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CHAPTER 18
Electrochemistry
OXIDATION–REDUCTION

Reactions where electrons are transferred from one atom
to another are called oxidation–reduction reactions


redox reactions for short
Atoms that lose electrons are being oxidized, atoms that
gain electrons are being reduced
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na → Na+ + 1 e–
oxidation
Cl2 + 2 e– → 2 Cl–
reduction
Oil
Rig
OXIDATION & REDUCTION

Oxidation is the process that occurs when
the oxidation number of an element increases
an element loses electrons
a compound adds oxygen
 a compound loses hydrogen
 a half-reaction has electrons as products




Reduction is the process that occurs when
the oxidation number of an element decreases
an element gains electrons
 a compound loses oxygen
 a compound gains hydrogen
 a half-reaction has electrons 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
4.
(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
EXAMPLE: DETERMINE THE OXIDATION STATES OF
ALL THE ATOMS IN A PROPANOATE ION, C3H5O2–
 There
are no free elements or free ions in
propanoate, so the first rule that applies is Rule 3b
(C3) + (H5) + (O2) = −1
 Because
all the atoms are nonmetals, the next rule
we use is Rule 5, following the elements in order
H = +1
 O = −2

Note: unlike
(C3) + 5(+1) + 2(−2) = −1 charges, oxidation
(C3) = −2
states can be
C = −⅔
fractions!
PRACTICE – ASSIGN AN OXIDATION STATE
TO EACH ELEMENT IN THE FOLLOWING
• Br2
Br = 0 (Rule 1)
• K+
K = +1 (Rule 2)
• LiF
Li = +1 (Rule 4a) & F = −1 (Rule 5)
• CO2
O = −2 (Rule 5) & C = +4 (Rule 3a)
• SO42−
O = −2 (Rule 5) & S = +6 (Rule 3b)
• Na2O2 Na = +1 (Rule 4a) & O = −1 (Rule 3a)
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
+1 −2
oxidation
reduction
Copyright © 2011 Pearson Education, Inc.
Oxidation–Reduction
• Oxidation and reduction must occur simultaneously
 if an atom loses electrons another atom must take them
• The reactant that reduces an element in another
reactant is called the reducing agent
 the reducing agent contains the element that is oxidized
• The reactant that oxidizes an element in another
reactant is called the 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
Copyright © 2011 Pearson Education, Inc.
Common Oxidizing Agents
Copyright © 2011 Pearson Education, Inc.
Common Reducing Agents
Copyright © 2011 Pearson Education, Inc.
Example: Assign oxidation states, determine
the element oxidized and reduced, and
determine the oxidizing agent and reducing
agent in the following reaction
Reducing
Agent
Oxidizing
Agent
Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O
0
+7
−2
+1
+3
+4 −2
+1 −2
Reduction
Oxidation
Copyright © 2011 Pearson Education, Inc.
Practice – Assign oxidation states, determine
the element oxidized and reduced, and
determine the oxidizing agent and reducing
agent in the following reactions
Sn4+ + Ca → Sn2+ + Ca2+
+4
0
+2
+2
Ca is oxidized, Sn4+ is reduced
Ca is the reducing agent, Sn4+ is the oxidizing agent
F2 + S → SF4
0
0
+4−1
S is oxidized, F is reduced
S is the reducing agent, F2 is the oxidizing agent
Copyright © 2011 Pearson Education, Inc.
18.2 Balancing Redox Reactions
• Some redox reactions can be balanced by
the method we previously used, but many are
hard to balance using that method
many are written as net ionic equations
many have elements in multiple compounds
• The main principle is that electrons are
transferred – so if we can find a method to
keep track of the electrons it will allow us to
balance the equation
Copyright © 2011 Pearson Education, Inc.
Half-Reactions
• We generally split the redox reaction into two
separate half-reactions – a reaction just
involving oxidation or reduction
the oxidation half-reaction has electrons as products
the reduction half-reaction has electrons as reactants
3 Cl2 + I− + 3H2O → 6 Cl− + IO3− + 6 H+
0
−1
+1 −2
−1
+5 −2
+1
oxidation: I− → IO3− + 6 e−
reduction: Cl2 + 2 e− → 2 Cl−
Tro: Chemistry: A Molecular Approach, 2/e
16
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Balancing Redox Reactions by the
Half-Reaction Method
• In this method, the reaction is broken down into
•
two half-reactions, one for oxidation and another
for reduction
Each half-reaction includes electrons
 electrons go on the product side of the oxidation halfreaction – loss of electrons
 electrons go on the reactant side of the reduction halfreaction – gain of electrons
• Each half-reaction is balanced for its atoms
• Then the two half-reactions are adjusted so that
the electrons lost and gained will be equal when
combined
Copyright © 2011 Pearson Education, Inc.
Balancing Redox Reactions
1. assign oxidation states
a) determine element oxidized and element reduced
2. write ox. & red. half-reactions, including electrons
a) ox. electrons on right, red. electrons on left of arrow
3. balance half-reactions by mass
a)
b)
c)
d)
first balance elements other than H and O
add H2O where need O
add H+ where need H
if reaction done in Base, neutralize H+ with OH−
4. balance half-reactions by charge
a) balance charge by adjusting electrons
5. balance electrons between half-reactions
6. add half-reactions
7. check by counting atoms and total charge
Copyright © 2011 Pearson Education, Inc.
Example 18.2: Balancing redox reactions
in acidic solution
1. assign oxidation
states and
determine element
oxidized and
element reduced
Fe2+ + MnO4– → Fe3+ + Mn2+
2. separate into
oxidation &
reduction halfreactions
ox: Fe2+ → Fe3+
+2
+7 −2
+3
+2
oxidation
reduction
red: MnO4– → Mn2+
Copyright © 2011 Pearson Education, Inc.
Example 18.2: Balancing redox reactions
in acidic solution
3. balance halfreactions by mass
a) first balance atoms
other than O and H
b) balance O by
adding H2O to side
that lacks O
c) balance H by
adding H+ to side
that lacks H
Fe2+ → Fe3+
MnO4– → Mn2+
MnO4– → Mn2+ + 4H2O
MnO4– + 8H+ → Mn2+ + 4H2O
Copyright © 2011 Pearson Education, Inc.
Example 18.2: Balancing redox reactions
in acidic solution
4.
balance each
half-reaction
with respect to
charge by
adjusting the
numbers of
electrons
a)
b)
electrons on
product side for
oxidation
electrons on
reactant side for
reduction
Fe2+ → Fe3+ + 1 e−
MnO4– + 8H+ → Mn2+ + 4H2O
+7
+2
MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O
Copyright © 2011 Pearson Education, Inc.
Example 18.2: Balancing redox reactions
in acidic solution
5. balance
electrons
Fe2+ → Fe3+ + 1 e− } x 5
between halfMnO4– + 8H+ + 5 e− → Mn2+ + 4H2O
reactions
6. add half2+ → 5 Fe3+ + 5 e−
5
Fe
reactions,
canceling
MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O
electrons and
common
5 Fe2+ + MnO4– + 8H+ → Mn2+ + 4H2O + 5 Fe3+
species
reactant side
Element
product side
7. Check that
5
Fe
5
numbers of
1
Mn
1
4
O
4
atoms and
8
H
8
total charge
+17
charge
+17
are equal
Copyright © 2011 Pearson Education, Inc.
Practice – Balance the following equation
in acidic solution
I– + Cr2O72− → Cr3+ + I2
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Practice – Balancing redox reactions
1. assign oxidation
states and
determine
element oxidized
and element
reduced
2. separate into
oxidation &
reduction halfreactions
I− + Cr2O72– → I2 +
−1
+6 −2
0
Cr3+
+3
oxidation
reduction
ox:
I− → I2
red: Cr2O72– → Cr3+
Copyright © 2011 Pearson Education, Inc.
Practice – Balancing redox reactions
3. balance halfreactions by
mass
a) first balance
atoms other than
O and H
b) balance O by
adding H2O to
side that lacks O
c) balance H by
adding H+ to side
that lacks H
ox:
I − → I2
ox:
2 I − → I2
red: Cr2O72– → Cr3+
red: Cr2O72– → 2 Cr3+
red: Cr2O72– → 2Cr3+ +7H2O
Cr2O72– +14H+ → 2Cr3+ +7H2O
Copyright © 2011 Pearson Education, Inc.
Practice – Balancing redox reactions
4. balance each
half-reaction
with respect to
charge by
adjusting the
numbers of
electrons
a)
b)
electrons on
product side
for oxidation
electrons on
reactant side
for reduction
2 I− → I2 + 2e−
Cr2O72– +14H+ → 2Cr3+ +7H2O
Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O
Copyright © 2011 Pearson Education, Inc.
Practice – Balancing redox reactions
5.
6.
7.
balance
2 I− → I2 + 2e−
electrons
between
6 I− → 3 I2 + 6e−} x 3
halfreactions
2– +14H+ + 6e−→ 2Cr3+ +7H O
Cr
O
2 7
2
add halfreactions,
canceling Cr2O72– +14H+ + 6 I−→ 2Cr3+ +7H2O + 3 I2
electrons
reactant side
Element
product side
and
2
Cr
2
common
6
I
6
species
7
O
7
check
14
H
14
+6
charge
+6
Copyright © 2011 Pearson Education, Inc.
EXAMPLE 18.3 Balancing Redox Reactions Occurring in Basic
Solution
Balance the equation occurring in basic solution:
SOLUTION
To balance redox reactions occurring in basic solution, follow the half-reaction
method outlined above, but add an extra step to neutralize the acid with OH– as
shown in step 3 below.
1. Assign oxidation states.
2. Separate the overall reaction into two halfreactions.
Oxidation:
Reduction:
3. Balance each half-reaction with respect
to mass.
• Balance all elements other than H
and O.
• Balance O by adding H2O.
• Balance H by adding H+.
• Neutralize H+ by adding enough OH–
to neutralize each H+. Add the same
number of OH– ions to each side of
the equation.
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continued…
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4. Balance each half-reaction with respect to
charge.
5. Make the number of electrons in both halfreactions equal.
6. Add the half-reactions together.
7. Verify that the reaction is balanced.
FOR PRACTICE 18.3
Balance the following redox reaction occurring in basic solution.
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18.3 VOLTAIC (OR GALVANIC) CELLS: GENERATING
ELECTRICITY FROM SPONTANEOUS CHEMICAL
REACTIONS
Electrical Current


When we talk about the current of a
liquid in a stream, we are discussing
the amount of water that passes by in a
given period of time.
When we discuss electric current, we
are discussing the amount of electric
charge that passes a point in a given
period of time.
 whether as electrons flowing
through a wire, or ions flowing
through a solution.
REDOX REACTIONS GENERATING
CURRENT



Redox reactions involve the
transfer of electrons from one
substance to another.
Therefore, redox reactions
have the potential to
generate an electric current.
To use that current, we need
to separate the place where
oxidation is occurring from
the place where reduction is
occurring.
A VOLTAIC (GALVANIC) CELL
ELECTROCHEMISTRY




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 .
COMPONENTS OF AN ELECTROCHEMICAL
CELL
• Oxidation and reduction half-reactions are kept separate in halfcells.
• Electron flow through a wire along with ion flow through a
solution constitutes an electric circuit.
• Requires a conductive solid electrode to allow the transfer of
electrons.
 through external circuit
 metal or graphite
• requires ion exchange between the two half-cells of the system

electrolyte (salt bridge for galvanic cell)
THE ELECTRODES

Anode
Oxidation occurs at anode
 In a voltaic cell, anode is labelled (-) with a negative
sign because the anode releases the electrons.


Cathode
Reduction occurs at cathode
 In a voltaic cell, cathode is labelled (+) with a positive
sign because cathode (attracts) receives the electrons.


Electrons are (-) and will like to travel to anything
with a (+) (cathode in this case). Electrons are
released from the oxidation process at the anode.
A VOLTAIC (GALVANIC) CELL
The salt bridge is
required to complete
the circuit and
maintain charge
balance.
CURRENT



Current is the number of electrons that flow through the
system per second
 unit = Ampere
1 A of current = 1 Coulomb of charge flowing by each
second
 1 A = 6.242 x 1018 electrons per second
Electrode surface area dictates the number of electrons
that can flow
 larger batteries produce larger currents
VOLTAGE



The difference in potential energy between the reactants
and products is the potential difference
 unit = Volt
1 V = 1 J of energy per 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 and
the cathode in a voltaic cell is called the cell potential.
The cell potential depends on the relative ease with which
the oxidation and reduction happens.
The cell potential under standard conditions is called the
standard emf, E°cell
 25 °C, 1 atm for gases, 1 M concentration of solution
 sum of the cell potentials for the half-reactions
ELECTROCHEMICAL CELL NOTATION

Shorthand description of a voltaic cell

Electrode | electrolyte || electrolyte | electrode

Oxidation half-cell on the left, reduction half-cell on the right


Single | = phase barrier
 if multiple electrolytes in same phase, a comma is used rather
than |
 often use an inert electrode
Double line || = salt bridge
VOLTAIC CELL
Anode = Zn(s)
the anode is
oxidized to Zn2+
Cathode = Cu(s)
Cu2+ ions are
reduced at the
cathode
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
ELECTRODES


Many times the anode is made of the metal that is oxidized and
the cathode is made of the same metal as is produced by the
reduction.
However, if the redox reaction we are running involves the
oxidation or reduction of an ion to a different oxidation state, or
the oxidation or reduction of a gas, we may use an inert electrode.
 an inert electrode is one that not does participate in the
reaction, but just provides a surface for the transfer of
electrons to take place on
Because the half-reaction
involves reducing the Mn
oxidation state from +7 to
+2, we use an electrode
that will provide a surface
for the electron transfer
without reacting with the
MnO4−.
Platinum works well
because it is extremely
nonreactive and conducts
electricity.
Fe(s) | Fe2+(aq) || MnO4-(aq), Mn2+(aq), H+(aq) | Pt(s)
18.4 STANDARD REDUCTION POTENTIALS


A half-reaction with a strong
tendency to occur has a large
positive half-cell potential.
When two half-cells are
connected, the electrons 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.
We select as a standard halfreaction the reduction of
H+ to H2 under standard
conditions, which we assign a
potential difference = 0 V
standard hydrogen
electrode, SHE
Zn(s) | Zn2+(1 M) || H+(1 M) | H2(g) | Pt(s)
HALF-CELL POTENTIALS





SHE reduction potential is defined to be exactly 0 V.
Standard Reduction Potentials compare the tendency for a
particular reduction half-reaction to occur relative to the
reduction of H+ to H2
 under standard conditions
Half-reactions with a stronger tendency toward reduction than
the SHE have a + value for E°red
Half-reactions with a weaker tendency toward reduction than the
SHE have a  value for E°red
For an oxidation half-reaction, E°oxidation = − E°reduction
CALCULATING CELL POTENTIALS UNDER
STANDARD CONDITIONS



E°cell = E°oxidation + E°reduction
When adding E° values for the half-cells, do not multiply
the half-cell E° values, even if you need to multiply the halfreactions to balance the equation
This is due to the fact that Volt = Joules/Coulomb (potential
energy per unit charge) (It is an intensive property like
density)
Example 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 ox (anode): Al(s)  Al3+(aq) + 3 e−
reaction into
the oxidation red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l)
and reduction
half-reactions
find the E for Eox of Al = −Ered of Al3+ = +1.66 V
each halfEred of NO3− = +0.96 V
reaction and E = (+1.66 V) + (+0.96 V) = +2.62 V
cell
sum to get
Ecell
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Practice – Calculate Ecell for the reaction at 25 C IO3–
(aq) + 6 H+(aq) + 5 I−(aq) → 3 I2(s) + 3 H2O(l)
Reduction Half-Reaction
Ered, V
F2(g) + 2e− 2 F−(aq)
+2.87
IO3−(aq) + 6 H++ 5e− ½I2(s) + 3H2O(l)
+1.20
Ag+(aq) + 1e− Ag(s)
+0.80
I2(s) + 2e− 2 I−(aq)
+0.54
Cu2+(aq) + 2e− Cu(s)
+0.34
Cr3+(aq) + 1e− Cr2+(aq)
−0.50
Mg2+(aq) + 2e− Mg(s)
−2.37
Copyright © 2011 Pearson Education, Inc.
Practice – Calculate Ecell for the reaction at 25 C
2 IO3– + 12 H+ + 10 I−→ 6 I2 + 6 H2O
separate the ox (anode): 2 I−(s)  I2(aq) + 2 e−
reaction into
the oxidation red: IO3−(aq) + 6 H+(aq) + 5 e−  ½ I2(s) + 3 H2O(l)
and reduction
half-reactions
find the E for Eox of I− = −Ered of I2 = −0.54 V
each halfEred of IO3− = +1.20 V
reaction and E = (−0.54 V) + (+1.20 V) = +0.66 V
cell
sum to get
Ecell
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PREDICTING THE SPONTANEOUS DIRECTION OF
AN OXIDATION-REDUCTION REACTION


A spontaneous reaction will take place when a
reduction half-reaction is paired with an oxidation
half-reaction lower on the table.
If paired the other way, the reverse reaction is
spontaneous.
Cu2+(aq) + 2 e−  Cu(s) Ered = +0.34 V
Zn2+(aq) + 2 e−  Zn(s) Ered = −0.76 V
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
spontaneous
Cu(s) + Zn2+(aq)  Cu2+(aq) + Zn(s)
nonspontaneous
Example 18.5: Predict if the following reaction is
spontaneous under standard conditions
Fe(s) + Mg2+(aq)  Fe2+(aq) + Mg(s)
separate the ox:
reaction into
the oxidation red:
and reduction
half-reactions
look up the
relative
positions of
the reduction
half-reactions
red:
red:
Fe(s)  Fe2+(aq) + 2 e−
Mg2+(aq) + 2 e−  Mg(s)
Fe2+(aq) + 2 e−  Fe(s)
Mg2+(aq) + 2 e−  Mg(s)
because Mg2+ reduction is below Fe2+
reduction, the reaction is NOT
spontaneous as written
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the reaction is
Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s)
spontaneous in
ox:
Mg(s)  Mg2+(aq) + 2 e−
the reverse
red: Fe2+(aq) + 2 e−  Fe(s)
direction
sketch the cell
and label the parts
– oxidation occurs
at the anode;
electrons flow
from anode to
cathode
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Practice – Decide whether each of the
following will be spontaneous as written or in
the reverse direction
F2(g) + 2 I−(aq)  I2(s) + 2 F−(aq)
Reduction Half-Reaction
F2(g) + 2e− 2 F−(aq)
spontaneous as written
Mg(s) + 2 Ag+(aq)  Mg2+(aq) + 2 Ag(s)
spontaneous as written
Cu2+(aq) + 2e− Cu(s)
Cr3+(aq) + 1e− Cr2+(aq)
spontaneous in reverse
+2
Cr2+(aq)
 Cu(s) + 2
Ag+(aq) + 1e− Ag(s)
I2(s) + 2e− 2 I−(aq)
Cu2+(aq) + 2 I−(aq)  I2(s) + Cu(s)
Cu2+(aq)
IO3−(aq) + 6 H++ 5e− I2(s) + 3H2O(l)
Cr3+(aq)
Mg2+(aq) + 2e− Mg(s)
spontaneous as written
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Practice – Which of the following materials can
be used to oxidize Cu without oxidizing Ag?
a)
F−
b)
I−
c)
I2
d)
Cr3+
Reduction Half-Reaction
F2(g) + 2e− 2 F−(aq)
IO3−(aq) + 6 H++ 5e− I2(s) + 3H2O(l)
Ag+(aq) + 1e− Ag(s)
I2(s) + 2e− 2 I−(aq)
Cu2+(aq) + 2e− Cu(s)
Cr3+(aq) + 1e− Cr2+(aq)
Mg2+(aq) + 2e− Mg(s)
Copyright © 2011 Pearson Education, Inc.
Practice – Sketch and label a voltaic cell
in which one half-cell has Ag(s) immersed in 1 M AgNO3, and the
other half-cell has a Pt electrode immersed in 1 M Cr(NO3)2 and 1 M
Cr(NO3)3 Write the half-reactions and overall reaction, and determine
the cell potential under standard conditions.
Reduction Half-Reaction
Ered, V
F2(g) + 2e− 2 F−(aq)
+2.87
IO3−(aq) + 6 H++ 5e− ½I2(s) + 3H2O(l)
+1.20
Ag+(aq) + 1e− Ag(s)
+0.80
I2(s) + 2e− 2 I−(aq)
+0.54
Cu2+(aq) + 2e− Cu(s)
+0.34
Cr3+(aq) + 1e− Cr2+(aq)
−0.50
Mg2+(aq) + 2e− Mg(s)
−2.37
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e− →
e− →
e− →
e− →
salt
bridge
anode = Pt
cathode = Ag
Cr2+
Ag+
Cr3+
ox:
Cr2+(aq)  Cr3+(aq) + 1 e−
red: Ag+(aq) + 1 e−  Ag(s)
E
= +0.50 V
E
tot: Cr2+(aq) + Ag+(aq)  Cr3+(aq) + Ag(s)
= +0.80 V
E
= +1.30 V
Copyright © 2011 Pearson Education, Inc.
PREDICTING WHETHER A METAL WILL
DISSOLVE IN AN ACID


Metals whose ion reduction reaction lies below H+
reduction on the table will dissolve in acid
 as a single displacement reaction
Almost all metals will dissolve in HNO3
 having N reduced rather than H
 Au and Pt dissolve in HNO3 + HCl
NO3−(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(l)
Practice – Which of the following metals will
dissolve in HC2H3O2(aq)? Write the reaction.
Reduction Half-Reaction
Au3+(aq) + 3e− Au(s)
a)
Ag
b)
Cu
Cu2+(aq) + 2e− Cu(s)
c)
Fe
2H+(aq) + 2e− H2(g)
d)
Cr
Cr3+(aq) + 3e− Cr(s)
Ag+(aq) + 1e− Ag(s)
Fe3+(aq) + 3e− Fe(s)
Mg2+(aq) + 2e− Mg(s)
c)
2 Fe(s) + 6 HC2H3O2(aq) → 2 Fe(C2H3O2)3(aq) + 3 H2(g)
d)
2 Cr(s) + 6 HC2H3O2(aq) → 2 Cr(C2H3O2)3(aq) + 3 H2(g)
Copyright © 2011 Pearson Education, Inc.
18.5 CELL POTENTIAL, FREE ENERGY AND
K (EQUILIBRIUM CONSTANT)


For a spontaneous reaction
 one that proceeds in the forward direction with the
chemicals in their standard states
 DG° < 1 (negative)
 E° > 1 (positive)
 K>1
DG° = −RTlnK = −nFE°cell
 n is the number of electrons
 F = Faraday’s Constant = 96,485 C/mol e−
EXAMPLE 18.6 Relating ∆G and Ecell
Use the tabulated electrode potentials to calculate ∆G for the reaction:
Is the reaction spontaneous?
SORT You are given a redox reaction and
asked to find ∆G.
STRATEGIZE Use the tabulated values
of electrode potentials to calculate Ecell.
Then use Equation 18.3 to calculate ∆G
from Ecell.
CONCEPTUAL PLAN
SOLVE Break the reaction up into
oxidation and reduction half-reactions and
find the standard electrode potentials for
each. Find Ecell by subtracting Ean from
Ecat.
SOLUTION
Oxidation (Anode) :
Reduction (Cathode) :


2 Br  (aq) 
 Br22 (l )  2 e 




I 22 ( s)  2 e 
 2 I (aq)



I 22 ( s)  2 Br (aq) 
 2 I  ( aq)  Br22 (l )
E   1.09 V
E   0.54 V



Ecell
cell  Ecat
cat  Ean
an
 0.55 V
Calculate ∆G from Ecell. The value of n
(the number of moles of electrons)
corresponds to the number of electrons
that are canceled in the half-reactions.
Remember that 1 V = 1 J/C.
Since ∆G is positive, the reaction is not
spontaneous under standard conditions.
continued…
© 2011 Pearson Education, Inc.
EXAMPLE 18.7 Relating Ecell and K
Use the tabulated electrode potentials to calculate K for the oxidation of copper by H+
(at 25 C).
SORT You are given a redox reaction and
asked to find K.
GIVEN:
FIND: K
STRATEGIZE Use the tabulated values
of electrode potentials to calculate Ecell.
Then use Equation 18.6 to calculate ∆G
from Ecell.
CONCEPTUAL PLAN
SOLVE Break the reaction up into
oxidation and reduction half-reactions and
find the standard electrode potentials for
each. Find Ecell by subtracting Ean from
Ecat.
SOLUTION
Calculate K from Ecell. The value of n (the
number of moles of electrons) corresponds
to the number of electrons that are
canceled in the half-reactions.
continued…
© 2011 Pearson Education, Inc.
18.6 CELL POTENTIAL AND
CONCENTRATION



We know there is a relationship between the reaction
quotient, Q, the equilibrium constant, K, and the free
energy change, DGº.
Changing the concentrations of the reactants and products
so they are not 1 M will affect the standard free energy
change, DGº .
Because DGº determines the cell potential, Ecell, the voltage
for the cell will be different when the ion concentrations are
not 1 M.
NERNST EQUATION
EXAMPLE 18.8 Calculating Ecell Under Nonstandard Conditions
Determine the cell potential for an electrochemical cell based on the following two
half-reactions:
Oxidation:
Reduction:
SORT You are given the half-reactions for
a redox reaction and the concentrations of
the aqueous reactants and products. You
are asked to find the cell potential.
GIVEN:
FIND: Ecell
STRATEGIZE Use the tabulated values
of electrode potentials to calculate Ecell.
Then use Equation 18.9 to calculate Ecell.
CONCEPTUAL PLAN
SOLVE Write the oxidation and reduction
half-reactions, multiplying by the
appropriate coefficients to cancel the
electrons. Find the standard electrode
potentials for each. Find Ecell.
SOLUTION
continued…
© 2011 Pearson Education, Inc.
Calculate Ecell from Ecell.The value of n
(the number of moles of electrons)
corresponds to the number of electrons (6
in this case) canceled in the half-reactions.
Determine Q based on the overall
balanced equation and the given
concentrations of the reactants and
products. (Note that pure liquid water,
solid MnO2, and solid copper are omitted
from the expression for Q.)
CHECK The answer has the correct units (V). The value of Ecell is larger than , as expected based on
Le Châtelier’s principle because one of the aqueous reactants has a concentration greater than
standard conditions and the one aqueous product has a concentration less than standard conditions.
Therefore the reaction has a greater tendency to proceed toward products and has a greater cell
potential.
© 2011 Pearson Education, Inc.
SUMMARIZING ELECTROCHEMICAL CELLS

In all electrochemical cells,
 Anode is where OXIDATION happens
 Cathode is where REDUCTION happens

In voltaic cells, the source of electrons is anode, and anode is –

In voltaic cells, the cathode draws electrons and has a + sign.


In electrolytic cells, anode = oxidation, but an external power
source is hooked to anode with the + terminal connected to anode
(to pull electrons away from anode)
In electrolytic cells, cathode = reduction, but an external power
source is hooked to cathode with the – terminal connected to
cathode (to force electrons towards cathode)
ELECTROLYSIS



Electrolysis is the process of
using electrical energy to
break a compound apart
Electrolysis is done in an
electrolytic cell
Electrolytic cells can be used to
separate elements from their
compounds
ELECTROLYSIS



In electrolysis we use electrical energy to overcome the energy
barrier of a non-spontaneous reaction, allowing it to occur
The reaction that takes place is the opposite of the spontaneous
process
2 H2(g) + O2(g)  2 H2O(l)
spontaneous
2 H2O(l)  2 H2(g) + O2(g)
electrolysis
Some applications are (1) metal extraction from minerals and
purification, (2) production of H2 for fuel cells, (3) metal plating
ELECTROLYTIC CELLS


The electrical energy is supplied by a Direct Current power supply
 AC alternates the flow of electrons so the reaction won’t be able
to proceed
Some electrolysis reactions require more voltage than Ecell predicts.
This is called the overvoltage
ELECTROLYTIC CELLS

The source of energy is a battery or DC power supply

The + terminal of the source is attached to the anode

The − terminal of the source is attached to the cathode



Electrolyte can be either an aqueous salt solution or a molten ionic
salt
Cations in the electrolyte are attracted to the cathode and anions
are attracted to the anode
Cations pick up electrons from the cathode and are reduced, anions
release electrons to the anode and are oxidized
ELECTROLYSIS OF PURE COMPOUNDS

The compound 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)
ELECTROLYSIS OF WATER
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.
80
MIXTURES OF IONS


When more than one cation is present, the cation that is
easiest to reduce will be reduced first at the cathode
 most positive E°red (on top)
When more than one anion is present, the anion that is
easiest to oxidize will be oxidized first at the anode
 most negative E°red (on bottom)
EXAMPLE 18.9A: PREDICT THE HALF-REACTION OCCURRING AT THE
ANODE AND CATHODE FOR THE ELECTROLYSIS OF A MIXTURE OF MOLTEN
ALBR3(L) AND MGBR2(L)
in the electrolysis of a
molten salt, the
anion is oxidized and
the cation is reduced
ox:
2 Br−(l) → Br2(g) + 2 e−
red:
Al3+(aq) + 3 e− → Al(s)
Mg2+(aq) + 2 e− → Mg(s)
use the values of E°
of the half-reactions
as a guide to
determine which halfreaction is easiest
red:
Al3+(aq) + 3 e− → Al(s)
Mg2+(aq) + 2 e− → Mg(s)
E°= −1.66 V
E°= −2.37 V
Top reduction is preferred, Al 3+ will reduce to
Al.
oxidation occurs at the anode:
anode and reduction
cathode:
at the cathode
2 Br−(l) → Br2(g) + 2 e−
Al3+(aq) + 3 e− → Al(s)
PRACTICE – PREDICT THE HALF-REACTION OCCURRING AT
THE ANODE AND CATHODE FOR THE ELECTROLYSIS OF A
MIXTURE OF MOLTEN MGCL2(L) AND MGBR2(L)
in the electrolysis of a
molten salt, the
anion is oxidized and
the cation is reduced
ox:
use the values of E°
of the half-reactions
as a guide to
determine which halfreaction is easiest
ox:
red:
2 Cl−(l)  Cl2(g) + 2 e−
2 Br−(l)  Br2(g) + 2 e−
Mg2+(aq) + 2 e−  Mg(s)
Cl2(g) + 2 e−  2Cl−(l)
E°= 1.36 V
Br2(g) + 2 e− 2 Br−(l)
E°= 1.09 V
Br is on bottom, Br - is preferred for oxidation
to Br.
oxidation occurs at the anode:
anode and reduction
cathode:
at the cathode
2 Br−(l)  Br2(g) + 2 e−
Mg2+(aq) + 2 e−  Mg(s)
ELECTROLYSIS OF AQUEOUS SOLUTIONS


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
 O2 + 4 e− + 4H+ 2 H2O
Ered° = −0.83 v @ stand. cond.
Ered° = −0.41 v @ pH 7
Ered° = 1.23 v @ stand. cond.
Ered° = 0.82 v @ pH 7
EXAMPLE 18.9B: PREDICT THE HALF-REACTION
OCCURRING AT THE ANODE AND CATHODE FOR THE
ELECTROLYSIS LII(AQ)
in the electrolysis of
an aqueous salt, the
anion or water (or
electrode) is oxidized,
and the cation or
water is reduced
ox: 2 I−(aq)  I2(s) + 2 e−
2 H2O(l)  O2(g) + 4e− + 4H+(aq)
red: Li+(aq) + 1 e−  Li(s)
2 H2O(l) + 2 e−  H2(g) + 2 OH−(aq)
use the values of E°
of the half-reactions
as a guide to
determine which halfreaction is easiest
ox: O2 + 4e− + 4H+2 H2O
I2(s) + 2 e−  2 I−(aq)
E°= 0.82 V
E°= 0.54 V
red: 2 H2O + 2 e−  H2 + 2 OH− E°= −0.41 V
Li+(aq) + 1 e−  Li(s)
E°= −3.04 V
oxidation occurs at the anode:
anode and reduction
cathode:
at the cathode
2 I−(aq)  I2(s) + 2 e−
2 H2O(l) + 2 e−  H2(g) + 2 OH−(aq)
PRACTICE – PREDICT THE HALF-REACTION OCCURRING AT
THE ANODE AND CATHODE FOR THE ELECTROLYSIS
NABR(AQ)
in the electrolysis of
an aqueous salt, the
anion or water (or
electrode) is oxidized,
and the cation or
water is reduced
ox: 2 Br−(aq)  Br2(l) + 2 e−
2 H2O(l)  O2(g) + 4 e− + 4H+(aq)
red: Na+(aq) + 1 e−  Na(s)
2 H2O(l) + 2 e−  H2(g) + 2 OH−(aq)
use the values of E°
of the half-reactions
as a guide to
determine which halfreaction is easiest
ox: Br2(s) + 2 e− 2 Br−(aq)
O2 + 4e− + 4H+2 H2O
red: 2 H2O + 2 e−  H2 + 2 OH−
Na+(aq) + 1 e−  Na(s)
oxidation occurs at the anode:
anode and reduction
cathode:
at the cathode
Tro: Chemistry: A Molecular Approach, 2/e
E°= 1.09 V
E°= 0.82 V
E°= −0.41 V
E°= −3.04 V
2 H2O(l)  O2(g) + 4 e− + 4H+(aq)
2 H2O(l) + 2 e−  H2(g) + 2 OH−(aq)
86
STOICHIOMETRY OF ELECTROLYSIS


In an electrolytic cell, the amount of product made is related to the
number of electrons transferred
 essentially, the electrons are a reactant
The number of moles of electrons that flow through the electrolytic
cell depends on the current and length of time
 1 Amp = 1 Coulomb of charge/second
 1 mole of e− = 96,485 Coulombs of charge
 Faraday’s constant
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:
3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min
mass Au, g
Conceptual
t(s), amp
Plan:
charge (C)
mol e−
mol Au
g Au
Relationships:
Solve:
Check:
units are correct, answer is reasonable because 10 A running
for 1 hr ~ 1/3 mol e−
PRACTICE – CALCULATE THE AMPERAGE REQUIRED TO
PLATE 2.5 G OF AU IN 1 HOUR (3600 SEC)
AU3+(AQ) + 3 E− → AU(S)
Given:
Find:
Conceptual
Plan:
3 mol e− : 1 mol Au, mass = 2.5 g, time = 3600 s
current, A
g Au
mol Au
mol e−
charge
amps
Relationships:
Solve:
Check:
units are correct, answer is reasonable because 10 A running
for 1 hr ~ 1/3 mol e−