to lower entropy (ΔS < 0) since the solid is more organized and

has less available microstates than the liquid. The reaction is spontaneous (ΔG < 0) since the freezing point of water is 0oC, and this

temperature, -10oC, is below the freezing point. Choice E

10. Ga(s) + In3+ → Ga3+ + In(s) observable reaction occurs This means that In3+ is more easily reduced than Ga3+ and that Ga(s) is more

easily oxidized than In(s). Based on just this observation, we can infer the following table:

In3+ + 3e- → In(s) Ga3+ + 3e- → Ga(s)

Bi(s) + In3+ → no observable reaction If an observable reaction occurred, it would be Bi(s) + In3+ → In(s) + Bi+3 Since no reaction occurs, the reverse reaction would be spontaneous. This implies that In3+ is less easily reduced than Bi3+, and that Bi(s) is less easily oxidized than In(s) Based on just this observation, we can infer the following table:

Bi3+ + 3e- → Bi(s) In3+ + 3e- → In(s) Based on these two tables, it should be clear that In3+ is between Bi3+ and Ga3+.

In3+ can react with Ga(s) but will not react with Bi(s) The overall table will then be: Bi3+ + 3e- → Bi(s) In3+ + 3e- → In(s) Ga3+ + 3e- → Ga(s) Bi3+ is the ion most easily reduced; Ga(s) is the element most easily oxidized. This is also consistent with the diagonal rule: the spontaneous reaction will occur

between upper left/lower right positions on the chart. In3+ will react with Ga(s) (upper left/lower right) and will not react with Bi(s) (upper right/lower left).

Choice A

11. [IrCl2(NH3)4]I2 Correct name is tetraamminedichloroiridium (IV) iodide. Na3[Co(CN)6] Correct name is sodium hexacyanocobaltate (III). Na[FeCl4] Correct name is sodium tetrachloroferrate (III). K4[Fe(CN)6] Correct name is potassium hexacyanoferrate (II) [Pt(CO)3(H2O)]Br2 Correct name is aquotricarbonylplatinum(II) bromide Choice D

12. Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) This reaction will be pushed to the right, making Cu(OH)2(s) more soluble, if we

add acid. OH- is a base. The H+ ions from the acid will react with the OH- ions making H2O, and the removal of OH- ions pushes the reaction to the right.

CaCO3(s) Ca2+(aq) + CO3

2-(aq) This reaction will be pushed to the right, making CaCO3(s) more soluble, if we

add acid. CO32- is a base. The H+ ions from the acid will react with the CO3

2- ions making HCO3

-, or excess acid can react with CO32- making H2CO3 (which becomes

CO2 and H2O resulting in bubbling). The removal of CO32- ions pushes the reaction

to the right. AgCl(s) Ag+(aq) + Cl-(aq) Adding acid will have no effect on this equilibrium, because Cl- is not a base. It

is the conjugate “base” of the strong acid HCl, meaning that Cl- has negligible base properties in aqueous solution. Since HCl is 100% ionized, Cl- has no ability to attract H+ ions.

Choice D

13. Ag3PO4(s) 3Ag+(aq) + PO43-(aq)

[PO4

3-] = 4.4 x 10-5 M [Ag+] = 3(4.4 x 10-5) = 1.32 x 10-4 M Ksp = [Ag+]3[PO4

3-] = (1.32 x 10-4)3 (4.4 x 10-5) = 1.0 x 10-16 Choice C 14. A precipitate will form when Q > Ksp

AgBr will form when [Ag+][Br-] > 7.7 x 10-13 AgCl will form when [Ag+][Cl-] > 1.6 x 10-10 Initially [Br-] = 0.010 M and [Cl-] = 0.010 M

AgBr(s) will form when [Ag+] > 7.7x10−13

0.010 = 7.7 x 10-11 M

AgCl(s) will form when [Ag+] > 1.6x10−10

0.010 = 1.6 x 10-8 M

AgBr(s) forms first since it is less soluble and requires a lower [Ag+] to start precipitation. When AgCl precipitates, [Ag+] = 1.6 x 10-8 M as shown above. At that point:

[Br-] = 7.7x10−13

[Ag+ ] = 7.7x10

−13

1.6x10−8 = 4.8 x 10-5 M Choice A

15. Based upon the data in the problem, the overall cell reaction is: Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) Eo = 0.46 volts 0.46 volts is the standard potential of this cell. Standard potential means that all

concentrations are 1.0 M. In this problem [Ag+] = 0.031 M and [Cu2+] = 0.10 M To calculate the potential when we have non-standard concentrations, we use the

Nernst equation.

E = Eo - 0.0592n

log Q

E = 0.46 - 0.05922

log 0.10(0.031)2

= 0.40 volts Choice B

16. I2(s) + 2Cl-(aq) → 2I-(aq) + Cl2(g) This is a non-spontaneous reaction. Based on the Eo data in the problem, Eo for

this reaction = (0.53 – 1.36) = -0.83 volts

log K = nEo

0.0592 = 2(−0.83)

0.0592 = -28.04

K = 10-28.04 = 9.1 x 10-29 Choice A

17. 2Au(s) + 3Ca2+(aq) → 2Au3+(aq) + 3Ca(s) This is a non-spontaneous reaction. Based on the Eo values given in the problem,

the Eo for this cell reaction would be -1.50 – 2.87 = -4.37 volts ΔG = -nFE = - 6 (96500) (-4.37) = 2.53 x 106 J or 2.53 x 103 kJ Choice B 18. AgBr(s) ⇄ Ag+ (aq) + Br-(aq) Ksp = 7.7 × 10-13 Ag+(aq) + 2NH3(aq) ⇄ [Ag(NH3)2]+(aq) Kf = 1.7 × 107 Adding the reactions, and multiplying the K’s, we get: AgBr(s) + 2NH3(aq) ⇄  Ag(NH3)2

+(aq) + Br- (aq) K = 1.3 x 10-5 solid 1.5-2x x x

x2

(1.5− 2x)2 = 1.3 x 10-5 Solving: x = 5.4 x 10-3 M Choice B

19. Anode: Al → Al3+ + 3e- Cathode: 2H+ + 2e- → H2 Multiply anode half reaction by 2 and cathode half reaction by 3 in order to

cancel the electrons. Then add: 2Al + 6H+ → 2Al3+ + 3H2 Choice E 20. A precipitate will occur if Q > Ksp For PbCl2, the expression for Q and Ksp is [Pb2+][Cl-]2 In all three cases 100 mL of a Pb(NO3)2 solution is mixed with 100 mL of a NaCl

solution. This means the final volume will be 200 mL, and each solution will be

diluted by a dilution factor: initialvolumefinalvolume

= 100200

or 0.50

This means that for the Q test, we must use half the original concentrations because

of this mutual dilution. For case X this will be ½ (0.10) = 0.050 For case Y, this will be ½ (0.05) = 0.025. For case Z, this will be ½(0.01) = 0.0050

X: Q = [Pb2+][Cl-]2 = (0.050)(0.050)2 = 1.2 x 10-4 Q > Ksp Precipitate forms. Y. Q = [Pb2+][Cl-]2 = (0.025)(0.025)2 = 1.6 x 10-5 Q < Ksp No precipitate forms. Z. Q = [Pb2+][Cl-]2 = (0.005)(0.005)2 = 1.2 x 10-7 Q < Ksp No precipitate forms. Choice A 21. ΔHo = 9.16 = 2(33.18) = -57.2 kJ ΔSo = 304.3 – 2(240.1) = -175.9 J K-1

ΔGo = ΔHo - TΔSo = -57.2 – 623( −175.91000

) = 52.4 kJ Choice D

22. ΔG = ΔH -TΔS A reaction is spontaneous if ΔG < 0. In an endothermic reaction ΔH > 0. If ΔH > 0, then in order to make ΔG < 0, ΔS must be positive and the temperature

must be sufficiently high. Then the -TΔS term will be numerically bigger than the ΔH term.

Choice A

23. [Cr(en)2Cl2]+ has two geometric isomers, cis and trans, where in the cis isomer the Cl atoms are 90o from each other, and in the trans isomers the Cl atoms are 180o from each other.

The cis isomer, where the Cl astoms are 90o from each other, has non-

superimposable mirror images. So the cis isomer is split into two optical isomers which are mirror images of each other, but not identical.

The trans isomer has a superimposable mirror image, meaning that there is only

one version of the trans isomer. Thus, there are total of 3 isomers. Two versions of the cis isomer (right/left pair of

enantiomers) and one version of the trans isomer. This is illustrated on the bottom of page 915 of the textbook. Choice C 24. HCOOH() HCOOH(g) At the normal boiling point, liquid and vapor are in equilibrium when the pressure of the vapor = 1.00 atm. At that temperature ΔGo = 0. ΔHo = 363 – (-410) = 47 kJ ΔSo = 251 – 130 = 121 J K-1 ΔGo = ΔHo - TΔSo = 0 at the boiling point (when pressure of vapor = 1.00 atm)

47 – T( 1211000

) = 0 Solving T = 388 K Choice B

25. Oxidation half reaction: Cu(s) → Cu2+(aq) + 2e- Reduction half reaction: NO3

-(aq) + 4H+(aq) + 3e- → NO(g) + 2H2O() Add the half reactions, multiplying the oxidation half reaction by 3 and the reduction

half reaction by 2 in order to cancel the electrons. The result is: 3Cu(s) + 8H+(aq) + 2NO3

-(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O() Choice C