Download (40 points)

Name____[KEY]_______________________
[KEY]
Date:______________
Period:______
Review Unit 12 Test (Chp 20): Electrochemistry
[KEY]
Section I Multiple Choice
1. The loss of electrons by an element is called __________.
A)
B)
C)
D)
E)
reduction
transduction
fractionation
oxidation
LEO says GER
potentiation
Fe2S3 + 12 HNO3  2 Fe(NO3)3 + 3 S + 6 NO2 + 6 H2O
2. Which substance oxidizes another substance in the reaction above?
A)
B)
C)
D)
E)
Fe2S3
S
NO2
HNO3
H2O
= gets reduced
N gets reduced from +5 (in HNO3) to +4 (in NO2)
–
–
MnO4 + Br  MnO2 + Br2
3. As the reaction represented above proceeds, the oxidation number of manganese changes from…
A)
B)
C)
D)
E)
+4 to 0
+7 to 0
+7 to +4
+8 to +4
+1 to +2
Mn in MnO4– :
(Mn) + 4(O)
(Mn) + 4(–2)
(Mn) + –8
Mn
=
=
=
=
Mn in MnO2 :
–1
–1
–1
+7
(Mn) + 2(O)
(Mn) + 2(–2)
(Mn) + –4
Mn
=
=
=
=
0
0
0
+4
4. In which of the following species does phosphorus have the same oxidation number as it does in
–
H2PO4 ?
P in H2PO4– :
P in P4O10 :
A) P O
4
B)
C)
D)
E)
6
P4O10
P8
HPO32–
H3PO3
2(H) + (P) + 4(O) = –1
2(+1) + (P) + 4(–2) = –1
2 + (P) + –8 = –1
P = +5
4(P) + 10(O) = 0
4(P) + 10(–2) = 0
4(P) + –20 = 0
P = +5
5. Which one of the following reactions is a redox reaction?
A)
B)
C)
D)
E)
H2O + NaCl  NaOH + HCl
Pb2+ + 2 Cl–  PbCl2
AgNO3 + HCl  HNO3 + AgCl
NaOH + HCl  NaCl + H2O
NONE of the above is a redox reaction.
1
…Pb + …H2O2 + …H+  …Pb4+
+ …H2O
6. What is the coefficient of the peroxide molecule when the equation above is balanced in acidic solution?
A)
B)
C)
D)
E)
1
2
3
4
8
0
+1 –1
+1
+4
+1 –2
…Pb + …H2O2 + …H+  …Pb4+ + …H2O
O
HE
BAL
RED:
2 e– + H2O2 + 2 H+  2 H2O
4 e– + 2 H2O2 + 4 H+  4 H2O
Pb  Pb4+ + 4 e–
OX:
Pb + 2 H2O2 + 4 H+  Pb4+ + 4 H2O
U
…Fe2+ + …O2 + …H+  …Fe3+ + …H2O
7. If 1 mole of O2 oxidizes Fe2+ in acidic solution according to the reaction represented above, how many
moles of Fe3+ ions can be formed?
A)
B)
C)
D)
E)
2
3
4
5
6
O
+2
0
+1
+3
+1 –2
2+
+
3+
…Fe + …O2 + …H  …Fe + …H2O
HE
RED: 4 e– + O2 + 4 H+  2 H2O
Fe+2  Fe+3 + e–
4 Fe+2  4 Fe+3 + 4 e–
OX:
BAL
4 Fe2+ + O2 + 4 H+  4 Fe3+ + 2 H2O
U
1 mol O2 x 4 mol Fe3+ = 4 mol Fe3+ ions
1 mol O2
Questions 8-9 refer to a galvanic cell constructed using two half-cells and using the two half-reactions
represented below.
Cu2+(aq) + e–  Cu+(aq)
Pb2+(aq) + 2 e–  Pb(s)
Eo = 0.15 V
Eo = –0.13 V
8. As the cell operates, ionic species that are found in the half-cell containing the anode include which of
the following?
I. Cu2+
A)
B)
C)
D)
E)
I only
II only
III only
I and III only
II and III only
II. Cu+
III. Pb2+
RED:
Cu2+(aq) + e–  Cu+(aq)
CAThode
OX:
ANode
Pb  Pb2+(aq) + 2 e–
9. What is the standard cell potential for the galvanic cell?
A)
B)
C)
D)
E)
–0.28 V
–0.02 V
0.02 V
0.28 V
0.43 V
E = Ered + Eox
+0.28 V = (+0.15) + (+0.13)
2
Eored = 0.15 V
°ox = +0.13 V
M(s) + 2 Cr3+(aq)  M2+(aq) + 2 Cr2+(aq)
Cr3+(aq) + e–  Cr2+(aq)
Eo = +0.15 V
Eo = –0.41 V
10. According to the information above, what is the standard reduction potential for the half-reaction
M2+(aq) + 2 e–  M(s)
A)
B)
C)
D)
E)
–0.97 V
–0.56 V
–0.26 V
0.26 V
0.56 V
Eo = ?
Cr3+(aq) + e–  Cr2+(aq)
RED:
M(s)  M2+(aq) + 2 e–
OX:
…OR…
Eo = Eored + Eoox
Eored = –0.41 V
Eoox = ?
OVR: M(s) + 2 Cr3+(aq)  M2+(aq) + 2 Cr2+(aq) Eo = +1.15 V
0.15 = (–0.41) + (x)
0.56 = x
Eoox (M) = +0.56
Eored(M) = –0.56 V
11. Which one of the following is the most effective choice to reduce another substance?
A)
B)
C)
D)
E)
O2
Li
Ca
F2
H2
Best at getting “oxidized”
Loses e–’s easily
Must be lithium (Group I metal)
LOWEST Eored (reduction potential)
LOWEST IE (ionization energy) , LOWEST Zeff
–
3 MnO4 (aq) + 24 H+(aq) + 5 Fe(s)  3 Mn2+(aq) + 5 Fe3+(aq) + 12 H2O(l)
12. True statements about the reaction represented above include which of the following?
–
I. MnO4 (aq) oxidizes another substance.
II. The oxidation state of hydrogen changes from +1 to 0.
III. Fe is oxidized.
A)
B)
C)
D)
E)
I only
III only
II and III only
I and III only
I, II, and III
I.
–
because Mn in MnO4 gets reduced from +7 to +2
so it must “oxidize” another substance
III. because Fe goes from 0 to +3 (loses e–’s)
Fe2+ + 2 Na(s)  Fe(s) + 2 Na+
13. If the ∆G° for the reaction above is +142 kJ/mol, which of the following correctly describes the
standard voltage, E°, and the equilibrium constant, K, for this reaction?
A)
B)
C)
D)
E)
E° is positive and K > 1
E° is negative and K < 1
E° is positive and K < 1
E° is negative and K > 1
E° is zero and K = 1
If ∆Go is positive, then…
If ∆Go is positive, then…
Eo is – and the reaction is
unfavorable.
∆Go = –nFEo
O
R
and Eo is – because…
(+) = –nF(–)
K must be < 1 because…
K must be < 1 because…
UNfavorable (forward) reactions
favor reactants at equilibrium.
∆Go = –RT ln K
and ∆Go is + so…
(+) = –RT ln K
and ln K must be –
3
14. A concentration cell using nickel electrodes as the anode and cathode was constructed, and the
observed voltage was found to be 1.00 volt instead of the standard cell potential, E°, of 0 volts.
Which of the following could correctly account for this observation?
A)
B)
C)
D)
E)
The nickel anode was larger than the nickel cathode
The electrolyte at the anode was Ni(NO3)2, while the electrolyte at the cathode was NiSO4
The Ni2+ solution at the anode was less concentrated than the Ni2+ solution at the cathode
The solutions in the half-cells had different volumes
The Ni2+ solution at the cathode was less concentrated than the Ni2+ solution at the anode
Q = [Ni2+]
[Ni2+]
If ° , then Q = (1.0)
(1.0)
.
Q = 1 and  = ° and ° = 0 V
But if E > Eo , then Q < 1 , and there must be more reactant so the reaction would shift right at a
faster rate to increase the voltage.
[Ni2+] must be lower at the anode so it increases as reaction proceeds to the right to make more Ni2+.
15. Consider an electrochemical cell based on the reaction:
Q = [Sn2+](PH2)
[H+]
2 H+(aq) + Sn(s)  Sn2+(aq) + H2(g)
Which of the following actions would change the measured cell potential?
A)
B)
C)
D)
E)
increasing the [Sn2+] in the anode compartment
increasing the pH in the cathode compartment
lowering the pH in the cathode compartment
increasing the pressure of hydrogen gas in the cathode compartment
ALL of the above will change the measure cell potential. if Q ≠ 1, then E ≠ °
16. In an electrolytic cell the sign of ∆G° is always __________ and is the reaction is _______________.
A)
B)
C)
D)
E)
positive, favorable
negative, favorable
negative, favorable
positive, unfavorable
zero, at equilibrium
17. A steady electric current is passed through molten MgCl2 for exactly 1.00 hour, producing 243 g of
Mg metal. If the same current is passed through molten AlCl3 for 1.00 hour, the mass of Al metal
produced is closest to
A)
B)
C)
D)
E)
27.0 g
54.0 g
120. g
180. g
270. g
A little tricky but…
243 g Mg x 1 mol Mg = 10 mol Mg
24.3 g Mg
10 mol Mg x
2 mol e– x 1 mol Al3+ x 27 g Al = 180 g Al
1 mol Mg2+
3 mol e–
1 mol
4
P
RED:
Cu2+(l) + 2 e–  Cu(s)
CAThode
Pt(s)
Cu(s)
Cl2(g)
OX:
ANode
Cl–(l)
Cu2+(l)
2 Cl–(l)  Cl2(g) + 2 e–
(answer choices show H2O is not a participant)
Questions 18-19 refer to an electrolytic cell that involves the electrolysis of CuCl2.
18. Which of the following occurs in the reaction?
A)
B)
C)
D)
E)
Cu is reduced at the cathode.
Cl– is oxidized at the anode.
Copper is converted from the 0 oxidation state to the +2 oxidation state.
CuCl2 is reduced at the cathode.
Cl– acts as an oxidizing agent.
19. Which of the following expressions is correct for the maximum mass of copper, in grams, that could
be plated out by electrolyzing aqueous CuCl2 for 16 hours at a constant current of 3.0 amperes?
A) (16)(3,600)(3.0)(63.55)
(96,485)(2)
B) (16)(3,600)(3.0)(63.55)
(96,485)
C) (16)(3,600)(3.0)(63.55)(2)
(96,485)
D) (16)(60)(3.0)(96,485)
(63.55)(2)
E) (16)(60)(3.0)(96,485)(2)
(63.55)
16 hr x 60 min x 60 s x 3.0 C x 1mol e– x 1 mol Cu2+ x 63.55 g Cu = ___ g Cu
1 hr 1 min
1s
96485 C
2 mol e–
1 mol Cu
20. If 0.060 faraday is passed through an electrolytic cell containing a solution of In3+ ions, the
maximum number of moles of In that could be deposited at the cathode is…
A)
B)
C)
D)
E)
0.010 mole
0.020 mole
0.030 mole
0.060 mole
0.18 mole
0.060 F = 0.060 mol e–
(at 96,485 C per mol e–)
0.060 mol e– x 1 mol In3+ = 0.020 mol In
3 mol e–
5
Questions 21-24 refer to the electrochemical cell and reaction below.
The spontaneous reaction that occurs when the cell above operates is:
2 Ag+ + Cd(s)  2 Ag(s) + Cd2+
Q = [Cd2+]
[Ag+]2
(A) Voltage increases.
(B) Voltage decreases but remains positive.
(C) Voltage becomes zero and remains at zero.
(D) No change in voltage occurs.
(E) Direction of voltage change cannot be predicted without additional information.
Which of the above occurs for each of the following circumstances?
21. A 50-milliliter sample of a 2-molar Cd(NO3)2 solution is added to the left beaker.
(B) Voltage decreases but remains positive.
Q = [Cd2+]
(1.0)2
.
Q > 1 and E < Eo There is more product
so the reaction shifts left at a faster rate
which decreases the voltage.
22. The silver electrode is made larger.
(D) No change in voltage occurs.
Ag(s) is not in the Q expression and does not affect °.
23. The salt bridge is replaced by a platinum wire.
(C) Voltage becomes zero and remains at zero.
Salt bridge is required to transfer ions and
maintain charge balance as electrons flow in the cell.
24. Current is allowed to flow for 5 minutes.
Q = (>
(B) Voltage decreases but remains positive.
1.0)
(< 1.0)
6
As rxn occurs, reactants decrease and
products increase, so…
Q > 1 and E < Eo There is more product
so the reaction shifts left at a faster rate
which decreases the voltage.
Questions 25-27 refer to galvanic cells made from different combinations of the three half–cells
described below.
Half–cell 1: strip of Cu(s) in 1.00 M Cu(NO3)2(aq)
Half–cell 2: strip of Ag(s) in 1.00 M Ag(NO3)2(aq)
Half–cell 3: strip of Cr(s) in 1.00 M Cr(NO3)3(aq)
Galvanic
Cell
Half–cells
Cell Reaction
E°cell (V)
A
1 and 2
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
0.46
B
1 and 3
2 Cr(s) + 3 Cu2+(aq)  2 Cr3+(aq) + 3 Cu(s)
?
C
2 and 3
Cr(s) + 3 Ag+(aq)  Cr3+(aq) + 3 Ag(s)
1.54
25. What is the standard cell potential of galvanic cell B?
(A)
(B)
(C)
(D)
2.00 V
1.08 V
–1.08 V
–2.00 V
No SRPs of elements given so
Eo = Eored + Eoox
is useless.
The rxn in Cell B can be made by combining the rxns in cells A & C:
Cell A:
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
° = 0.46 V
Cu (aq) + 2 Ag(s)  Cu(s) + 2 Ag (aq)
° = –0.46 V
2+
Cell C:
Cell B:
+
Cr(s) + 3 Ag+(aq)  Cr3+(aq) + 3 Ag(s)
2 Cr(s) + 3 Cu2+(aq)  2 Cr3+(aq) + 3 Cu(s)
° = 1.54 V
° = ?
26. In galvanic cells A and B, which of the following takes place in half–cell 1 ?
(A)
(B)
(C)
(D)
Reduction occurs in both cell A and cell B.
Cu(s) in 1.00 M Cu(NO3)2(aq)
Oxidation occurs in both cell A and cell B.
Reduction occurs in cell A, and oxidation occurs in and cell B.
Oxidation occurs in cell A, and reduction occurs in and cell B.
Cell A:
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
(Cu gets oxidized to Cu2+)
Cell B: 2 Cr(s) + 3 Cu2+(aq)  2 Cr3+(aq) + 3 Cu(s)
(Cu2+ gets reduced to Cu)
27. If the half-cell containing 1.00 M Ag(NO3)2(aq) in galvanic cells A and C is replaced with a half-cell
containing 0.10 M Ag(NO3)2(aq), what will be the effect on the cell voltage of the two galvanic cells?
(A)
(B)
(C)
(D)
The voltage will increase in both cells.
The voltage will decrease in both cells.
The voltage will increase in cell A and decrease in cell C.
The voltage will decrease in cell A and increase in cell C.
Cell A:
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
Q = [Cu2+]
[Ag+]2
Q = (1.0)
(0.10)2
Q > 1 ,  < °
Cell C:
Cr(s) + 3 Ag+(aq)  Cr3+(aq) + 3 Ag(s)
Q = [Cr3+]
[Ag+]3
Q = (1.0)
(0.10)3
Q > 1 ,  < °
7
Section II
Free Response Questions
YOU MAY USE YOUR CALCULATOR FOR SECTION II
Directions: Read each question carefully and write your response in the space provided following each
question. Your responses to these questions will be scored on the basis of the accuracy and relevance of
the information cited. Explanations should be clear and well organized. Specific answers are preferable to
broad, diffuse responses. For calculations, clearly show the method used and the steps involved in
arriving at your answers. It is to your advantage to do this, since you may obtain partial credit if you do
and you will receive little or no credit if you do not.
1.
An unknown metal M forms a soluble compound, M(NO3)2.
(a) A solution of M(NO3)2 is electrolyzed. When a constant current of 2.50 amperes is
applied for 35.0 minutes, 3.06 grams of the metal M is deposited. Calculate the molar
mass of M and identify the metal.
35.0 min x 60 s x 2.50 C x 1 mol e– x 1 mol M = 0.0272 mol M
3.06 g M = 112 g/mol = Cd
1 min
1s
96,485 C 2 mol e–
0.0272 mol M
(b) The metal identified in (a) is used with zinc to construct a galvanic cell shown below.
Write the net ionic equation for the cell reaction and calculate the cell potential, E.
K
+
Cl
-
M e ta l
M
1 .0 M
Z in c
1 .0 M
M (N O 3)2
RED: Cd2+ + 2e–  Cd
OX:
Zn  Zn2+ + 2e–
Half-Reaction
Eo (V)
M2+(aq) + 2e–  M(s)
–0.40
Zn2+(aq) + 2 e–  Zn(s)
–0.76
Z n S O4
Ered = –0.40 V
Eox = +0.76 V
Cd2+ + Zn  Cd + Zn2+
E = Ered + Eox
+0.36 V = (–0.40) + (+0.76)
(c) Predict the sign of the standard free energy change, G, at 25C for this reaction. Explain.
  
G is – because the galvanic cell described is a favorable reaction under standard reactions with
° that is + which produces a voltage.
(d) How would the cell potential, cell , differ from the cell shown in (b) if the initial concentration
of ZnSO4 is 0.10-molar, but the concentration of the M(NO3)2 solution remains unchanged.
Q = [Zn2+]
[Cd2+]
Q = (0.10)
(1.0)
.
Q < 1, therefore E > Eo because there is more reactant so the reaction would shift right at a faster rate.
8
2. An electrochemical cell consists of a tin electrode in an acidic solution of 1.0 molar Sn2+ connected by a
salt bridge to a second compartment with a silver electrode in an acidic solution of 1.0 molar Ag+. The
two reduction half-reactions for the overall reaction that occurs in the cell are shown in the table below.
Half-Reaction
Eo (V)
Ag+(aq) + e–  Ag(s)
+0.80
Sn2+(aq) + 2 e–  Sn(s)
–0.14
(a) Write the equation for the half–cell reaction occurring at the anode.
RED: Ag+ + e–  Ag
OX:
Sn  Sn2+ + 2e–
(anode reaction)
(b) Write the balanced chemical equation for the overall favorable cell reaction that
occurs when the circuit is complete. Calculate the standard voltage, E, for this cell.
RED:
Ag+ + e–  Ag
2 Ag+ + 2 e–  2 Ag
Ered = +0.80 V
Sn  Sn2+ + 2 e–
OX:
+
Eox = +0.14 V
2+
2 Ag + Sn  2 Ag + Sn
E = Ered + Eox
+0.94 V = (0.80) + (+0.14)
(c) Determine the value of the standard free energy change, Go, for the reaction at 25oC.

G = –nFE
= –(2)(96,485)(0.94) = –180,000 J/molrxn or –180 kJ/molrxn
(per molrxn b/c 2 mol e– per molrxn)
(d) Calculate the equilibrium constant for this cell reaction at 298 K.
ΔGo = −RT ln K
o/ RT
Keq = e–G
ln K = ΔGo
−RT
e–(–180,000/(8.314 x 298)) = 3.57 1031
(e) A cell similar to the one described above is constructed with solutions that have initial
concentrations of 1.0 molar Sn2+ and 0.020 molar Ag+. How would the cell potential, cell ,
differ from the standard cell potential, °cell ? Justify your answer in terms of LeChatelier’s
principle.
Q = [Sn2+]
[Ag+]2
Q = (1.0)
(0.020)2
.
Q > 1, therefore E < Eo because there is more product so the reaction would shift left at a faster rate.
9
Collected H2(g)
3.
Collected O2(g)
Water was electrolyzed, as shown in the diagram above, for 5.61 minutes using a constant current of
0.513 ampere. A small amount of nonreactive electrolyte was added to the container before the
electrolysis began. The temperature was 298 K and the atmospheric pressure was 1.00 atm.
(a) Write the balanced equation for the half reaction that took place at the anode.
2 H2O(l)  O2(g) + 4 H+(aq) + 4 e–
(ANode is OXidation so H2O is oxidized from O–2 to O20)
(b) Write the balanced equation for the half reaction that took place at the cathode. Why is the volume
of O2(g) collected different from the volume of H2(g) collected as shown in the diagram?
2 e– + 2 H2O(l)  H2(g) + 2 OH–(aq)
(CAThode is REDuction so H2O is reduced from H+ to H20)
When water decomposes according to the balanced equation 2 H2O(l)  2 H2(g) + O2(g) twice as
many moles of hydrogen are produced than moles of oxygen.
(c) Calculate the amount of electric charge, in coulombs, that passed through the solution.
5.61 min x 60 s x 0.513 C = 173 C
1 min
1s
(d) Calculate the number of moles of H2(g) produced during the electrolysis.
173 C x 1 mol e– x 1 mol H2 = 8.96 x 10–4 mol H2
96,485 C 2 mol e–
OR:
173 C x 1 mol e– x 2 mol H2 = 8.96 x 10–4 mol H2
96,485 C 4 mol e–
(e) Calculate the volume, in liters, at 298 K and 1.00 atm of dry H2(g) produced during the
electrolysis.
PV = nRT
(1.00)V = (8.96 x 10–4)(0.08206)(298)
10
V = 0.0219 L
–
5 Fe2+(aq) + MnO4 (aq) + 8 H+(aq)  5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)
4.
The mass percent of iron in a soluble iron(II) compound is measured using a titration based on the
balanced equation above.
–
(a) What is the oxidation number of manganese in the permanganate ion, MnO4 (aq)?
+7
Mn + 4(–2) = –1
(b) Identify the species that reduces another species in the reaction represented above.
Fe2+ reduces by getting oxidized from Fe2+ to Fe3+
The mass of a sample of the iron(II) compound is carefully measured before the sample is dissolved
in distilled water. The resulting solution is acidified with H2SO4(aq). The solution is then titrated
with MnO4–(aq) until the end point is reached.
(c) Let the variables g , M , and V be defined as follows:
g = the mass, in grams, of the sample of the iron(II) compound
–
M = the molarity of the MnO4 (aq) used as the titrant
–
V = the volume, in liters, of MnO4 (aq) added to reach the end point
–
In terms of these variables, the number of moles of MnO4 (aq) added to reach the end point of
the titration is expressed as M x V. Using the variables defined above, the molar mass of iron
(55.85 g∙mol–1), and the coefficients in the balanced chemical equation, write the expression for
each of the following quantities.
(i)
The number of moles of iron in the sample
–
–
L MnO4 x mol MnO4 x 5 mol Fe2+ = mol Fe2+
–
–
1 L MnO4
1 mol MnO4
(V)
x
(M)
x
5
= VxMx5
(ii)
The mass of iron in the sample, in grams
mol Fe2+ x 55.85 g Fe = mass Fe
1 mol Fe
V x M x 5 x 55.85 OR mol Fe x 55.85
(iii)
The mass percent of iron in the compound
mass Fe x 100 = mass % Fe
mass sample
V x M x 5 x 55.85 x 100 = mass % Fe
g
(d) What effect will adding too much titrant have on the experimentally determined value of the
mass percent of iron in the compound? Justify your answer.
The experimentally determined mass percent of the iron in the compound will be too large.
V is too large  expression in part (c)(iii) above will be too large.
11
 e– flow
cathode
5.
An electrochemical cell is constructed with an open switch, as shown in the diagram above. A strip
of Sn and a strip of unknown metal, X are used as electrodes. When the switch is closed, the mass of
the Sn electrode increases. The half-reactions are shown below.
Sn2+ (aq) + 2 e–  Sn(s)
X3+(aq) + 3 e–  X(s)
E˚ = –0.14 V
E˚ = ?
(a) In the diagram above, label the electrode that is the cathode. Justify your answer.
Sn electrode is the cathode because the cathode is the site of reduction (gain in electrons) and will convert
Sn2+ ions into neutral Sn metal increasing the mass of the electrode.
(b)In the diagram above, draw an arrow indicating the direction of electron flow in the external
circuit when the switch is closed.
(see diagram)
(c) If the standard cell potential E˚cell is +0.60 V, what is the standard potential, in volts for the
X3+/X electrode?
2+
0.60 = (–0.14) + x
x = +0.74
–
RED: Sn (aq) + 2 e  Sn(s)
E = Ered + Eox
OR
3+
E˚red = –0.14 V
X(s)  X (aq) + 3 e
OX:
–
E˚ox = x
E˚cell = +0.60 V
Eox(X) = +0.74 V
Ered(X) = –0.74 V
E˚ox (X) = +0.74 V
so… Ered(X) = –0.74 V
(d)Write balanced net-ionic equation for the overall chemical reaction in the cell.
RED: Sn2+ + 2 e–  Sn
3 Sn2+ + 6 e–  3 Sn
X  X3+ + 3 e–
2 X  2 X3+ + 6 e–
OX:
2+
3 Sn + 2 X  3 Sn + 2 X
3+
12
e) In the cell, the concentrations of Sn2+ and X3+ are changed from 1.0 M to 0.50 M.
(i)
Substitute all appropriate values into a Q expression. (Do not do any calculations.)
Q = [X3+]2
[Sn2+]3
(ii)
Q = (0.50)2
(0.50)3
.
On the basis of your response in (e)(i), will the cell potential be greater than, less than, or
equal to E˚cell? Explain.
Q > 1, therefore E < Eo because there is more product so the reaction would shift left at a faster rate.
6.
A galvanic cell is constructed using a chromium electrode in a 1.00 M solution of Cr(NO3)3 and a
silver electrode in a 1.00 M solution of Ag(NO3)2. Both solutions are at 25C.
The half-reactions are shown below.
Ag+ (aq) + e–  Ag(s) E˚ = +0.80 V
Cr3+(aq) + 3 e–  Cr(s) E˚ = –0.74 V
(a) Write a balanced net ionic equation for the favorable reaction that occurs as the cell
operates. Identify the species responsible for reducing and the species responsible for oxidizing.
RED:
Ag+ + e–  Ag
3 Ag+ + 3 e–  3 Ag
Cr  Cr3+ + 3 e–
OX:
+
3 Ag + Cr  3 Ag + Cr
3+
(b) A partial diagram of the cell is shown here.
(i)
Which metal is the anode?
Cr (gets oxidized)
Ag
Cu
Cr
(ii)
What additional component is
necessary to make the cell operate?
1 .0 M
salt bridge
(iii)
C r(N O 3)3
1 .0 M
CAg
u (N O 3)3
What function does the component in (ii) serve?
transfer of ions or charge (not electrons) to maintain charge balance
(c) How does the potential of this cell change if the concentration of AgNO3 is changed to 3.00 M at
25C? Justify your answer in terms of LeChatelier’s principle.
Q = [Cr3+]
[Ag+]3
Q = (1.0)
(3.0)3
.
Q < 1, therefore E > Eo because there is more reactant so the reaction would shift right at a faster rate.
13
Answer Key
1.
2.
3.
4.
2007
#3
e– flow, overall rxn, ∆G, electrolysis (PV=nRT)
Review
2005B #2(a)-(e)
Review
2004 #6
2000 #2
½ rxn, overall rxn, E, K, Ered (X), electrolysis
Review
1988 B (ELECTRO) ½ rxns, overall rxn, E, K, Nernst
2012 #6
activity series, AN ½ rxn, describe change in mass of electrodes, E & Q
Review
1993 D (ELECTRO) overall rxn, AN/CAT, salt bridge, Nernst
2010B #3
identify Ox#, redox titration stoich
Review
2007 #5
5.
Review
6.
Review
½ rxn, AN/CAT, electrolysis(H2O) (PV=nRT)
AN/CAT, e– flow, Ered (X), overall rxn, Nernst
identify Ox#, redox titration stoich (w/ variables not #’s)
1993 D (ELECTRO) overall rxn, AN/CAT, salt bridge, Nernst
14