Chemistry: Structure and Properties
Third Edition
Lecture Presentation
Chapter 19
Electrochemistry
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Oxidation–Reduction (Redox)
• 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.
• Increase in oxidation state is oxidation; decrease in oxidation state is
reduction.
2 Na(s ) + Cl2 (g ) ¾¾
® 2 Na +Cl- (s )
Na ¾¾
® Na+ + 1 e –
oxidation,
Na is going from 0 to +1 oxidation state
Cl2 + 2 e – ¾¾
® 2Cl– reduction,
Cl is going from 0 to -1 oxidation state
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Oxidation and 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; or
– 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; or
– a half-reaction has electrons as reactants.
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Half-Reactions
• The redox reaction is split into two separate half-reactions.
– An oxidation reaction
The oxidation half-reaction has electrons as products.
– A reduction reaction
The reduction half-reaction has electrons as reactants.
3 Cl2 + I- + 3 H2 O ¾¾
® 6 Cl- + I O3 - + 6 H+
0
-1
+1 -2
-1
+5 -2
Oxidation:
I- ¾¾
® IO3 - + 6 e - (unbalanced)
Reduction:
Cl2 + 2 e – ¾¾
® 2Cl–
+1
(unbalanced)
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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
half-reaction—loss of electrons.
– Electrons go on the reactant side of the reduction
half-reaction—gain of electrons.
• Each half-reaction is balanced for its atoms first, then for
charge via electrons.
• Then, the two half-reactions are adjusted so that the electrons
lost and gained will be equal when combined.
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Balancing Redox Reactions
1.
Assign oxidation states by determining the element oxidized and the element reduced.
2.
Write oxidation and reduction half-reactions, including electrons.
a) Oxidation electrons should be on the right and reduction electrons on the left
of the arrow.
3.
Balance half-reactions by mass.
a) First balance elements other than H and O.
b) Add H2O where O is needed.
c) Add H+ where H is needed.
d) If the reaction is carried out in a basic solution, neutralize H+ with OH- .
4.
Balance half-reactions:
a) Balance charge by adjusting electrons.
b) Balance electrons between half-reactions.
5.
Add half-reactions together.
6.
Check by counting atoms and total charge on both sides.
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Practice Problem: Balancing Redox HalfReactions in an Acidic Environment (1 of 9)
Example 19.1 Half-Reaction Method of Balancing Aqueous Redox Equations in
Acidic Solution
Balance the redox equation.
Al(s ) + Cu2+ (aq ) ® Al3+ (aq ) + Cu(s )
Procedure For…
Half-Reaction Method of Balancing Aqueous Redox Equations in Acidic Solution
General Procedure
Solution
Step 1 Assign oxidation states to all atoms and identify the substances being oxidized
and reduced.
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Practice Problem: Balancing Redox HalfReactions in an Acidic Environment (2 of 9)
Step 2 Separate the overall reaction into two half-reactions: one for oxidation and one
for reduction.
Oxidation:
Al(s ) ® Al3+ (aq )
Reduction:
Cu2+ (aq ) ® Cu(s )
Step 3 Balance each half-reaction with respect to mass in the following order:
• Balance all elements other than H and O.
• Balance O by adding
H2O.
• Balance H by adding
H+ .
All elements are balanced, so proceed to the next step.
Step 4 Balance each half-reaction with respect to charge by adding electrons. (Make
the sum of the charges on both sides of the equation equal by adding as many electrons
as necessary.)
Al(s ) ® Al3+ (aq ) + 3 e 2 e - Cu2+ (aq ) ® Cu(s )
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Practice Problem: Balancing Redox HalfReactions in an Acidic Environment (3 of 9)
Step 5 Make the number of electrons in both half-reactions equal by multiplying one
or both half-reactions by a small whole number.
2[Al(s ) ® Al3+ (aq ) + 3 e - ]
2 Al(s ) ® 2 Al3+ (aq ) + 6 e 3[2 e - Cu2+ (aq ) ® Cu(s )]
6 e - + 3 Cu2+ (aq ) ® 3 Cu(s )
Step 6 Add the two half-reactions together, canceling electrons and other species as
necessary.
2 Al(s ) ¾¾
® 2 Al3 + (aq ) + 6e 6e - + 3 Cu2 + (aq ) ¾¾
® 3 Cu(s)
2+
2 Al(s) + 3 Cu (aq ) ¾¾
® 2 Al3 + (aq ) + 3 Cu(s )
Step 7 Verify that the reaction is balanced with respect to both mass and charge.
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Practice Problem: Balancing Redox HalfReactions in an Acidic Environment (4 of 9)
Reactants
Products
2 AI
2 AI
3 Cu
3 Cu
6+ Charge
6+ Charge
For Practice 19.1
Balance the redox reaction in acidic solution.
H+ (aq ) + Cr(s ) ® H2 (g ) + Cr 2+ (aq )
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Practice Problem: Balancing Redox HalfReactions in an Acidic Environment (5 of 9)
Example 19.3 Balancing Redox Reactions Occurring in Basic Solution
Balance the equation occurring in basic solution.
I- (aq ) + MnO4- (aq ) ® I2 (aq ) + MnO2 (s )
Solution
To balance redox reactions occurring in basic solution, follow the half-reaction method
outlined in Examples 19.1 and 19.2, but add an extra step to neutralize the acid with
OH-
as shown in Step 3 of this example.
1.
Assign oxidation states.
2.
Separate the overall reaction into two half-reactions.
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Practice Problem: Balancing Redox HalfReactions in an Acidic Environment (6 of 9)
Oxidation: I (aq ) ® I2 (aq )
Reduction: MnO-4 (aq ) ® MnO2 (s )
3. Balance each half-reaction with respect to mass:
(a) Balance all elements other than H and O.
(b) Balance O by adding H2O.
(c) Balance H by adding H+ .
(d) Neutralize H+ by adding enough OH- to neutralize each
H+ . Add the same number of OH- ions to each side of the
equation. Adjust the mole of H2O.
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Practice Problem: Balancing Redox HalfReactions in an Acidic Environment (7 of 9)
(a)
ìï2 I- (aq ) ¾¾
® I2 (aq )
í
® MnO2 (s )
ïîMnO4 (aq ) ¾¾
(b)
ìï2 I- (aq ) ¾¾
® I2 (aq )
í
® MnO2 (s ) + 2 H2O( l )
ïîMnO4 (aq ) ¾¾
(c)
ìï2 I- (aq ) ¾¾
® I2 (aq )
í
+
® MnO2 ( s) + 2 H2O( l )
ïî4 H (aq ) + MnO 4 (aq ) ¾¾
(d)
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Practice Problem: Balancing Redox HalfReactions in an Acidic Environment (8 of 9)
4.
Balance each half-reaction with respect to charge.
2 I- (aq ) ® I2 (aq ) + 2 e 4 H2O(I ) + MnO 4- (aq ) + 3 e - ® MnO2 (s ) + 2 H2O(I ) + 4 OH- (aq )
5.
Make the number of electrons in both half-reactions equal.
3[2 I- (aq ) ® I2 (aq ) + 2 e - ]
6 I- (aq ) ® 3 I2 (aq ) + 6 e 2 [4 H2O(I ) + MnO 4- (aq ) + 3 e - ® MnO2 (s ) + 2 H2O(I ) + 4 OH- (aq )]
8 H2O(I ) + 2 MnO 4- (aq ) + 6 e - ® 2 MnO2 (s ) + 4 H2O(I ) + 8 OH- (aq )
6.
Add the half-reactions together.
6I- (aq ) ¾¾
® 3I2 (aq ) + 6e 4 8 g-H2O(l ) + 2MnO -4 (aq ) + 6e - ¾¾
® 2MnO2 (s ) + 4H2q ( l ) + 8OH- (aq )
6 I- (aq ) + 4H2O(l ) + 2MnO 4- (aq ) ¾¾
® 3I2 (aq ) + 2MnO2 ( s) + 8OH- (aq )
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Practice Problem: Balancing Redox HalfReactions in an Acidic Environment (9 of 9)
7. Verify that the reaction is balanced.
Reactants
Products
6I
6I
8H
8H
2 Mn
2 Mn
12 O
12 O
-8 Charge
-8 Charge
minus 8 charge
8 minus charge
For Practice 19.3
Balance the following redox reaction occurring in basic solution.
CIO- (aq ) + Cr(OH)-4 (aq ) ® CrO24- (aq ) + CI- (aq )
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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.
– Also known as a galvanic cell.
• Nonspontaneous redox reactions can be made to occur
in an electrolytic cell by the addition of
electrical energy.
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A Spontaneous Redox Reaction:
Zn(s ) + Cu2+ (aq ) ® Zn2+ (aq ) + Cu(s )
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Voltaic (Galvanic) Cells: Spontaneous
Redox Reactions
• Electrical current: The amount of electric charge that passes a point in a
given period of time
– Whether as electrons flowing through a wire or as ions flowing through
a solution.
• Redox reactions involve the movement of electrons from one substance to
another.
– Therefore, redox reactions have the potential to generate an electric
current.
• A spontaneous redox reaction does not require external energy to proceed.
– The reaction’s DG
is negative.
• Voltaic (galvanic) cells produce an electrical current from spontaneous redox
reactions.
– To use that current, we need to separate the place where oxidation is
occurring from the place where reduction is occurring.
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Electrochemical Cells
• Oxidation and reduction half-reactions are kept separate in
half-cells of an electrochemical cell.
• To constitute an electrical circuit:
– Electrons flow through a wire along with ions (electrolyte)
flowing through a solution via the salt bridge.
• The flow of electrons requires a conductive solid electrode to allow
the transfer of electrons either through:
– An external circuit or
– Metal or graphite electrode
• An electrochemical cell requires the exchange of ions between the
two half-cells of the system via a salt bridge.
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Voltaic Cell (1 of 2)
The salt bridge is required to complete the circuit and maintain charge
balance.
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Voltage and Current
• Voltage is the difference in potential energy between the reactants and products. It is
also called the potential difference.
– Unit = volt = potential energy per unit of charge
•
1 V = 1J / C
– The voltage needed to drive electrons through the external circuit
• The amount of force pushing the electrons through the wire is called the
electromotive force, emf.
• Current is the number of electrons that flow through the system per second.
Unit = ampere
• 1 A of current = 1 coulomb of charge flowing per second
• Electrode surface area dictates the number of electrons that can flow.
– Larger batteries produce larger currents.
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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
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
– Sum of the cell potentials for the half-reactions
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Voltaic Cell (2 of 2)
Cathode = Cu(s )
Anode = Zn(s )
The anode is oxidized to
Zn2+ ions.
Cu2+ ions are reduced at the cathode.
Zn(s ) | Zn2+ (aq ) || Cu2+ (aq ) | Cu(s )
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Conceptual Connection 19.1 (1 of 2)
Which statement best captures the difference between volts and
amps?
(a) The volt is a unit that quantifies the difference in electrical
potential energy, and the amp is a unit that quantifies the
flow of electrical current.
(b) The amp is a unit that quantifies the difference in electrical
potential energy, and the volt is a unit that quantifies the flow
of electrical current.
(c) The volt and amp are two different units to measure the same
thing, the flow of electrical current
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Conceptual Connection 19.1 (2 of 2)
Which statement best captures the difference between volts and
amps?
(a) The volt is a unit that quantifies the difference in
electrical potential energy, and the amp is a unit that
quantifies the flow of electrical current.
(b) The amp is a unit that quantifies the difference in electrical
potential energy, and the volt is a unit that quantifies the flow
of electrical current.
(c) The volt and amp are two different units to measure the same
thing, the flow of electrical current
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Electrodes of an Electrochemical Cell
• Anode
– Electrode where oxidation occurs
– Anions attracted to it
– Connected to positive end of battery in an electrolytic cell
– Loses weight in electrolytic cell
• Cathode
– Electrode where reduction occurs
– Cations attracted to it
– Connected to negative end of battery in an
electrolytic cell
– Gains weight in electrolytic cell
Electrode where plating takes place in electroplating
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Electrodes
• Typically,
– 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.
• 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 does not participate in
the reaction but just provides a surface for the
transfer of electrons to take place on.
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Conceptual Connection 19.2 (1 of 2)
In a voltaic cell, in which direction do electrons flow?
(a) From higher potential energy to lower potential energy
(b) From the cathode to the anode
(c) From lower potential energy to higher potential energy
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Conceptual Connection 19.2 (2 of 2)
In a voltaic cell, in which direction do electrons flow?
(a) From higher potential energy to lower potential energy
(b) From the cathode to the anode
(c) From lower potential energy to higher potential energy
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Cell Notation
• Shorthand description of a voltaic cell is written as follows:
electrode electrolyte
electrolyte electrode
– Oxidation half-cell is on the left; reduction half-cell is on
the right.
– Single line vertical (|) indicates a phase barrier
§ If multiple electrolytes are in same phase, a comma is
used rather than a single vertical line
§ Often use an inert electrode
(|).
– Double vertical line (||) separate the two half-reactions.
• Example:
Zn(s ) | Zn2+ (aq ) || Cu2+ (aq ) | Cu(s )
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Electrochemical Cell Notation with an
Inert Electrode
The half-reaction involves reducing Mn oxidation state from +7 to +2. An inert
electrode (Pt) will provide a surface for the electron transfer without reacting
with the MnO-4 .
Fe(s )| Fe2+ (aq ) || MnO-4 (aq ), Mn2+ (aq ), H+ (aq ) |Pt(s )
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Standard Reduction Potential
• The potential of a half-reaction cannot be measured directly.
– Only the potential relative to another half-reaction can be measured.
• To overcome this limitation, a standard half-reaction for the reduction of
H+ to H2
is selected and assigned a potential difference of 0 V.
– Standard hydrogen electrode, SHE
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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 .
• Half-reactions with a stronger tendency toward reduction than the
SHE have a positive
E°.
• Half-reactions with a stronger tendency toward oxidation than the
SHE have a negative E °.
• Cell potential for any electrochemical cell is
°
°
°
Ecell
= Ecat
- Ean
• Cell potential is positive for spontaneous reactions and negative for
nonspontaneous reactions.
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Cell Potential (E °)
E naught
• 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.
– Electrons flow from anode to cathode.
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Which Way Will Electrons Flow?
• Under standard conditions, zinc has a stronger tendency to oxidize than copper.
– Electrons flow from anode to cathode.
– Therefore, the electrons flow from zinc, making zinc the anode.
Zn ® Zn2+ + 2 e -
E °ox = + 0.76 V
Cu ® Cu2+ + 2 e-
E°ox = - 0.34 V
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Tendencies from the Table of Standard
Reduction Potentials
• A redox reaction will be spontaneous when there is a
strong tendency for the oxidizing agent to be reduced
and the reducing agent to be oxidized.
– Higher on the table of standard reduction potentials =
stronger tendency for the reactant to be reduced.
– Lower on the table of standard reduction potentials =
stronger tendency for the product to be oxidized.
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Table of Standard Reduction Potentials
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Calculating Cell Potentials under
Standard Conditions
• Cell potentials are intensive properties of matter.
– Because cell potentials are intensive physical properties, when
determining the cell potential, do not multiply the half-cell
E°
values, even if you need to multiply the half-reactions to balance
the redox equation.
• The cell potential of an electrochemical cell (E°cell ) is the difference
between the electrode potential of the cathode and that of the anode.
• Cell potentials can be determined using the following equation:
E °cell = E °cathode (reduction) - E °anode (oxidation)
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Practice Problem: Calculating Standard Potentials
for Electrochemical Cells from Half-Reactions (1 of 4)
Example 19.4 Calculating Standard Potentials for Electrochemical Cells from
Standard Electrode Potentials of the Half-Reactions
Use tabulated standard electrode potentials to calculate the standard cell potential for this
reaction occurring in an electrochemical cell at 25 °C. (The equation is balanced.)
Al(s ) + NO3 - (aq ) + 4 H+ (aq ) ® Al3+ (aq ) + NO(g ) + 2 H2O(l )
Solution
Begin by separating the reaction into oxidation and reduction half-reactions. (In this case,
you can readily see that Al(s) is oxidized. In cases where it is not so apparent, you may
want to assign oxidation states to determine the correct half-reactions.)
Oxidation:
Al(s ) ® Al3+ (aq ) + 3 e -
Reduction:
NO3 - (aq ) + 4 H+ (aq ) +3 e - ® NO(g ) + 2 H2O(l )
Look up the standard electrode potentials for each half-reaction in Table 19.1. Add the
half-cell reactions together to obtain the overall redox equation. Calculate the standard
cell potential by subtracting the electrode potential of the anode from the electrode
potential of the cathode.
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Practice Problem: Calculating Standard Potentials
for Electrochemical Cells from Half-Reactions (3 of 4)
Oxidation
(Anode):
Reduction
(Cathide);
Al(s ) ® Al3+ (aq ) + 3 e- E° = -1.66 V
NO3 - (aq ) + 4 H+ (aq ) + 3 e - ® NO(g ) + 2 H2O(l ) E ° = 0.96 V
Al(s ) + NO3 - (aq ) + 4 H+ (aq ) ® Al3+ (aq ) + NO(g ) + 2 H2O(l )
E °cell = E °cat - E °an
= 0.96 V - ( -1.66 V)
= 2.62 V
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Practice Problem: Calculating Standard Potentials
for Electrochemical Cells from Half-Reactions (4 of 4)
For Practice 19.4
Use tabulated standard electrode potentials to calculate the standard
cell potential for this reaction occurring in an electrochemical cell at
25 °C. (The equation is balanced.)
3 Pb2+ (aq ) + 2 Cr(s ) ® 3Pb(s ) + 2 Cr 3+ (aq )
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Conceptual Connection 19.3 (1 of 2)
An electrode has a negative electrode potential. Which
statement is correct regarding the potential energy of an
electron at this electrode?
(a) An electron at this electrode has a lower potential energy
than it has at a standard hydrogen electrode.
(b) An electron at this electrode has a higher potential energy
than it has at a standard hydrogen electrode.
(c) An electron at this electrode has the same potential energy
as it has at a standard hydrogen electrode.
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Conceptual Connection 19.3 (2 of 2)
An electrode has a negative electrode potential. Which
statement is correct regarding the potential energy of an
electron at this electrode?
(a) An electron at this electrode has a lower potential energy
than it has at a standard hydrogen electrode.
(b) An electron at this electrode has a higher potential
energy than it has at a standard hydrogen electrode.
(c) An electron at this electrode has the same potential energy
as it has at a standard hydrogen electrode.
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Predicting Spontaneity of Redox
Reactions
• 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.
• When
E°cell is
– positive, the redox reaction of the cell is spontaneous ( DG°
will be negative).
Zn(s ) + Cu2+ (aq ) ® Zn2+ (aq ) + Cu(s )
• When
spontaneous reaction
E°cell is
– negative, the redox reaction of the cell is nonspontaneous ( DG°
will be positive).
Cu(s ) + Zn2+ (aq ) ® Cu2+ (aq ) + Zn(s )
Cu2+ (aq ) + 2 e - ® Cu(s )
Zn2+ (aq ) + 2 e - ® Zn(s )
nonspontaneous reaction
E °red = + 0.34 V
E °red = - 0.76 V
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Practice Problem: Predicting Spontaneous Redox
Reactions and Sketching an Electrochemical Cell (1 of 5)
Example 19.5 Predicting Spontaneous Redox Reactions and Sketching
Electrochemical Cells
Without calculating
E °cell , predict whether each of these redox reactions is spontaneous
(when the reactants and products are in their standard states). If the reaction is
spontaneous as written, make a sketch of the electrochemical cell in which the reaction
could occur. If the reaction is not spontaneous as written, write an equation for the
direction in which the spontaneous reaction occurs and sketch the corresponding
electrochemical cell. In your sketches, make sure to label the anode (which should be
drawn on the left), the cathode, and the direction of electron flow.
a.
Fe(s ) + Mg2+ (aq ) ® Fe2+ (aq ) + Mg(s )
b.
Fe(s ) + Pb2+ (aq ) ® Fe2+ (aq ) + Pb(s )
Solution
a.
Fe(s ) + Mg2+ (aq ) ® Fe2+ (aq ) + Mg(s )
This reaction involves the reduction of
Mg2+ .
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Practice Problem: Predicting Spontaneous Redox
Reactions and Sketching an Electrochemical Cell (3 of 5)
Mg2+ (aq ) + 2 e - ® Mg(s )
E ° = -2.37V
Fe(s ) ® Fe2+ (aq ) + 2 e -
E ° = -0.45V
and the oxidation of Fe:
The magnesium half-reaction has the more negative electrode potential and
therefore repels electrons more strongly and undergoes oxidation. The iron
half-reaction has the more positive electrode potential and therefore attracts
electrons more strongly and undergoes reduction. So the reaction as written is
not spontaneous. (The reaction pairs the reduction of Mg2+ with the reverse
of a half-reaction above it in Table 19.1—such pairings are not spontaneous.)
For Practice 19.5
Are these redox reactions spontaneous under standard conditions?
(a)
Zn(s ) + Ni2 + (aq ) ¾¾
® Zn2 + (aq ) + Ni(s )
(b)
Zn(s) + Ca2 + (aq ) ¾¾
® Zn2 + (aq ) + Ca(s )
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Practice Problem: Predicting Spontaneous Redox
Reactions and Sketching an Electrochemical Cell (4 of 5)
However, the reverse reaction is spontaneous:
Fe2+ (aq ) + Mg(s ) ® Fe(s ) + Mg2+ (aq )
The corresponding electrochemical cell is shown in Figure 19.9.
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Practice Problem: Predicting Spontaneous Redox
Reactions and Sketching an Electrochemical Cell (5 of 5)
This reaction involves the reduction of
Pb2 + and the oxidation of Fe:
Pb2 + (aq ) + 2e - ¾¾
® Pb(s) E ° = -0.13 V
Fe(s ) ® Fe2+ (aq ) + 2 e -
E ° = -0.45V
The iron half-reaction has the more negative electrode potential and therefore
repels electrons more strongly and undergoes oxidation. The lead half-reaction
has the more positive electrode potential and therefore attracts electrons more
strongly and undergoes reduction. So the reaction as written is not
spontaneous.
For Practice 19.5
Are these redox reactions spontaneous under standard conditions?
(a)
Zn(s ) + Ni2 + (aq ) ¾¾
® Zn2 + (aq ) + Ni(s )
(b)
Zn(s) + Ca2 + (aq ) ¾¾
® Zn2 + (aq ) + Ca(s )
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Conceptual Connection 19.4 (1 of 2)
A solution contains both NaI and NaBr. Which oxidizing agent could
you add to the solution to selectively oxidize I- (aq ) but not
(a)
Cl2
(b)
H2O2
(c)
CuCl2
Br - (aq )?
(d) HNO3
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Conceptual Connection 19.4 (2 of 2)
A solution contains both NaI and NaBr. Which oxidizing agent could
you add to the solution to selectively oxidize I- (aq ) but not
(a)
Cl2
(b)
H2O2
(c)
CuCl2
Br - (aq )?
(d) HNO3
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Predicting whether a Metal Will Dissolve
in an Acid
• Metals dissolve in acids
– If the reduction of H+ (aq )
is easier than the reduction of
the metal
– If their ion reduction reaction
lies below H+ reduction on
the table
• Almost all metals will dissolve in
HNO3 .
Zn(s ) + 2H+ (aq ) ® Zn2 + (aq ) + H2 (g )
– Having N reduced rather than
H+
– Au and Pt dissolve in HNO3 + HCI.
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Conceptual Connection 19.5 (1 of 2)
Which metal dissolves in HNO3 but not in HCl?
(a) Fe
(b) Au
(c) Ag
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Conceptual Connection 19.5 (2 of 2)
Which metal dissolves in HNO3 but not in HCl?
(a) Fe
(b) Au
(c) Ag
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E °cell , DG °, and K
E naught sub cell, delta G naught, and K
• For a spontaneous reaction, one
that proceeds in the forward
direction with all substances in
their standard states
•
–
DG° < 1 (negative)
–
E ° > 1 (positive)
–
K >1
DG° = -RT ln K = -nFE°cell
– n = the number of moles of
electrons
– F = Faraday’s constant =
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Practice Problem: Relating DG ° and E °cell (1 of 4)
delta G naught and E naught sub ce
Example 19.6 Relating
DG °and E °cell
Use the tabulated electrode potentials to calculate
DG° for the reaction.
I2 (s ) + 2 Br - (aq ) ® 2 I- (aq ) + Br2 (l )
Is the reaction spontaneous under standard conditions?
Sort You are given a redox reaction and asked to find
Given
I2 (s ) + 2 Br - (aq ) ® 2 I- (aq ) + Br2 (l )
Find
DG°
DG°.
Strategize Refer to the values of electrode potentials in Table 19.1 to calculate
Then use Equation 19.3 to calculate
E°cell.
DG° from E °cell.
Conceptual Plan
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Practice Problem: Relating DG ° and E °cell (2 of 4)
delta G naught and E naught sub ce
Table 19.1 Standard Electrode Potentials at 25 °C
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Practice Problem: Relating DG ° and E °cell (3 of 4)
delta G naught and E naught sub ce
Solve Separate the reaction into oxidation and reduction half-reactions and find the
standard electrode potentials for each. Determine
E°cell by subtracting
E °an from E °cat.
Solution
Calculate DG° 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.
DG° = -nFE °cell
æ 96,485 C ö æ
Jö
= -2 mol e- ç
0.55
÷
ç
C ÷ø
è mol e ø è
= +1.1´ 105 J
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Practice Problem: Relating DG ° and E °cell (4 of 4)
delta G naught and E naught sub ce
Since DG° is positive, the reaction is not spontaneous under standard conditions.
Check The answer is in the correct units (joules) and seems reasonable in magnitude
( » 110 kJ). You have seen (in Chapter 18) that values of
DG° typically range from
plus or minus tens to hundreds of kilojoules. The sign is positive, as expected for a
reaction in which
E °cell is negative.
For Practice 19.6
Use tabulated electrode potentials to calculate
DG° for the reaction.
2Na(s ) + 2 H2O(l ) ® H2 (g ) + 2 OH- (aq ) + 2 Na+ (aq )
Is the reaction spontaneous under standard conditions?
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Conceptual Connection 19.6 (1 of 2)
Consider the result of Example 19.6. The calculation revealed
that the reaction is not spontaneous. Based on conceptual
reasoning which statement best explains why I2 does not oxidize
Br -?
(a) Br is more electronegative than I; therefore we do not expect
Br - to give up an electron to I2 .
(b) I is more electronegative than Br; therefore we do not expect
I2 to give up an electron to Br - .
(c) Br - is in solution and I2 is a solid. Solids do not gain
electrons from substances in solution.
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Conceptual Connection 19.6 (2 of 2)
Consider the result of Example 19.6. The calculation revealed
that the reaction is not spontaneous. Based on conceptual
reasoning which statement best explains why I2 does not oxidize
Br -?
(a) Br is more electronegative than I; therefore we do not
expect Br - to give up an electron to I2 .
(b) I is more electronegative than Br; therefore we do not expect
I2 to give up an electron to Br - .
(c) Br - is in solution and I2 is a solid. Solids do not gain
electrons from substances in solution.
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Practice Problem: Relating K and E °cell (1 of 4)
K and E naught sub cell
Example 19.7 Relating
E °cell and K
Refer to tabulated electrode potentials to calculate K for the oxidation of copper by
H+ (at 25 ° C).
Cu(s ) + 2 H+ (aq ) ® Cu2+ (aq ) + H2 (g )
Sort You are given a redox reaction and asked to find K.
Given
Cu(s ) + 2 H+ (aq ) ® Cu2+ (aq ) + H2 (g )
Find K
Strategize Refer to the values of electrode potentials in Table 19.1 to calculate
Then use Equation 19.6 to calculate
E°cell.
DG° from E °cell.
Conceptual Plan
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Practice Problem: Relating K and E °cell (2 of 4)
K and E naught sub cell
Table 19.1 Standard Electrode Potentials at
25 °C
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Practice Problem: Relating K and E °cell (3 of 4)
K and E naught sub cell
Solve Separate the reaction into oxidation and reduction half-reactions and find the
standard electrode potentials for each. Find
E°cell by subtracting E °an from E °cat.
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.
E °cell =
0.0592 V
log K
n
log K = E °cell
log K = -0.34 V
n
0.0592 V
2
= -11.48
0.0592 V
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Practice Problem: Relating K and E °cell (4 of 4)
K and E naught sub cell
K = 10-11.48 = 3.3 ´ 10 -12
Check The answer has no units, as expected for an equilibrium constant. The magnitude
of the answer is small, indicating that the reaction lies far to the left at equilibrium, as
expected for a reaction in which
E°cell is negative.
For Practice 19.7
Use the tabulated electrode potentials to calculate K for the oxidation of iron by
H+ (at 25 ° C).
2 Fe(s ) + 6 H+ (aq ) ® 2Fe3+ (aq ) + 3 H2 (g )
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Conceptual Connection 19.7 (1 of 2)
A redox reaction has an equilibrium constant of K = .1.2 ´ 103
°
°
and Ecell
for the
Which statement is true regarding DGrxn
reaction?
°
°
(a) Ecell
is positive and DGrxn
is positive.
(b)
°
°
Ecell
is negative.
is negative and DGrxn
°
°
(c) Ecell
is positive and DGrxn
is negative.
°
°
(d) Ecell
is positive.
is negative and DGrxn
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Conceptual Connection 19.7 (2 of 2)
A redox reaction has an equilibrium constant of K = .1.2 ´ 103
°
°
and Ecell
for the
Which statement is true regarding DGrxn
reaction?
°
°
(a) Ecell
is positive and DGrxn
is positive.
(b)
°
°
Ecell
is negative.
is negative and DGrxn
°
°
(c) Ecell
is negative.
is positive and DGrxn
°
°
(d) Ecell
is positive.
is negative and DGrxn
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Cell Potential at Concentration
• There is a relationship between the reaction quotient, Q; the
equilibrium constant, K; and the standard free energy change,
DG°.
• Changing the concentrations of the reactants and products so that
they are not 1 M will affect the standard free energy change,
DG°.
• Because DG° determines the cell potential, Ecell , the voltage for the
cell will be different when the ion concentrations are not 1 M.
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Deriving the Nernst Equation
DG = DG° + RT ln Q
( -) nFEcell = ( -) nFE °cell + RT ln Q
RT
Ecell = E °cell In Q
nF
This is the Nernst equation.
At 25°C (T = 298 K),
R = 8.314 J / m ol·K
Faraday’s constant (F) = 96,500 C / mol e n = the number of moles of electrons
Converting from ln to log (2.303), the Nernst equation becomes
Ecell = E °cell -
0.0592 V
log Q
n
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Practice Problem: Calculating Ecell (1 of 4)
E sub cell
Example 19.8 Calculating Ecell under Nonstandard Conditions
Determine the cell potential for an electrochemical cell based on the following two halfreactions:
Oxidation:
Cu(s ) ® Cu2+ (aq, 0.010 M) + 2 e-
Reduction: MnO4 - (aq, 2.0 M) + 4 H+ (aq, 1.0 M) + 3 e - ® MnO2 (s ) + 2 H2O(l )
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
[MnO4 - ] = 2.0 M; [H+ ] = 1.0 M; [ Cu2+ ] = 0.010 M
Find
Ecell
Strategize Use the tabulated values of electrode potentials to calculate Ecell .
Then use the Nernst equation to calculate
Ecell .
Conceptual Plan
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Practice Problem: Calculating Ecell (2 of 4)
E sub cell
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
Ecell .
Solution
Calculate
Ecell from Ecell . The value of n (the number of moles of electrons) corresponds
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Practice Problem: Calculating Ecell (3 of 4)
E sub cell
to the number of electrons canceled in the half-reactions, six in this case. 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.)
Ecell = E °cell -
0.0592 V
log Q
n
0.0592 V
[Cu2+ ]3
= E °cell log
n
[MnO4 - ]2 [H+ ]8
0.0592 V
[0.010]3
= 1.34 V log
6
[2.0]2 [1.0]8
= 1.34 V - ( -0.065 V)
= 1.41 V
Check The answer has the correct units (V). The value of
Ecell is larger than
E°cell ,
as expected, based on Le Châtelier’s principle, because one of the aqueous reactants
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Practice Problem: Calculating Ecell (4 of 4)
E sub cell
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 a greater cell potential.
For Practice 19.8
Determine the cell potential of an electrochemical cell based on the following two halfreactions:
Oxidation:
Ni(s ) ® Ni2+ (aq, 2.0 M) + 2 e -
Reduction:
VO2+ (aq, 0.010 M) + 2 H+ (aq, 1.0 M) + e- ® VO2+ (aq, 2.0 M) + H2O(l )
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Conceptual Connection 19.8 (1 of 2)
In an electrochemical cell, Q = 0.0010 and K = 0.10. What can
°
you conclude about Ecell and Ecell
?
°
(a) Ecell is positive and Ecell
is negative.
°
is positive.
(b) Ecell is negative and Ecell
°
are positive.
(c) Both Ecell and Ecell
°
(d) Both Ecell and Ecell
are negative.
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Conceptual Connection 19.8 (2 of 2)
In an electrochemical cell, Q = 0.0010 and K = 0.10. What can
°
you conclude about Ecell and Ecell
?
°
is negative.
(a) Ecell is positive and Ecell
°
is positive.
(b) Ecell is negative and Ecell
(c) Both
°
are positive
Ecell and Ecell
°
(d) Both Ecell and Ecell
are negative
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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 solution.
• 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.
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Concentration Cell
When the cell concentrations are
equal, there is no difference in
energy between the half-cells, and
no electrons flow.
When the cell concentrations are different,
electrons flow from the side with the less
concentrated solution (anode) to the side with
the more concentrated solution (cathode).
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Batteries: Dry Cell
• Electrolyte in paste form
–
NH4CI
• Anode = Zn
Zn(s) ® Zn2+ (aq ) + 2 e-
• Cathode = graphite rod
•
MnO2 is reduced.
2 MnO2 (s ) + 2 NH4 + (aq ) + 2 e® Mn2O3 (s ) + 2 NH3 (g ) + H2O(l )
• Cell voltage = 1.5 V
• Expensive, nonrechargeable,
heavy, easily corroded
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Batteries: Alkaline Dry Cell
• Same basic cell as acidic dry cell,
except electrolyte is alkaline KOH
paste
• Anode = Zn (or Mg)
Zn(s ) + 2 OH- (aq ) ® Zn ( OH)2 (s ) + 2 e-
• Cathode = graphite or brass rod
•
MnO2 is reduced.
2 MnO2 (s ) + 2 H2O(l ) + 2 e -
® 2MnO ( OH) (s ) + 2 OH- (aq )
• Cell voltage = 1.5 V
• Longer shelf life than acidic dry
cells, with little corrosion of zinc
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Batteries: Lead Storage Battery
• Six electrochemical cells in series
• Electrolyte = 30% H2SO4
• Anode = Pb
Pb(s ) + HSO4 - (aq ) ® PbSO 4 (s ) + H+ (aq ) + 2 e-
• Cathode = Pb coated with PbO2
• PbO2 is reduced.
PbO2 (s ) + HSO4 - + 3 H+ (aq ) + 2 e - ® PbSO4 (s ) + 2 H2O(l )
• Individual cell voltage = 2.09 V
• Rechargeable, heavy
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NiCad Battery
• Electrolyte is concentrated KOH solution.
• Anode = Cd
Cd(s ) + 2 OH- (aq ) ® Cd ( OH)2 (s ) + 2 e -
E °ox = 0.81 V
• Cathode = Ni coated with NiO2
•
NiO(OH) is reduced.
2NiO(OH)(s ) + 2 H2O(l ) + 2 e - ® 2 Ni ( OH)2 (s ) + 2OH- (aq )
E °red = 0.45 V
• Cell voltage = 1.3 V
• Rechargeable, long life, light;
however, recharging incorrectly can
lead to battery breakdown.
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NiMH 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 °ox = 0.89 V
• Cathode = Ni coated with
•
NiO2
NiO(OH) is reduced.
NiO(OH)(s ) + H2O(l ) + e - ® Ni ( OH)2 (s ) + OH- (aq ) E °red = 0.45 V
• Cell voltage = 1.3 V
• Rechargeable, long life, light, more environmentally friendly
than NiCad, greater energy density than NiCad
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Lithium Ion Battery
• Anode is graphite
impregnated with Li ions.
• Cathode is Li-transition
metal oxide.
– Reduction of transition
metal
• Li ion migration from anode
to cathode causes a
corresponding migration of
electrons from anode to
cathode.
• Rechargeable, long life,
very light, greater energy
density.
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Fuel Cells
• They are like batteries in which reactants are constantly being added.
– So they never run down!
• Anode and cathode are both Pt coated metal.
• Electrolyte is OH- solution.
• Anode reaction:
2 H2 (g ) + 4 OH- (aq ) ® 4 H2O(l ) + 4 e -
• Cathode reaction:
O2 (g ) + 2 H2O(l ) + 4 e - ® 4 OH- (aq )
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Electrochemical Cells Overview
• In all electrochemical cells, oxidation occurs at the anode and
reduction occurs at the cathode.
• In voltaic cells (spontaneous reactions;
E°cell is positive),
– the anode is the source of electrons and has a ( -) charge;
– the cathode draws electrons and has a (+) charge.
• In electrolytic cells (nonspontaneous reactions; E°cell is negative),
– electrons are drawn off the anode, so there must be a place to
release the electrons—the positive terminal of the battery;
– electrons are forced toward the anode, so there must be a
source of electrons—the negative terminal of the battery.
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Voltaic versus Electrolytic Cells
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Electrolysis (1 of 2)
• Electrolysis is the
process of using
electrical energy to
break a compound
apart.
• Electrolysis occurs in
an electrolytic cell.
• Electrolytic cells can
be used to separate
elements from their
compounds.
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Electrolysis (2 of 2)
• In electrolysis, we use electrical energy to overcome the
energy barrier of a nonspontaneous 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 )
2 H2O(l ) ® 2 H2 (g ) + O2 (g )
spontaneous
electrolysis
• Some applications of electrolysis are the following:
1. Metal extraction from minerals and purification
2. Production of H2 for fuel cells
3. Metal plating
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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.
• The source of energy is a battery or DC power supply.
– The positive terminal of the source is attached to the anode.
– The negative 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.
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Electrolysis of Aqueous Solutions
• Possible cathode reactions:
– Reduction of cation to metal
– Reduction of water to H2
§
2 H2O + 2 e - ® H2 + 2 OH-
E °red = -0.83 V at standard conditions
E °red = -0.41 V at pH 7
• Possible anode reactions:
– Oxidation of anion to element
– Oxidation of
§
H2O to O2
2 H2O ® O2 + 4 e - + 4H+
E °ox = -1.23 V at standard conditions
E °ox = -0.82 V at pH 7
– Oxidation of electrode:
§ Particularly Cu
§ Graphite doesn’t oxidize
• Half-reactions that lead to least negative
Ecell will occur.
– Unless overvoltage changes the conditions
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Electrolysis of Water
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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.
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Conceptual Connection 19.9 (1 of 2)
Which statement is true for both electrolytic and voltaic
cells?
(a) The cell spontaneously produces a positive charge.
(b) Electrons flow from the anode to the cathode.
(c) Oxidation occurs at the cathode.
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Conceptual Connection 19.9 (2 of 2)
Which statement is true for both electrolytic and voltaic
cells?
(a) The cell spontaneously produces a positive charge.
(b) Electrons flow from the anode to the cathode.
(c) Oxidation occurs at the cathode.
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Electrolysis of Pure Compounds
• The compound must be in
molten (liquid) state.
• Electrodes are normally
graphite.
• Cations are reduced at the
cathode to metal element.
• Anions are oxidized at the
anode to nonmetal
element.
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Mixtures of Ions and Electrolysis
• 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
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Practice Problem: Predicting the
Products of Electrolysis Reactions (1 of 4)
Example 19.9 Predicting the Products of Electrolysis Reactions
Predict the half-reaction occurring at the anode and the cathode for electrolysis for each
reaction.
a. a mixture of molten
AIBr3 and MgBr2
b. an aqueous solution of LiI
Solution
a. In the electrolysis of a molten salt, the anion is oxidized and the cation is reduced.
However, this mixture contains two cations. Start by writing the possible oxidation and
reduction half-reactions that might occur.
Since Br - is the only anion, write the equation for its oxidation, which occurs at the anode.
At the cathode, both the reduction of
AI3+ and the reduction of
Mg2+ are possible.
The one that actually occurs is the one that occurs more easily. Since the reduction of
AI3+
has a more positive electrode potential in aqueous solution, this ion is more easily
reduced.
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Practice Problem: Predicting the
Products of Electrolysis Reactions (2 of 4)
Therefore, the reduction of
AI3+ occurs at the cathode.
b. Because LiI is in an aqueous solution, two different oxidation half-reactions are possible
at the anode, the oxidation of I- and the oxidation of water. Write half-reactions for each,
including the electrode potential. Remember to use the electrode potential of water under
conditions in which [H+ ] = 10-7 M. Since the oxidation of
I- has the more negative
electrode potential, it will be the half-reaction to occur at the anode.
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Practice Problem: Predicting the
Products of Electrolysis Reactions (3 of 4)
Similarly, write half-reactions for the two possible reduction half-reactions that
might occur at the cathode: the reduction of Li+ and the reduction of water.
Since the reduction of water has the more positive electrode potential (even
when considering overvoltage, which would raise the necessary voltage by
about 0.4 – 0.6 V), it is the half-reaction that occurs at the cathode.
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Practice Problem: Predicting the
Products of Electrolysis Reactions (4 of 4)
For Practice 19.9
Predict the half-reactions occurring at the anode and the
cathode for the electrolysis of aqueous Na2SO4 .
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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
• Conceptual plan:
– time (in seconds) ® coulombs ® moles of electrons ®
moles of metal ® grams of metal
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Conceptual Connection 19.10 (1 of 2)
® Ag(s ).
Silver plating uses the reaction Ag+ (aq ) + e - ¾¾
How many moles of electrons must pass through an
electrolytic cell for silver plating in order to plate 3 moles of
Ag?
(a) 1 mol e(b) 2 mol e(c) 3 mol e(d) 4 mol e-
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Conceptual Connection 19.10 (2 of 2)
® Ag(s ).
Silver plating uses the reaction Ag+ (aq ) + e - ¾¾
How many moles of electrons must pass through an
electrolytic cell for silver plating in order to plate 3 moles of
Ag?
(a) 1 mol e(b) 2 mol e(c) 3 mol e (d) 4 mol e-
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Practice Problem: Stoichiometry of
Electrolysis (1 of 3)
Example 19.10 Stoichiometry of Electrolysis
Gold can be plated out of a solution containing
Au3+ according to the half-reaction:
Au3+ (aq ) + 3 e - ® Au(s )
What mass of gold (in grams) is plated by a 25-minute flow of 5.5 A current?
Sort You are given the half-reaction for the plating of gold, which shows the stoichiometric
relationship between moles of electrons and moles of gold. You are also given the current
and duration. You must find the mass of gold that is deposited in that time.
Given 3mol e - : 1 mol Au
5.5amps
25 min
Find g Au
Strategize You need to find the amount of gold, which is related stoichiometrically to the
number of electrons that have flowed through the cell. Begin with time in minutes and
convert to seconds. Then, since current is a measure of charge per unit time, use the
given current and the time to find the number of coulombs.
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Practice Problem: Stoichiometry of
Electrolysis (2 of 3)
You can use Faraday’s constant to calculate the number of moles of electrons and the
stoichiometry of the reaction to find the number of moles of gold. Finally, use the molar
mass of gold to convert to mass of gold.
Conceptual plan
Solve Follow the conceptual plan to solve the problem, canceling units to arrive at the
mass of gold.
Solution
1 mol Au 196.97 g Au
60 s 5.5 C 1 mol e
25 min ´
´
´
´
´
= 5.6 g Au
1s
1 min
96,485 C 3 mol e 1 mol Au
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Practice Problem: Stoichiometry of
Electrolysis (3 of 3)
Check The answer has the correct units (g Au). The magnitude of the answer
is reasonable if you consider that 10 amps of current for 1 hour is the
equivalent of about 1/ 3 mol of electrons (check for yourself), which would
produce 1/ 9 mol (or about 20 g) of gold.
For Practice 19.10
Silver can be plated out of a solution containing Ag+ according to the
half-reaction:
Ag+ (aq ) + e - ® Ag(s )
How much time (in minutes) does it take to plate 12 g of silver using a current
of 3.0 A?
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Corrosion: Undesirable Redox Reaction
• Corrosion is the spontaneous
oxidation of a metal by chemicals in
the environment.
– Mainly O2
• Because many materials used are
active metals, corrosion can be a
very big problem.
– Metals are often used for their
strength and malleability, but
these properties are lost when
the metal corrodes.
– For many metals, the product
of corrosion does not adhere to
the metal, and as it flakes off
more metal can corrode.
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Reduction of O2
o sub 2
• O2 is very easy to reduce in moist conditions.
O2 (g ) + 2 H2O(l ) + 4 e - ® 4 OH- (aq )
E ° = 0.40 V
• O2 is even easier to reduce under acidic conditions.
O2 (g ) + 4 H+ (aq ) + 4 e - ® 2 H2O(l )
E ° = 1.23 V
• Because the reduction of most metal ions lies below O2 on
the table of standard reduction potentials, the oxidation of
those metals by O2 is spontaneous.
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Rusting (1 of 2)
• At the anodic regions, Fe(s) is oxidized to Fe2+ .
• The electrons travel through the metal to a cathodic region where O2 is
reduced.
– In acidic solution from gases dissolved in the moisture
• The Fe2+ ions migrate through the moisture to the cathodic region, where
they are further oxidized to Fe3+ , which combines with the oxygen and water
to form rust.
• Rust is hydrated iron(III) oxide, Fe2O3·nH2O.
The exact composition depends on the conditions.
– Moisture must be present, where water is a reactant.
It is required for ion flow between cathodic and anodic regions.
• Electrolytes promote rusting, because they enhance current flow.
• Acids promote rusting by lowering pH which will lower E°red of O2 .
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Rusting (2 of 2)
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Preventing Corrosion (1 of 2)
• 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, such as Al, form an oxide that strongly attaches to the
metal surface, preventing the rest of the metal from corroding.
• Another method to protect one metal is to attach it to a more reactive metal
that is cheap.
– Sacrificial electrode
Galvanized nails
– If a metal more active than iron, such as magnesium or aluminum, is in
electrical contact with iron, the metal (i.e. magnesium or aluminum)
rather than the iron, will be oxidized.
– This is the principle that underlies the use of sacrificial electrodes to
prevent the corrosion of iron.
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Preventing Corrosion (2 of 2)
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Conceptual Connection 19.11 (1 of 2)
Which of these metals does not act as a sacrificial
electrode for iron?
(a) Cu
(b) Mg
(c) Zn
(d) Mn
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Conceptual Connection 19.11 (2 of 2)
Which of these metals does not act as a sacrificial
electrode for iron?
(a) Cu
(b) Mg
(c) Zn
(d) Mn
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