Semester 131 Exam I All Answers are A:

Useful Physical constants and Formula 

PHYSICAL CONSTANTS:	Useful Equations:	

R = 8.314 J mol-1 K-1		L = k/c ,  G =1/R= k (A/l)       	
R = 0.08314 L bar K-1 mol-1		= Fu+ ; u+ = speed L/v	
kB = 1.381 x 10-23 J K-1		 =  u- ; u+ = speed L/v	
1 J = 1 kg m2 s-2		L = l+ + l-	
1 atm = 1.101325 x 105 Pa		a = L/Lo	
b = 0.51 mol-1/2 dm3/2    (in H2O, 25oC)		       	
1 atm = 1.01325 x 105 Pa; 1 bar = 105 Pa; 		
1 Torr = 1 mmHg = 133.322 Pa		
RT/F= 0.0257 at 298 K
g = 9.81 m s-2
0 °C= 273.15 K	DF = RT/zF ln (c1/c2)	
L = 6.022 x 1023 mol-1 (Avogadro's number)	DG = - zFE	
h = 6.626 x 10-34 J s	E = E° – (RT/z F)ln(k)         	
F = 96500 C mol-1	DS = z F (¶E/¶T)p =-(¶ΔG/¶T)p	
c = 2.998 x 108 m s-1               (speed of light)	Ln x = 2.303 log x	
DG=RT ln (c1/c2)
DG= DH-TDS	pH = - Log [H+]
Kw= [OH-] [ H+]	
E° = (RT/z F)ln(ko)         	t+ = u+/(u++u-); t+ = I+/(I++I-)	
1/k = (eoekBT/e2Siciz2iL)1/2                                    
                  (thickness of ionic atmosphere)	eo = 8.854x10-12C2N-1m-2
e = 78 at 25oC
e = 1.602x10-19 C	
aa' = t+ Q/FcA               (travelled distance)		

1.	Write the overall reaction of the following electrochemical cell: 

Ag½AgCl(s) ½KCl(aq) ½Hg2Cl2(s) ½Hg
	
A)	2Ag(s) + Hg2Cl2(s) ® 2AgCl(s) +2Hg(l)	
B)	Ag(s) + Hg2Cl2(s) ® AgCl(s) +Hg(l)	
C)	Ag(s) + AgCl(s) ® Hg(s) + Hg2Cl2(s)	
D)	2Ag(s) + 2AgCl(s) ® 2Hg(s) + Hg2Cl2(s)	
E)	2AgCl(s) ® Hg2Cl2(s)	

2.	The solubility product (Ksp) of BaSO4 is 9.2 x 10-11 mol2dm-6. Calculate the solubility s (M) of BaSO4 in a solution of 0.10 M in KNO3 and 0.10 M KCl assuming DHLL? (B = 0.51)
	
A)	7.83 x 10-5	
B)	1.24 x 10-5	
C)	2.74 x 10-5	
D)	3.07 x 10-5	
E)	4.04 x 10-5

3.	Two solutions of equal volume are separated by a membrane which is permeable to K+ and Cl- ions but not to P- ions. The initial concentrations are:   
  
[K+] =[Cl-] =  0.025 M½ [K+] = [P-] = 0.100 M

Calculate the equilibrium concentration of the chloride Cl- ion (M) on the right hand-side of the membrane.
	
A)	0.0042	
B)	0.0011	
C)	0.040	
D)	0.010	
E)	0.020	

4.	The standard glass electrode used in the pH measurements is:
	
A)	Permeable to H+ ion	
B)	Permeable to Na+ and K+ ions	
C)	Permeable to Li+ and Cs+ ions 	
D)	Permeable to acetate CH3COO- ion	
E)	Permeable to all metal ions	

5.	At 25oC, a solution containing compound A and its reduced form AH2 has a standard electrode potential of -0.52 V. A solution containing B and BH2 has a standard potential of -0.23 V. A cell is constructed with these systems as half-cells where the overall reaction is spontaneous in the forward direction. What is the emf in volts of the cell?
	
A)	0.29	
B)	0.75	
C)	0.98	
D)	0.50	
E)	0.20	

6.	According to Faraday Law, the amount of material generated at the electrode surface during electrolysis is NOT: 
	
A)	related to the geometric shape of the electrode	
B)	proportional to the increase in the mass of the electrode	
C)	related to the quantity of electricity passed	
D)	related to the stoichiometric coefficients of the reactants	
E)	related to its molecular weight	

7.	The Arrhenius theory for the variation of molar conductivity with concentration is applicable to:
	
A)	Weak electrolytes	
B)	Strong electrolytes	
C)	Both weak and strong electrolytes	
D)	Non electrolytes	
E)	Both strong electrolytes and non electrolytes	

8.	Calculate the equilibrium constant of 1.00% dissociated acetic acid at a concentration of 0.10 M
	
A)	1.01x10-5 	
B)	1.10x10-4 	
C)	2.32x10-5 	
D)	0.01x10-5 	
E)	3.10x10-5 

9.	The relationship of the ionic atmosphere of 0.01 M CuCl to 0.01 M CuCl2 at 250C and eH2O = 78 can be stated as follows:
	
A)	The ionic atmosphere thickness of CuCl is 1.7 times that for CuCl2	
B)	The ionic atmosphere thickness for CuCl2 is 1.7 times that for CuCl	
C)	The ionic atmosphere thickness for CuCl2 is 3 times that for CuCl	
D)	The ionic atmosphere thickness for CuCl is 3 times that for CuCl2	
E)	The ionic atmosphere thickness for both electrolytes is the same	

10.	A very weak acid (HA) exhibits a molar conductivity of 2.45 mS m2 mol-1 at 298K for a 0.015 M concentration. Given the limiting ionic conductivity of H+ and A- as 33.76 mS m2 mol-1 and 27.50 m S m2 mol-1, calculate the equilibrium dissociation constant of HA.
	
A)	2.50 x 10-5 	
B)	3.97 x 10-5 	
C)	4.15 x 10-5 	
D)	5.27 x 10-5 	
E)	1.38 x 10-5 	

11.	Given,

the potential of the cell,

at 25 oC, will be greater than zero if the activity x of Ni2+ is greater than,
	
A)	0.122	
B)	0.616	
C)	1.61	
D)	2.10	
E)	0.476	

12.	Given the following Eo values,

Eo for the process:          
                        
is:	
A)	–0.912  V	
B)	 0.336  V	
C)	 1.824  V	
D)	–1.824  V	
E)	–0.336  V	

13.	Given,

	calculate, at 25 °C, the equilibrium constant for the reaction,

A)	2.3 x 10-32	
B)	3.8 x 10-27	
C)	4.7 x 1031	
D)	3.5 x 1026	
E)	0.97 x 10-23	

14.	The electrode potential E of the cell:   

is given by,

where mu refers to the molality of HCl (without units) and is its mean activity coefficient at that concentration.
Use the information given in the plot below to obtain the mean activity coefficient,, for a 0.1000 m HCl solution.:
	
A)	0.9156	
B)	1.0003	
C)	0.4492	
D)	0.004530	
E)	0.8163	

15.	At 25 °C the measured emf of the cell:

is -1.0137 V.  Calculate the solubility product Kosp for AgBr given that,

	 
	
A)	5.35 x 10-13	
B)	1.90 x 10-13	
C)	1.27 x 10-32	
D)	6.33 x 10-12	
E)	3.45 x 10-22	

16.	The electrolytic conductivity is 3.82 S m-1 for a solution 0.1 M in KCl and 0.2 M in NaCl (a strong electrolyte). Calculate the ionic molar conductivity (lNa+) for Na+ ion in S m2mol-1.

               [Note: lK+ = 7.4 × 10-3  and lCl- = 7.6 × 10-3 S m2mol-1]
	
A)	4 × 10-3  S m2 mol-1	
B)	5 × 10-3  S m2 mol-1	
C)	6 × 10-3  S m2 mol-1	
D)	7 × 10-3  S m2 mol-1	
E)	8 × 10-3  S m2 mol-1	

17.	In the moving boundary experiment a 10.0 mA current was passed through a column of a 0.10 M LiCl solution. Given that the cross-sectional area of the column is 2.0 x 10-4 m2 and that of the ionic molar conductivities of Li+ and Cl- are 4.0 x 10-3 and 7.5 x 10-3 S m2 mol-1, respectively, calculate the speed (m sec-1) of the Li+ ion. 
 	
A)	1.8 x 10-6 	
B)	3.2x 10-6	
C)	7.1 x 10-6	
D)	8.3 x 10-6	
E)	9.0 x 10-6	

18.	After correction for the water conductivity, the conductivity of a saturated aqueous solution of AgCl at 25 oC was found to be 0.1887 
mS m-1. What is the solubility of silver chloride at this temperature?    
[lo of Ag+ = 61.9 S cm2 mol-1 and lo of Cl- = 76.31 S cm2 mol-1]
	
A)	1.36 x 10-5 mol L-1	
B)	2.14 x 10-3 mol L-1	
C)	5.12 x 10-5 mol L-1	
D)	2.65 x 10-4 mol L-1	
E)	7.21 x 10-7 mol L-1	

19.	The mobilities of H+ and Cl- at 25 oC in water are 3.623 x 10-7 m2 s-1 V-1 and 7.91 x 10-8 m2 s-1 V-1, respectively. What fraction of the current is carried by the protons in 1.0 x 10-3 M HCl(aq)?
	
A)	0.82	
B)	0.72	
C)	0.54	
D)	0.14	
E)	0.05	

20.	A solution of MnO42- is electrolytically reduced to Mn3+. A current of 6.74 amp is passed through the solution for 15.0 minutes. What is the number of moles of Mn3+ produced in this process? (1 faraday = 96,486 coulombs)
	
A)	0.0211 mol	
B)	0.0512 mol	
C)	0.0900 mol	
D)	0.0176 mol	
E)	0.0748 mol	

Answer Key

1.	A	
2.	A	
3.	A	
4.	A	
5.	A	
6.	A	
7.	A	
8.	A	
9.	A	
10.	A	
11.	A	
12.	A	
13.	A	
14.	A	
15.	A	
16.	A	
17.	A	
18.	A	
19.	A	
20.	A	

Semester 122:
PHYSICAL CONSTANTS:	Useful Equations:	

R = 8.314 J mol-1 K-1		 =     , G =1/R=  A/L       	
R = 0.08314 L bar K-1 mol-1		= F u+ ; 
u+ = speed x Length/Voltage	
kB = 1.381 x 10-23 J K-1		 = F u- ; u = speed L/V	
1 J = 1 kg m2 s-2		 = + 	
1 atm = 1.101325 x 105 Pa		 = 	
B = 0.51 mol-1/2 dm3/2 (in H2O, 25oC)			
1 atm = 1.01325 x 105 Pa; 1 bar = 105 Pa; 		
1 Torr = 1 mmHg = 133.322 Pa		
RT/F= 0.0257 at 298 K
g = 9.81 m s-2
0 °C= 273.15 K	 = RT/zFln (c1/c2)	
L = 6.022 x 1023 mol-1 Avogadro's number	G- zFE	
h = 6.626 x 10-34 J s	E = E – (RT/z F)ln(K)         	
F = 96485 C mol-1	S = z F (E/T)p=-(ΔG/T)p	
c = 2.998 x 108 m s-1	Ln x = 2.303 log x	
G=RT ln (c1/c2)
G= H-TS	pH = - Log [H+]
Kw= [OH-] [ H+]	
E = (RT/z F)ln(Ko)         	t+ = u+/(u++u-); t+ = I+/(I++I-)	

Hydrogen Electrode: H+ (aq) + e- (g) →½H2			Eo= 0 V
Silver-Silver Chloride: e- + AgCl (s) →Ag (s) + Cl- (aq)	Eo=0.22233 V
Calomel Electrode: e- + ½Hg2Cl2(s) → Cl- (aq) + Hg (l)	Eo=0.3337 V

1)	How many grams of aluminum nitrate, Al(NO3)3, must be dissolved in 3.0 L of water to obtain a solution that has the same ionic strength as a 0.10 M sodium sulfate, Na2SO4, solution? Assume that both electrolytes are strong.
	(Molar mass of Al(NO3)3 = 213.01)

a) 88 g
	b) 27 g
	c) 13 g
	d) 32 g
	e) 43 g

2)	Which of the following solutions of strong electrolytes has the highest  conductivity()?

a) 0.0200 M.
b) 0.0100 M.
c) 0.0200 M.
d) 0.0200 MHCl.
e) 0.0100 M. 

3)	The solubility of strontium sulfate (SrSO4) in pure water is 3.1 x 10-5 M at 25 oC. What is the solubility of strontium sulfate in an aqueous solution that is 0.30 M in NaCl, and 0.50 M in NaBr? Is this change in solubility a salting in or a salting out effect?  Select the completely correct answer from the following choices.
	
Choice	Solubility in a solution
that is 0.3 M in NaCl
and 0.5 M in NaBr
	This effect
is called	
a)	2.1x10-3 M	salting in	
b)	1.0x10-3 M	salting in	
c)	2.1x10-3 M	salting out	
d)	1.0x10-3 M	salting out	
e)	6.5x10-8 M	salting in	

4)	What is the value of the constant current needed to deposit 0.061 g Au at the cathode when passed for 3.0 h through a solution of Au2(SO4)3.
	(Atomic mass of Au: 196.9665 g mol-1)

a)	2.4 mA 
b)	8.3 mA
c)	30 A
d)	8.6 A
e)	4.2 mA

5) 	At 25 oC a weak acid H2A has molar conductivities  at a concentration of moldm3 and at infinite dilution.   The equilibrium constant for this acid is given by:

a)	
b)	
c)	
d)	
e)	

6)	How, in accordance with the Debye-Huckel limiting law, will the pH of a 0.1 M aqueous solution of a weak acid, HX, change if enough NaCl is added to make its solution 0.01 M in NaCl?
	
a)	No change because NaCl is a neutral salt.The pH increases because the ionic strength increases.
b)	The pH decreases because NaCl is acidic.
c)	The pH decreases because NaCl decreases the ionic strength.
d)	The pH decreases because NaCl increases the ionic strength.

7)	Knowing that EoZn2+/Zn = - 0.7618 V and EoCr3+/Cr = - 0.744 V we expect that in the cell,
Zn(s)|Zn2+(1.0 M), Cr3+(2.0 M )|Cr(s)
	
	25oC,

a)	the zinc electrode dissolves.
b)	the chromium electrode dissolves.
c)	chromium is deposited at the chromium electrode.
d)	the zinc ion concentration increases.
e)	the chromium ion concentration decreases.

8)	A 0.0100 M solution of a weak acid AH has a molar conductivity of 300 S cm2 mol-1.  Given (HX)=425 S cm2mol-1,(AM)= 75 S cm2 mol-1 and (MX)= 100 S cm2 mol-1 calculate Kc for this acid.  (at 25 oC).

a)	1.63x10-2
b)	1.29x10-3
c)	2.25x10-2
d)	1.29x10-2
e)	8.00x10-3
9)	The ionic atmosphere model explains that the asymmetry or relaxation effect that a central ion experiences under the influence of an external electrical field 
	is a result of:
	a) A dragging effect caused by solvent viscosity.
	b) A dragging effect caused by oppositely charged ions.
	c) Dragging caused by solvent molecules moving in an opposite direction.
d) The neutralization of ions.
e) An excluded volume effect.

10)	What is the emfin volts of the cell,

		Pt(s), Cl2 (g, 1 bar)| NaCl (aq, 0.01 M)|Hg2Cl2(s)|Pt(s)

at 25oCgiven that the solubility product of mercurous chloride, Hg2Cl2, is 3.1 x 10-10 and Eo = -1.0257 V?

a)	1.3   
b)	-1.3   
c)	0.28
d)	-0.28 
e)	2.82 

11) 	A system with 2.000 dm3 of 0.0150 M of the polyelectrolyte Na3P (aq) in compartment 1 is separated from 2.000 dm3 of 0.0500 M NaCl (aq) in compartment 2 by a membrane permeable to Na+ and Cl only is allowed to reach equilibrium.  The equilibrium concentration of Cl in compartment 2 is:

a) 0.0145 M
b) 0.1000 M
c) 0.0500 M
d) 0.0122 M
e) 0.0328 M
12)	The thickness (r) of the ionic atmosphere for strong electrolytes in a solvent with a dielectric constant  at temperature T is given by:

	whereCi is the concentration of ion i in mol/L and Ziis its charge. According to the equation, which of the two strong aqueous electrolyte solutions; 0.001 M NaCl or 0.001M PbF3 at the same temperature would have a larger average separation between its oppositely charged ions at 25 C and why?

a) The PbF3 solution because it has the small flouride ions. 
b) The PbF3 solution because it has ions of higher valence.
c) The NaCl solution because it has the large chloride ion.
d) The PbF3 solution because it has more ions and ions with higher valence.  
e) The  NaCl solution because its has fewer ions and the ions have lower valence.

13) 	What is the standard emf, Eo, for the reaction

	Cr2O72- + 14 H+ + 12 e- → 2 Cr + 7 H2O
using
	Cr2O72- + 14 H+ + 6 e- → 2 Cr3+ + 7 H2O 	Eo = 1.33 V
and
	Cr3+ + 3 e- → Cr	Eo = -0.744 V

a)	-0.293   V
b)	1.037    V
c)	0.293    V
d)	0.586    V
e)	-0.586   V

14)	Given,
	
	Calculate the standard solubility product of at 25 oC.
	a)	
	b)	
	c)	
	d)	
	e)		

15)	At 25 oC the cell

	Cd(s)|CdCl2(aq, 0.05M)|AgCl(s)|Ag(s)

has an emf of 0.7041 V and Eo = 0.5717 V.  Calculate the mean activity coefficient()of the Cd2+ and Cl-in the 0.05 M CdCl2 solution.

a)	56 %
b)	30 %
c)	9 %
d)	0.406
e)	5 %

16)	The potential Eo of the standard cell,

is 0.5385 V at 278 K and 0.5355 V at 298 K.  If in this temperature range Eo varies linearly with temperature, calculate H o for the reaction,

at 288 K.

a) -51.9 KJ mol-1
b) -47.7 KJ mol-1
c) 104 KJ mol-1
d) 95 KJ mol-1	
	e) -95 KJ mol-1

17) 	Given the following information

Claim (or statement)	Reason	
Eo for the cell 
is negative thus the silver electrode is the negative electrode.	At the electrode atoms are oxidized to ions and electrons are released	
	
	which of the following choices is correct?	
Choice	Claim/Statement	Reason	
a)	False	False	
b)	True	False	
c)	True	True and correct explanation of the statement	
d)	False	True	
e)	True	True but not a correct explanation of the statement	

18)	The conductivity of a 0.1 M solution of NaOH is 0.0221 S cm-1. When an equal volume of 0.1 M HCl is added, the conductivity falls to 0.0056 S cm-1. Calculate  of NaCl.
a)	351 S cm2 mol-1
b)	2215 S cm2 mol-1
c)	265 S cm2 mol-1
d)	65.5 S cm2 mol-1
e)	112 S cm2 mol-1

19) 	Calculate the velocity of Li+ ions in m sec-1 in a 0.1 M LiCl, when a current of 1 A is passed through a column of the solution. The cross-sectional area of the column is 2 x 10-4 m2, and the molar conductivities of Li+ and Cl- ions are 4 x 10-3 S cm2 mol-1and 7.5 x 10-3 S cm2 mol-1, respectivity.
a)	1.1 x 10-5 m sec-1
b)	2.1 x 10-4 m sec-1
c)	4.3 x 10-4 m sec-1
d)	1.8 x 10-4 m sec-1
e)	6.8 x 10-5 m sec-1

20)	Use the reduction half-reaction: 

CrS (s) + 2e- Cr(s) + S(aq) 	with E0= - 0.400 V
	
and assume DHLL to apply andthat H2SO4dissociatesto completely dissociate in water,to calculate the EMF in volts forof the following cell at 25 C:

				Cr(s)|CrSO4(s)|H2SO4 (aq, 0.00100 m)|H2(g,1bar), Pt

a)	-0.653 V0.336
b)	0.147  V
c)	0.653  V0.648
d)	0.556  V
e)	-0.147 V

The student is asked to change molalities to activities. This requires him to calculate the ionic strength then use the Debye Huckel limiting law to get and then multiply these by their molalities to get and  then use these in the Nernst equation to get E. This does not seem to me like a 6-minute-question.  Here we have four concepts in one question.  I suggest removal of “DHLL to apply and” as I show above and in this case the correct answer would be 0.648  V.   The concentration is so low that the activities are very close in value to the molalities.

Answer Key 001.
1	D	
2	C	
3	A	
4	B	
5	D	
6	E	
7	A	
8	C	
9	B	
10	A	
11	E	
12	E	
13	C	
14	B	
15	D	
16	B	
17	A	
18	E	
19	D	
20	C