Ch17.1 – Galvanic Cells
Ch4: Redox involves the transfer of electrons
(OIL RIG)
Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)
Half rxns:
Ox:
Red:
Red agent: _____
Ox agent: _____
lose 2e- (oxidized)
Zn(s) + CuSO4(aq)
ZnSO4(aq) + Cu(s)
gained 2e- (reduced)
Ox ½ Reaction: Zn(s) Zn(aq)+2 +2eRed ½ Reaction: Cu+2(aq) + 2e- Cu(s)
Red agent: Zn
Ox agent: Cu+2
lose 2e- (oxidized)
Zn(s) + CuSO4(aq)
ZnSO4(aq) + Cu(s)
gained 2e- (reduced)
Oxidized ½ Reaction: Zn(s) Zn(aq)+2 +2eReduced ½ Reaction: Cu+2(aq) + 2e- Cu(s)
e-
Cu
CuSO4 soln
e-
V
e-
Red agent: Zn
Ox agent: Cu+2
e-
Zn
ZnSO4 soln
lose 2e- (oxidized)
Zn(s) + CuSO4(aq)
ZnSO4(aq) + Cu(s)
gained 2e- (reduced)
Oxidized ½ Reaction: Zn(s) Zn(aq)+2 +2eReduced ½ Reaction: Cu+2(aq) + 2e- Cu(s)
e-
Cu
CuSO4 soln
e-
V
e-
Red agent: Zn
Ox agent: Cu+2
e-
Zn
ZnSO4 soln
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)
Cu
Cu
Cu
V eeeeelectrons transfer
Na+ Cl- Na+ Cl- Na+ Cl- Na+ Cl- Na+
thru the wire
Cl(salt bridge)
ClNa+
(necessary to maintain
Na+
Zn
Clan ion charge balance)
ClNa+
Na+
Zn
ClClZn
CuSO4 soln
ZnSO4 soln
V
Cu
Cu
Na+ Cl- Na+ Cl- Na+ Cl- Na+ Cl- Na+
Cl(salt bridge)
ClNa+
(necessary to maintain
Na+
Clan ion charge balance)
ClNa+
Na+
ClCl-
Cu
electrons transfer
thru the wire
Zn
Zn
Zn
Cu+2 SO4-2
SO4-2
Cu+2
CuSO4 soln
cathode cell
ε
Zn+2 SO4-2
SO4-2
Zn+2
ZnSO4 soln
anode cell
Cell Potential, cell
Reduction Half Cell
Cu+2 + 2e– → Cu
Zn → Zn+2 + 2eCu+2 + 2e- → Cu
E˚ → standard conditions
25˚C, I Molar
Oxidation Half Cell
Zn → Zn+2 + 2e–
E˚ = + .76V
E˚ = + .34 V
Ecell=+1.10 V
Cell Potentials
Ex1)
Al+3(aq) + Mg(s) Al(s) + Mg+2(aq)
Give the balanced cell rxn and EMF for the cell.
Ex1)
Al+3(aq) + Mg(s) Al(s) + Mg+2(aq)
Give the balanced cell rxn and EMF for the cell.
Al+3 + 3e- Al
Mg+2 + 2e- Mg
Eo = –1.66 V
Eo = –2.37 V
Al+3(aq) + Mg(s) Al(s) + Mg+2(aq)
Ex1)
Give the balanced cell rxn and EMF for the cell.
Al+3 + 3e- Al
Mg+2 + 2e- Mg
Eo = –1.66 V
Eo = –2.37 V
2Al+3 + 6e- 2Al
Eo = –1.66 V Don’t multiply!
3Mg 3Mg+2 + 6eEo = +2.37 V
2Al+3(aq) + 3Mg(s) 2Al(s) + 3Mg+2(aq) Eo = +0.71 V
eMg metal
lost e-‘s so
it is oxidized.
Anode
V
e-
Mg
Al
Mg+2
Al+3
Mg | Mg+2 || Al+3 | Al
Anode cell
Cathode cell
Al+3 gained e-‘s so
it was reduced
in charge. Formed
more Al on the electrode.
The metal the
e’s head to is the cathode
Ex2) MnO4-(aq) + H+(aq) + ClO3-(aq) ClO4-(aq) + Mn+2(aq) + H2O(l)
Give the balanced cell rxn and EMF for the cell.
MnO4- + 5e- + 8H+ Mn+2 + 4H2O
ClO4- + 2e- + 2H+ ClO3- + H2O
.
Eo = +1.51 V
Eo = +1.19 V
Mn+7 gains 5e- to become Mn+2: reduced in charge → cathode cell
Ex2) MnO4-(aq) + H+(aq) + ClO3-(aq) ClO4-(aq) + Mn+2(aq) + H2O(l)
Cl+5 loses 2e- to become Cl+7: oxidized → anode cell
MnO4- + 5e- + 8H+ Mn+2 + 4H2O
ClO4- + 2e- + 2H+ ClO3- + H2O
Eo = +1.51 V
Eo = +1.19 V
2MnO4- + 10e- + 16H+ 2Mn+2 + 8H2O
Eo = +1.51 V
5ClO3- + 5H2O 5ClO4- + 10e- + 10H+ Eo = –1.19 V
2MnO4-(aq) + 6H+(aq) + 5ClO3-(aq) 5ClO4-(aq) + 2Mn+2(aq) + 3H2O(l)
Eo = +0.32 V
e-
V
e-
?
?
?
?
.
|
Anode cell
||
|
Cathode cell
Mn+7 gains 5e- to become Mn+2: reduced in charge → cathode cell
Ex2) MnO4-(aq) + H+(aq) + ClO3-(aq) ClO4-(aq) + Mn+2(aq) + H2O(l)
Cl+5 loses 2e- to become Cl+7: oxidized → anode cell
MnO4- + 5e- + 8H+ Mn+2 + 4H2O
ClO4- + 2e- + 2H+ ClO3- + H2O
Eo = +1.51 V
Eo = +1.19 V
2MnO4- + 10e- + 16H+ 2Mn+2 + 8H2O
Eo = +1.51 V
5ClO3- + 5H2O 5ClO4- + 10e- + 10H+ Eo = –1.19 V
2MnO4-(aq) + 6H+(aq) + 5ClO3-(aq) 5ClO4-(aq) + 2Mn+2(aq) + 3H2O(l)
Eo = +0.32 V
ePt ClO 3
ClO4H+
V
eMn+2 Pt
MnO4H+
Pt | ClO3- , ClO4- , H+ || Mn+2, MnO4Anode
Cathode
Since REDOX occurs w
aqueous ions, we need
a non-reactive metal as
electrodes, so Pt makes
a great choice.
H2 gas bubbles off
, H+ | Pt of cathode
Ex3)
Ag+ + e- Ag
Fe+3 + e- Fe+2
Eo = +0.80V
Eo = +0.77V
Setup a galvanic cell, give the balanced cell rxn and EMF for the cell.
Ex3)
Ag+ + e- Ag
Fe+3 + e- Fe+2
Eo = +0.80V
Eo = +0.77V
Setup a galvanic cell, give the balanced cell rxn and EMF for the cell.
Ag+ + e- Ag
RIG
Fe+2 Fe+3 + e- OIL
Ag+ + Fe+2 Ag + Fe+3
e-
Pt
Fe+3
Fe+2
V
e-
Ag
Ag+
Pt | Fe+2(aq) ,Fe+3(aq) || Ag+(aq) | Ag
Anode
Cathode
Eo = +0.80V
Eo = -0.77V
Eo = +0.03V
Since there’s no Fe(s),
we need another metal
to be the electrode. Pt
is non-reactive, so makes
a great choice.
Ex4)
Fe+2 + 2e- Fe
MnO4- + 5e- Mn+2 + 4H2O
Eo = –0.44V
Eo = +1.51V
Setup a galvanic cell, give the balanced cell rxn and EMF for the cell.
e-
V
e-
?
?
?
Ch17 HW#1 p880 29,31,33
?
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. Cl2 + 2e- → 2ClBr2 + 2e- → 2Br-
E˚= 1.36 V
E˚= 1.09 V
b. MnO4 + 8H+ + 5e- → Mn2 + 4H2O
IO4- + 2H+ +2e- → IO3- + H2O
E˚= 1.51 V
E˚= 1.60
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. Cl2 + 2e- → 2ClBr2 + 2e- → 2Br-
E˚= 1.36 V
E˚= 1.09 V
Cl2 + 2e- → 2Cl2Br- → Br2 + 2eCl2 + 2Br- → Br2 + 2Cle-
Pt
Br2
Br-
V
e-
Pt
Cl-
E˚= 1.36 V
E˚= –1.09 V
E˚=
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. MnO4- + 8H+ + 5e- → Mn+2 + 4H2O E˚= 1.51 V
E˚= 1.60 V
IO4- + 2H+ +2e- → IO3- + H2O
Mn+7 gains 5e- to become Mn+2: reduced in charge → cathode cell
b. MnO4- + 8H+ + 5e- → Mn+2 + 4H2O E˚= 1.51 V
IO4- + 2H+ +2e- → IO3- + H2O
E˚= 1.60 V
I+7 loses 2e- to become I+5: oxidized → anode cell
5(IO4- + 2H+ +2e- → IO3- + H2O)
2(Mn+2 + 4H2O → MnO4- + 8H+ + 5e-)
E˚= +1.60 V
E˚= –1.51 V
Mn+7 gains 5e- to become Mn+2: reduced in charge → cathode cell
b. MnO4- + 8H+ + 5e- → Mn+2 + 4H2O E˚= 1.51 V
IO4- + 2H+ +2e- → IO3- + H2O
E˚= 1.60 V
I+7 loses 2e- to become I+5: oxidized → anode cell
5IO4- + 10H+ + 10e- → 5IO3- + 5H2O
E˚= +1.60 V)
2 Mn+2 + 8H2O → 2MnO4- + 16H+ + 10e- E˚= –1.51 V)
e-
Pt
IO4H+
IO3-
Anode
V
e-
MnO4
Mn+2
-
Pt
H2 gas bubbles out
H+
Cathode
Pt | IO4-(aq), IO3-(aq) ,H+ || MnO4-(aq), Mn+2(aq) ,H+ | Pt
31. Give the standard notation for each cell in Exercise 29.
a. Cl2 + 2e- → 2ClBr2 + 2e- → Br-
E˚= 1.36 V
E˚= 1.09 V
Cl2 + 2e- → 2Cl2Br- → Br2 + 2eCl2 + 2Br- → Br2 + 2Cl-
E˚= 1.36 V
E˚= –1.09 V
Pt | Br- , Br2 || Cl2 | Cl- | Pt
Cl2
e-
Pt
Br2
Br-
V
e-
Pt
Cl-
31. Give the standard notation for each cell in Exercise 29.
a. Cl2 + 2e- → 2ClBr2 + 2e- → Br-
E˚= 1.36 V
E˚= 1.09 V
Pt | Br- , Br2 || Cl2 | Cl- | Pt
b. MnO4 + 8H+ + 5e- → Mn2 + 4H2O
IO4- + 2H+ +2e- → IO3- + H2O
E˚= 1.51 V
E˚= 1.60 V
Pt | IO4-(aq), IO3-(aq) ,H+ || MnO4-(aq), Mn+2(aq) ,H+ | Pt
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. Cu2+ + e-→ Cu+
Au3+ + 3e- → Au
E0 = +0.16
E0 = +1.50
Au3+ + 3e- → Au
E0 = +1.50
3Cu+ → 3Cu2+ + 3e- E0 = –0.16
Au3+ + 3Cu+ → 3Cu2+ + Au
b.
Cd2+ + 2e- → Cd
VO2+ + 2H+ + e- → VO2+ +H2O
E0 = –0.40
E0 = +1.00
2VO2+ + 4H+ + 2e- → 2VO2+ + 2H2O
Cd → Cd2+ + 2e2VO2+ + 4H+ + Cd → Cd2+ + 2VO2+ + 2H2O
E0 = +1.00
E0 = +0.40
Ch17.1 cont
Ch17 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. H2O2 + 2H+ + 2e- → 2H2O
O2 + 2H+ + 2e- → H2O2
b. Mn2+ + 2e- → Mn
Fe3+ + 3e- → Fe
E˚= 1.78 V
E˚= 0.68 V
E˚= -1.18 V
E˚= -0.036 V
32. Give the standard line notation for each cell in Exercise 30.
30/32a)
a. H2O2 + 2H+ + 2e- → 2H2O
O2 + 2H+ + 2e- → H2O2
e-
V
e-
E˚= 1.78 V
E˚= 0.68 V
30/32a)
a. H2O2 + 2H+ + 2e- → 2H2O
O2 + 2H+ + 2e- → H2O2
E˚= 1.78 V
E˚= 0.68 V
H2O2 + 2H+ + 2e- → 2H2O
H2O2 → O2 + 2H+ + 2e-
E˚= +1.78 V
E˚= –0.68 V
O2
e-
Pt
H 2 O2
H+
Anode
V
e-
Pt
H 2O
H 2 O2
+
H
Cathode
Pt | H2O2 , H+ || H2O2 , H+ , H2O | Pt
30/32b)
b. Mn2+ + 2e- → Mn
Fe3+ + 3e- → Fe
e-
Anode
E˚= -1.18 V
E˚= -0.036 V
V
e-
Cathode
30/32b)
b. Mn2+ + 2e- → Mn
Fe3+ + 3e- → Fe
E˚= -1.18 V
E˚= -0.036 V
2Fe3+ + 6e- → 2Fe
3Mn → 3Mn2+ + 6e-
E˚= -0.036 V
E˚= +1.18 V
e-
V
e-
Fe
Mn
Fe+3
Anode
Mn+2
Cathode
Fe | Fe+3 || Mn+2 | Mn
35. Calculate εo 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) ↔ Cu2+(aq) + 2Ag(s)
b. Zn2+ (aq) + Ni(s) ↔ Ni2+ (aq) + Zn(s)
35. Calculate εo 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) ↔ Cu2+(aq) + 2Ag(s)
2Ag+(aq) + 2e- ↔ 2Ag(s)
Cu(s) ↔ Cu2+(aq) + 2e-
b. Zn2+ (aq) + Ni(s) ↔ Ni2+ (aq) + Zn(s)
E˚= +0.80 V
E˚= –0.16 V
35. Calculate εo 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) ↔ Cu2+(aq) + 2Ag(s)
2Ag+(aq) + 2e- ↔ 2Ag(s)
Cu(s) ↔ Cu2+(aq) + 2e-
E˚= +0.80 V
E˚= –0.16 V
b. Zn2+ (aq) + Ni(s) ↔ Ni2+ (aq) + Zn(s)
Zn2+ (aq) + 2e- ↔ Zn(s)
Ni(s) ↔ Ni2+ (aq) + 2e-
E˚= –0.76 V
E˚= +0.23 V
Ch17.2 – Electric Work and Free Energy
EMF
work (J)
potential difference (V) =
charge (C)
−w
ε=
In your text, work is defined
q
as flowing out of a system.
Equations you might remember:
∆G = ∆H – T∆S
enthalpy
entropy
−w
ε=
q
∆E = q + w
internal
energy
heat
work
potential
(voltage)
charge
Equations you might remember:
∆G = ∆H – T∆S
enthalpy
−w
ε=
q
∆E = q + w
entropy
internal
energy
heat
work
∆G = ∆H – T∆S
These equations are ideally related
w = ∆E – q
∆G = w
∆G = -qεmax
potential
(voltage)
charge
Equations you might remember:
∆G = ∆H – T∆S
∆E = q + w
enthalpy
internal
energy
entropy
heat
−w
ε=
q
work
potential
(voltage)
charge
∆G = ∆H – T∆S
These equations are ideally related
w = ∆E – q
∆G = w
∆G = -qεmax = -nFεmax
If you feel so inclined
u may wanna look at ex 17.3,
pg850 to c this in action.
moles
Farady: the charge of
one mole of electrons
96,485 Coulombs / 1 mole
Dependence of Cell Potential on Concentration
Ex1) For the cell rxn:
2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s)
predict whether Ecell is larger or smaller than Eocell if:
a) [Al+3] = 2.0M, [Mn+2] = 1.0M
b) [Al+3] = 1.0M, [Mn+2] = 3.0M
Eocell = 0.48V
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.
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)
Cd2+, IO3-, K+, H2O, AuCl4-, I2
Reminder: ox agents get reduced in charge (They add electrons).
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)
Cd2+, IO3-, K+, H2O, AuCl4-, I2
Reminder: ox agents get reduced in charge (They add electrons).
Cd2+ + 2e- → Cd
IO3- + 6H+ + 5e- → ½I2 + 3H2O
K+ + e- → K
2H2O + 2e- → 2H2 + OHAuCl4- + 3e- → Au + 4ClI2 + 2e- → 2I-
E˚= –0.40 V
E˚= +1.20 V strongest
E˚= –2.92 V weakest
E˚= –0.83 V
E˚= +0.99 V
E˚= +0.54 V
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 Cu2+ ? The lower will oxidize.
b. Is H+ (aq) capable of oxidizing Mg(s) ?
c. Is Fe3+ (aq) capable of oxidizing I-?
d. Is Fe3+ (aq) capable of oxidizing Br -?
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 Cu2+ ? The lower will oxidize.
No
b. Is H+ (aq) capable of oxidizing Mg(s) ?
Yes
c. Is Fe3+ (aq) capable of oxidizing I-?
Yes
d. Is Fe3+ (aq) capable of oxidizing Br -?
No
51. Consider only the species (at standard conditions)
Na+, Cl-, Ag+, Zn2+, Zn, Pb In answering the following questions,
give reasons for you answers
(use data from table 17.1)
a. Which is the strongest
+:
Na
oxidizing agent?
Itself gets reduced:
Cl- :
b. Which is the strongest
reducing agent?
Itself gets oxidized:
c. Which species can be
oxidized by SO42- (aq)
in acid?
Ag+:
Zn2+:
Zn :
Pb :
d. Which species can be reduced by Al(s)?
Al itself gets oxidized (flipped), everything above it will reduce.
51. Consider only the species (at standard conditions)
Na+, Cl-, Ag+, Zn2+, Zn, Pb In answering the following questions,
give reasons for you answers
(use data from table 17.1)
a. Which is the strongest
+:–2.70
Na
oxidizing agent?
Itself gets reduced:
Cl- :–1.36
Ag+
+:+0.80
Ag
b. Which is the strongest
Zn2+:–0.76
reducing agent?
Itself gets oxidized:
Zn :+0.76
Zn
Pb :+0.13
c. Which species can be
oxidized by SO42- (aq)
in acid?
Zn and Pb
their reduction potentials are less than SO42d. Which species can be reduced by Al(s)?
Al itself gets oxidized (flipped), everything above it will reduce.
Ag+, Zn2+ are willing to reduce, not Na+
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 Br2 but not oxidize Cl- to Cl2
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 Mn2+ but not oxidize Ni to Ni2+
55. A galvanic cell is based on the following half-reactions at 25˚C
Ag+ + e- → Ag
H2O2 + 2H+ + 2e- → 2H2O
Eo = +0.80 V
Eo = +1.78 V
Predict whether Gcell is larger or smaller than Gocell for the following cases
H2O2 + 2H+ + 2e- → 2H2O
Eo = +1.78 V
Ag → Ag+ + eEo = –0.80 V
H2O2 + 2H+ + 2Ag → 2Ag+ + 2H2O Eo = +0.98 V
a. [Ag+] = 1.0 M, [H2O2] = 2.0 M, [H+] = 2.0 M
b. [Ag+] = 2.0 M, [H2O2] = 1.0 M, [H+ ] = 1 x 10 -7 M
55. A galvanic cell is based on the following half-reactions at 25˚C
Ag+ + e- → Ag
H2O2 + 2H+ + 2e- → 2H2O
Eo = +0.80 V
Eo = +1.78 V
Predict whether Gcell is larger or smaller than Gocell for the following cases
H2O2 + 2H+ + 2e- → 2H2O
Eo = +1.78 V
Ag → Ag+ + eEo = –0.80 V
H2O2 + 2H+ + 2Ag → 2Ag+ + 2H2O Eo = +0.98 V
a. [Ag+] = 1.0 M, [H2O2] = 2.0 M, [H+] = 2.0 M
[H2O2] and [H+] increased favors products, Gcell is _________
b. [Ag+] = 2.0 M, [H2O2] = 1.0 M, [H+ ] = 1 x 10 -7 M
[Ag+] increased, favors reactants AND
[H2O2] and [H+] decreased favors reactants,
Gcell is ________
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 + HSO4- PbSO4 + H+ + 2eCathode rxn: PbO2 + HSO4- + 3H+ + 2e- PbSO4 + 2H2O
Cell reaction:
Lead storage battery cell reaction:
Pb(s) + PbO2(s) + 2HSO4-(aq) + 2H+(aq) 2PbSO4(s) + 2H2O(l)
Dry cell battery (alkaline version):
Anode reaction:
Zn + 2OH- ZnO + H2O + 2eCathode rxn: 2MnO2 + H2O + 2e- Mn2O3 + 2OHCell reaction:
Corrosion - the oxidation of metals.
Oxidation of Iron
Fe Fe+2 + 2eO2 + 2H2O + 4e- 4OH-
Prevention
Galvanization:
Ch17 HW#4 p883 48,50,52,73a
Fe Fe+2 + 2e- –EMF = 0.44V
Zn Zn+2 + 2e- –EMF = 0.76V
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
Cr3+, H2, Zn, Li, F-, Fe2+
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
Cr3+, H2, Zn, Li, F-, Fe2+
5. 2Cr3+ + 7H2O → Cr2O72- + 14H+ + 6e-
–1.33
3. H2 → 2H+ + 2e-
0.00
2. Zn → Zn+2 + 2e-
+0.76
1. Li → Li+ + e-
+3.05
6. 2F- → F2 + 2e-
–2.87
4. Fe2+ → Fe3+ + e-
–0.77
50. Answer the following questions using data from table 17.1
(all under standard conditions)
a. Is H2(g) capable of reducing Ag+ (aq)?
H2 → 2H+ + 2e0.00
b. Is H+(g) capable of reducing Ni2+ (aq)?
H2 → 2H+ + 2e-
0.00
c. Is Fe2+(aq) capable of reducing VO2+ (aq)?
Fe2+ → Fe3+ + e-
–0.77
d. Is Fe2+(aq) capable of reducing Cr3+ (aq) to Cr2+(aq)?
Fe2+ → Fe3+ + e- –0.77
52. Consider only the species (at standard condition)
Ce4+, Ce3+, Fe2+, Fe3+, Fe, Mg2+, Mg, Ni2+, 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 Ce4+ ?
d. Which of the species can be oxidized by H+(aq)?
e. Which of the species can be reduced by H2(g)?
52. Consider only the species (at standard condition)
Ce4+, Ce3+, Fe2+, Fe3+, Fe, Mg2+, Mg, Ni2+, 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 Ce4+ ?
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 H2(g)?
Any rxn with a positive voltage (E > 0.00) in the forward direction
will be reduced:
73. Consider a galvanic cell based on the following half reactions
Zn2+ + 2e- → Zn
Fe2+ + 2e- → Fe
E˚ = -0.76 V
E˚ = -0.44 V
a. Determine the overall cell reaction and calculate E˚cell.
Zn → Zn2+ + 2eFe2+ + 2e- → Fe
E˚ = +0.76 V
E˚ = -0.44 V
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
Ex of galvanic cell:
Anode:
Zn Zn+2 + 2eCathode: Cu+2 + 2e- Cu
current
q
I=
t
+0.76 V
+0.34 V
+1.10 V
charge
time
(coulombs)
Units: (amperes) =
(sec)
1 mol of electrons = 1 farady of charge = 96,485 coulombs
Ex1) How long must a current of 5.00A be applied to a soln of Ag+
to produce 10.5g of silver?
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 Al3+
Electrolysis of water
2H2O O2 + 4H+ + 4e4H2O + 4e- 2H2 + 4OH-
–EMF = –1.23V
–EMF = –0.83V
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?
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
Ex2) An acidic soln contains the ions: Ce+4, VO2+, Fe+3.
Using EMF’s from Table 17.1, give the order of oxidizing ability.
(The order they arte reduced.)
Ex2) An acidic soln contains the ions: Ce+4, VO2+, Fe+3.
Using EMF’s from Table 17.1, give the order of oxidizing ability.
(The order they arte reduced.)
Ce+4 + e- Ce+3
VO2+ + 2H+ + e- VO2+2 + H2O
Fe+3 + e- Fe+2
Oxidizing ability:
Ce+4 > VO2+ > Fe+3
Ce+4 is reduced at the lowest voltage.
Ch17 HW#5 p883 79,91,95 + Ch17 Rev
E = +1.70V
E = +1.00V
E = +0.77V
Ch17 HW#5 p883 79,91,95 + Ch17 Rev
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 Al3+ (In class?)
b. 1.0g Ni from aqueous Ni2+
c. 5.0 mol Ag from aqueous Ag+
91. A solution at 25°C contains 1.0 M Cd2+, 1.0 M Ag+, 1.0 M Au3+,
and 1.0 M Ni2+ in the cathode compartment of an electrolytic cell.
Predict the order in which the metals will plate out as the voltage
is gradually increased.
95. In the electrolysis of an aqueous solution of Na2SO4, what reactions
occur at the anode and the cathode? (Assume standard conditions.)
S2O82- + 2e- → 2SO42O2 + 4H+ = 4e- → 2H2O
2H2O + 2e- → H2 + 2OHNa+ + e- → Na
E°= 2.01 V
E°= 1.23 V
E°= -0.83 V
E°= -2.71 V
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. Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
H2O2 + 2H+ + 2e- → 2H2O
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. MnO4-(aq) + I-(aq) ↔ I2(aq) + Mn2+(aq)
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 Cu2+ to Cu but not reduce Cu2+ to Cu+
b. Reduce Br2 to Br- but not to reduce I2 to I-
59. Consider the concentration cell shown below. Calculate the cell potential
at 25°C when the concentration of Ag+ in the compartment on the right
is the following.
a. 1.0 M
b. 2.0 M
c. 0.10 M
d. 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
[Ag+]
= 1.0M
AP Chemistry – Ch16 FRQ Review
1. Use principles of thermodynamics to answer the following questions.
(a) The gas N2O4 decomposes to form the gas NO2 according to the equation:
(i) Predict the sign of ∆H°for the reaction. Justify your answer.
(ii) Predict the sign of ∆S°for the reaction. Justify your answer.
(b) One of the diagrams best represents the relationship between ∆G°
and temperature for the reaction given in part (a).
Assume that ∆H°and ∆S°are independent of temperature.
Draw a circle around the correct graph.
Explain why you chose that graph in terms of the relationship ∆G°= ∆H°– T∆S°.
(c) A reaction mixture of N2O4 and NO2 is at equilibrium.
N2O4(g)
↔
2NO2(g)
Heat is added to the mixture while the mixture is maintained at constant pressure
(i) Explain why the concentration of N2O4 decreases.
(ii) The value of Keq at 25°C is 5.0 x 10-3.
Will the value of Keq at 100°C be greater than, less than, or equal to this value?
(d) Using the value of Keq at 25°C given in part (c-ii), predict whether the value of ∆H°
is expected to be greater than, less than, or equal to the value of T∆S°. Explain.
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 CuSO₄(ag) at 25°C,
as shown in the diagram. As the cell operates, copper metal is deposited
onto one electrode and O₂(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.
Half-reaction
O₂(g) + 4H⁺(ag) + 4e⁻ → 2H₂O(l)
Cu²⁺(ag) + 2e⁻ → Cu(s)
E° (V)
+1.23
+0.34
(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.
…a beaker containing 1.0 M CuSO₄(ag) at 25°C. As the cell operates, copper metal
is deposited onto one electrode and O₂(g) is produced as the other electrode.
The two reduction half-reactions for the overall reaction:
Half-reaction
E° (V)
O₂(g) + 4H⁺(ag) + 4e⁻ → 2H₂O(l)
+1.23
Cu²⁺(ag) + 2e⁻ → Cu(s)
+0.34
(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 25°C and 1.16 atm,
of the O₂(g) that is produced.
Ch16,17 FRQ Reviews from 2015 AP Test
1. Metal-air cells a relatively new type of portable energy source consisting of a metal
anode, an alkaline electrolyte paste that contains water, and a porous
cathode membrane that lets in oxygen from the air. A schematic of the cell is shown.
Reduction potentials for the cathode and 3 possible metal anodes are given below.
Half Reaction
(at pH 11 and 298K)
E
O2(g) + 2 H2O (l) + 4e- → 4OH-(aq)
+0.34 V
ZnO(s) + H2O(l) + 2e- → Zn(s)+ 2OH-(aq)
-1.31 V
Na2O(s) + H2O(l) + 2e- → 2Na(s) + 2OH-(aq) -1.60 V
CaO(s) + H2O(l) + 2e- → Ca(s) + 2OH-(aq)
-2.78 V
(a) Early forms of metal-air cells used zinc as the anode.
Zinc oxide is produced as the cell operates according
to the overall equation: 2Zn(s) + O2(g) → 2ZnO(s)
(i) Using the data in the table above, calculate
the cell potential for the zinc-air cell.
(ii) The electrolyte paste contains OH- ions.
On the diagram of the cell, draw an arrow
to indicate the direction of migration of OH- ions
through the electrolyte as the cell operates.
(b) A fresh zinc-air cell is weighted on an analytical balance before being placed
in a hearing aid for use.
(i) As the cell operates, does the mass of the cell increase, decrease
or remains the same?
(ii) Justify your answer to part (b-i) in terms of the equation for the overall cell reaction.
(c) The zinc-air cell is taken to the top of mountain where the air pressure is lower.
(i) Will the cell potential be higher, lower, or the same as the cell potential
at the lower elevation?
(ii) Justify your answer to part (c-i) based on the equation for the overall cell reaction
and the information above.
(d) Metal-air cells need to be lightweight for many applications. In order to transfer
more electrons with a smaller mass, Na and Ca are investigated as potential anodes.
A 1.0g anode of which of these metals would transfer more electrons, assuming that
the anode is totally consumed during the lifetime of a cell?
Justify your answer with calculations.
(e) The only common oxide of zinc has the formula ZnO.
(i) Write the electron configuration for a Zn atom in the ground state.
(ii) From which sublevel are electrons removed when a Zn atom
in the ground state is oxidized.
1. Answer the following questions about the solubility of Ca(OH)2
(Ksp = 1.3 x 10-6).
(a) Write a balanced chemical equation for the dissolution of Ca(OH)2(s)
in pure water.
(b) Calculate the molar solubility of Ca(OH)2 in 0.10 M Ca(OH3)2.
(c) In the box below, complete a particle representation diagram that includes
four water molecules with proper orientation around the Ca2+ ion.
Represent water molecules as
Ca2+
V
1.0M CuSO4(aq)
1.0M H+(aq)