Electrochemistry
John W. Moore
Conrad L. Stanitski
Peter C. Jurs
http://academic.cengage.com/chemistry/moore
Chapter 18
Electrochemistry and Its
Applications
Stephen C. Foster • Mississippi State University
Electrochemistry is the study and use of e- flow in
chemical reactions.
Redox reactions generate (and use) e• Those e- can be harnessed (batteries).
• Corrosion is an electrochemical reaction.
Applied e- flow:
• Can drive reactant-favored reactions toward products.
• Rechargeable batteries, electrolysis, and
electroplating…
Redox Reactions
Redox Reactions
Oxidation Number Refresher
• Pure element = 0.
• Monatomic ion = charge of ion.
• (ox. numbers in a species) = overall charge.
Oxidation & reduction (Redox) always occur together.
Element
F
Cl, Br, I
H
O
ox. no.
−1
−1
+1
−2
Exceptions?
None
Interhalogens
Metal hydrides = -1
Metal peroxides = -1
Halogen oxides
Redox Reactions
• Oxidation = loss of e- = increase in ox. no.
+2 e-
2 HCl(aq) + Mg(s)
+1 -1
H2(g) + MgCl2(aq)
0
0
-2
+2 -1
e-
H+ is reduced, Mg is oxidized.
Using Half-Reactions to Understand Redox
Give oxidation numbers for each atom. Identify the oxidizing
and reducing agents:
6 Fe2+ + Cr2O72- + 14 H3O+
6 Fe3+ + 2 Cr3+ + 21 H2O
Species
Fe2+
Cr2O72H3O+
+1
Fe3+
Cr3+
H2O
• Reduction = gain of e- = decrease in ox. no.
Redox reactions split into half reactions:
Ox. number
Explanation
+2
charge on ion
O = -2; Cr = +6
O is usually -2; 2(Cr) + 7(-2) = -2
O = -2; H = +1
O is usually -2; H is usually
+3
+3
O = -2; H = +1
Fe2+ → Fe3+ oxidation
Cr(+6) → Cr3+ reduction
charge on ion
charge on ion
O is usually -2, H is usually +1
Fe2+ = reducing agent
Cr2O72- = oxidizing agent
1
Using Half-Reactions to Understand Redox
Balancing Redox Equations
Half-reactions may include different numbers of e-:
Redox in acidic or basic solutions are harder
(H2O, H3O+ or OH- are often omitted…).
Al3+(aq) + 3 e-
Al(s)
Zn2+(aq) + 2 e-
Zn(s)
e- must balance in the full reaction.
2[ Al(s)
3[ Zn2+(aq) + 2 e2 Al(s) + 3 Zn2+(aq)
Al3+(aq) + 3 e- ]
Zn(s) ]
H3AsO4 + I2 →
HAsO2 + IO3-
which occurs in aqueous acidic solution.
2 Al3+(aq) + 3 Zn(s)
Balancing Redox Equations in Acidic Solution
H3AsO4 + I2
Balance:
HAsO2 + IO3-
(acidic solution)
(i) What is oxidized? Reduced?
I2 (I = 0) → IO3- (I = +5) oxidation
H3AsO4 (As = +5) → HAsO2 (As = +3) reduction
(ii) Write unbalanced half-reactions:
H3AsO4 → HAsO2
I2 → IO3(iii) Balance atoms (except H and O).
H3AsO4 → HAsO2
I2 → 2 IO3-
Balancing Redox Equations in Acidic Solution
(vii) Equalize e- and add.
5 [ H3AsO4 + 2 H+ + 2 e- → HAsO2 +2 H2O ]
1[
I2 + 6 H2O → 2 IO3- + 12 H+ + 10 e- ]
5 H3AsO4 + 10 H+ + 10 e- + I2 + 6 H2O
2H+
→ 5 HAsO2 + 10 H2O + 2 IO3- + 12 H+ + 10 e4H2O
5 H3AsO4 + I2 → 5 HAsO2 + 4 H2O + 2 IO3- + 2 H+
(viii) Make H3O+ (H2O + H+). Add H2O if needed.
5 H3AsO4 + I2 → 5 HAsO2 + 2 H2O + 2 IO3- + 2 H3O+
Balancing Redox Equations in Acidic Solution
(iv) Balance O (add H2O as needed).
H3AsO4 → HAsO2 + 2 H2O
I2 + 6 H2O → 2 IO3(v) Balance H (add H+ as needed).
H3AsO4 + 2 H+ → HAsO2 +2 H2O
I2 + 6 H2O → 2 IO3- + 12 H+
(vi) Balance charges (add e- ).
H3AsO4 + 2 H+ + 2 e- → HAsO2 +2 H2O
I2 + 6 H2O → 2 IO3- + 12 H+ + 10 ezero charge 2(-1) + 12(+1)}
10(-1)
0
=
10
+ -10
Balancing Redox Equations in Basic Solution
Balance the following (basic conditions):
N2 + S2S + N2H4
(i) Oxidized? Reduced?
N2 (N = 0) → N2H4 (N = -2) reduction
S2- (S = -2) → S (S = 0)
oxidation
(ii) Unbalanced half-reactions:
N2 → N2H4
S2- → S
(iii) Balance (except H and O).
N2 → N2H4
S2- → S
2
Balancing Redox Equations in Basic Solution
Balancing Redox Equations in Basic Solution
(vi) Equalize e- and add.
1 [ N2 + 4 H+ + 4 e- → N2H4 ]
2 [ S2- → S + 2 e- ]
+
N2 + 4 H + 4 e- + 2 S2- → N2H4 + 2 S + 4 eN2 + 4 H+ + 2 S2- → N2H4 + 2 S
(iv) Balance O (add H2O as needed).
N2 → N2H4
S2- → S
(v) Balance H (add H+ as needed).
N2 + 4 H+ → N2H4
S2- → S
(vii) Make H2O (H+ + OH-). Add OH-.
N2 + 4 H+ + 4 OH- + 2 S2- → N2H4 + 2 S + 4 OH-
(vi) Balance charges (add e- ).
N2 + 4 H+ + 4 e- → N2H4
S2- → S + 2 e-
N2 + 4 H2O + 2 S2- → N2H4 + 2 S + 4 OH-
Electrochemical Cells
Electrochemical Cells
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
e-
e-
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
Linked oxidation and reduction
reactions.
e-
e-
e- move across an external
conductor.
Also called a voltaic cell or a battery.
battery
• A battery is strictly a series of linked voltaic cells.
Electrochemical Cells
Electrodes (anode & cathode)
Allow e- to pass in and out of
solution.
A salt bridge (or porous barrier)
is required...
Anode
(oxidation)
Salt
bridge
Cathode
(reduction)
Electrochemical Cells
Salt bridge
Contains a salt solution (e.g. K2SO4 ).
Ions pass into the cells (restricts bulk flow).
Stops charge buildup.
porous
plug
Zn
SO42Zn2+
SO42- released
as Zn → Zn2+
K2SO4
Cu
K+
Cu2+
2 K+ released
as Cu2+ → Cu
3
Electrochemical Cells
Electrochemical Cells
Zinc is removed:
Zn
A compact notation:
Zn2+ + 2 e-
• Oxidation at the anode (both vowels).
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s)
anode cell
• Zn supplies e-.
cathode cell
• Anode has “-” charge.
Current flows from anode to cathode.
Copper is deposited:
Cu2+ + 2 e-
Cu
• Reduction at the cathode (both consonants).
• Cu2+ accepts e-.
| = phase boundary.
|| = salt bridge.
Details (e.g. concentration) listed after each species.
• Cathode has “+” charge.
Electrochemical Cells & Voltage
Electrical work
= charge x ΔEp
= (number of e-) ΔEp
SI Units
Charge: 1 coulomb (C) = 1 ampere x second = 1 As
Potential:
1 volt (V) = 1 J C-1
Electrochemical Cells & Voltage
Cell voltage varies if conditions vary.
A standard voltage ( E° ) occurs if:
• All [solute] = 1 M.
 or saturated if the solubility < 1 M.
• All gases have P = 1 bar.
Voltage depends on cell
chemistry ≠ size.
• All solids are pure.
Charge depends on nreactants ≈
size.
Electrochemical Cells & Voltage
Electrochemical Cells & Voltage
Standard hydrogen electrode (SHE)
E°cell is positive = product favored reaction
(E°cell < 0 is reactant favored)
favored
Absolute voltages cannot be measured.
Pt | H2(1bar), 1M H3O+ ||
E° = 0 V (oxidation & reduction).
• They are measured relative to a standard electrode.
2 H3O+(aq, 1M) + 2 e-
H2(g, 1 bar) + 2 H2O (ℓ)
4
Using Standard Cell Potentials
Reduction Half Reaction
E° (V)
→ 2 F-(aq)
+2.87
H2O2(aq) + 2 H3O + 2 e- → 4 H2O(ℓ)
+1.77
F2(g) + 2 e-
Electrochemical Cells & Voltage
Cu2+(aq) + 2 eCu(s)
Zn(s)
Zn2+(aq) + 2 e-
reduction
oxidation
MnO4-(aq)+8 H3O+ + 5 e- → Mn2+(aq) + 12 H2O(ℓ) +1.51
Tabulated as
reductions.
Cl2(g) + 2 e-
→ 2 Cl-(aq)
+1.358
Br2(g) + 2 e-
→ 2 Br-(aq)
+1.066
Ag+(aq) + e-
→ Ag(s)
+0.799
Cu2+(aq) + 2 e-
→ Cu(s)
+0.337
2 H3O+(aq) + 2 e-
→ H2(g) + 2 H2O(ℓ)
Ni2+(aq)
e-
→ Ni(s)
-0.25
Fe2+(aq) + 2 e-
→ Fe(s)
-0.44
Zn2+(aq) + 2 e-
→ Zn(s)
-0.763
Al3+(aq) + 3 e-
→ Al(s)
-1.66
Li+(aq)
→ Li(s)
-3.045
+2
+
e-
0.00
Electrochemical Cells & Voltage
Reaction
Process
E°red (Table)
E°
Zn → Zn2+ + 2 e-
oxidation
-0.76 V
+0.76 V
→ Cu
reduction
+0.34 V
+0.34 V
Cu2+
+2
e-
E°cell = E°cathode - E°anode
E°cell = 0.34 – (-0.76) V
The overall voltage:
E°cell = E°Zn2+, reduction + E°Cu, oxidation
If an equation is reversed, E° → -1 x E°
E°oxidation = - E°reduction.
E° tables are reduction values, so:
E°cell = E°reduction + E°oxidation
E°cell = E°cathode - E°anode
Electrochemical Cells & Voltage
What is E° for a Ni(s) | Ni2+|| Ag+ | Ag(s) cell?
Reduction Half Reaction
Ag+(aq)
.
.
.
+
e-
→
Ag(s)
E° (V)
+0.799
.
.
.
Ni2+(aq) + 2 e- → Ni(s)
-0.25
= 1.10 V
Or
E°cell = E°oxid + E°red
= 0.34 + 0.76 V = 1.10 V
Anode written on the left; cathode the right.
E° = E°cathode - E°anode = 0.799 – (-0.25) V = 1.05 V
Electrochemical Cells & Voltage
Using Standard Cell Potentials
If Ni(s)| Ni2+(aq, 1M) is connected to SHE, the Ni
electrode loses mass over time. E°cell = 0.25 V. Is Ni
oxidized or reduced? What is E° for the Ni half cell?
1. Tabulated half-cell E° are reductions.
Ni loses mass:
Ni(s) → Ni2+(aq) + 2 eNi is oxidized (a
anode).
3. More positive E° = easier reduction.
Since:
E°cell = E°cathode - E°anode
0.25 V = E°SHE - E°anode = 0 - E°anode
E°anode = -0.25 V
(Note: this is the tabulated reduction value)
2. Reactions can be reversed.
4. Less positive E° = easier oxidation (for the reverse
reaction).
5. A “left” species, will oxidize any “right” below it.
6. E° depends on [reactant] and [product], but not on
nreactant or nproduct (i.e not stoichiometric coefficient).
5
Using Standard Cell Potentials
Using Standard Cell Potentials
Will Zn(s) react with a 1 M iron(III) solution? If so what
is E° for the reaction?
Will Zn react with a 1 M iron(III) solution? If so what is E° for the reaction.
Reduction Half Reaction
F2(g) + 2
→ 2
e-
E° (V)
F-(aq)
+2.87
→ Fe2+(aq)
Fe3+(aq) + e-
Yes!
+2
→ Zn(s)
e-
E° = +0.771 V
E° = -1(-0.763 V)
2 Fe3+ + Zn → 2 Fe2+ + Zn2+
E°cell = +1.534 V
+0.771
2 H3O+(aq) + 2 e- → H2(g) + 2 H2O(ℓ)
Zn2+(aq)
2 Fe3+ + 2 e- → Fe2+
Zn
→ Zn2+ + 2 e-
+0.0
-0.763
Fe3+ (“left”) is higher than Zn (“right”).
Note
2 x (Fe3+ reaction) to balance e-.
E° (Fe3+) is not doubled.
Using Standard Cell Potentials
Using Standard Cell Potentials
a) Will Al(s) react with a 1 M tin(IV) solution?
b) Will 1 M Fe2+ react with Sn(s)?
What’s the voltage of: Al(s)|Al3+,1M || Sn2+,1M|Sn(s) ?
Sn4+ + 2 eSn2+ + 2 eFe2+ + 2 eAl3+ + 3 e-
→
→
→
→
Sn2+ (s)
Sn (s)
Fe (s)
Al (s)
+0.15 V
-0.14 V
-0.44 V
-1.66 V
(a) Yes – “left” (Sn4+) above “right” (Al).
(b) No – “left” (Fe2+) below “right” (Sn).
E° and Gibbs Free Energy
Product-favored reactions: ΔG° < 0.
Spontaneous cell reactions: E°cell > 0.
ΔG° = –n F E°
E°cell
with n = moles of e- transferred,
F = Faraday constant = charge/(mol of e-).
= (e- charge) x (Avogadro’s number).
= (1.60218 x 10-19 C)(6.02214 x 1023 mol-1).
F = 96,485 C/mol = 96,500 C/mol (3 sig. fig.)
Sn2+ + 2 e- → Sn
Al
→ Al3+ + 3 eBalance e-:
3(Sn2+ + 2 e- → Sn)
2(Al
→ Al3+ + 3 e- )
2+
3 Sn + 2 Al → 3 Sn + 2 Al3+
E° = -0.14 V
E° = -1(-1.66 V)
E° = -0.14 V
E° = +1.66 V
E°cell = +1.52 V
E° and Gibbs Free Energy
Cu2+ + Zn(s) → Cu(s) + Zn2+
E°cell = 1.10 V
Spontaneous.
ΔG° = − nFE°cell = −2 mol (96500 C/mol)(1.10 V)
= −2 .12 x105 J
(1 J = 1 C V)
= −212 kJ
6
ΔG°, E°cell, and K°
Since:
ΔG°, E°cell, and K°
ΔG° = − RT ln K° = −nFE°cell
Determine K° for:
Cu2+ + Zn(s) → Cu(s) + Zn2+
RT
E°cell = R T ln K° = (2.303)
log K°
nF
nF
E°cell = 0.0257 V ln K°
n
At 298 K:
E°cell = 1.10 V
E°cell = 0.0592 V log K°
n
or,
1.10 V =
0.0592 V log K°
2
2 mol e
log K° = 37.16
0.0592 V
E°cell =
log K°
n
K° = 1037.16 = 1.5 x 1037
Effect of Concentration on Cell Potential
Effect of Concentration on Cell Potential
E° values apply if [solute] = 1 M (or saturated).
What is the voltage for:
Cu2+ + Zn(s) → Cu(s) + Zn2+
2+
if [Cu ] = 0.1 M and [Zn2+] = 5.0 M. E°cell = 1.10 V.
Other conditions:
Ecell = E°cell −
RT
ln Q
nF
Nernst equation
E = E° −
At 298 K:
Ecell = E°cell − 0.0592 V log Q
n
E = 1.10 −
Ecell = E°cell − 0.0257 V ln Q
n
5.0
0.1
Concentration Cells
Concentration dependence leads to:
Zn | Zn2+ (dilute) || Zn2+ (conc.) | Zn
Ecell ≠ 0 V
Zn | Zn2+ (0.01M) || Zn2+ (1M) | Zn
anode = oxidation
0.0592
log
2
= 1.05 V
Concentration Cells
Example
[Zn2+]
0.0592
log
[Cu2+]
2
E = E° −
E = 0.0 −
0.010
0.0592
log
1
2
E = 0.0 −
0.0592
log 10-2
2
cathode = reduction
Zn(s) → Zn2+ (0.01M) + 2 eZn2+ (1M) + 2 e- → Zn(s)
Zn2+(1M) → Zn2+(0.01M)
(net reaction)
[Zn2+]dilute
0.0592
log
[Zn2+]conc
2
E = 0.0592 V
7
Common Batteries
Primary Batteries
Alkaline Battery:
Primary battery
One time use. Not easily rechargeable
Zn(s) + 2 OH-(aq) → ZnO(aq) + H2O(ℓ) + 2 eMnO2(s) + H2O(ℓ) + e- → MnO(OH)(s) + OH-(aq)
Overall
Secondary battery
Rechargeable battery.
Zn(s) + H2O(ℓ) + 2 MnO2(s) → ZnO(aq) + 2 MnO(OH)(s)
Ecell = 1.54 V when new.
Primary Batteries
Secondary Batteries
Mercury battery
Lead-Acid Battery (high capacity, high current).
Pb(s) + HSO4-(aq) + H2O(ℓ) → PbSO4(s) + H3O+(aq) + 2 ePbO2(s) + 3 H3O+(aq) + HSO4-(aq) + 2 e- → PbSO4(s) + 5 H2O
Pb + PbO2(s) + 2 H3O+ + 2 HSO4-(aq) → 2 PbSO4(s) + 4 H2O
Net E° = +2.041 V
Zn(s) + 2 OH- → ZnO(aq) + H2O(ℓ) + 2 eHgO(s) + H2O(ℓ) + 2 e- → Hg(ℓ) + 2 OH-(aq)
Zn(s) + HgO(s) → Hg(ℓ) + ZnO(aq)
• Insoluble PbSO4 stays on the electrodes
• The reaction is reversed by recharging.
Ecell = 1.35 V
Lead-Acid Storage Battery
6 cells in series (12 V).
Secondary Batteries
Nickel-Cadmium (Nicad).
Cd(s) + 2 OH-(aq) → Cd(OH)2(s) + 2 e2[NiO(OH)(s) + H2O(ℓ) + e- → Ni(OH)2(s) + OH-(aq)]
Cd(s) + 2 NiO(OH)(s) + 2 H2O → Cd(OH)2(s) + 2 Ni(OH)2(s)
Insoluble
(Rechargable…)
net: E° = +1.299 V
8
Secondary Batteries
Secondary Batteries
Nickel-metal hydride (NiMH). Doesn’t use toxic Cd.
Lithium Ion. Low mass, high energy density.
MH(s) + OH-(aq) → M(s) + H2O(ℓ) + e-
Li(s) (in polymer) → Li+ (in polymer) + eLi+ ( in CoO2 ) + e- + CoO2 → LiCoO2
NiO(OH)(s) + H2O(ℓ) + e- → Ni(OH)2(s) + OH-(aq)
MH(s) + NiO(OH)(s) → M(s) + Ni(OH)2(s)
E°cell =+1.4 V
Li(s) + CoO2(s) → LiCoO2(s)
M is a metal alloy in KOH
Ecell = 3.4 V
Fuel Cells
Electrolysis
Convert bond energy into electricity.
Electrolytic cell:
cell Applied voltage forces a reaction to
occur. e.g. electrolysis of molten NaCl:
ProtonProton-Exchange
Membrane (PEM) fuel cell.
H2 → 2 H+ + 2 e½ O2 + 2 H+ + 2 e- → H2O
V
e-
e-
Pt catalyst
2 Na+ + 2 e- → 2Na (ℓ)
O2 in
H2 in
2
→ Cl2 (g) + 2
+2
Cl-
→ Cl2 (g) + 2 Na(ℓ)
H+
gases flow
through
channels
2
Na+
E° = -2.714 V
Cl-
e-
E° = -1.358 V
E° = -4.072 V
H+
Graphite electrodes.
Pt catalyst coated on both
sides of the membrane.
H2 out
H2O out
Anode
Cathode
Na(ℓ) and Cl2(g) produced if > 4.1 V is applied.
However, melting NaCl takes lots of energy…
H+ exchange
membrane
Electrolysis
Electrolysis
reductions
K+(aq) + e- → K(s)
oxidations
Aqueous solutions? Other reactions can occur.
Consider KI(aq):
2 I-(aq) → I2(aq) + 2 e-
E° = -2.925 V
2 H2O(ℓ) + 2 e- → H2(g) + 2 OH-(aq) E° = -0.828 V
E° = -0.535 V
6 H2O(ℓ) → O2(g) + 4 H3O+(aq) + 4 e- E° = -1.229 V
E°ox → -1 x E°red
The most positive E half reactions occur…
2 H2O + 2 I- → H2 + I2 + 2 OH-
Ecell = -1.363 V
Brown = I2 ; Purple = OH- (phenolphthalein)
9
Electrolysis
Electrolysis
Reduction Half Reaction
Aqueous
oxidation will
not occur.
Aqueous
reduction will
not occur.
F2(g) +2 e-
E° (V)
→ 2 F-(aq)
+2.87
→ 4 H2O(ℓ)
+1.77
→ 2 Cl-(aq)
+1.358
O2(g) + 4 H3O+(aq) + 4 e- → 6 H2O(ℓ)
+1.229
Br2(g) +2 e-
→ 2 Br-(aq)
+1.066
Ag+(aq) + e-
→ Ag(s)
+0.799
2 H3O+(aq) + 2 e-
→ H2(g) + 2 H2O(ℓ)
Ni2+(aq) + 2 e-
→ Ni(s)
-0.25
Fe2+(aq) + 2 e-
→ Fe(s)
-0.44
Zn2+(aq) + 2 e-
→ Zn(s)
-0.763
2 H2O(ℓ) + 2 e-
→ H2(g) + 2 OH-(aq)
-0.8277
Al3+(aq) + 3 e-
→ Al(s)
-1.66
Na+(aq) + e-
→ Na(s)
-2.714
K+(aq) + e-
→ K(s)
-2.925
H2O2(aq) + 2 H3O+2
e-
Cl2(g) +2 e-
0.00
Electrolysis of Brine
Summary
Metal ions are reduced if E°red > −0.8 V
Aqueous Na+, K+, Mg2+, Al3+ … cannot be reduced.
Anions can be oxidized if E°red < +1.2 V
Aqueous F- … cannot be oxidized.
In practice Erequired > E calculated.
Overvoltage is needed
• Cl-(aq) can be oxidized to Cl2(g).
Counting Electrons
Chlorine is produced from NaCl(aq) by the chlor-alkali
process.
2 Cl-(aq) → Cl2(g) + 2 e2 H2O(ℓ) + 2 e- → 2 OH-(aq) + H2(g)
Cu2+(aq) + 2 e-
Cu(s)
2 mol e- ≡ 1 mol Cu.
Also
charge = current x time
1 coulomb = 1 ampere x 1 second
1C=1As
NaOH(aq) is 21- 30% NaOH by weight.
Counting Electrons
Counting Electrons
Determine the mass of Cu plated onto an electrode from
a Cu2+ solution by the application of a 10. A current for
10. minutes.
What is the cost to produce 14. g of Al (mass of a soda
can) by the reduction of Al3+. Assume V = 4.0 V and 1
kWh of electricity costs 10 cents.
Cu2+(aq) + 2 e-
Charge = 10. A x 10. min x (60 s/min) = 6.0 x
6.0 x103 C
Al3+ + 3 e-
Cu(s)
103
C
1 mol e- 1 mol Cu 63.55 g
= 2.0 g
96500 C 2 mol e- 1 mol Cu
14. g 1 mol Al
26.98 g
e-
3 mol
1 mol Al
Al
96500 C
1 mol e-
= 1.5 x 105 C
1J
Energy used = (1.5 x105 C)(4.0V) 1 C x 1V
1 kWh
3.60 x 106 J
= 0.17 kWh (or 1.7 cents)
10
Corrosion: Product-Favored Reactions
Corrosion: Product-Favored Reactions
Oxidation of metals by the environment.
Iron “rusts”
Anode:
M(s)
Mn+ + n e-
Cathode: often involve water and/or O2
O2(g) + 2 H2O(ℓ) + 4 e- → 4 OH-(aq)
2 H2O(ℓ) + 2 e- → 2 OH-(aq) + H2(g)
Anode:
2[Fe(s)
Cathode: O2(g) + 2 H2O(ℓ) + 4 e2 Fe(s) + O2(g) + 2 H2O(ℓ)
Iron(II) hydroxide is converted to rust by O2 and H2O:
Rust
(Fe2O3·xH2O(s); x = 2 - 4)
Anode and cathode must be electrically connected.
(The metal itself acts as the conductor).
Corrosion: Product-Favored Reactions
Fe2+ + 2 e-]
4 OH-(aq)
2 Fe(OH)2(s)
Corrosion Protection
Anodic inhibition
Iron nails corroding:
• Blue = Fe2+ indicator
• Purple = OH- indicator
• Paint or coat the surface.
• Form thin metal oxide coat:
2 Fe(s) + 2 Na2CrO4(aq) + 2 H2O(ℓ)
→ Fe2O3(s) + Cr2O3(s) + 4 NaOH(aq)
Stress points corrode
quickly.
Insoluble coating;
impervious to O2 and H2O
Cathodic protection – Attach a more reactive metal
which will corrode first.
Corrosion Prevention
Iron can also be galvanized (coated with Zn):
Zn(OH)2 (insoluble film) forms on the
surface.
Zn → Zn2++ 2 e- E° = 0.763 V
Fe → Fe2+ + 2 e- E° = 0.44 V
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