solutions to examples
Post on 30-Dec-2015
104 Views
Preview:
DESCRIPTION
TRANSCRIPT
SOLUTIONS TO EXAMPLES
Solubility of AgCl(s) in water at 25oC is 1.274 x 10-5 mol kg-1.
Calculate the solubility of AgCl(s) in 0.010 mol kg-1 Na2SO4(aq).
i
2iizm
2
1I 030.0)2)(010.0()1)(020.0(
2
1 22
I < 0.05 Use Debye-Hückel law
IzzAlog
176.0030.0)2)(1()509.0(log
666.0
Ksp = aAg+ aCl- For AgCl dissolving in H2O assume = 1 since m 0
Ksp = (1.27410-5)2 = 1.62310-10
For I = 0.030 mol kg-1 Ignore Ag+ and Cl- in solution as conc’s v. low
Ksp = aAg+ aCl- = mAg+ mCl-
1.62310-10 = (0.666)2 m2
m = 1.9110-5 mol kg-1 = solubility
In the presence of Na2SO4 the solution is no longer ideal calculate activity coeff’s
AgCl(s) Ag+(aq) + Cl-(aq) calculate [Ag+] or [Cl-]
Calculate Ksp for the ideal soln and assume it be the same for the non-ideal soln
Example 0Example 0
Consider the Daniell cell:
Using the standard reduction potentials, calculate the equilibrium constant at 25 C.
Standard reduction potentials at 25 C:
Cu2+ + 2e- Cu(s) Eo = +0.34 V
Zn2+ + 2e- Zn(s) Eo = −0.76 V
Overall reaction:
Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)
The cell potential at standard conditions:
)s(Cu)aq(CuSO)aq(ZnSO)s(Zn 44
Since K is large reactiongoes to completion
oanode
ocathode
ocell EEE = +1.10 V
ERT
nFKln
At equilibrium: Ecell = 0 V
K = 1.6 1037
Example 1Example 1
Reduction rxn
Oxidation rxn
)V10.1()K298)(molKJ315.8(
)molC10649.9)(2(11
14
Units: KmolKJ
VmolC11
1
J
VC But J = C V w = V q
E°cell > 0 cell reaction is spontaneous expect K > 1
Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)
Electrochemical reaction involved: 2H+(aq) + 2e H2(g)
Let E1 and E2 be potentials for the initial and final states, i.e. for 5 and 20 mmol kg–1
The mean activity coefficients of HBr in 5.0 and 20.0 mmol kg–1 are 0.930 and 0.879, respectively. Consider a hydrogen electrode in HBr(aq) solution at 25 °C operating at 1.15 atm.
Calculate the change in the electrode potential when the molality of the acid solution is changed from 5.0 and 20.0 mmol kg–1.
QlnnF
RTEE o 2
H
2Ho
a
Pln
nF
RTEE
mV2.34V0342.0)100.20)(879.0(
)100.5)(930.0(log05916.0E
3
3
232
o1 )100.5()930.0(
)15.1(log
2
05916.0EE
232
o2 )100.20()879.0(
)15.1(log
2
05916.0EE
)15.1(
)100.5()930.0(
)100.20()879.0(
)15.1(log
2
05916.0EEE
232
23212
and
Example 2Example 2
Ecell > 0 Galvanic cell spontaneous reaction written as above
Reduction: Cl2(g) + 2e− 2Cl–(aq) E° = +1.36 V
Suggest a Pt electrode as cathode.
Oxidation: Mn(s) Mn2+(aq) + 2e−
Mn2+(aq) + 2e− Mn(s) E° = ? V
Suggest a Mn electrode as anode.
E°cell = E°cathode – E°anode = 1.36 V – x = 2.54 V
Mn(s) | Mn2+(aq), Cl–(aq) | Cl2(g) | Pt(s)
Devise a cell in which the cell reaction is:
Mn(s) + Cl2(g) MnCl2(aq)
Give the half reactions at the electrodes and from the standard cell potential of 2.54 V deduce the standard potential for the Mn2+/Mn(s) redox couple.
Given: E°(Cl2/Cl-) = +1.36 V
E°anode = -1.18 V or E° for the redox couple Mn2+ / Mn(s) = -1.18 V
Example 3Example 3
Half reactions:
Cell:
Standard potential for the Mn2+/Mn(s) redox couple:
(R-H): Cd2+(aq) + 2e Cd(s) E° = –0.40 V
(L-H): 2AgBr(s) + 2e 2Ag(s) + 2Br–(aq) E° = +0.07 V
Cell reaction: Cd2+(aq) + 2Ag(s) + 2Br–(aq) Cd(s) + 2AgBr(s)
E°cell = E°cathode – E°anode = E°R-H – E°L-H = – 0.47 V
Need to calculate activities:
ICd(NO3)2 = 0.010 and IKBr = 0.050
hence we can use the Debye-Hückel limiting law
Estimate the cell potential at 25°C for
Ag(s)|AgBr(s)|KBr(aq, 0.050 mol kg–1)||Cd(NO3)2(aq,0.0034 mol kg–1)|Cd(s)
E°(R-H) = –0.40 V E°(L-H) = +0.07 V (assume non-ideal solutions)
Write the spontaneous electrochemical reaction.
2
BrCdaa
1Q
2
QlnnFRT
EE cello
cell
Example 4Example 4
Ecell < 0 non-spontaneous electrochemical reaction
Cd2+(aq) + 2Ag(s) + 2Br–(aq) Cd(s) + 2AgBr(s)
The spontaneous electrochemical reaction:
Cd(s) + 2AgBr(s) Cd2+(aq) + 2Ag(s) + 2Br–(aq)
-0.076 -0.084Ecell = –0.63 V
2BrCd
ocellcell
aa
1log
2
05916.0EE BrCd alog05916.0alog
2
05916.047.0
Brmlog05916.0
010.0)1)(2(509.0log
050.0)1)(1(509.0log
791.0
769.0
Cdmlog2
05916.0 )0034.0)(791.0(log
2
05916.0
)050.0)(769.0(log05916.0
= -0.076
= -0.084
-0.47
Note: the cell is considered at standard conditions and Ecell > 0
(R-H) Ag+(aq) + e- Ag(s) E° = +0.80 V
(L-H) AgI(s) + e- Ag(s) + I–(aq) E° = –0.15 V
Spontaneous electrochemical reaction:
Ag+(aq) + I–(aq) AgI(s) E°cell = +0.95 V
K = 1.11016
Therefore Ksp = K–1 = 8.710–17
Ksp = [Ag+] [I-] = [Ag+]2 = 8.7 10–17
Solubility = [AgI(aq)] = 9.410–9 mol kg–1
The standard potential of the cell below at 25 °C is 0.95 V.
Ag(s) |AgI(s) | AgI(aq) | Ag(s)
Calculate: a) its solubility constant and b) the solubility of AgI .
ERT
nFKln
AgI(s) Ag+(aq) + I–(aq)
Example 5Example 5
Calculate the degree of ionization and the acid dissociation constant at 298 K for a 0.010 M acetic acid solution that has a resistance of 2220 . The resistance of a 0.100 M potassium chloride solution was also found to be 28.44 . m(0.1 M KCl) = 129 S cm2 mol-1
o(H+) = 349.6 S cm2 mol-1 o(CH3COO-) = 40.9 S cm2 mol-1
omm o
m
m
om
12om molcmS5.390)6.349)(1()9.40)(1(
cm
R
C
To find the cell constant (C), we can use the data for the KCl solution.
Degree of ionisation, :
where
Example 6Example 6
om
m
12om molcmS5.390
1235
14
m molcmS5.16cmmol100.1
cmS1065.1
c
141
cmS1065.12220
cm366.0
R
C
Degree of ionisation, :
R
C
cm
Using KCl data to find the cell constant (C):
11234m cmS0129.0)molcmS129)(cmmol1000.1(c
c = 0.100 M = 0.100 mol dm-3 = 1.0010-4 mol cm-3
11 cm366.0)44.28)(cmS0129.0(RC
Finding m:
0423.05.390
5.16om
m
or 4.23%
Acid dissociation constant, Ka:
]COOHCH[
]COOCH][OH[K
3
33a
CH3COOH + H2O CH3COO- + H3O+
c]OH[ 3
c]COOCH[ 3
c1]COOHCH[ 3
c)1(
)c)(c(Ka
5
22
1087.1)0423.0(1
)010.0()0423.0(
1
c
Also pKa = -log(1.8710-5) = 4.73
The molar conductivity of a strong electrolyte in water at 25 °C was found to be 109.9 S cm2 mol-1 for a concentration of 6.2 10-3 mol L-1 and 106.1 S cm2 mol-1 for a concentration of 1.5 10-3 mol L-1.
Estimate the limiting molar conductivity of the electrolyte.
Note: Strong electrolyte Kohlrausch law
Using K, calculate limiting molar conductivity from Kohlrausch law.
Subtract two equations, solve for K.
Note: is the same in for both eqn’som
cKomm
121m2m ccK
2/112
33MmolcmS0.95
105.1102.6
1.1069.109
12
1m2m
ccK
cKomm
M102.6)MmolcmS0.95(molcmS9.109 32/112om
12 12o
m molcmS4.102
= -0.095 S cm-2
)Lmol102.6(K)molcmS9.109( 13om
12
)Lmol105.1(K)molcmS1.106( 13om
12
2 equations, 2 unknowns!
Example 7Example 7
The limiting molar conductivities of KCl, KNO3, and AgNO3 are 149,9, 145.0, and 133.4 S cm2 mol-1, respectively (all at 25 °C).
What is the limiting molar conductivity of AgCl at this temperature?
To solve: manipulate the 3 equations above to obtain the one for AgCl
We can apply the Kohlrausch law of independent migration of ions.
)Cl()Ag()AgCl(om
Recall:
om
12om molcmS9.149)Cl()K()KCl(
1233
om molcmS0.145)NO()K()KNO(
1233
om molcmS4.133)NO()Ag()AgNO(
)KNO()KCl()AgNO()AgCl( 3om
om3
om
om
1212 molcmS3.138molcmS0.1459.1494.133
Example 8Example 8
The mobility of the NO3- ion in aqueous solution at 25 °C is 7.4010-8 m2 s-1 V-1.
(Viscosity of water is 0.89110-3 kg m-1 s-1).
Calculate its diffusion coefficient and the effective radius at this temperature.
Calculate the hydrodynamic radius:
Use the Einstein relation between the mobility and diffusion coefficient:
Having “a” and the radius of a simple ion (without coordinated water) plus the dimension of a single water molecule, you would be able to predict the number of molecules in the hydrated share of the ion.
Remember: in the calculations you have to show the work on units.
Without that, the work might be considered as not done at all.
m1029.1)sm1090.1)(smkg10891.0(6
)K298)(KJ10381.1( 10129113
123
Or use the equation: a6
zeu
J = kg m2 s-2
J = V C
zF
RTuD 129
1
111128sm1090.1
)molC48596)(1(
)K298)(molKJ315.8)(Vsm1040.7(
D6
kTa
Example 9Example 9
zF
RTuD
Considering units:
)molC(
)K)(molKJ)(Vsm(1
11112
CJVsm 112
BUT w = qV J = C V
CVCVsm 112
12 sm
)sm)(smkg(
)K)(KJ(1211
1
D6
kTa
2smkg
J
BUT J = kg m2 s-2
2
22
smkg
smkg
m
Also: V = IR V = A and q = It C = As
Note that we deal with large and positive overpotential.
Hence the Tafel equation for anodic current can be applied.
= 0.138 V
j0 = 2.82 mA cm–2
The electron transfer coefficient of a certain electrode in contact with the redox couple M3+ / M4+ in aqueous solution at 25 °C is 0.39. The current density is found to be 55.0 mA cm–2 when the overvoltage is 125 mV.
What is the overvoltage required for a current density 75.0 mA cm–2?
What is the exchange current density?
RT
nF)1(
o jej
)(RT
nF)1(
j
jln 12
1
2
M3+ M4+ + e-
j1 = 55.0 mA cm–2 1 = 0.125 V j2 = 75.0 mA cm–2 2 = ?
F = 96485 C mol-1 R = 8.315 J K-1 mol-1 T = 298 K n = 1 = 0.39
Example 10Example 10
RT
nF)1(
oejj
The relatively high positive overpotential applied results in very little reduction taking place.
for a 2 electron process
The exchange current density and the electron transfer coefficient for the reaction 2H+ + 2e H2(g) on nickel at 25 °C are 6.3 10–6 A cm–2 and 0.58, respectively.
Determine what current density would be required to obtain an overpotential of 0.20 V as calculated from the Butler-Volmer equation and the Tafel equation.
RT
nF
RT
nF)1(
o eejjButler-Volmer:
Tafel:
00012.046.693)103.6(j 6
23 cmA1037.4
RT
nF)1(
oejjPositive potential Anodic current
46.693)103.6(j 623 cmA1037.4
F = 96485 C mol-1 R = 8.315 J K-1 mol-1 T = 298 K n = 2
= 0.58 = 0.20 V Jo = 6.3 10–6 A cm–2
Example 11Example 11
top related