CHM 3410 – Problem Set 7
Due date: Wednesday, November 3rd Do the following problems. Show your work.
1) Using the half-cell potential data in Table 6.2 in the Appendix of Atkins (and not the thermochemical data table) find the numerical value for K for each of the following reactions
a) PbSO4(s) D Pb2+(aq) + SO42-(aq) (solubility product for lead (II) sulfate)
b) 2 Cu+(aq) D Cu(s) + Cu2+(aq) (disproportionation reaction for Cu+ ion)
2) The Debye-Huckel limiting law (equn 5.75) gives a theoretical relationship between log10 g± and I1/2, where I, the ionic strength, is defined by eq 5.76.
Using the data in Table 5.5 in the appendix of Atkins, test the validity of the Debye-Huckel limiting law by plotting log10 g± vs I1/2for KCl (a 1-1 electrolyte) and CaCl2 (a 1-2 electrolyte). Include the theoretical values on your plot. Limit your plot to data with b £ 0.5 .
EXTRA CREDIT. The extended Debye-Huckel law (equn 5.78) can be used to find an improved fit to experimental data for the mean activity coefficient (g±). Find a method for determining the values for B and C in equn 5.78 that gives the best agreement between the equation and the experimental data for KCl analyzed in part a (use A = 0.509). Plot log10 g± vs I1/2for KCl, and include a curve representing the best fit theoretical values from the extended Debye-Huckel theory.
3) The two dimensional Maxwell-Boltzmann distribution, which would apply, for example, to the free motion of particles adsorbed onto a surface, is
f(v) dv = N v exp(-Mv2/2RT) dv (3.1)
where N is a constant.
a) Find the value for N that makes f(v) dv a normalized distribution. This means finding the value for N that makes the distribution satisfy the following requirement
ò0¥ f(v) dv = 1 (3.2)
b) Find c, crms, and cmp for the two dimensional Maxwell-Boltzmann distribution.
4) Find crms (rms average speed), l (mean free path), and z (average number of collisions a molecule makes per unit time) for benzene vapor at T = 500. K and p = 1.000 atm. Also find f, the fraction of benzene molecules that have a speed between 500. m/s and 600. m/s, for the above conditions.
Exercises
6.20 b Write the cell reaction and electrod half-reactions and calculate the standard potential for each of the following cells:
a) Pt|Cl2(g)|HCl(aq)||K2CrO4(aq)|Ag2CrO4(s)|Ag(s)
b) Pt|Fe3+(aq),Fe2+(aq)||Sn4+(aq),Sn2+(aq)|Pt
c) Cu|Cu2+(aq)||Mn2+(aq),H+(aq)|MnO2(s)|Pt
Problems
6.16 Consider the cell Zn(s)|ZnCl2(0.0050 mol/kg)|Hg2Cl2(s)|Hg() for which the cell reaction is Hg2Cl2(s) + Zn(s) ® 2 Hg() + 2 Cl-(aq) + Zn2+(ag). Given that E°(Zn2+,Zn) = - 0.7628 V, E°(Hg2Cl2,Hg) = + 0.2676 V, and that the cell potential is + 1.2272 V (a) write the Nernst equation for the cell. Determine (b) the standard cell potential, (c) DGrxn, DG°rxn, and K for the cell reaction, (d) the mean ionic activity and activity coefficient of ZnCl2 from the measured cell potential, and (e) the mean ionic activity coefficient of ZnCl2 from the Debye-Huckel limiting law. (f) Given that (¶Ecell/¶T)p = - 4.52 x 10-4 V/K, calculate DSrxn and DHrxn.
Solutions.
1) a) ox Pb(s) ® Pb2+(aq) + 2 e- E° = + 0.13 V
red PbSO4(s) + 2 e- ® Pb(s) + SO42-(aq) E° = - 0.36 V
cell PbSO4(s) D Pb2+(aq) + SO42-(aq) E°cell = - 0.23 V
ln K = nFE°cell = (2) (96485. C/mol) (- 0.23 V) = - 17.91 so K = e-17.91 = 1.7 x 10-8
RT (8.314 J/mol.K) (298. K)
b) ox Cu+(aq) ® Cu2+(aq) + e- E° = - 0.16 V
red Cu+(aq) + e- ® Cu(s) E° = + 0.52 V
cell 2 Cu+(aq) D Cu(s) + Cu2+(aq) E°cell = + 0.36 V
ln K = nFE°cell = (1) (96485. C/mol) (+ 0.36 V) = + 14.02 so K = e14.02 = 1.2 x 106
RT (8.314 J/mol.K) (298. K)
Note that the large value for K in the above reaction is the reason why Cu+(aq) ion is unstable in aqueous solution.
2) a) From Debye-Huckel theory
log10g± = - | z+ z- | A I1/2 At T = 25. °C, A = 0.509 .
For KCl, I = (1/2) [ (1)2 bKCl + (-1)2 bKCl ] = bKCl
For CaCl2, I = (1/2) [ (2)2 bCaCl2 + (-1)2 (2 bCaCl2) ] = 3 bCaCl2
Therefore, for KCl, log10g± = - (1) (0.509) (bKCl)1/2 = - 0.509 (bKCl)1/2
and for CaCl2, log10g± = - (2) (0.509) (3 bCaCl2)1/2 = - 1.763 (bCaCl2)1/2
The data from Table 5.5 of the Appendix of Atkins is tabulated below. The data are plotted on the next page, along with the line representing the predicted values for log10g± based on Debye-Huckel theory.
b b1/2 g± (KCl) log10g±(KCl) g±(CaCl2) log10g±(CaCl2)
0.001 0.0316 0.966 - 0.0150 0.888 - 0.0516
0.005 0.0707 0.929 - 0.0320 0.789 - 0.1029
0.01 0.100 0.905 - 0.0434 0.732 - 0.1355
0.05 0.223 0.830 - 0.0809 0.584 - 0.2336
0.10 0.316 0.798 - 0.0980 0.524 - 0.2807
0.50 0.707 0.769 - 0.1141 0.510 - 0.2924
Based on the above plot it appears that Debye-Huckel theory works well at low molalityes for KCl, but not for CaCl2.
EXTRA CREDIT
log10g± = - | z+ z- | A I1/2 + C I
1 + B I1/2
If we apply this to KCl, with A = 0.509, C = 0, we get
log10g± = - 0.509 b1/2
1 + B b1/2
The easiest way to find the best value for B is by trial and error
b1/2 log10g±(KCl) B = 1 B = 2 B = 1.9 B = 1.8
0.0316 - 0.0150 - 0.0156 - 0.0151 - 0.0152 - 0.0152
0.0707 - 0.0320 - 0.0336 - 0.0315 - 0.0317 - 0.0319
0.100 - 0.0434 - 0.0463 - 0.0424 - 0.0428 - 0.0431
0.223 - 0.0809 - 0.0785 - 0.0797 - 0.0810
0.316 - 0.0980 - 0.0986 - 0.1005 - 0.1025
0.707 - 0.1141 - 0.1491 - 0.1536 - 0.1583
Based on the above I would say B = 1.8 gives the best agreement between the experimental values for log10g±(KCl) and the values calculated from extended Debye-Huckel theory. Note that the calculated values begin to deviate from the experimental values at b1/2 > 0.25
To get the best value for B and C at the same time, the above procedure would be repeated using different values of C, until a best fitting value for B and C were obtained.
3) a) Normalization means
1 = ò0¥ N v exp(- Mv2/2RT) dv = N ò0¥ v exp(- Mv2/2RT) dv
The general solution for this integral is
ò0¥ x2n+1 exp(-ax2) dx = n!/2an+1 ; n = 0, 1, 2, ...
In this case n = 0, a = M/2RT, and so
1 = N 2RT = NRT , and so N = M/RT
2 M M
b) c = ò0¥ v (M/RT) v exp(- Mv2/2RT) dv = (M/RT) ò0¥ v2 exp(- Mv2/2RT) dv
The general solution for this integral is
ò0¥ x2n exp(-ax2) dx = (pa)1/2 (1.3.5.....2n-1)/(2a)n+1 ; n = 0, 1, 2, ...
In this case n = 1, a = M/2RT, and so
c = (M/RT) (pM/2RT)1/2 = (pRT/2M)1/2
(M/RT)2
crms = { ò0¥ v2 (M/RT) v exp(- Mv2/2RT) dv }1/2 = { (M/RT) ò0¥ v3 exp(- Mv2/2RT) dv }1/2
We have given the general form of this integral above. In this case n = 1, a = M/2RT
crms = { (M/RT) (1/2) (2RT/M)2 }1/2 = (2RT/M)1/2
For cmp, we need
d f(v)/dv = 0 = d/dv [ (M/RT) v exp(- Mv2/2RT) ]
= (M/RT) [ exp(- Mv2/2RT) + v exp(- Mv2/2RT) (- 2vM/2RT) ]
= (M/RT) exp(- Mv2/2RT) [ 1 - v2M/RT ] = 0
For the above to be equal to zero the term in the square brackets must be zero, and so
1 - v2M/RT = 0 , v = (RT/M)1/2
4) For benzene, s = 0.88 x 10-18 m2.
p = 1.000 atm = 1.013 x 105 N/m2
T = 500. K
M = 78.12 x 10-3 kg/mol
crms = (3RT/M)1/2 = [3(8.3145 J/mol.K)(500.0 K)/(0.07812 kg/mol)]1/2 = 400. m/s
z = 4sNA p
(pMRT)1/2
So z = 4 (0.88 x 10-18 m2) (6.022 x 1023 mol-1) (1.013 x 105. N/m2) = 6.72 x 109 collisions/s
[ p (0.07812 kg/mol) (8.3145 J/mol.K) (500.0 K) ]1/2
l = RT = (8.314 J/mol.K) (500.0 K)
21/2 s NA p 21/2 (0.88 x 10-18 m2) (6.022 x 1023) (1.013 x 105. N/m2)
= 5.48 x 10-8 m = 54.8 nm
Note that we could also find l using the relationship l = z/c, where c = average speed = (8/3p)1/2 crms. The value for l found in this way is the same as above.
For f (500 m/s < v < 600 m/s) we will use the approximation
f(a < v < b) = òab f(v) dv @ f(vave) Dv vave = (a + b)/2 Dv = b - a
f(v) = 4p (M/2pRT)3/2 v2 exp(-Mv2/2RT)
vave = (500 + 600)/2 = 550 m/s Dv = (600 - 500) = 100 m/s
M = 78.12 x 10-3 kg/mol T = 500.0 K
f(vave) = 4p [0.07812/2p (8.3145) (500.0 ) ]3/2 (550)2 exp[ - (0.07812) (550)2/2 (8.3145) (500.0) ]
= 0.001146
So f(a < v < b) @ f(vave) Dv = (0.001146) (100) = 0.1146 (or 11.5 %)
For a more precise value we could divide the range into two rectangles, 500 – 550 m/s and 550 – 600 m/s. In fact, by using a larger number of smaller intervals we can find the fraction to an arbitrary degree of precision.
Exercise 6.20b
We use the cell convention that the anode is the left electrode and the cathode is the right electrode. Note that if E°cell < 0, that simply means the cell operates spontaneously in the reverse direction for standard conditions.
a) ox 2 Cl-(aq) ® Cl2(g) + 2 e- E° = - 1.36 V
red Ag2CrO4(s) + 2 e- ® 2 Ag(s) + CrO42-(aq) E° = + 0.45 V
cell Ag2CrO4(s) + 2 Cl-(aq) D 2 Ag(s) + Cl2(g) + CrO42-(aq) E°cell = - 0.91 V
b) ox 2 Fe2+(aq) ® 2 Fe3+(aq) + 2 e- E° = - 0.77 V
red Sn4+(aq) + 2 e- ® Sn2+(aq) E° = + 0.15 V
cell 2 Fe2+(aq) + Sn4+(aq) D 2 Fe3+(aq) + Sn2+(aq) E°cell = - 0.62 V
c) ox Cu(s) ® Cu2+(aq) + 2 e- E° = - 0.34 V
red MnO2(s) + 4 H+(q) + 2 e- ® Mn2+(aq) + 2 H2O() E° = + 1.23 V
cell Cu(s) + MnO2(s) + 4 H+(aq) D Cu2+(aq) + Mn2+(aq) + 2 H2O() E°cell = + 0.89 V
Problem 6.16
ox Zn(s) ® Zn2+(aq) + 2 e- E° = + 0.7628 v
red Hg2Cl2(s) + 2 e- ® 2 Hg() + 2 Cl-(aq) E° = + 0.2676 v
net Zn(s) + Hg2Cl2(s) ® 2 Hg() + Zn2+(aq) + 2 Cl-(aq) E°cell = + 1.0304 v
a) Ecell = E°cell - (RT/nF) ln Q Q = (aZn2+) (aCl-)2
Ecell = 1.0304 v - [ (8.3145 J/mol.K) (298.15 K) / 2 (96485 C/mol) ] ln Q
= 1.0304 v - (0.01285 v) ln Q
b) From above, E°cell = 1.0304 v
c) DG°rxn = - n F E°cell = - (2) (96485. C/mol) (1.0304 v) = - 198.8 kJ/mol
DGrxn = - n F Ecell = - (2) (96485 C/mol) (1.2272 v) = - 236.8 kJ/mol
lnK = n F E°cell = (2) (96485. C/mol) (1.0304 v) = 80.21 , K = exp(80.21) = 6.9 x 1034
RT (8.3145 J/mol.K) (298.15 K)
d) If we solve the Nernst equation in part a for ln Q, we get
ln Q = E°cell - Ecell = (1.0304 v - 1.2272 v) = - 15.32 , Q = exp( - 15.32 ) = 2.23 x 10-7
(RT/nF) (0.01285 v)
However, Q = (aZn2+) (aCl-)2 = [(bZnCl2) g± ] [(2 bZnCl2) g± ]2 = 4 (bZnCl2)3 g±3
g± =[ Q/4 (bZnCl2)3 ]1/3 Since bZnCl2 = 0.0050 mol/kg
g± =[(2.23 x 10-7) / 4 (0.0050)3 ]1/3 = 0.764
e) In problem 2 we gave the expression for log10g± for an ionic compound with formula MX2. So
log10g± = - (2) (0.509) (3 bZnCl2)1/2 = - 1.763 (bZnCl2)1/2 = - 1.763 (0.0050)1/2 = - 0.1247
g± = 10-0.1247 = 0.750
f) DSrxn = n F (¶Ecell/¶T)p = 2 (96485. C/mol) (- 4.52 x 10-4 v/K) = - 87.22 J/mol.K
DHrxn = DGrxn + T DSrxn = - 236.8 kJ/mol + (298. K) (- 87.22 x 10-3 kJ/mol.K) = -262.8 kJ/mol