CHM 3410 – Problem Set 8
Due date: Wednesday, October 31st
Do all of the following problems. Show your work.
"V'ant Hoff was the first chemist to make the connection between free energy and voltage, but Nernst was the one
who ended up with his name on the equation." - Patrick Coffey, Cathedral of Science.
1) Data for the equilibrium constant for a chemical reaction is often fit to a power series expansion in powers of
(1/T). Consider the following expression for ln K
ln K = a + (b/T) + (c/T 2)
(1.1)
where a, b, and c are constants.
Find an expression for Grxn, Hrxn, and Srxn for a reaction whose equilibrium constant is given by
equn. 1.1.
2) For the galvanic cell
Pt(s)|H2(g)|H+(aq)||Cl-(aq)|Cl2(g)|Pt(s)
(2.1)
it is found (at T = 25.0 C) that Ecell = 1.362 v and (Ecell/T)p = - 1.20 x 10-3 v/K.
a) Give the half-cell oxidation reaction, the half-cell reduction reaction, and the net cell reaction
corresponding to the above galvanic cell. The net cell reaction should be the same as reaction 2.2.
b) Using only information given in the problem, find Grxn, Hrxn, and Srxn for the reaction
H2(g) + Cl2(g)  2 HCl(aq)
(2.2)
3) Using only half-cell reduction data (Table 7.2 of the appendix of Atkins) find the numerical value for the
equilibrium constant for the following reactions at T = 25. C.
a) Cd(OH)2(s)  Cd2+(aq) + 2 OH-(aq)
b) Sn(s) + Sn4+(aq)  2 Sn2+(aq)
4) Europium has two common cationic forms, Eu2+ and Eu3+. Half-cell potentials involving these two ions inclusde
the following (values for E are at T = 25. C).
Eu2+(aq) + 2 e-  Eu(s)
E = - 3.442 v
Eu3+(aq) + 3 e-  Eu(s)
E = - 2.407 v
(4.1)
(4.2)
Based on these half-cell potentials, find the value for E at T = 25. C for the process
Eu3+(aq) + e-  Eu2+(aq)
HINT: E4.3 + E4.1  E4.2, but G4.3 + G4.1 = G4.2.
(4.3)
5) Self-ionization of water is the reaction
H2O()  H+(aq) + OH-(aq)
(5.1)
At T = 25.0 C, the equilibrium constant for the above reaction is Kw = 1.011 x 10-14.
If we assume that aH2O = 1, as is approximately true for dilute solutions, then
Kw = (aH+) (aOH-)  [H+] [OH-] ()2 = KM ()2
(aH2O)
KM = [H+] [OH-]
(5.2)
where KM is the product of the concentrations of hydrogen and hydroxide ions and () = ( H+ OH-)1/2 is the mean
activity coefficient for hydrogen and hydroxide ion.
The value for KM has been measured for a series of aqueous solutions of potassium chloride, KCl, a 1:1
strong electrolyte (Harned, H. S. and W. J. Hamer, J.Amer.Chem.Soc. 55, 2194 (1933)). Some of the data are given
below.
bKCl (mol/kg)
KM
bKCl (mol/kg)
KM
0.01
0.02
0.03
0.04
1.24 x 10-14
1.33 x 10-14
1.39 x 10-14
1.44 x 10-14
0.06
0.11
0.21
0.51
1.51 x 10-14
1.64 x 10-14
1.76 x 10-14
1.89 x 10-14
a) For each of the above solutions find the experimental value for , the mean activity coefficient.
b) Theoretical vales for the mean activity coefficient can be found using Debye-Huckel theory
log10() = - A |z+ z-| I1/2
(3.3)
I = (1/2) i zi2 bi
(3.4)
where
For water at T = 25.0 C, A = 0.509. For solutions of potassium chloride (ignoring the small contribution to I from
H+ and OH- ions), I = bKCl.
Find the value for  predicted using Debye-Huckel theory for each of the above solutions.
EXTRA CREDIT - Plot the experimental values for log10() vs I1/2. Also indicate in the plot the line that
represents the values for log10() predicted using Debye-Huckel theory. Briefly discuss the agreement (or lack of
agreement) of Debye-Huckel theory with the experimental results.
Solutions.
1)
Grxn = - RT ln K = - RT { a + (b/T) + (c/T2) } = - R { aT + b + (c/T) }
Hrxn = - R (d(ln K)/d(1/T)) = - R (d/d(1/T)) { a + (b/T) + (c/T 2) }
= - R { b + (2c/T) }
Srxn = - ((d Grxn)/dT) = R (d/dT) { (aT + b + (c/T) } = R { a – (c/T2) }
As a check, we note that
Grxn = Hrxn - TSrxn, as expected.
2)
a) The half-cell and net cell reactions are
ox
H2(g)  2 H+(aq) + 2 ered
Cl2(g) + 2 e-  2 Cl-(aq)
_____________________________
net
H2(g) + Cl2(g)  2 HCl(aq)
b) Grxn = - FEcell = - (2) (96485. C/mol) (1.362 v) = - 262.8 kJ/mol
Srxn = F(dEcell/dT) = (2) (96485. C/mol) (- 1.20 x 10-3 v/K) = - 231.6 J/mol.K
Finally, since Grxn = Hrxn - T Srxn
Hrxn = Grxn + T Srxn = ( - 262.8 kJ/mol) + (298.15 K) ( - 0.2316 kJ/mol.K) = - 331.9 kJ/mol
3)
a)
ox
Cd(s)  Cd2+(aq) + 2 e-
red
Cd(OH)2(s) + 2 e-  Cd(s) + 2 OH-(aq)
_________________________________________
net
Cd(OH)2(s)  Cd2+(aq) + 2 OH-(aq)
E = + 0.40 v
E = - 0.81 v
____________
Enet = - 0.41 v
K = (aCd2+)(aOH-)2
(aCd(OH)2)
ln K = FEnet = (2)(96485 C/mol)(- 0.41 v) = - 31.92 , so K = e-31.92 = 1.4 x 10-14
RT
(8.3145 J/mol.K)(298.15 K)
a)
ox
Sn(s)  Sn2+(aq) + 2 e-
red
Sn4+(aq) + 2 e-  Sn2+(aq)
_______________________________
net
Sn(s) + Sn4+(aq)  2 Sn2+(aq)
K=
E = + 0.14 v
E = + 0.15 v
____________
Enet = + 0.29 v
(aSn2+)2
(aSn)(aSn4+)
ln K = FEnet = (2)(96485 C/mol)(+ 0.29 v) = + 22.57 , so K = e22.57 = 6.4 x 109
RT
(8.3145 J/mol.K)(298.15 K)
4)
To do this problem we need to work with free energy, and use the the relationship
G = - FE
Eu3+(aq) + e-  Eu2+(aq)
G4.3
Eu2+(aq) + 2 e-  Eu(s)
_____________________
Eu3+(aq) + 3 e-  Eu(s)
G4.1 = 664.2 kJ/mol
______
G4.2 = 696.7 kJ/mol
where we have used the numbering of reactions in the problem, and the given half-cell potentials to find G
So
G4.3 + G4.1 = G4.2
G4.3 = G4.2 - G4.1 = (696.7 kJ/mol) – (664.2 kJ/mol) = 32.5 kJ/mol
Since
G = - FE ,
E - G/F =
5)
a) To find experimental values for  we used the relationship
-
(32500. J/mol)
= - 0.337 v
(1) (96485. C/mol)
Kw = KM ()2
 = (Kw/KM)1/2
b) For theoretical values for  we use the Debye-Huckel theory.
log10() = - A |z+ z- | I1/2
At T = 25.0 C, A = 0.509. For H+ and OH- ions | z+ z- | = 1. Finally, if we ignore the small concentration of ions
generated by the self ionization reaction, I = bKCl.
The information calculated from the above and used to carry out the plot is given on the next page. Note
that in a plot of log10() vs I1/2 the theoretical values of log10() will fall on a line with y-intercept equal to 0.000
and slope equal to - 0.509.
bKCl
I1/2
()exp
log10()exp
()DH
log10()DH
0.0100
0.0200
0.0300
0.0400
0.1000
0.1414
0.1732
0.2000
0.9030
0.8719
0.8528
0.8379
- 0.0443
- 0.0596
- 0.0691
- 0.0768
0.8894
0.8473
0.8163
0.7910
- 0.0509
- 0.0720
- 0.0882
- 0.1018
0.0600
0.1100
0.2100
0.5100
0.2449
0.3317
0.4583
0.7141
0.8183
0.7852
0.7579
0.7314
- 0.0871
- 0.1050
- 0.1204
- 0.1359
0.7504
0.6779
0.5844
0.4330
- 0.1247
- 0.1688
- 0.2333
- 0.3635
Notice that there are significant deviations even at relatively low values of I1/2. If we do a best fit to a second order
polynomial, we get
log10() = - 0.0045 - 0.4158 I1/2 + 0.3228 I
which gives a better fit to the experimental data.