Electrochemistry
Chapter 20
1
Half-reaction method…remember?
• Example
•
•
Al(s) + Cu2+(aq)  Al3+(aq) + Cu(s)
Oxidation: Al(s)  Al3+(aq) + 3eReduction: 2e- + Cu2+(aq)  Cu(s)
– Use lowest common multiple to make both equivalent in number of electrons
• Oxidation  multiply by 2
– 2Al(s)  2Al3+(aq) + 6e-
• Reduction  multiply by 3
– 6e- + 3Cu2+(aq)  3Cu(s)
– Collate (electrons cross out)
• Net reaction:
2Al(s) + 3Cu2+(aq)  2Al3+(aq) + 3Cu(s)
DOES EVERYTHING BALANCE?
(Make sure to balance after every step!)
2
In acidic milieu
• Oxidation:
• Reduction:
Fe2+(aq) + MnO4-(aq)  Fe3+(aq) + Mn2+(aq)
Fe2+(aq)  Fe3+(aq) + e8H+(aq) + MnO4-(aq)  4H2O(l) + Mn2+(aq)
– What did I do in the above half-rxn?
– Is it fully balanced?
•
•
•
5e- + 8H+(aq) + MnO4-(aq)  4H2O(l) + Mn2+(aq)
Balance both half-reactions:
5Fe2+(aq)  5Fe3+(aq) + 5e- (multiply by 5; why?)
5e- + 8H+(aq) + MnO4-(aq)  4H2O(l) + Mn2+(aq)
Collate
Net rxn:
5Fe2+(aq) + 8H+(aq) + MnO4-(aq)  5Fe3+(aq) +4H2O(l) + Mn2+(aq)
3
Solve
VO2+(aq) + Zn(s)  VO2+(aq) + Zn2+(aq)
4
Answer
VO2+(aq) + Zn(s)  VO2+(aq) + Zn2+(aq)
• Oxidation:
Zn(s)  Zn2+(aq) + 2e• Reduction:
e- + 2H+(aq) + VO2+(aq)  VO2+(aq) + H2O(l)
• Balancing both half-reactions:
Zn(s)  Zn2+(aq) + 2e2e- + 4H+(aq) + 2VO2+(aq)  2VO2+(aq) + 2H2O(l)
• Collate
• Net reaction:
Zn(s) + 4H+(aq) + 2VO2+(aq)  2VO2+(aq) + 2H2O(l) + Zn2+(aq)
5
In basic milieu
• Oxidation:
• Reduction:
I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s)
I-(aq)  I2(aq) + e2I-(aq)  I2(aq) + 2eMnO4-(aq)  2OH-(aq) + MnO2(s)
2H+(aq) + 2OH-(aq) + MnO4-(aq)  2OH-(aq)+ MnO2(s)
2H2O(l) + MnO4-(aq)  2OH-(aq)+ MnO2(s)
2H2O(l) + MnO4-(aq)  4OH-(aq)+ MnO2(s)
3e- + 2H2O(l) + MnO4-(aq)  4OH-(aq)+ MnO2(s)
• Balance both half-reactions:
• Collate
• Net rxn:
6I-(aq)  3I2(aq) + 6e-
6e- + 4H2O(l) + 2MnO4-(aq)  8OH-(aq)+ 2MnO2(s)
6I-(aq) + 4H2O(l) + 2MnO4-(aq)  3I2(aq) + 8OH-(aq)+ 2MnO2(s)
6
Solve
• Al(s) + H2O(l)  Al(OH)4-(aq) + H2(g)
7
Answer
• Al(s) + H2O(l)  Al(OH)4-(aq) + H2(g)
• Oxidation:
•
•
•
•
Al(s) + 4OH-(aq)  Al(OH)4-(aq) + 3eReduction:
2e- + 2H2O(l)  2OH-(aq) + H2(g)
Balance each half-reaction:
2Al(s) + 8OH-(aq)  2Al(OH)4-(aq) + 6e6e- + 6H2O(l)  6OH-(aq) + 3H2(g)
Collate
Net-reaction:
2Al(s) + 2OH-(aq) + 6H2O(l)  2Al(OH)4-(aq) +3H2(g)
8
Electricity
• Movt of electrons
• Movt of electrons through wire connecting 2
•
•
half-reactions  electrochemical cell
Also called voltaic or galvanic cell
Cell produces current from spontaneous rxn
– Example: copper in solution of AgNO3 is spontaneous
• On the other hand, an electrolytic cell uses
electrical current to drive a non-spontaneous
chemical rxn
9
Voltaic cell
• Solid Zn in zinc ion solution =
•
•
•
•
•
•
half-cell
Likewise, Cu/Cu-ion solution
Wire attached to each solid
Salt bridge =
1. contains electrolytes,
2. connects 2 half-cells,
3. anions flow to neutralize
accumulated cations at anode
and cations flow to neutralize
accumulated anions at cathode
(completes circuit)
• “An Ox” = anode oxidation
• Has negative charge because
releases electrons
• “Red Cat” = reduction
cathode
• Has positive charge because
takes up electrons
10
Electrical current
•
•
•
•
•
•
•
•
Measured in amperes (A)
1 A = 1 C/s
Coulomb = unit of electric charge
e- = 1.602 x 10-19 C
1 A = 6.242 x 1018 e-/s
Electric current driven by difference in potential
energy per unit of charge: J/C
Potential difference (electromotive force or emf)
= volt (V)
Where 1 V = 1 J/C
11
More…
• In the voltaic cell, potential difference (emf)
between cathode and anode is referred to as
– Cell potential (Ecell)
• Under standard conditions (1 M, 1 atm, 25°C),
cell potential is
Standard cell potential = E°cell
•
• Cell potential = measure of overall tendency of
redox rxn to occur spontaneously
• Thus, the higher the E°cell, the greater the
spontaneity
12
Electrochemical notation
• Cu(s)|Cu2+(aq)||Zn2+(aq)|Zn(s)
• Notation describes voltaic cell
• An ox on left
• Red cat on right
• Separated by double vertical line (salt
bridge)
• Single vertical line separates diff phases
13
Electrochemical notation
• Some redox rxns reactants & products in
same phase
• Mn doesn’t precipitate out  uses Pt at
cathode
– Pt is inert, but provides area for electron
gain/loss
• Fe(s)|Fe2+(aq)||MnO4-(aq), H+(aq),
Mn2+(aq)|Pt(s)
• Write out net reaction
14
Answer
• Fe(s)|Fe2+(aq)||MnO4-(aq), H+(aq),
Mn2+(aq)|Pt(s)
• Oxidation:
Fe(s)  Fe2+(aq) + 2e• Reduction:
5e- + MnO4-(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l)
• Net-reaction:
5Fe(s) + 2MnO4-(aq) + 16H+(aq)  5Fe2+(aq) + 2Mn2+(aq) + 8H2O(l)
15
Standard reduction potentials
• One half-cell must have a potential of zero
to serve as reference
– Standard hydrogen electrode (SHE) halfcell
• Comprises Pt electrode in 1 M HCl w/ H2
bubbling at 1 atm:
• 2H+(aq) + 2e-  H2(g); E°red = 0.00 V
16
Example
• Throw zinc into 1M HCl
• Zn(s)|Zn2+(aq)||2H+(aq)|H2(g)
• E°cell = E°ox + E°red = 0.76 V
• If E°red = 0.00 V (as the reference)
• Then E°ox = 0.76 V (= oxid of Zn half-rxn)
• Reduction of Zn-ion
– Is = -0.76 V (non-spontaneous)
17
Problem
• Cr(s)|Cr3+(aq)||Cl-(aq)|Cl2(g)
• What is the std cell pot (E°cell) given oxid
of Cr = 0.73 V and Cl red = 1.36V?
• Hint: standard electrode potentials are
intensive properties; e.g., like density
– Stoichiometry irrelevant!
18
Solution
• E°cell = E°ox + E°red = 0.73V + 1.36V
= 2.09V
19
Appendix M, pages A-33-35
• Standard reduction potentials in aqueous
solution @ 25°C
• Also, pg. 967, Table 20.1 (gives increasing
strengths of ox/red agents)
– Let’s take a look at it
• Does increasing strengths of ox/red agents make
sense?
• What happens to oxidizing agent, reducing agent?
20
Problem
• Calculate the standard cell potential for
the following:
Al(s) + NO3-(aq) + 4H+(aq)  Al3+(aq) + NO(g) + 2H2O(l)
21
Answer
• Oxidation
Al(s)  Al3+(aq) + 3e-; E°ox = 1.66V
• Reduction
NO3-(aq) + 4H+(aq) + 3e-  NO(g) + 2H2O(l);
E°red = 0.96V
• E°cell = E°ox + E°red =1.66V + 0.96V =
2.62V
22
Predicting the spontaneous
direction of a redox rxn
• Generally, any reduction half-rxn is
spontaneous when paired w/reverse of
half-rxn below it in table of standard
reduction potentials
• Let’s look at table
• Predict the exact value and spontaneity for
the following:
Fe(s) + Mg2+(aq)  Fe2+(aq) + Mg(s)
23
Answers
•
Fe(s) + Mg2+(aq)  Fe2+(aq) + Mg(s)
Oxidation
• Reduction
Fe(s)  Fe2+(aq) + 2e-; E°ox = 0.45V
Mg2+(aq) + 2e-  Mg(s); E°red = -2.37V
• E°cell = E°ox + E°red =0.45V + -2.37V =
-1.92V
nonspontaneous
24
Will metal X dissolve in acid?
• Metals whose reduction half-rxns lie below
reduction of proton to hydrogen gas will
dissolve in acids
• Why?
– Just look at the table!
• Nitric acid is exception
– Let’s take a look
25
E°cell, G°, K
• What must the values for E°cell, G°, & K
be in order to have a spontaneous rxn?
• G°<0
• E°cell>0
• K>1
– Product-favored
26
Relationship between G° & E°cell
• Faraday’s Constant (F) = 96,485 C/mol e• G° =-ne-FE°cell
• Problem:
• Calculate G° for
I2(s) + 2Br-(aq)  2I-(aq) + Br2(l)
Is it spontaneous?
27
Solution: it’s nonspontaneous!
I2(s) + 2Br-(aq)  2I-(aq) + Br2(l)
• Oxidation
2Br-(aq)  Br2(l) + 2e-; E°ox = -1.09V
• Reduction
I2(s) + 2e-  2I-(aq); E°red = 0.54V
• E°cell = -1.09V + 0.54V = -0.55V
 G ° =-n e- FE ° cell =-(2m ol e )  (
-
96, 485 C
m ol e
-
)  (  0.55V )  1.1  10 J (w he re V =
5
J
)
C
28
Problem
2Na(s) + 2H2O(l)  H2(g) + 2OH-(aq) + 2Na+(aq)
Is it spontaneous?
29
Solution: it’s spontaneous!
• Oxidation
2Na(s)  2Na+(aq) + 2e-; E°ox = 2.71V
• Reduction
2H2O(l) + 2e-  H2(g) + 2OH-(aq) E°red = -0.83V
• E°cell = 2.71V + -0.83V = 1.88V
 G ° =-n e- FE ° cell =-(2m ol e )  (
-
96, 485 C
m ol e
-
)  (1.88V )  -3.63  10 J (w here V =
5
J
C
30
)
Relationship between E°cell & K
E ° cell =
RT
 lnK
nF
W here R = 8.314 J
m ol  K
, T = 298.15 K , F = 96,4 85 C
m ol e
-
lnK = 2.303logK
 E ° cell =
0.0592V
log K
n e-
31
,&
Problem
• Calculate K for
2Cu(s) + 2H+(aq)  Cu2+(aq) + H2(g)
32
Solution: is it product-favored?
• Oxidation
2Cu(s)  Cu2+(aq) + 2e-; E°ox = -0.34V
• Reduction
2H+(aq) + 2e-  H2(g); E°red = 0.00V
E°cell = -0.34V
E ° cell =
0.0592V
 0.34V 
log K
n e0.0592V
log K
2
K  3.3  10
12
33
Cell potential & concentration:
Nernst Equation
• Concentration ≠ 1M
– Non-standard
conditions
• Under standard
•
conditions, Q = 1
 Ecell = E°cell
E cell = E° cell -
0.0592 V
logQ
n e-
34
Problem
• Compute the cell potential, given
Cu(s)  Cu2+(aq, 0.010 M) + 2eMnO4-(aq, 2.0 M) + 4H+(aq, 1.0M) + 3e-  MnO2(s) + 2H2O(l)
35
Solution
• Balance the equation!
• Oxidation
3Cu(s)  3Cu2+(aq) + 6e-; E°ox = -0.34V
• Reduction
2MnO4-(aq)+ 8H+(aq) + 6e-  2MnO2(s) + 4H2O(l) ; E°red = 1.68V
E cell = E ° cell -
E cell = 1.34V -
E cell = 1.34V -
0.0592 V
logQ
n e0.0592 V
log
6
0.0592 V
6
[C u
2+ 3
]
- 2
+ 8
[M nO 4 ] [H ]
log
[0.010]
2
3
[2.0] [1.0]
E cell = 1.34V -(-0.065V ) = 1.41V
8
36
To summarize
• If Q<1, rxn goes to products
– Ecell > E°cell
• If Q>1, rxn goes to reactants
– Ecell < E°cell
• If Q = K, @ eq.,
– E°cell = 0 (& Ecell = 0)
• Explains why all batteries die
37
Concentration cells
• Voltaic cells can be constructed from two similar half-rxns where
difference in concentration drives current flow
Cu(s) + Cu2+(aq, 2.0M)  Cu2+(aq, 0.010M) + Cu(s)
– E°cell = 0 since both half-rxns are the same
• However, using Nernst equation, different concentrations yield
0.068V
– Let’s take a look
• Flow is from lower Cu-ion concentration half-cell to higher
one
– Down the concentration gradient
• The electrons will flow to the concentrated cell where they dilute the Cu-ion
concentration
• Results in  Cu-ion concentration in dilute cell &  Cu-ion
concentration in concentrated cell
38
Batteries
• Dry-cell batteries
– Don’t contain large amounts of water
• Anode
• Cathode
Zn(s)  Zn2+(aq) + 2e-
2MnO2(s) + 2NH4+(aq) + 2e-  Mn2O3(s) + 2NH3(g) + H2O(l)
– Cathode is carbon-rod immersed in moist (acidic)
paste of MnO2 that houses NH4Cl
• 1.5 V
39
Batteries
• More common dry-cell type
– Alkaline battery
• Anode
Zn(s) + 2OH-(aq)  Zn(OH)2(s) + 2e• Cathode
2MnO2(s) + 2H2O(l) + 2e-  2MnO(OH)(s) + 2OH-(aq)
• Longer shelf-life, “live” longer
• Cathode in basic paste
40
Car Batteries
• Lead-acid storage batteries
• 6 electrochemical cells (2V) in series
• Anode
•
Pb(s) + HSO4-(aq)  PbSO4(s) + H+(aq) + 2eCathode
PbO2(s) + HSO4-(aq) + 3H+(aq) + 2e-  PbSO4(s) + 2H2O(l)
• In 30% soln of sulfuric acid
• If dead due to excess PbSO4 covering electrode surfaces
• Re-charge (reverse rxn)  converts PbSO4 to Pb and
PbO2
41
Rechargeable batteries
• Ni-Cd
• Anode
• Cathode
Cd(s) + 2OH-(aq)  Cd(OH)2(s) + 2e-
2NiO(OH)(s) + 2H2O(l) + 2e-  2Ni(OH)2(s) + 2OH-(aq)
•
•
•
•
•
•
KOH, usually, used
1.30 V
Reverse rxn recharges battery
Excess recharging  electrolysis of water
EXPLOSION!!!
Muhahahaha!
42
43
Rechargeable batteries
• Since Cd is toxic
– Developed safer alternative
• Ni-MH
• Hybrid car batteries: high energy density
• Same cathode rxn as previous
• Anode
MH(s) + OH-(aq)  M(s) + H2O(l) + e-
• Commonly, M = AB5, where A is rare earth mixture of La,
Ce, Nd, Pr, and B is Ni, Co, Mn, and/or Mn
• Very few use AB2, where A = Ti and/or V
44
45
Rechargeable batteries
• Anode made of graphite w/incorporated
Li-ions between carbon layers
• Ions spontaneously migrate to cathode
• Cathode = LiCoO2 or LiMn2O4
• Transition metal reduced
• Used in laptop computers, cell phones,
digital cameras
• Light weight and high E density
46
Fuel cell
• Reactants flow through battery
– Undergo redox rxn
• Generate electricity
• Hydrogen-oxygen fuel cell
• Anode
• Cathode
2H2(g) + 4OH-(aq)  4H2O(l) + 4eO2(g) + 2H2O(l) + 4e-  4OH-(aq)
• Used in space-shuttle program
– And Arnold’s Hummah
47
Electrolysis
• Electrical current used
•
•
to drive
nonspontaneous
redox rxn
In electrolytic cells
Used in
– Electrolysis of water
– Metal plating: silver
coated on metal,
jewelry, etc.
48
Electrolytic cells: using electricity to
run a rxn
• Anode is “+”  gives electrons, connected
to positive terminal of power source
• Cathode is “-”  takes electrons,
connected to negative terminal of power
source
• Opposite scheme of voltaic cell!
49
Predicting the products of
electrolysis
1. Pure molten salts
–
Anion oxidized/cation reduced
• Obtain 2Na(s) and Cl2(g) from electrolysis of NaCl
2. Mixture of cations or anions
–
K+/Na+ and Cl-/Br- present
• Look at page 967 & compare half-cell
potentials
–
Cation/anion preferably reduced that has least
negative, or most positive, half-cell potential
50
Example
• Predict the half-rxn occurring at the anode and the cathode for
electrolysis of
• Oxidation
• Reduction
AlBr3 & MgBr2
Br-(l)  Br2(g) + 2e-; E°ox = -1.09V
Bromide will be oxidized at the anode
Al3+(l) + 3e-  Al(s); E°red = -1.66V
Mg2+(l) + 2e-  Mg(s) ; E°red = -2.37V
Reduction of Al will occur at the cathode since its potential is greater
than Mg’s
51
Predicting the products of
electrolysis
3. aqueous solns: same as #2
– Water redox might occur simultaneously
2H2O(l)  O2(g) + 4H+(aq) + 4e2H2O(l) + 2e-  H2(g) + 2OH-(aq)
E°ox = -0.82 V & E°red = -0.41 V
E°cell = -1.23 V
52
Problem
• Given oxidation of I- = -0.54 V & reduction
of Li+ = -3.04 V, which, if any, gases
would be formed and where; i.e., at
cathode/anode?
53
Solution
• Oxidation
2I-(aq)  2I2(aq) + 2e-; E°ox = -0.54V
2H2O(l)  O2(g) + 4H+(aq) + 4e-; E°ox = -0.82V
•
I- will be oxidized at the anode
Reduction
2Li+(aq) + 2e-  2Li(s); E°red = -3.04 V
2H2O(l) + 2e-  H2(g) + 2OH-(aq); E°red = -0.41 V
Water will be reduced at the cathode
54
Stoichiometry of electryolysis
• Can use e- stoichiometric relations to
predict moles and/or grams of substances
• Remember, unit of current = ampere = A
= 1 C (magnitude of current)/s (time of
current flow)
• Also, F = 96,485 C/mole e-
55
Problem
• Gold can be plated out of a soln
containing the Au3+ according to
Au3+(aq) + 3e-  Au(s)
• What mass of gold (in grams) will be
plated by the flow of 5.5 A of current for
25 mins?
56
Solution
25 m in 
60 s
1 m in

5.5 C
1s

m ol e

96, 485 C

1m ol A u
3 m ol e


196.97 g
 5.6 g A u
1m olA u
57