Electrochemistry
Chapter 20
1
Half-reaction method…remember?
• Example
2+
3+
Al(s)
(s) + Cu (aq) ⇒ Al (aq) + Cu(s)
(s)
3+
• Oxidation: Al(s)
(s) ⇒ Al (aq) + 3e
• Reduction: 2e- + Cu2+(aq) ⇒ Cu(s)
(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
1
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 halfhalf-rxn?
rxn?
– Is it fully balanced?
•
•
•
5e- + 8H+(aq) + MnO4-(aq) ⇒ 4H2O(l) + Mn2+(aq)
Balance both halfhalf-reactions:
5Fe2+(aq) ⇒ 5Fe3+(aq) + 5e- (multiply by 5; why?)
5e- + 8H+(aq) + MnO4-(aq) ⇒ 4H2O(l) + Mn2+(aq)
Collate
Net rxn:
rxn:
5Fe2+(aq) + 8H+(aq) + MnO4-(aq) ⇒ 5Fe3+(aq) +4H2O(l) + Mn2+(aq)
3
Solve
2+
2+
VO2+(aq) + Zn(s)
(s) ⇒ VO (aq) + Zn (aq)
4
2
Answer
2+
2+
VO2+(aq) + Zn(s)
(s) ⇒ VO (aq) + Zn (aq)
• Oxidation:
2+
Zn(s)
(s) ⇒ Zn (aq) + 2e
• Reduction:
e- + 2H+(aq) + VO2+(aq) ⇒ VO2+(aq) + H2O(l)
• Balancing both halfhalf-reactions:
2+
Zn(s)
(s) ⇒ Zn (aq) + 2e
+
+
2+
2e + 4H (aq) + 2VO2 (aq) ⇒ 2VO (aq) + 2H2O(l)
• Collate
• Net reaction:
+
+
2+
2+
Zn(s)
(s) + 4H (aq) + 2VO2 (aq) ⇒ 2VO (aq) + 2H2O(l) + Zn (aq)
5
In basic milieu
• Oxidation:
• Reduction:
I-(aq)
aq) + MnO4 (aq) ⇒ I2(aq) + MnO2(s)
I-(aq)
aq) ⇒ I2(aq) + e
2I-(aq) ⇒ I2(aq) + 2e-
MnO4-(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 halfhalf-reactions:
• Collate
• Net rxn:
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
3
Solve
• Al(s)
(s) + H2O(l) ⇒ Al(OH)4 (aq) + H2(g)
7
Answer
• Al(s)
(s) + H2O(l) ⇒ Al(OH)4 (aq) + H2(g)
• Oxidation:
•
•
•
•
Al(s)
(s) + 4OH (aq) ⇒ Al(OH)4 (aq) + 3e
Reduction:
2e- + 2H2O(l) ⇒ 2OH-(aq) + H2(g)
Balance each halfhalf-reaction:
2Al(s) + 8OH-(aq) ⇒ 2Al(OH)4-(aq) + 6e6e- + 6H2O(l) ⇒ 6OH-(aq) + 3H2(g)
Collate
NetNet-reaction:
2Al(s) + 2OH-(aq) + 6H2O(l) ⇒ 2Al(OH)4-(aq) +3H2(g)
8
4
Electricity
• Movt of electrons
• Movt of electrons through wire connecting 2
•
•
halfhalf-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 nonnon-spontaneous
chemical rxn
9
Voltaic cell
• Solid Zn in zinc ion solution =
halfhalf-cell
•
•
•
•
•
•
Likewise, Cu/CuCu/Cu-ion solution
Wire attached to each solid
Salt bridge =
1. contains electrolytes,
2. connects 2 halfhalf-cells,
3. anions flow to neutralize
accumulated cations at anode
and cations flow to neutralize
accumulated anions at cathode
(completes circuit)
• “An Ox”
Ox” = anode oxidation
• Has negative charge because
•
•
releases electrons
“Red Cat”
Cat” = reduction
cathode
Has positive charge because
takes up electrons
10
5
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)
emf)
= volt (V)
Where 1 V = 1 J/C
11
More…
• In the voltaic cell, potential difference (emf
(emf))
between cathode and anode is referred to as
– Cell potential (E
(Ecell)
• Under standard conditions (1 M, 1 atm,
atm, 25°
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
6
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
2+
• Fe(s)|Fe2+(aq)||MnO4-(aq), H+(aq)
(aq), Mn (aq)|Pt(s)
• Write out net reaction
14
7
Answer
2+
• Fe(s)|Fe2+(aq)||MnO4-(aq), H+(aq)
(aq), Mn (aq)|Pt(s)
• Oxidation:
2+
Fe(s)
(s) ⇒ Fe (aq) + 2e
• Reduction:
5e- + MnO4-(aq) + 8H+(aq) ⇒ Mn2+(aq) + 4H2O(l)
• Net-reaction:
5Fe(s) + 2MnO4-(aq) + 16H+(aq) ⇒ 5Fe2+(aq) + 2Mn
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) halfhalfcell
• Comprises Pt electrode in 1 M HCl w/ H2
bubbling at 1 atm:
• 2H+(aq) + 2e- ⇒ H2(g); E°red = 0.00 V
16
8
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(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
9
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’
Let’s take a look at it
• Does increasing strengths of ox/red agents make
sense?
• What happens to oxidizing agent, reducing agent?
20
10
Problem
• Calculate the standard cell potential for
the following:
+
3+
Al(s)
(s) + NO3 (aq) + 4H (aq) ⇒ Al (aq) + NO(g)
(g) + 2H2O(l)
21
Answer
• Oxidation
3+
°
Al(s)
(s) ⇒ Al (aq) + 3e ; E ox = 1.66V
• Reduction
NO3-(aq) + 4H+(aq) + 3e- ⇒ NO(g)
(g) + 2H2O(l);
E°red = 0.96V
• E°cell = E°ox + E°red =1.66V + 0.96V = 2.62V
22
11
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:
2+
2+
Fe(s)
(s) + Mg (aq) ⇒ Fe (aq) + Mg(s)
(s)
23
Answers
•
2+
2+
Fe(s)
(s) + Mg (aq) ⇒ Fe (aq) + Mg(s)
(s)
Oxidation
2+
°
Fe(s)
(s) ⇒ Fe (aq) + 2e ; E ox = 0.45V
• Reduction
°
Mg2+(aq) + 2e- ⇒ Mg(s)
(s); E red = -2.37V
• E°cell = E°ox + E°red =0.45V + -2.37V =
-1.92V
nonspontaneous
24
12
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’
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
– ProductProduct-favored
26
13
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 =-(2mol e- ) × (
96, 485C
J
) × ( −0.55V) = 1.1× 105 J (where V= )
mol e
C
28
14
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 =-(2mol e- ) × (
96, 485C
J
) × (1.88V) = -3.63 × 105 J (where V= )
mol e
C
30
15
Relationship between E°cell & K
E°cell =
RT
• lnK
nF
Where R = 8.314 J
mol • K
, T = 298.15 K, F = 96,485 C
mol e-
,&
lnK = 2.303logK
0.0592V
∴ E°cell =
log K
n e-
31
Problem
• Calculate K for
2Cu(s) + 2H+(aq) ⇒ Cu2+(aq) + H2(g)
32
16
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
log K
n e-
0.0592V
log K
2
K = 3.3 × 10−12
−0.34V =
33
Cell potential & concentration:
Nernst Equation
• Concentration ≠ 1M
– NonNon-standard
conditions
• Under standard
•
conditions, Q = 1
∴ Ecell = E°cell
E cell = E°cell -
0.0592 V
logQ
n e-
34
17
Problem
• Compute the cell potential, given
2+
Cu(s)
(s) ⇒ Cu (aq, 0.010 M) + 2e
MnO4-(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 -
0.0592 V
logQ
n e-
E cell = 1.34V-
0.0592 V
[Cu 2+ ]3
log
6
[MnO 4 - ]2 [H + ]8
0.0592 V
[0.010]3
log
6
[2.0]2 [1.0]8
= 1.34V-(-0.065V) = 1.41V
E cell = 1.34VE cell
36
18
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 halfhalf-rxns where
difference in concentration drives current flow
2+
2+
Cu(s)
(s) + Cu (aq, 2.0M) ⇒ Cu (aq, 0.010M) + Cu(s)
(s)
– E°cell = 0 since both halfhalf-rxns are the same
• However, using Nernst equation, different concentrations yield
0.068V
– Let’
Let’s take a look
• Flow is from lower CuCu-ion concentration halfhalf-cell to higher
one
– Down the concentration gradient
• The electrons will flow to the concentrated cell where they dilute
dilute the CuCu-ion
concentration
• Results in ⇑ CuCu-ion concentration in dilute cell & ⇓ CuCu-ion
concentration in concentrated cell
38
19
Batteries
• DryDry-cell batteries
– Don’
Don’t contain large amounts of water
• Anode
• Cathode
2+
Zn(s)
(s) ⇒ Zn (aq) + 2e
2MnO2(s) + 2NH4+(aq) + 2e- ⇒ Mn2O3(s) + 2NH3(g) + H2O(l)
– Cathode is carboncarbon-rod immersed in moist (acidic)
paste of MnO2 that houses NH4Cl
• 1.5 V
39
Batteries
• More common drydry-cell type
– Alkaline battery
• Anode
Zn(s)
(s) + 2OH (aq) ⇒ Zn(OH)2(s) + 2e
• Cathode
2MnO2(s) + 2H2O(l) + 2e- ⇒ 2MnO(OH)(s) + 2OH-(aq)
• Longer shelfshelf-life, “live”
live” longer
• Cathode in basic paste
40
20
Car Batteries
•
•
•
•
LeadLead-acid storage batteries
6 electrochemical cells (2V) in series
Anode
+
Pb(s)
(s) + HSO4 (aq) ⇒ PbSO4(s) + H (aq)
(aq) + 2e
Cathode
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
• ReRe-charge (reverse rxn)
rxn) ⇒ converts PbSO4 to Pb and
PbO2
41
Rechargeable batteries
• NiNi-Cd
• Anode
• Cathode
Cd(s)
(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!
Muhahahaha!
42
21
43
Rechargeable batteries
• Since Cd is toxic
– Developed safer alternative
• NiNi-MH
• Hybrid car batteries: high energy density
• Same cathode rxn as previous
• Anode
MH(s)
(s) + OH (aq)
aq) ⇒ M(s)
(s) + H2O(l) + e
• Commonly, M = AB5, where A is rare earth mixture of
La, Ce,
Ce, Nd,
Nd, Pr, and B is Ni, Co, Mn,
Mn, and/or Mn
• Very few use AB2, where A = Ti and/or V
44
22
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
23
Fuel cell
• Reactants flow through battery
– Undergo redox rxn
• Generate electricity
• HydrogenHydrogen-oxygen fuel cell
• Anode
• Cathode
2H2(g) + 4OH-(aq) ⇒ 4H2O(l) + 4eO2(g) + 2H2O(l) + 4e- ⇒ 4OH-(aq)
• Used in spacespace-shuttle program
– And Arnold’
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
24
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
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 halfhalf-cell
potentials
–
Cation/anion
Cation/anion preferably reduced that has least
negative, or most positive, halfhalf-cell potential
50
25
Example
• Predict the halfhalf-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)
(s); E red = 1.66V
2+
°
Mg (l) + 2e ⇒ Mg(s)
(s) ; E red = -2.37V
Reduction of Al will occur at the cathode since its potential is greater
than Mg’
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
26
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
27
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)
(s)
• What mass of gold (in grams) will be
plated by the flow of 5.5 A of current for
25 mins?
56
28
Solution
60 s 5.5C mol e− 1mol Au 196.97 g
25 min×
×
×
×
×
= 5.6 g Au
1min 1s
96, 485C 3mol e− 1molAu
57
29