Transcript Chapter

Chemistry II
Chapter 18
Electrochemistry
Redox Reaction
• one or more elements change oxidation number
 all single displacement, and combustion,
 some synthesis and decomposition
Redox Reaction
• always have both oxidation and reduction
 split reaction into oxidation half-reaction and a reduction
half-reaction
• aka e- transfer reactions
 half-reactions include eoxidizing agent is reactant molecule that causes oxidation
 contains element reduced
reducing agent is reactant molecule that causes reduction
 contains the element oxidized
Oxidation & Reduction
• oxidation:
 ox number of an element increases
 element loses e compound adds O
 compound loses H
 half-reaction has e- as products
• reduction:
 ox number of an element decreases
 element gains e compound loses O
 compound gains H
 half-reactions have e- as reactants
Rules for Assigning Oxidation States
• rules are in order of priority
1. free elements have an oxidation state = 0
 Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)
2. monatomic ions have an oxidation state equal to their
charge
 Na = +1 and Cl = -1 in NaCl
3. (a) the sum of the oxidation states of all the atoms in a
compound is 0
 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0
Rules for Assigning Oxidation States
3.
(b) the sum of the oxidation states of all the atoms in a polyatomic
ion equals the charge on the ion
 N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1
4.
(a) Group I metals have an oxidation state of +1 in all their
compounds
 Na = +1 in NaCl
(b) Group II metals have an oxidation state of +2 in all their
compounds
 Mg = +2 in MgCl2
Rules for Assigning Oxidation States
5.
in their compounds, nonmetals have oxidation states according to the
table below
 nonmetals higher on the table take priority
Nonmetal
Oxidation State
Example
F
-1
CF4
H
+1
CH4
O
-2
CO2
Group 7A
-1
CCl4
Group 6A
-2
CS2
Group 5A
-3
NH3
Oxidation and Reduction
• oxidation occurs when an atom’s oxidation state increases
•
during a reaction
reduction occurs when an atom’s oxidation state
decreases during a reaction
CH4 + 2 O2 → CO2 + 2 H2O
-4 +1
0
+4 –2
oxidation
reduction
+1 -2
Oxidation–Reduction
• oxidation and reduction must occur simultaneously
 if an atom loses electrons another atom must take them
• reactant that reduces an element in another reactant = reducing agent
 the reducing agent contains the element that is oxidized
• reactant that oxidizes element in another reactant = oxidizing agent
 the oxidizing agent contains the element that is reduced
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2 is the oxidizing agent
Identify the Oxidizing and Reducing Agents in
Each of the Following
3 H2S + 2 NO3– + 2 H+ → 3 S + 2 NO + 4 H2O
MnO2 + 4 HBr → MnBr2 + Br2 + 2 H2O
Identify the Oxidizing and Reducing Agents in
Each of the Following
red ag
ox ag
+1 -2
+5 -2
3 H2S + 2 NO3– + 2 H+ 3 S + 2 NO + 4 H2O
+1
0
+2 -2
oxidation
reduction
ox ag
red ag
+4 -2
+1 -1
MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O
+2 -1
oxidation
reduction
0
+1 -2
+1 -2
Common Oxidizing Agents
Oxidizing Agent
Product when Reduced
O2
O-2
H2O2
H2O
F2, Cl2, Br2, I2
F-, Cl-, Br-, I-
ClO3- (BrO3-, IO3-)
Cl-, (Br-, I-)
H2SO4 (conc)
SO2 or S or H2S
SO3-2
S2O3-2, or S or H2S
HNO3 (conc) or NO3-1
NO2, or NO, or N2O, or N2, or NH3
MnO4- (base)
MnO2
MnO4- (acid)
Mn+2
CrO4-2 (base)
Cr(OH)3
Cr2O7-2 (acid)
Cr+3
Common Reducing Agents
Reducing Agent
Product when Oxidized
H2
H+
H2O2
O2
I-
I2
NH3, N2H4
N2
S-2, H2S
S
SO3-2
SO4-2
NO2-
NO3-
C (as coke or charcoal)
CO or CO2
Fe+2 (acid)
Fe+3
Cr+2
Cr+3
Sn+2
Sn+4
metals
metal ions
Balancing Redox Reactions
1. assign oxidation numbers
a) determine element oxidized and element reduced
2. write ox. & red. half-reactions, including ea) ox. electrons on right, red. electrons on left of arrow
3. balance half-reactions by mass
a) first balance elements other than H and O
b) add H2O where need O
c) add H+ where need H
d) neutralize H+ with OH- in base
4. balance half-reactions by charge
a) balance charge by adjusting e5. balance e- between half-reactions
6. add half-reactions
7. check
Ex 18.3 – Balance the equation:
I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Assign
Oxidation
States
Separate into
half-reactions
I(aq) + MnO4(aq)  I2(aq) + MnO2(s)
ox: I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s)
Ex 18.3 – Balance the equation:
I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Balance
Balancehalf-ox: ox:
2 I(aq)
I(aq)
2
I

I2(aq)
I2(aq)
I2(aq)
(aq)

 MnO
reactions
half- by red: red:
4 H+MnO
+
MnO

MnO
+ 22(s)H2+O2(l)H2O(l)
(aq) 4 (aq) 4 (aq)  2(s)
mass
reactions
+



4
H
+
4
OH
+
MnO

MnO
+
2
H
O
+
4
OH
by
mass
(aq)
(aq)
4
(aq)
2(s)
2
(l)
(aq)
then O by
adding
then
in
base,
HHby
2O 4 H2O(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq)
adding H+
neutralize
MnO4(aq) + 2 H2O(l)  MnO2(s) + 4 OH(aq)
+
the H
with OH-
Ex 18.3 – Balance the equation:
I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Balance
Halfreactions
by
charge
Balance
electrons
between
halfreactions
ox: 2 I(aq)  I2(aq) + 2 e
red: MnO4(aq) + 2 H2O(l) + 3 e  MnO2(s) + 4 OH(aq)
ox: 2 I(aq)  I2(aq) + 2 e } x3
red: MnO4(aq) + 2 H2O(l) + 3 e  MnO2(s) + 4 OH(aq) }x2
ox: 6 I(aq)  3 I2(aq) + 6 e
red: 2 MnO4(aq) + 4 H2O(l) + 6 e  2 MnO2(s) + 8 OH(aq)
Ex 18.3 – Balance the equation:
I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Add the ox: 6 I(aq)  3 I2(aq) + 6 e
Halfred: 2 MnO4(aq) + 4 H2O(l) + 6 e  2 MnO2(s) + 8 OH(aq)
reactions tot: 6 I(aq)+ 2 MnO4(aq) + 4 H2O(l)  3 I2(aq)+ 2 MnO2(s) + 8 OH(aq)
Check
Reactant
Count
Element
Product
Count
6
I
6
2
Mn
2
12
O
12
8
H
8
8
charge
8
Practice - Balance the Equation
H2O2 + KI + H2SO4 ® K2SO4 + I2 + H2O
Practice - Balance the Equation
H2O2 + KI + H2SO4 ® K2SO4 + I2 + H2O
+1 -1
+1 -1 +1 +6 -2
+1 +6 -2
0
+1 -2
oxidation
reduction
ox:
red:
2 I- I2 + 2eH2O2 + 2e- + 2 H+  2 H2O
tot
2 I- + H2O2 + 2 H+  I2 + 2 H2O
H2O2 + 2 KI + H2SO4  K2SO4 + I2 + 2 H2O
Practice - Balance the Equation
ClO3- + Cl-  Cl2 (in acid)
Practice - Balance the Equation
ClO3- + Cl+5 -2
-1
→
Cl2 (in acid)
0
oxidation
reduction
ox:
red:
2 Cl- → Cl2 + 2 e- } x5
2 ClO3- + 10 e- + 12 H+ → Cl2 + 6 H2O} x1
tot
10 Cl- + 2 ClO3- + 12 H+ → 6 Cl2 + 6 H2O
ClO3- + 5 Cl- + 6 H+ → 3 Cl2 + 3 H2O
Electrical Current
• current of a liquid in a stream, = amount
of water that passes by in a given period
of time
• electric current = amount of electric
charge that passes a point in a given
period of time
 whether as e- flowing through a wire
or ions flowing through a solution
Redox Reactions & Current
• redox reactions involve the transfer of e- from one substance to
another
• therefore, redox reactions have the potential to generate an electric
current
• in order to use that current, we need to separate the place where
oxidation is occurring from the place that reduction is occurring
Electric Current Flowing
Directly Between Atoms
Electric Current Flowing
Indirectly Between Atoms
Electrochemical Cells
• electrochemistry is the study of redox reactions that
produce or require an electric current
• the conversion between chemical energy and electrical
energy is carried out in an electrochemical cell
• spontaneous redox reactions take place in a voltaic cell
 aka galvanic cells
• nonspontaneous redox reactions can be made to occur in
an electrolytic cell by the addition of electrical energy
Electrochemical Cells
• redox reactions kept separate
•
•
•
 half-cells
e- flow in a wire along and ion flow in solution constitutes
an electric circuit
requires a conductive metal or graphite electrode to
allow the transfer of e through external circuit
ion exchange between the two halves of the system
 electrolyte
Electrodes
• Anode
 electrode where oxidation occurs
 anions attracted to it
 connected to positive end of battery in electrolytic cell
 loses weight in electrolytic cell
• Cathode
 electrode where reduction occurs
 cations attracted to it
 connected to negative end of battery in electrolytic cell
 gains weight in electrolytic cell
electrode where plating takes place in electroplating
Voltaic Cell
the salt bridge is
required to complete
the circuit and
maintain charge
balance
Current and Voltage
• # e- that flow through the system per second is
the current
 unit = Ampere
 1 A of current = 1 Coulomb of charge per second
 1 A = 6.242 x 1018 e-/sec.
 Electrode surface area dictates the number of e- that
can flow
Current and Voltage
• the difference in potential energy between the reactants
and products is the potential difference
 unit = Volt
 1 V of force = 1 J of energy/Coulomb of charge
 the voltage needed to drive electrons through the
external circuit
 amount of force pushing the electrons through the wire
is called the electromotive force, emf
Cell Potential
• the difference in potential energy between the anode the
cathode in a voltaic cell is called the cell potential
• cell potential depends on the relative ease with which the
oxidizing agent is reduced at the cathode and the
reducing agent is oxidized at the anode
• the cell potential under standard conditions is called the
standard emf, E°cell
 25°C, 1 atm for gases, 1 M concentration of solution
 Ecell = EOX + ERED
Cell Notation
• shorthand description of Voltaic cell
electrode | electrolyte || electrolyte | electrode
• oxidation half-cell on left, reduction half-cell on the right
• single | = phase barrier, double line || = salt bridge
 if multiple electrolytes in same phase, a comma is
used rather than |
 often use an inert electrode
Fe(s) | Fe2+(aq) || MnO4(aq), Mn2+(aq), H+(aq) | Pt(s)
Standard Reduction Potential
• a half-reaction with a strong tendency to
•
•
•
occur has a large + half-cell potential
two half-cells are connected, e- will flow so
that the half-reaction with the stronger
tendency will occur
we cannot measure the absolute tendency of
a half-reaction, we can only measure it
relative to another half-reaction
select as a standard half-reaction the
reduction of H+ to H2 under standard
conditions, which we assign a potential
difference = 0 v
 standard hydrogen electrode, SHE
Half-Cell Potentials
• SHE reduction potential is defined to be exactly 0 v
• half-reactions with a stronger tendency toward reduction than the SHE
have a + value for E°red
• half-reactions with a stronger tendency toward oxidation than the SHE
have a  value for E°red
• E°cell = E°oxidation + E°reduction
 E°oxidation = E°reduction
 when adding E° values for the half-cells, do not multiply the halfcell E° values, even if you need to multiply the half-reactions to
balance the equation
 When E°cell > 0 reaction may be spontaneous
Ex 18.4 – Calculate Ecell for the reaction at 25C
Al(s) + NO3−(aq) + 4 H+(aq)  Al3+(aq) + NO(g) + 2 H2O(l)
Separate the reaction
into the oxidation and
reduction halfreactions
ox:
Al(s)  Al3+(aq) + 3 e−
red:
NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l)
find the E for each
half-reaction and sum
to get Ecell
Eox = −Ered = +1.66 v
Ered = +0.96 v
Ecell = Eox + Ered
Ecell = (+1.66 v) + (+0.96 v) = +2.62 v
Ex 18.4a – Predict if the following reaction is spontaneous
under standard conditions
Fe(s) + Mg2+(aq)  Fe2+(aq) + Mg(s)
Separate the reaction
into the oxidation and
reduction halfreactions
ox:
Fe(s)  Fe2+(aq) + 2 e−
red:
Mg2+(aq) + 2 e−  Mg(s)
look up the E halfreactions
Ecell = Eox + Ered = +0.45 + -2.37 = -1.92
since Ecell = -ve the reaction is NOT spontaneous as
written
[Mg2+ reduction is below Fe2+ reduction, the reaction is NOT spontaneous as written]
the reaction is spontaneous
in the reverse direction
Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s)
ox:
red:
sketch the cell and label the
parts – oxidation occurs at
the anode; electrons flow
from anode to cathode
Mg(s)  Mg2+(aq) + 2 e−
Fe2+(aq) + 2 e−  Fe(s)
Practice - Sketch and Label the Voltaic Cell
Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s) , Write the HalfReactions and Overall Reaction, and Determine the
Cell Potential under Standard Conditions.
Fe(s)  Fe2+(aq) + 2 e−
E = +0.45 V
red: Pb2+(aq) + 2 e−  Pb(s)
E = −0.13 V
ox:
tot: Pb2+(aq) + Fe(s)  Fe2+(aq) + Pb(s)
Spontaneous
E = +0.32 V
Predicting Whether a Metal Will Dissolve in
an Acid
•
•
acids dissolve in metals if the reduction of the metal ion is
easier than the reduction of H+(aq)
metals whose ion reduction reaction lies below H+
reduction on the table will dissolve in acid
All
have
+ve Eox
Ecell = Eox + 0 = +ve
E°cell, ΔG° and K
• for a spontaneous reaction
 one that proceeds in the forward direction with the
chemicals in their standard states
 ΔG° < 1 (negative)
 E° > 1 (positive)
K > 1
• ΔG° = −RTlnK = −nFE°cell
 n is the number of electrons
 F = Faraday’s Constant = 96,485 C/mol e−
Example 18.6- Calculate ΔG° for the reaction
I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
Given: I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
Find: DG°, (J)
Concept Plan:
E°ox, E°red
E°cell
Ecell  Eox  Ered
DG°
DG   nFE cell
Relationships:
2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 v
Solve: ox:
DG  nFE

cell

 2 I−(aq)
red: I2(l) + 2 e− →
DG  2 mol e 96,485
 0.55 
C = +0.54 v
E°
mol e 
J
C
tot:  I2(l) + 2Br−(aq)5 → 2I−(aq) + Br2(l) E° = −0.55 v
DG  1.110 J
Answer: since DG° is +, the reaction is not spontaneous in
the forward direction under standard conditions
E°cell, ΔG° and K
• ΔG° = −RTlnK = −nFE°cell
E°cell =
RT x ln K
nF
R = 8.314 J/mol.K
lnK = 2.303log K
F = 96,485 C/mol e-
E°cell = 0.0592 logK
n
Example 18.7- Calculate K at 25°C for the reaction
Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq)
Given: Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq)
Find: K
Concept Plan:
E°ox, E°red
E°cell
E

cell
 E E

ox

red
E

cell
K
0.0592 V

log K
n
Relationships:
0.0592
V 2+ + 2 e−

Cu
→
Cu
E° = −0.34 v
Solve: Eox:
log(aq)K
cell  (s)
n

+
red: 2 H (aq) + 2 e−2 →
E° = +0.00 v
molHe2(aq)
log K   0.34 V 
 11.5
0.05922+
V
+
tot: Cu11
Cu (aq) + H2(g) E° = −0.34 v
(s).5+ 2H (aq) →
12
K  10
 3.2 10
Answer: since K < 1, the position of equilibrium lies far to
the left under standard conditions


Nonstandard Conditions the Nernst Equation
•
Relationship between Ecell (nonstandard) and E°cell (standard)
ΔG = ΔG° + RT ln Q
•
Subs. ΔG° = -nFE°cell into above eqn.
-nFEcell = -nFE°cell + -RTlnQ
(divide by -nF)
Ecell = E°cell - (RT/nF) log Q
R = 8.314 J/mol.K, (RT/nF)lnQ = (0.0592/n)logQ
Ecell = E° - (0.0592/n) logQ
Called the Nernst equation
Nonstandard Conditions the Nernst Equation
Ecell = E°cell - (0.0592/n) log Q at 25°C
1. when Q = 1 (std. conditions) Ecell = E°cell
2. At equilibrium,Q = K,
Ecell = E°cell - (0.0592/n) log K
and (0.0592/n) log K = E°cell
Ecell = 0
•
•
Potential reaches zero as concentrations approach equilibrium
Used to calculate E when concentrations not 1 M
E at Nonstandard Conditions
Reactant conc. > standard conditions
Product conc. < standard conditions … reaction shifts right
Example 18.8- Calculate Ecell at 25°C for the reaction
3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + 3 Cu2+(aq) + 4 H2O(l)
3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)
Given:
[Cu2+] = 0.010 M, [MnO4−] = 2.0 M, [H+] = 1.0 M
Ecell
Find:
Concept Plan:
E°ox, E°red
Ecell  Eox  Ered
Relationships:
Solve:
ox:
Cu(s)V→ Cu2+E(aq)
cell + 2
0.0592
E cell  E cell
− 
red: MnO4
E°cell
E cell  E cell 
Ecell
0.0592 V
log Q
n
2 3
0
.
0592
V
[
Cu
]
e− }x3E° =log−0.34 v 3  8
n
[MnO 4 ] [H ]
E cell
log Q
+
−
2 H2V
O(l) }x2
(aq) + 4 H
n (aq) + 3 e → MnO2(s) +0.0592
[0.010E°
]3 = +1.68 v
E
 1.34 V 
log
tot: 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) →cell2 MnO2(s) + Cu62+(aq) + [42.0]
H23O
[1(l))
.0]8E° = +1.34 v
E cell  1.41 V
Check:
units are correct, Ecell > E°cell as expected because
[MnO4−] > 1 M and [Cu2+] < 1 M
Concentration Cells
• it is possible to get a spontaneous reaction when the oxidation and
reduction reactions are the same, as long as the electrolyte
concentrations are different
• the difference in energy is due to the entropic difference in the
solutions
 the more concentrated solution has lower entropy than the less
concentrated
• electrons will flow from the electrode in the less concentrated solution
to the electrode in the more concentrated solution
 oxidation of the electrode in the less concentrated solution will
increase the ion concentration in the solution – the less
concentrated solution has the anode
 reduction of the solution ions at the electrode in the more
concentrated solution reduces the ion concentration – the more
concentrated solution has the cathode
Concentration Cell
when the cell concentrations
are equal there is no difference
in energy between the half-cells
and no electrons flow
Cu2+(aq) + 2e- → Cu(s) 0.34V
Cu(s) → Cu2+(aq) + 2e- -0.34V
Cu2+(aq) + Cu(s) → Cu(s) + Cu2+(aq)
E°cell = E°red + E°ox = 0V
Cu(s) Cu2+(aq) (1 M)  Cu2+(aq) (1 M) Cu(s)
Concentration Cell
when the cell concentrations are
different, e- flow from the side with
the less concentrated solution
(anode) to the side with the more
concentrated solution (cathode)
Cell potential Ecell calculated using
Nernst eqn.
Ecell = E° - (0.0592/n) logQ
Ecell = E° - (0.0592/n) log([OX]/[RED])
= 0.068V
Cu(s) Cu2+(aq) (0.010 M)  Cu2+(aq) (2.0 M) Cu(s)
LeClanche’ Acidic Dry Cell
• electrolyte in paste form
 ZnCl2 + NH4Cl
 or MgBr2
• anode = Zn (or Mg)
Zn(s) → Zn2+(aq) + 2 e-
• cathode = graphite rod
• MnO2 is reduced
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e→ 2 NH4OH(aq) + 2 Mn(O)OH(s)
• cell voltage = 1.5 v
• expensive, nonrechargeable, heavy, easily corroded
Alkaline Dry Cell
•
same basic cell as acidic dry cell, except electrolyte is alkaline
KOH paste
•
anode = Zn (or Mg)
Zn(s) → Zn2+(aq) + 2 e-
•
cathode = brass rod
•
MnO2 is reduced:
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e→ 2 NH4OH(aq) + 2 Mn(O)OH(s)
•
cell voltage = 1.54 v
•
longer shelf life than acidic dry cells and rechargeable, little
corrosion of zinc
Lead Storage Battery
• 6 cells in series
• electrolyte = 30% H2SO4
• anode = Pb
Pb(s) + SO42-(aq) → PbSO4(s) + 2 e-
• cathode = Pb coated with PbO2
• PbO2 is reduced:
PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e→ PbSO4(s) + 2 H2O(l)
• cell voltage = 2.09 v
• rechargeable, heavy
NiCad Battery
• electrolyte is concentrated KOH solution
• anode = Cd
Cd(s) + 2 OH-(aq) → Cd(OH)2(s) + 2 e- E0 = 0.81 v
• cathode = Ni coated with NiO2
• NiO2 is reduced:
NiO2(s) + 2 H2O(l) + 2 e- → Ni(OH)2(s) + 2OH-
E0 = 0.49 v
• cell voltage = 1.30 v
• rechargeable, long life, light – however recharging incorrectly can
lead to battery breakdown
Ni-MH Battery
• electrolyte is concentrated KOH solution
• anode = metal alloy with dissolved hydrogen
 oxidation of H from H0 to H+
M∙H(s) + OH-(aq) → M(s) + H2O(l) + e-
E° = 0.89 v
• cathode = Ni coated with NiO2
• NiO2 is reduced:
NiO2(s) + 2 H2O(l) + 2 e- → Ni(OH)2(s) + 2OH-
E0 = 0.49 v
• cell voltage = 1.30 v
• rechargeable, long life, light, more environmentally friendly than
NiCad, greater energy density than NiCad
Lithium Ion Battery
• electrolyte is concentrated KOH solution
• anode = graphite impregnated with Li
ions
• cathode = Li - transition metal oxide
 reduction of transition metal
• work on Li ion migration from anode to
cathode causing a corresponding
migration of electrons from anode to
cathode
• rechargeable, long life, very light, more
environmentally friendly, greater energy
density
Fuel Cells
• like batteries in which reactants
are constantly being added
 so it never runs down!
• Anode and Cathode both Pt
coated metal
• Electrolyte is OH– solution
• Anode Reaction:
2 H2 + 4 OH–
→ 4 H2O(l) + 4 e-
• Cathode Reaction:
O2 + 4 H2O + 4 e→ 4 OH–
Electrolytic Cell
• uses electrical energy to overcome the energy barrier and
•
•
•
•
•
cause a non-spontaneous reaction
 must be DC source
the + terminal of the battery = anode
the - terminal of the battery = cathode
cations attracted to the cathode, anions to the anode
cations pick up electrons from the cathode and are
reduced, anions release electrons to the anode and are
oxidized
some electrolysis reactions require more voltage than Etot,
called the overvoltage
electroplating
In electroplating, the work piece
is the cathode.
Cations are reduced at
cathode and plate to the
surface of the work piece.
The anode is made of the
plate metal. The anode
oxidizes and replaces the
metal cations in the solution
Electrochemical Cells
• in all electrochemical cells, oxidation occurs at the anode, reduction
occurs at the cathode
• in voltaic cells,
 anode is the source of electrons and has a (−) charge
 cathode draws electrons and has a (+) charge
• in electrolytic cells
 electrons are drawn off the anode, so it must have a place to
release the electrons, the + terminal of the battery
 electrons are forced toward the anode, so it must have a source of
electrons, the − terminal of the battery
Electrolysis
• electrolysis is the process of using
electricity to break a compound apart
• electrolysis is done in an electrolytic cell
• electrolytic cells can be used to separate
elements from their compounds
 generate H2 from water for fuel cells
 recover metals from their ores
Electrolysis of Water
Electrolysis of Pure Compounds
•
•
•
•
must be in molten (liquid) state
electrodes normally graphite
cations are reduced at the cathode to metal element
anions oxidized at anode to nonmetal element
Electrolysis of NaCl(l)
Mixtures of Ions
• when more than one cation is present, the cation that is easiest to
reduce will be reduced first at the cathode
 least negative or most positive E°red
• when more than one anion is present, the anion that is easiest to
oxidize will be oxidized first at the anode
 least negative or most positive E°ox
Electrolysis of Aqueous Solutions
• Complicated by more than one possible oxidation and reduction
• possible cathode reactions
 reduction of cation to metal
 reduction of water to H2
 2 H2O + 2 e- → H2 + 2 OH-
• possible anode reactions
 oxidation of anion to element
 oxidation of H2O to O2
 2 H2O → O2 + 4e- + 4H+
 oxidation of electrode
 particularly Cu
 graphite doesn’t oxidize
E° = -0.83 v @ stand. cond.
E° = -0.41 v @ pH 7
E° = -1.23 v @ stand. cond.
E° = -0.82 v @ pH 7
• half-reactions that lead to least negative Etot will occur
 unless overvoltage changes the conditions
Electrolysis of NaI(aq)
with Inert Electrodes
possible oxidations
2 I- → I2 + 2 e2 H2O → O2 + 4e- + 4H+
E° = −0.54 v
E° = −0.82 v
possible reductions
Na+ + e- → Na0
E° = −2.71 v
2 H2O + 2 e- → H2 + 2 OH- E° = −0.41 v
overall reaction
2 I−(aq) + 2 H2O(l) → I2(aq) + H2(g) + 2 OH-(aq)
Faraday’s Law
• the amount of metal deposited during electrolysis is
directly proportional to the charge on the cation, the
current, and the length of time the cell runs
 charge that flows through the cell = current x time
Example 18.10- Calculate the mass of Au that can be plated in
25 min using 5.5 A for the half-reaction
Au3+(aq) + 3 e− → Au(s)
Given:
Find:
Concept Plan:
Relationships:
3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min
mass Au, g
t(s), amp
charge (C)
5.5 C
1s
mol e−
1 mol e 
96,485 C
mol Au
1 mol Au
3 mol e 
g Au
196.97 g
1 mol Au
Solve:
60 s 5.5 C 1 mol e  1 mol Au 196.97 g
25 min 





1 min 1 s
96,485 C 3 mol e 1 mol Au
 5.6 g Au
Check:
units are correct, answer is reasonable since 10 A
running for 1 hr ~ 1/3 mol e−
Corrosion
• corrosion is the spontaneous oxidation of a metal by
chemicals in the environment
• since many materials we use are active metals,
corrosion can be a very big problem
Rusting
• rust is hydrated iron(III) oxide
• moisture must be present
 water is a reactant
 required for flow between cathode and anode
• electrolytes promote rusting
 enhances current flow
• acids promote rusting
 lower pH = lower E°red
Preventing Corrosion
• one way to reduce or slow corrosion is to coat the metal surface to
keep it from contacting corrosive chemicals in the environment
 paint
 some metals, like Al, form an oxide that strongly attaches to the
metal surface, preventing the rest from corroding
• another method to protect one metal is to attach it to a more reactive
metal that is cheap
 sacrificial electrode
 galvanized nails
Sacrificial Anode