Ch17.1 Galvanic Cells Ch4: Redox involves the transfer of
electrons (OIL RIG) Zn (s) + CuSO 4(aq) ZnSO 4(aq) + Cu (s) Half
rxns: Ox:Red agent: _____ Red: Ox agent: _____
Slide 2
lose 2e- (oxidized) Zn (s) + CuSO 4(aq) ZnSO 4(aq) + Cu (s)
gained 2e- (reduced) Ox Reaction: Zn (s) Zn (aq) +2 +2e - Red
agent: Zn Red Reaction: Cu +2 (aq) + 2e - Cu (s) Ox agent: Cu
+2
Slide 3
lose 2e- (oxidized) Zn (s) + CuSO 4(aq) ZnSO 4(aq) + Cu (s)
gained 2e- (reduced) Oxidized Reaction: Zn (s) Zn (aq) +2 +2e - Red
agent: Zn Reduced Reaction: Cu +2 (aq) + 2e - Cu (s) Ox agent: Cu
+2 e - e - e - e - Cu Zn CuSO 4 soln ZnSO 4 soln V
Slide 4
lose 2e- (oxidized) Zn (s) + CuSO 4(aq) ZnSO 4(aq) + Cu (s)
gained 2e- (reduced) Oxidized Reaction: Zn (s) Zn (aq) +2 +2e - Red
agent: Zn Reduced Reaction: Cu +2 (aq) + 2e - Cu (s) Ox agent: Cu
+2 e - e - e - e - Cu Zn CuSO 4 soln ZnSO 4 soln V
Slide 5
Galvanic cell (Voltaic cell, wet cell battery) - convert
chemical potential energy into electrical energy Electrodes -
metals in voltaic cells Anode - negative electrode, electrons
produced here (Reducing agent - OIL) Cathode - positive electrode,
electrons head here (Ox agent - RIG) e - e - e - e - electrons
transfer Na + Cl - Na + Cl - Na + Cl - Na + Cl - Na + thru the wire
Cl - (salt bridge) Cl - CuNa + (necessary to maintain Na + Zn Cl -
an ion charge balance) Cl - CuNa + Na + Zn Cl - Cu Zn CuSO 4 soln
ZnSO 4 soln V
Slide 6
Cell Potential, cell Reduction Half Cell Oxidation Half Cell Cu
+2 + 2e Cu Zn Zn +2 + 2e Zn Zn +2 + 2e - E = +.76V Cu +2 + 2e - CuE
= +.34 V E cell =+1.10 V electrons transfer Na + Cl - Na + Cl - Na
+ Cl - Na + Cl - Na + thru the wire Cl - (salt bridge) Cl - CuNa +
(necessary to maintain Na + Zn Cl - an ion charge balance) Cl -
CuNa + Na + Zn Cl - Cu Zn Cu +2 SO 4 -2 Zn +2 SO 4 -2 SO 4 -2 Cu +2
SO 4 -2 Zn +2 CuSO 4 soln ZnSO 4 soln cathode cell anode cell E
standard conditions 25C, I Molar V
Slide 7
Cell Potentials
Slide 8
Ex1) Al +3 (aq) + Mg (s) Al (s) + Mg +2 (aq) Give the balanced
cell rxn and EMF for the cell.
Slide 9
Ex1) Al +3 (aq) + Mg (s) Al (s) + Mg +2 (aq) Give the balanced
cell rxn and EMF for the cell. Al +3 + 3e - AlE o = 1.66 V Mg +2 +
2e - Mg E o = 2.37 V
Slide 10
Ex1) Al +3 (aq) + Mg (s) Al (s) + Mg +2 (aq) Give the balanced
cell rxn and EMF for the cell. Al +3 + 3e - AlE o = 1.66 V Mg +2 +
2e - Mg E o = 2.37 V 2Al +3 + 6e - 2AlE o = 1.66 V 3Mg 3Mg +2 + 6e
- E o = +2.37 V 2Al +3 (aq) + 3Mg (s) 2Al (s) + 3Mg +2 (aq) E o =
+0.71 V Mg metalAl +3 gained e - s so lost e - s so it was reduced
it is oxidized. in charge. Formed Anode more Al on the electrode.
The metal the Mg | Mg +2 || Al +3 | Al es head to is the cathode
Anode cell Cathode cell V e-e- e-e- Mg Al Al +3 Mg +2 Dont
multiply!
Slide 11
Ex2) MnO 4 - (aq) + H + (aq) + ClO 3 - (aq) ClO 4 - (aq) + Mn
+2 (aq) + H 2 O (l) Give the balanced cell rxn and EMF for the
cell. MnO 4 - + 5e - + 8H + Mn +2 + 4H 2 O E o = +1.51 V ClO 4 - +
2e - + 2H + ClO 3 - + H 2 O E o = +1.19 V.
Slide 12
Ex2) MnO 4 - (aq) + H + (aq) + ClO 3 - (aq) ClO 4 - (aq) + Mn
+2 (aq) + H 2 O (l) MnO 4 - + 5e - + 8H + Mn +2 + 4H 2 O E o =
+1.51 V ClO 4 - + 2e - + 2H + ClO 3 - + H 2 O E o = +1.19 V 2MnO 4
- + 10e - + 16H + 2Mn +2 + 8H 2 O E o = +1.51 V 5ClO 3 - + 5H 2 O
5ClO 4 - + 10e - + 10H + E o = 1.19 V 2MnO 4 - (aq) + 6H + (aq) +
5ClO 3 - (aq) 5ClO 4 - (aq) + 2Mn +2 (aq) + 3H 2 O (l) E o = +0.32
V | || | Anode cell Cathode cell. V e-e- e-e- ? ? ? ? Mn +7 gains
5e - to become Mn +2 : reduced in charge cathode cell Cl +5 loses
2e - to become Cl +7 : oxidized anode cell
Slide 13
Ex2) MnO 4 - (aq) + H + (aq) + ClO 3 - (aq) ClO 4 - (aq) + Mn
+2 (aq) + H 2 O (l) MnO 4 - + 5e - + 8H + Mn +2 + 4H 2 O E o =
+1.51 V ClO 4 - + 2e - + 2H + ClO 3 - + H 2 O E o = +1.19 V 2MnO 4
- + 10e - + 16H + 2Mn +2 + 8H 2 O E o = +1.51 V 5ClO 3 - + 5H 2 O
5ClO 4 - + 10e - + 10H + E o = 1.19 V 2MnO 4 - (aq) + 6H + (aq) +
5ClO 3 - (aq) 5ClO 4 - (aq) + 2Mn +2 (aq) + 3H 2 O (l) E o = +0.32
V Since REDOX occurs w aqueous ions, we need a non-reactive metal
as electrodes, so Pt makes a great choice. H 2 gas bubbles off Pt |
ClO 3 -, ClO 4 -, H + || Mn +2, MnO 4 -, H + | Pt of cathode Anode
Cathode V e-e- e-e- Pt Mn +2 MnO 4 - H + ClO 3 - ClO 4 - H + Mn +7
gains 5e - to become Mn +2 : reduced in charge cathode cell Cl +5
loses 2e - to become Cl +7 : oxidized anode cell
Slide 14
Ex3) Ag + + e - AgE o = +0.80V Fe +3 + e - Fe +2 E o = +0.77V
Setup a galvanic cell, give the balanced cell rxn and EMF for the
cell.
Slide 15
Ex3) Ag + + e - AgE o = +0.80V Fe +3 + e - Fe +2 E o = +0.77V
Setup a galvanic cell, give the balanced cell rxn and EMF for the
cell. Ag + + e - Ag RIGE o = +0.80V Fe +2 Fe +3 + e - OILE o =
-0.77V Ag + + Fe +2 Ag + Fe +3 E o = +0.03V Since theres no Fe (s),
we need another metal to be the electrode. Pt is non-reactive, so
makes a great choice. Pt | Fe +2 (aq),Fe +3 (aq) || Ag + (aq) | Ag
Anode Cathode V e-e- e-e- Pt Ag Ag + Fe +2 Fe +3
Slide 16
Ex4) Fe +2 + 2e - FeE o = 0.44V MnO 4 - + 5e - Mn +2 + 4H 2 OE
o = +1.51V Setup a galvanic cell, give the balanced cell rxn and
EMF for the cell. Ch17 HW#1 p880 29,31,33 V e-e- e-e- ? ? ? ?
Slide 17
29. Sketch the galvanic cells based on the following
half-reactions. Show the direction of electron flow, show the
direction of electron flow, show the direction of ion migration
through the salt bridge, and identify the cathode and anode. Give
the overall balanced reaction, and determine E for the galvanic
cells. Assume that all concentrations are 1.0 M and that all
partial pressures are 1.0 atm. a. Cl 2 + 2e - 2Cl - E= 1.36 V Br 2
+ 2e - 2Br - E= 1.09 V b. MnO 4 + 8H + + 5e - Mn 2 + 4H 2 OE= 1.51
V IO 4 - + 2H + +2e - IO 3 - + H 2 OE= 1.60
Slide 18
Ch17 HW#1 p880 29,31,33 29. Sketch the galvanic cells based on
the following half-reactions. Show the direction of electron flow,
show the direction of electron flow, show the direction of ion
migration through the salt bridge, and identify the cathode and
anode. Give the overall balanced reaction, and determine E for the
galvanic cells. Assume that all concentrations are 1.0 M and that
all partial pressures are 1.0 atm. a. Cl 2 + 2e - 2Cl - E= 1.36 V
Br 2 + 2e - 2Br - E= 1.09 V Cl 2 + 2e - 2Cl - E= 1.36 V 2Br - Br 2
+ 2e - E= 1.09 V Cl 2 + 2Br - Br 2 + 2Cl - E= +0.27 V V e-e- e-e-
Pt Cl - Br - Br 2
Slide 19
Ch17 HW#1 p880 29,31,33 29. Sketch the galvanic cells based on
the following half-reactions. Show the direction of electron flow,
show the direction of electron flow, show the direction of ion
migration through the salt bridge, and identify the cathode and
anode. Give the overall balanced reaction, and determine E for the
galvanic cells. Assume that all concentrations are 1.0 M and that
all partial pressures are 1.0 atm. b. MnO 4 - + 8H + + 5e - Mn +2 +
4H 2 OE= 1.51 V IO 4 - + 2H + +2e - IO 3 - + H 2 OE= 1.60 V
Slide 20
b. MnO 4 - + 8H + + 5e - Mn +2 + 4H 2 OE= 1.51 V IO 4 - + 2H +
+2e - IO 3 - + H 2 OE= 1.60 V 5(IO 4 - + 2H + +2e - IO 3 - + H 2 O)
E= +1.60 V 2(Mn +2 + 4H 2 O MnO 4 - + 8H + + 5e - ) E= 1.51 V Mn +7
gains 5e - to become Mn +2 : reduced in charge cathode cell I +7
loses 2e - to become I +5 : oxidized anode cell
Slide 21
b. MnO 4 - + 8H + + 5e - Mn +2 + 4H 2 OE= 1.51 V IO 4 - + 2H +
+2e - IO 3 - + H 2 OE= 1.60 V 5IO 4 - + 10H + + 10e - 5IO 3 - + 5H
2 O E= +1.60 V) 2 Mn +2 + 8H 2 O 2MnO 4 - + 16H + + 10e - E= 1.51
V) 5IO 4 - + 2 Mn +2 + 3H 2 O 5IO 3 - + 2MnO 4 - + 6H + E= +0.09 V)
H 2 gas bubbles out Anode Cathode Pt | IO 4 - (aq), IO 3 - (aq),H +
|| MnO 4 - (aq), Mn +2 (aq),H + | Pt V e-e- e-e- Pt IO 4 - Mn +7
gains 5e - to become Mn +2 : reduced in charge cathode cell I +7
loses 2e - to become I +5 : oxidized anode cell IO 3 - MnO 4 - Mn
+2 H+H+ H+H+
Slide 22
31. Give the standard notation for each cell in Exercise 29. a.
Cl 2 + 2e - 2Cl - E= 1.36 V Br 2 + 2e - Br - E= 1.09 V Cl 2 + 2e -
2Cl - E= 1.36 V 2Br - Br 2 + 2e - E= 1.09 V Cl 2 + 2Br - Br 2 + 2Cl
- E= +0.27 V Pt | Br -, Br 2 || Cl 2 | Cl - | Pt V e-e- e-e- Pt Cl
- Br - Br 2 Cl 2
Slide 23
31. Give the standard notation for each cell in Exercise 29. a.
Cl 2 + 2e - 2Cl - E= 1.36 V Br 2 + 2e - Br - E= 1.09 V Pt | Br -,
Br 2 || Cl 2 | Cl - | Pt b. MnO 4 + 8H + + 5e - Mn 2 + 4H 2 OE=
1.51 V IO 4 - + 2H + +2e - IO 3 - + H 2 OE= 1.60 V Pt | IO 4 -
(aq), IO 3 - (aq),H + || MnO 4 - (aq), Mn +2 (aq),H + | Pt
Slide 24
33. Give the balanced cell reaction and determine E for the
galvanic cells based on the following half-reactions. Standard
reaction potentials are found in Table 17.1. a. Cu 2+ + e - Cu + E
0 = +0.16 Au 3+ + 3e - AuE 0 = +1.50 3Cu + 3Cu 2+ + 3e - E 0 = 0.16
Au 3+ + 3Cu + 3Cu 2+ + AuE 0 = +1.34 b. Cd 2+ + 2e - CdE 0 = 0.40
VO 2 + + 2H + + e - VO 2+ +H 2 OE 0 = +1.00 2VO 2 + + 4H + + 2e -
2VO 2+ + 2H 2 OE 0 = +1.00 Cd Cd 2+ + 2e - E 0 = +0.40 2VO 2 + + 4H
+ + Cd Cd 2+ + 2VO 2+ + 2H 2 OE 0 = +1.40
Slide 25
Ch17.1 contCh17 HW#2 p880 30,32,35 30. Sketch the galvanic
cells based on the following half-reactions. Show the direction of
electron flow, show the direction of ion migration through the salt
bridge, and identify the cathode and anode. Give the overall
balanced reaction, and determine E for the galvanic cells. Assume
that all concentrations are 1.0 M and that all partial pressures
are 1.0 atm. a. H 2 O 2 + 2H + + 2e - 2H 2 O E= 1.78 V O 2 + 2H + +
2e - H 2 O 2 E= 0.68 V b. Mn 2+ + 2e - MnE= -1.18 V Fe 3+ + 3e -
FeE= -0.036 V 32. Give the standard line notation for each cell in
Exercise 30.
Slide 26
30/32a) a. H 2 O 2 + 2H + + 2e - 2H 2 O E= 1.78 V O 2 + 2H + +
2e - H 2 O 2 E= 0.68 V V e-e- e-e-
Slide 27
30/32a) a. H 2 O 2 + 2H + + 2e - 2H 2 O E= 1.78 V O 2 + 2H + +
2e - H 2 O 2 E= 0.68 V H 2 O 2 + 2H + + 2e - 2H 2 O E= +1.78 V H 2
O 2 O 2 + 2H + + 2e - E= 0.68 V 2H 2 O 2 2H 2 O + O 2 E= +1.10 V
AnodeCathode Pt | H 2 O 2, H + || H 2 O 2, H +, H 2 O | Pt V e-e-
e-e- Pt H2O2H2O2 H+H+ H2O2H2O2 O2O2 H2OH2O H+H+
Slide 28
30/32b) b. Mn 2+ + 2e - MnE= -1.18 V Fe 3+ + 3e - FeE= -0.036 V
Anode Cathode V e-e- e-e-
Slide 29
30/32b) b. Mn 2+ + 2e - MnE= -1.18 V Fe 3+ + 3e - FeE= -0.036 V
2Fe 3+ + 6e - 2FeE= -0.036 V 3Mn 3Mn 2+ + 6e - E= +1.18 V 2Fe 3+ +
3Mn 3Mn 2+ + 2FeE= +1.14 V Anode Cathode Fe | Fe +3 || Mn +2 | Mn V
e-e- e-e- Fe Mn Mn +2 Fe +3
Slide 30
35. Calculate o values for the following cells. Which reactions
are spontaneous as written (under standard conditions)? Balance the
reactions that are not already balanced. Standard reduction
potentials are found in Table 17.1. a. 2Ag + (aq) + Cu(s) Cu 2+
(aq) + 2Ag(s) b. Zn 2+ (aq) + Ni(s) Ni 2+ (aq) + Zn(s)
Slide 31
35. Calculate o values for the following cells. Which reactions
are spontaneous as written (under standard conditions)? Balance the
reactions that are not already balanced. Standard reduction
potentials are found in Table 17.1. a. 2Ag + (aq) + Cu(s) Cu 2+
(aq) + 2Ag(s) 2Ag + (aq) + 2e - 2Ag(s) E= +0.80 V Cu(s) Cu 2+ (aq)
+ 2e - E= 0.16 V E= +0.64 V spontaneous b. Zn 2+ (aq) + Ni(s) Ni 2+
(aq) + Zn(s)
Slide 32
35. Calculate o values for the following cells. Which reactions
are spontaneous as written (under standard conditions)? Balance the
reactions that are not already balanced. Standard reduction
potentials are found in Table 17.1. a. 2Ag + (aq) + Cu(s) Cu 2+
(aq) + 2Ag(s) 2Ag + (aq) + 2e - 2Ag(s) E= +0.80 V Cu(s) Cu 2+ (aq)
+ 2e - E= 0.16 V E= +0.64 V spontaneous b. Zn 2+ (aq) + Ni(s) Ni 2+
(aq) + Zn(s) Zn 2+ (aq) + 2e - Zn(s) E= 0.76 V Ni(s) Ni 2+ (aq) +
2e - E= +0.23 V E= 0.53 V not spontaneous
Slide 33
Ch17.2 Electric Work and Free Energy EMF In your text, work is
defined as flowing out of a system.
Slide 34
Equations you might remember: G = H TS E = q + w
enthalpyentropy internal energy heat workpotential (voltage)
charge
Slide 35
Equations you might remember: G = H TS E = q + w G = H TS w = E
q G = w G = -q max enthalpy entropy internal energy heat
workpotential (voltage) charge These equations are ideally
related
Slide 36
Equations you might remember: G = H TS E = q + w G = H TS w = E
q G = w G = -q max = -nF max enthalpy entropy internal energy heat
workpotential (voltage) charge These equations are ideally related
moles Farady: the charge of one mole of electrons 96,485 Coulombs /
1 mole If you feel so inclined u may wanna look at ex 17.3, pg850
to c this in action.
Slide 37
Dependence of Cell Potential on Concentration Ex1) For the cell
rxn: 2Al (s) + 3Mn +2 (aq) 2Al +3 (aq) + 3Mn (s) E o cell = 0.48V
predict whether E cell is larger or smaller than E o cell if: a)
[Al +3 ] = 2.0M, [Mn +2 ] = 1.0M b) [Al +3 ] = 1.0M, [Mn +2 ] =
3.0M
Slide 38
Concentration Cells - nature will try to equalize the
concentration in the 2 cells, but only a small voltage will be
produced. Ex2) Determine the direction of electron flow and
designate the anode and cathode.
Slide 39
Ch17 HW#3 p881 47,49,51,53,55 (Do 47 in class) 47. Using data
from Table 17.1, place the following in order of increasing
strength as oxidizing agents. (all under standard conditions) Cd
2+, IO 3 -, K +, H 2 O, AuCl 4 -, I 2 Reminder: ox agents get
reduced in charge (They add electrons).
Slide 40
Ch17 HW#3 p881 47,49,51,53,55 (Do 47 in class) 47. Using data
from Table 17.1, place the following in order of increasing
strength as oxidizing agents. (all under standard conditions) Cd
2+, IO 3 -, K +, H 2 O, AuCl 4 -, I 2 Reminder: ox agents get
reduced in charge (They add electrons). Cd 2+ + 2e - CdE= 0.40 V IO
3 - + 6H + + 5e - I 2 + 3H 2 O E= +1.20 V strongest K + + e - K E=
2.92 V weakest 2H 2 O + 2e - 2H 2 + OH - E= 0.83 V AuCl 4 - + 3e -
Au + 4Cl - E= +0.99 V I 2 + 2e - 2I - E= +0.54 V
Slide 41
49. Answer the following questions using the data in table 17.1
(all under standard conditions) The higher potential will reduce.
a. Is H + (aq) capable of oxidizing Cu(s) to Cu 2+ ? The lower will
oxidize. b. Is H + (aq) capable of oxidizing Mg(s) ? c. Is Fe 3+
(aq) capable of oxidizing I - ? d. Is Fe 3+ (aq) capable of
oxidizing Br - ?
Slide 42
49. Answer the following questions using the data in table 17.1
(all under standard conditions) The higher potential will reduce.
a. Is H + (aq) capable of oxidizing Cu(s) to Cu 2+ ? The lower will
oxidize. No b. Is H + (aq) capable of oxidizing Mg(s) ? Yes c. Is
Fe 3+ (aq) capable of oxidizing I - ? Yes d. Is Fe 3+ (aq) capable
of oxidizing Br - ? No
Slide 43
Na + : Cl - : Ag + : Zn 2+ : Zn : Pb : 51. Consider only the
species (at standard conditions) Na +, Cl -, Ag +, Zn 2+, Zn, Pb In
answering the following questions, give reasons for you answers
(use data from table 17.1) a. Which is the strongest oxidizing
agent? Itself gets reduced: b. Which is the strongest reducing
agent? Itself gets oxidized: c. Which species can be oxidized by SO
4 2- (aq) in acid? d. Which species can be reduced by Al(s)? Al
itself gets oxidized (flipped), everything above it will
reduce.
Slide 44
Na + :2.70 Cl - :1.36 Ag + :+0.80 Zn 2+ :0.76 Zn :+0.76 Pb
:+0.13 51. Consider only the species (at standard conditions) Na +,
Cl -, Ag +, Zn 2+, Zn, Pb In answering the following questions,
give reasons for you answers (use data from table 17.1) a. Which is
the strongest oxidizing agent? Itself gets reduced: Ag + b. Which
is the strongest reducing agent? Itself gets oxidized: Zn c. Which
species can be oxidized by SO 4 2- (aq) in acid? Zn and Pb their
reduction potentials are less than SO 4 2- d. Which species can be
reduced by Al(s)? Al itself gets oxidized (flipped), everything
above it will reduce. Ag +, Zn 2+ are willing to reduce, not Na
+
Slide 45
53. Use the table of standard reduction potential (table 17.1)
to pick a reagent that is capable of each of the following
oxidations (under standard conditions in acidic solutions). a.
Oxidize Br to Br 2 but not oxidize Cl - to Cl 2
Slide 46
53. Use the table of standard reduction potential (table 17.1)
to pick a reagent that is capable of each of the following
oxidations (under standard conditions in acidic solutions). b.
Oxidize Mn to Mn 2+ but not oxidize Ni to Ni 2+
Slide 47
55. A galvanic cell is based on the following half-reactions at
25C Ag + + e - AgE o = +0.80 V H 2 O 2 + 2H + + 2e - 2H 2 OE o =
+1.78 V Predict whether G cell is larger or smaller than G o cell
for the following cases H 2 O 2 + 2H + + 2e - 2H 2 OE o = +1.78 V
Ag Ag + + e - E o = 0.80 V H 2 O 2 + 2H + + 2Ag 2Ag + + 2H 2 O E o
= +0.98 V a. [Ag + ] = 1.0 M, [H 2 O 2 ] = 2.0 M, [H + ] = 2.0 M b.
[Ag + ] = 2.0 M, [H 2 O 2 ] = 1.0 M, [H + ] = 1 x 10 -7 M
Slide 48
55. A galvanic cell is based on the following half-reactions at
25C Ag + + e - AgE o = +0.80 V H 2 O 2 + 2H + + 2e - 2H 2 OE o =
+1.78 V Predict whether G cell is larger or smaller than G o cell
for the following cases H 2 O 2 + 2H + + 2e - 2H 2 OE o = +1.78 V
Ag Ag + + e - E o = 0.80 V H 2 O 2 + 2H + + 2Ag 2Ag + + 2H 2 O E o
= +0.98 V a. [Ag + ] = 1.0 M, [H 2 O 2 ] = 2.0 M, [H + ] = 2.0 M [H
2 O 2 ] and [H + ] increased favors products, G cell is larger b.
[Ag + ] = 2.0 M, [H 2 O 2 ] = 1.0 M, [H + ] = 1 x 10 -7 M [Ag + ]
increased, favors reactants AND [H 2 O 2 ] and [H + ] decreased
favors reactants, G cell is much smaller
Slide 49
Ch17.3 Batteries A connection of galvanic cells in series, such
that the potentials of the individual cells add together to give a
total potential. Lead storage battery Anode reaction: Pb + HSO 4 -
PbSO 4 + H + + 2e - Cathode rxn: PbO 2 + HSO 4 - + 3H + + 2e - PbSO
4 + 2H 2 O Cell reaction:
Slide 50
Lead storage battery cell reaction: Pb (s) + PbO 2(s) + 2HSO 4
- (aq) + 2H + (aq) 2PbSO 4(s) + 2H 2 O (l)
Slide 51
Dry cell battery (alkaline version): Anode reaction: Zn + 2OH -
ZnO + H 2 O + 2e - Cathode rxn: 2MnO 2 + H 2 O + 2e - Mn 2 O 3 +
2OH - Cell reaction:
Slide 52
Corrosion - the oxidation of metals.
Slide 53
Oxidation of Iron Fe Fe +2 + 2e - O 2 + 2H 2 O + 4e - 4OH -
Prevention Galvanization: Fe Fe +2 + 2e - EMF = 0.44V Zn Zn +2 + 2e
- EMF = 0.76V Ch17 HW#4 p883 48,50,52,73a
Slide 54
48. Using the data from table 17.1, place the following in
order of increasing strength as reducing agents (all under standard
conditions). A reducing agent itself will get oxidized Cr 3+, H 2,
Zn, Li, F -, Fe 2+
Slide 55
Ch17 HW#4 p883 48,50,52,73a 48. Using the data from table 17.1,
place the following in order of increasing strength as reducing
agents (all under standard conditions). A reducing agent itself
will get oxidized Cr 3+, H 2, Zn, Li, F -, Fe 2+ 5. 2Cr 3+ + 7H 2 O
Cr 2 O 7 2- + 14H + + 6e - 1.33 3. H 2 2H + + 2e - 0.00 2. Zn Zn +2
+ 2e - +0.76 1. Li Li + + e - +3.05 6. 2F - F 2 + 2e - 2.87 4. Fe
2+ Fe 3+ + e - 0.77
Slide 56
50. Answer the following questions using data from table 17.1
(all under standard conditions) a. Is H 2 (g) capable of reducing
Ag + (aq)? H 2 2H + + 2e - 0.00 b. Is H + (g) capable of reducing
Ni 2+ (aq)? H 2 2H + + 2e - 0.00 c. Is Fe 2+ (aq) capable of
reducing VO 2 + (aq)? Fe 2+ Fe 3+ + e - 0.77 d. Is Fe 2+ (aq)
capable of reducing Cr 3+ (aq) to Cr 2 +(aq)? Fe 2+ Fe 3+ + e -
0.77
Slide 57
50. Answer the following questions using data from table 17.1
(all under standard conditions) a. Is H 2 (g) capable of reducing
Ag + (aq)? H 2 2H + + 2e - 0.00 No. b. Is H + (g) capable of
reducing Ni 2+ (aq)? H 2 2H + + 2e - 0.00 Yes. c. Is Fe 2+ (aq)
capable of reducing VO 2 + (aq)? Fe 2+ Fe 3+ + e - 0.77 No. d. Is
Fe 2+ (aq) capable of reducing Cr 3+ (aq) to Cr 2 +(aq)? Fe 2+ Fe
3+ + e - 0.77 No.
Slide 58
52. Consider only the species (at standard condition) Ce 4+, Ce
3+, Fe 2+, Fe 3+, Fe, Mg 2+, Mg, Ni 2+, Sn in answering the
following questions, give reasons for your answers (use data from
table 17.1) a. Which is the strongest oxidizing agent? b. Which is
the strongest reducing agent? c. Will iron dissolve in a 1.0 M
solution of Ce 4+ ? d. Which of the species can be oxidized by H +
(aq)? e. Which of the species can be reduced by H 2 (g)?
Slide 59
52. Consider only the species (at standard condition) Ce 4+, Ce
3+, Fe 2+, Fe 3+, Fe, Mg 2+, Mg, Ni 2+, Sn In answering the
following questions, give reasons for your answers (use data from
table 17.1) a. Which is the strongest oxidizing agent? Itself gets
reduced, look for largest V b. Which is the strongest reducing
agent? Itself gets oxidized, look for largest V when rxn flips c.
Will iron dissolve in a 1.0 M solution of Ce 4+ ? d. Which of the
species can be oxidized by H + (aq)? Any rxn when flipped over,
that has a negative voltage (E < 0.00), will be oxidized: e.
Which of the species can be reduced by H 2 (g)? Any rxn with a
positive voltage (E > 0.00) in the forward direction will be
reduced:
Slide 60
52. Consider only the species (at standard condition) Ce 4+, Ce
3+, Fe 2+, Fe 3+, Fe, Mg 2+, Mg, Ni 2+, Sn In answering the
following questions, give reasons for your answers (use data from
table 17.1) a. Which is the strongest oxidizing agent? Itself gets
reduced, look for largest V Ce 4+ (+1.70V) b. Which is the
strongest reducing agent? Itself gets oxidized, look for largest V
when rxn flips Mg (+2.37V) c. Will iron dissolve in a 1.0 M
solution of Ce 4+ ? Yes. Ce 4+ wants to reduce (+1.70V), Fe will
oxidize (+0.44V) d. Which of the species can be oxidized by H +
(aq)? Any rxn when flipped over, that has a negative voltage (E
< 0.00), will be oxidized: Ce 3+ (-1.70V), Fe 2+ (-0.77) e.
Which of the species can be reduced by H 2 (g)? Any rxn with a
positive voltage (E > 0.00) in the forward direction will be
reduced: Ce 4+ (+1.70V), Fe 3+ (+0.77V)
Slide 61
73. Consider a galvanic cell based on the following half
reactions Zn 2+ + 2e - Zn E = -0.76 V Fe 2+ + 2e - Fe E = -0.44 V
a. Determine the overall cell reaction and calculate E cell. Zn Zn
2+ + 2e - E = +0.76 V Fe 2+ + 2e - Fe E = -0.44 V
Slide 62
73. Consider a galvanic cell based on the following half
reactions Zn 2+ + 2e - Zn E = -0.76 V Fe 2+ + 2e - Fe E = -0.44 V
a. Determine the overall cell reaction and calculate E cell. Zn Zn
2+ + 2e - E = +0.76 V Fe 2+ + 2e - Fe E = -0.44 V Zn (aq) + Fe 2+
(aq) Fe (aq) + Zn 2+ (aq) E = +0.32 V
Slide 63
Ch17.4 Electrolysis Electrolytic Cell uses electrical energy to
produce a chemical change. Electrolysis forces a current thru a
cell to produce chemical change for which the cell potential is
negative. (Use what we know to develop a new twist): Ex of galvanic
cell: Anode: Zn Zn +2 + 2e - +0.76 V Cathode: Cu +2 + 2e - Cu +0.34
V +1.10 V
Slide 64
Ex of galvanic cell: Anode: Zn Zn +2 + 2e - +0.76 V Cathode: Cu
+2 + 2e - Cu +0.34 V +1.10 V charge current time Units: 1 mol of
electrons = 1 farady of charge = 96,485 coulombs
Slide 65
Ex1) How long must a current of 5.00A be applied to a soln of
Ag + to produce 10.5g of silver?
Slide 66
HW#79a) How long will it take to plate out each of the
following with a current of 100.0A? a. 1.0kg Al from aqueous Al
3+
Slide 67
Electrolysis of water 2H 2 O O 2 + 4H + + 4e - EMF = 1.23V 4H 2
O + 4e - 2H 2 + 4OH - EMF = 0.83V
Slide 68
Electroplating metals In an electrolytic cell, a soln contains
the following metal ions: Ag +, Cu +2, Zn +2. The voltage is
increased gradually. In which order will the metals be plated onto
the cathode?
Slide 69
Electroplating metals In an electrolytic cell, a soln contains
the following metal ions: Ag +, Cu +2, Zn +2. The voltage is
increased gradually. In which order will the metals be plated onto
the cathode? The higher the (+), the greater the tendency to occur.
Ag + + e - Ag E = +0.80V Cu +2 + 2e - Cu E = +0.34V Zn +2 + 2e - Zn
E = 0.76V The order of oxidizing ability (the order they are
reduced): Ag + > Cu +2 > Zn +2
Slide 70
Ex2) An acidic soln contains the ions: Ce +4, VO 2 +, Fe +3.
Using EMFs from Table 17.1, give the order of oxidizing ability.
(The order they arte reduced.)
Slide 71
Ex2) An acidic soln contains the ions: Ce +4, VO 2 +, Fe +3.
Using EMFs from Table 17.1, give the order of oxidizing ability.
(The order they arte reduced.) Ce +4 + e - Ce +3 E = +1.70V VO 2 +
+ 2H + + e - VO 2 +2 + H 2 O E = +1.00V Fe +3 + e - Fe +2 E =
+0.77V Oxidizing ability: Ce +4 > VO 2 + > Fe +3 Ce +4 is
reduced at the lowest voltage. Ch17 HW#5 p883 79,91,95 + Ch17
Rev
Slide 72
79. How long will it take to plate out each of the following
with a current of 100.0A? a. 1.0kg Al from aqueous Al 3+ (In
class?) b. 1.0g Ni from aqueous Ni 2+ c. 5.0 mol Ag from aqueous Ag
+
Slide 73
91. A solution at 25C contains 1.0 M Cd 2+, 1.0 M Ag +, 1.0 M
Au 3+, and 1.0 M Ni 2+ in the cathode compartment of an
electrolytic cell. Predict the order in which the metals will plate
out as the voltage is gradually increased.
Slide 74
95. In the electrolysis of an aqueous solution of Na 2 SO 4,
what reactions occur at the anode and the cathode? (Assume standard
conditions.) S 2 O 8 2- + 2e - 2SO 4 2- E = 2.01 V O 2 + 4H + = 4e
- 2H 2 O E = 1.23 V 2H 2 O + 2e - H 2 + 2OH - E = -0.83 V Na + + e
- Na E = -2.71 V
Slide 75
Ch17 Rev p880+ 34a,36a,54,59 + Bonus FRQ!!! 34. a. Give the
balanced cell reaction and determine the E for the galvanic cells
based on the following half-reactions. Standard reduction
potentials are found in table 17.1. a. Cr 2 O 7 2- + 14H + + 6e -
2Cr 3+ + 7H 2 O H 2 O 2 + 2H + + 2e - 2H 2 O
Slide 76
36. a. Calculate E values for the following cells. Which
reactions are spontaneous as written (under standard conditions)?
Balance the reactions. Standard reduction potentials are found in
Table 17.1. a. MnO 4 - (aq) + I - (aq) I 2 (aq) + Mn 2+ (aq)
Slide 77
54. Use the table of standard reduction potentials (Table 17.1)
to pick a reagent that is capable of each of the following
reductions (under standard conditions in acidic solution). a.
Reduce Cu 2+ to Cu but not reduce Cu 2+ to Cu + b. Reduce Br 2 to
Br - but not to reduce I 2 to I -
Slide 78
59. Consider the concentration cell shown below. Calculate the
cell potential at 25C when the concentration of Ag + in the
compartment on the right is the following. a. 1.0 Mb. 2.0 Mc. 0.10
Md. 4.0 x 10 -5 M e. Calculate the potential when both solutions
are 0.10 M in Ag +. For each case, also identify the cathode, the
anode, and the direction in which electrons flow. V Ag [Ag + ] =
1.0M
Slide 79
AP Chemistry - Ch17 FRQ Review 1. An external direct-current
power supply is connected to two platinum electrodes immersed in a
beaker containing 1.0 M CuSO (ag) at 25C, as shown in the diagram.
As the cell operates, copper metal is deposited onto one electrode
and O (g) is produced as the other electrode. The two reduction
half-reactions for the overall reaction that occurs in the cell are
show in the table below. (a) On the diagram, indicate the direction
of electron flow in the wire. (b) Write a balanced net ionic
equation for the electrolysis reaction that occurs in the cell. (c)
Predict the algebraic sign of G for the reaction. Justify your
answer. An electric current of 1.50 Amps passes through the cell
for 40.0 minutes. (d) Calculate the mass, in grams, of the Cu(s)
that is deposited on the electrode. (e) Calculate the dry volume,
in liters measured at 25C and 1.16 atm, of the O (g) that is
produced. V Cu Half-reactionE (V) O (g) + 4 H (ag) + 4 e 2 H O(l)
+1.23 Cu (ag) + 2 e Cu(s) +0.34