SOLUTIONS TO EXAMPLES. Solubility of AgCl(s) in water at 25 o C is 1.274 x 10 -5 mol kg -1. Calculate the solubility of AgCl(s) in 0.010 mol kg -1 Na.

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