Electrochemistry (Ch 17) - AP Chemistry

4/1/2011
ELECTROCHEMISTRY
Chapter 17
Ch 16 QUIZ due MON
LAB 17 Prelab due THURS
Ch 17 QUIZ from WED 4/13 – MON 4/18
LAB 17 due THURS 4/14
AP CHEMISTRY Practice Exam on SAT 4/16
EXAM VII (Ch 16 & 17) on WED 4/20
CH 17 – ELECTROCHEMISTRY
• Read pp 790-829
Suggested Problems (pp 830-834)
• Redox Review (p830):
13 – 16
• 25, 31, 33, 37, 43,
45, 51, 55, 57, 61,
65, 71, 73, 75
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ELECTROCHEMISTRY
• Many applications for electrochemistry
– Batteries
– Production of important industrial materials
– Electrodes (pH probes, blood/heart monitors)
• Electrochemistry is the study of the interchange of chemical and
electrical energy by oxidation-reduction reactions
– Focus on Two Opposite Processes:
• Generation of electric current by a spontaneous chemical reaction
• Use of electric current to induce a chemical change
GALVANIC CELLS
Section 17.1
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OXIDATION-REDUCTION REVIEW
AKA: “REDOX REDUX”
• An oxidation-reduction reaction involves the transfer of electrons
from the reducing agent to the oxidizing agent
– Oxidation is the loss of electrons (increase in oxidation number)
Fe 2+ Fe 3+
– Reduction is the gain of electrons (decrease in oxidation number)
– “LEO goes GER” or “OIL RIG”
MnO 4- Mn 2+
REDOX REDUX
• Consider the reaction between potassium permanganate and iron(II)
chloride in acidic solution: MnO 4- + Fe 2+ Mn 2+ + Fe 3+
– Reduction half-reaction: 8 H + + MnO 4- + 5 e - Mn 2+ + 4 H 2 O
– Oxidation half-reaction: 5 Fe 2+ 5 Fe 3+ + 5 e -
• Overall:
8 H + + MnO 4- + 5 Fe 2+ Mn 2+ + 5 Fe 3+ + 4 H 2 O
– In the same solution, MnO 4- and Fe 2+ transfer electrons but no useful work is
obtained from the chemical energy (released as heat)
– We can harness this energy (to do useful work) by separating the oxidizing
agent from the reducing agent, requiring that the electrons pass through a
wire (the flow of electrons is known as electric current)
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A GALVANIC CELL
A GALVANIC CELL
• Unfortunately, this set up won’t work.
– Electrons will flow from Fe 2+ to MnO 4- , but charge will build up in each
compartment and stop the flow of electrons
• A simple solution is to connect the solutions so that ions can flow to
keep the net charge in each compartment zero.
– A salt bridge is a U-tube filled with an electrolyte that will allow charges to
flow from one compartment to another to keep the net charge zero
– A porous disk (or cup in LAB 17) can also be used to do the same thing
• A galvanic cell is a device in which chemical energy is changed to
electrical energy
– Uses a spontaneous redox reaction to produce a current that can be used to
do work
– The reverse non-spontaneous process (converting electrical energy into
chemical energy) is called electrolysis (Section 17.7)
• You remember LEO goes GER, but do you remember what color he is
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A GALVANIC CELL
CELL POTENTIAL
• “LEO is a RED CAT”
– The electrode compartment in which reduction occurs is called the cathode
– The electrode compartment in which oxidation occurs is called the anode
• Reduction occurs at the cathode. Oxidation occurs at the anode.
– Electrons are pulled from the anode to the cathode
• The “pull” on the electrons is called the cell potential (E cell ) or the
electromotive force (emf) of the cell.
– Unit of electric potential is the volt (V)
• 1 V = 1 J/C (joule per coulomb)
– Measured with a voltmeter, but frictional heating of
the resisters wastes some of the useful energy of the cell,
so the reading will always measure less than the maximum cell potential
– A potentiometer measures the cell potential under conditions of zero
current with a variable-voltage device powering the reverse reaction
– Modern digital voltmeters draw only a negligible amount of current
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STANDARD REDUCTION POTENTIALS
Section 17.2
STANDARD REDUCTION POTENTIALS
• A galvanic cell’s redox reaction can always be broken down into two halfreactions.
The sum of the potential (E ) of each half-reaction equals the cell
potential (E cell ).
– For example, this cell reaction is:
2 H + (aq) + Zn(s) Zn 2+ (aq) + H 2 (g)
– The anode’s reaction (oxidation) is:
Zn Zn 2+ + 2 e -
– The cathode’s reaction (reduction) is:
Ch 17 - Electrochemistry 11
2 H + 2 e - H 2
• This cathode uses a Pt electrode in contact with 1 M H + bathed in H 2 (g) at 1 atm and is called the
standard hydrogen electrode
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STANDARD REDUCTION POTENTIALS
• The total potential (E cell ) can be measured (0.76 V), but there is no way to
measure the individual electrode potentials.
– To calculate potentials for each half-reaction, the reaction
2 H + 2 e - H 2
• where [H + ] = 1 M and P H2 = 1 atm
– has been assigned a potential of 0V
– E anode can be calculated because
E° cell = E° anode + E° cathode
• As a result, the reaction, ZnZn 2+ + 2 e - has a potential of 0.76 V
• Using the standard hydrogen potential as a reference, all other potentials
can be determined experimentally.
Ch 17 - Electrochemistry 13
STANDARD REDUCTION POTENTIALS
• What is the potential of the cathode
– Anode: Zn Zn 2+ + 2 e -
– Cathode: Cu 2+ + 2 e - Cu
• Since E° cell = E° anode + E° cathode
– E° cat = 1.10 V – 0.76 V = 0.34 V
• By convention, the potentials of
half reactions are given as reduction
potentials and are tabulated in
standard states in a list of standard reduction potentials
Ch 17 - Electrochemistry 14
STANDARD REDUCTION POTENTIALS
STANDARD REDUCTION POTENTIALS
• Combining two half-reactions to obtain a balanced oxidationreduction
reaction often requires two manipulations:
– One of the reduction half-reactions must be reversed (since redox reactions
must involve a substance being oxidized and a substance being reduced).
• The half-reaction with the largest positive potential will run as written (as a
reduction)
• The other half-reaction will be forced to run in reverse (as an oxidation)
• The net potential of the cell will be the difference between the two:
E° cell = E° cathode – E° anode
– Or “change the sign and add” (by changing the sign of the oxidation reaction)
Ch 17 - Electrochemistry 15
STANDARD REDUCTION POTENTIALS
• Combining two half-reactions to obtain a balanced oxidationreduction
reaction often requires two manipulations:
– Since the number of electrons lost must equal the number gained, the halfreactions
must be multiplied by integers as necessary to achieve the
balanced equation. However, the value of E° is not changed when a halfreaction
is multiplied by an integer.
• Since a standard reduction potential is an intensive property, the potential is not
multiplied by the integer.
STANDARD REDUCTION POTENTIALS
• Consider a galvanic cell based on the redox reaction
Fe 3+ (aq) + Cu(s) Cu 2+ (aq) + Fe 2+ (aq)
– The half reactions are:
Fe 3+ + e - Fe 2+ E° = 0.77 V (1)
Cu 2+ + 2 e - Cu E° = 0.34 V (2)
– To balance the cell reaction and calculate the standard cell potential, reaction (2)
must be reversed and multiply reaction (1) by two:
Cu Cu 2+ + 2 e -
– The cell reaction becomes:
-E° = -0.34 V
2 Fe 3+ + 2 e - 2 Fe 2+ E° = 0.77 V
Cu(s) + 2 Fe 3+ (aq) Cu 2+ (aq) + 2 Fe 2+ (aq)
E° cell = 0.43 V
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GALVANIC CELLS
• Consider a galvanic cell based on the reaction
Al 3+ (aq) + Mg(s) Al(s) + Mg 2+ (aq)
– The half-reactions are
• Al 3+ + 3 e - Al
E° = -1.66 V
• Mg 2+ + 2 e - Mg
E° = -2.37 V
– Give the balanced cell reaction and calculate E° for the cell.
• 2 Al 3+ (aq) + 3 Mg(s) 2 Al(s) + 3 Mg 2+ (aq)
• E° cell = 0.71V
LINE NOTATION
• Line notation is handy for representing electrochemical cells.
– The anode components are listed on the left and the cathode components are
listed on the right, separated by double vertical lines (indicating the salt bridge
or porous disk).
• For example, the cell on the previous slide would be:
Mg(s) | Mg 2+ (aq) || Al 3+ (aq) | Al(s)
– Phase difference (boundary) is indicated by a single vertical line
• Notice that the substance constituting the anode is far left and cathode is far right
– Commas are used to separate ionic species in the same solution
Pt(s)|ClO 3- (aq), ClO 4- (aq), H + (aq) ||H + (aq), MnO 4- (aq), Mn 2+ (aq)|Pt(s)
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COMPLETE DESCRIPTION OF A GALVANIC CELL
• NOTE: A cell will always run spontaneously in the direction that
produces a positive cell potential (E° cell )
• A complete description of a galvanic cell includes:
– The cell potential (always positive for a galvanic cell where E° cell = E° cat – E° an )
and the balanced cell reaction.
– The direction of electron flow, obtained by inspecting the half-reactions and
using the direction that gives a positive E cell .
– Designation of the anode and cathode
– The nature of each electrode and the ions present in each compartment. A
chemically inert conductor is required if none of the substances participating in
the half-reaction is a conducting solid.
DESCRIPTION OF A GALVANIC CELL
• Describe completely the galvanic cell based on the following halfreactions
under standard conditions:
Ag + + e - Ag
Fe 3+ + e - Fe 2+
E° = 0.80 V
E° = 0.77 V
– Ag + (aq) + Fe 2+ (aq) Fe 3+ (aq) + Ag(s) E° cell = 0.03 V
– Electrons flow from Fe 2+ to Ag +
– Anode: Fe 2+ compartment Cathode: Ag + compartment
– Pt(s)|Fe 2+ (aq), Fe 3+ (aq)||Ag + (aq)|Ag(s)
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CELL POTENTIAL, ELECTRICAL WORK, AND
FREE ENERGY
Section 17.3
CELL POTENTIAL AND WORK
• The work that can be done by the electrons transferred
through a wire depends on the “push” (thermodynamics)
behind the electrons.
– This electromotive force (emf) is defined in terms of a potential
difference (in volts) between two points in the circuit (1 V = 1 J/C)
– Cell potential (E ) and work (w) have opposite signs:
w max = –qE max
• In any real, spontaneous process some energy is always wasted – the
actual work realized is always less than the calculated maximum
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CELL POTENTIAL AND WORK
• Suppose a certain galvanic cell has a maximum potential (at zero
current) of 2.50 V
– In a particular experiment, 1.33 moles of electrons were passed through this
cell at an average actual potential of 2.10 V. The actual work done is: w = -qE
• 1 mole of electrons has a charge of 96,485 C (called a “faraday”, F = 96,485 C/mol)
– Actual work: w = -(1.33 mol)(96485 C/mol)(2.10 J/C) = -2.69 x 10 5 J
– Maximum work: w max = - (1.33)(96485)(2.50) = -3.21 x 10 5 J
– The efficiency of this cell is:
WORK AND FREE ENERGY
• Since
w max = ∆G
– For a galvanic cell,
w max = -qE max = ∆G
– Since q = nF
– Then,
∆G = -qE max = -nFE max
– Or
∆G = -nFE
– For standard conditions: ∆G° = -nFE°
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∆G = -nFE°
• This equation states that the maximum cell potential is directly
related to the free energy difference between the reactants and the
products in the cell.
– Provides an experimental means to obtain ∆G (as you will do in LAB)
– Confirms that a galvanic cell will run in the direction that gives a positive
value for E cell (positive E cell corresponds to a negative ∆G )
CALCULATING ∆G°
• Using the Standard Reduction Potentials Table, calculate ∆G° for the
reaction:
• Is this reaction spontaneous
Cu 2+ (aq) + Fe(s) Cu(s) + Fe 2+ (aq)
– E cell = 0.78 V
– ∆G = -1.5 x 10 5 J (SPONTANEOUS)
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PREDICTING SPONTANEITY
• Using the Standard Reduction Potentials Table, predict whether 1 M
HNO 3 will dissolve gold metal to form a 1 M Au 3+ solution.
DEPENDENCE OF CELL POTENTIAL ON
CONCENTRATION
Section 17.4
– E° cell = -0.54 V (not spontaneous under standard conditions)
– General rule: Oxidizing agents (reactants) higher on the list
will oxidize reducing agents (products) lower on the list
• Ex: H + will react with Fe but not Ag
• The best REDUCING AGENT is Li, the best OXIDIZING AGENT is F 2
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CONCENTRATION DEPENDENCE
• For the reaction,
Cu(s) + 2 Ce 4+ (aq) Cu 2+ (aq) + 2 Ce 3+ (aq)
has a potential of 1.36 V under standard conditions (1 M
concentrations all)
– What if [Ce 4+ ] is greater than 1 M
• Le Châtelier's principle predicts that the forward reaction will be favored,
increasing the cell potential.
– What if [Cu 2+ ] or [Ce 3+ ] is increased
• Cell potential is decreased
THE EFFECTS OF CONCENTRATION ON E
• For the cell reaction,
2 Al(s) + 3 Mn 2+ (aq) 2 Al 3+ (aq) + 3 Mn(s) E = 0.48 V
predict whether E cell is larger or smaller than E cell for the
following cases
– [Al 3+ ] = 2.0 M, [Mn 2+ ] = 1.0 M
• E cell < 0.48 V
– [Al 3+ ] = 1.0 M, [Mn 2+ ] = 3.0 M
• E cell > 0.48 V
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CONCENTRATION CELLS
• A galvanic cell in which both
compartments have the same
components but at different
concentrations is called a
concentration cell (typically
with small voltages)
– What is the potential of this cell
and what is the direction of
electron flow
– The relevant half-reaction is,
Ag + + e - Ag
E = 0.80 V
CONCENTRATION CELLS
• The electrons will flow in the
direction that will equalize the
concentrations of Ag + in the
two compartments
– Accomplished by transferring
electrons from the 0.1 M Ag +
compartment to the 1 M Ag +
compartment
– This produces more Ag + in the
left compartment and consumes
Ag + in the right compartment
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Ch 17 - Electrochemistry 34
CONCENTRATION CELLS
• Determine the direction of electron flow and designate the anode and
cathode for the cell represented below.
THE NERNST EQUATION
• The concentration dependence of cell potential results directly from the
dependence of free energy on concentration
– Recall, ∆G = ∆G + RT lnQ where Q is the reaction quotient
– Also recall, ∆G = -nFE and ∆G = -nFE
– So, -nFE = -nFE + RT lnQ
• Solving for cell potential,
– Fe 2+ + 2 e - Fe
– Transferring electrons from left (anode) to right (cathode)
– This equation gives the relationship between the cell potential and the concentrations
of the cell components is commonly called the Nernst equation (German chemist
Walther Hermann Nernst, 1864-1941)
– At 25C,
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CALCULATING E FOR NON-STANDARD CELLS
• A galvanic cell based on the reaction,
2 Al(s) + 3 Mn 2+ (aq) 2 Al 3+ (aq) + 3 Mn(s) E cell = 0.48 V
– If [Mn 2+ ] = 0.50 M and [Al 3+ ] = 1.50 M, at 25C
• The cell potential slightly decreases, as predicted by Le Chatelier’s principle
THE NERNST EQUATION
• The potential calculated from the Nernst equation is the maximum
potential before any current flow has occurred
– As the electrons flow from the anode to the cathode, the concentrations will
change and E cell will change as a result
• The cell will spontaneously discharge until it reaches equilibrium
– Where Q = K and E cell = 0
– A “dead” battery is a galvanic cell which has reached equilibrium (∆G = 0)
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THE NERNST EQUATION
• Describe the cell based on the following half reactions:
VO 2+ + 2 H + + e - VO 2+ + H 2 O
Zn 2+ + 2 e - Zn
E = 1.00 V
E = -0.76 V
where T = 25C, [VO 2+ ] = 2.0 M, [H + ] = 0.50 M, [VO 2+ ] = 1.0 x 10 -2 M,
[Zn 2+ ] = 1.0 x 10 -1 M
– 2 VO 2+ (aq) + 4 H + (aq) + Zn(s) 2 VO 2+ (aq) + 2 H 2 O(l) + Zn 2+ (aq)
– E cell = 1.76 V
– E = 1.76 – -0.13 = 1.89 V
ION-SELECTIVE ELECTRODES
• Electrodes that are
sensitive to the
concentration of a
particular ion are called
ion-selective
electrodes
– Example: pH meter
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CALCULATION OF EQUILIBRIUM CONSTANTS
FOR REDOX REACTIONS
• For a cell at equilibrium,
–E cell = 0 and Q = K
• At 25C, the Nernst equation, with E = 0, can be
rearranged to
EQUILIBRIUM CONSTANTS FROM CELL POTENTIALS
• For the oxidation-reduction reaction
S 4 O
2-
6 (aq) + Cr 2+ (aq) Cr 3+ (aq) + S 2 O
2-
3 (aq)
– the appropriate half-reactions are
S 4 O
2-
6 + 2 e - 2 S 2 O
2-
3 E = 0.17 V
Cr 3+ + e - Cr 2+ E = -0.50 V
Balance the redox reaction and calculate E and K (at 25C)
– 2 Cr 2+ (aq) + S 4 O
2-
6 (aq) 2 Cr 3+ (aq) + 2 S 2 O
2-
3 (aq)
– E = 0.67 V
Ch 17 - Electrochemistry 41
– K = 10 22.6 = 4 x 10 22 Ch 17 - Electrochemistry 42
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BATTERIES
Section 17.5
BATTERIES
• A battery is a galvanic cell or, more commonly, a group of galvanic
cells connected in series, where the potentials of the individual cells
add to give the total battery potential
– Direct current (DC)
– Common types:
• lead-acid
• acid dry cell
• alkaline dry cell
• nickel-cadmium
• fuel cell
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LEAD STORAGE BATTERY
• Lead storage batteries (lead-acid batteries) are commonly used in cars,
boats, and scooters because they are durable, have long life, and can
withstand temperatures between -30F and 120F
– Anode: lead
– Cathode: lead dioxide
• Overall:
Pb + HSO 4- PbSO 4 + H + + 2 e -
PbO 2 + HSO 4- + 3 H + + 2 e - PbSO 4 + 2 H 2 O
Pb(s) + PbO 2 (s) + 2 H + (aq) + 2 HSO 4- (aq) 2 PbSO 4 (s) + 2 H 2 O(l)
LEAD STORAGE BATTERY
• Six 2V cells provide
the 12 V potential of
a car battery
• The car’s alternator
charges the battery
by forcing the reverse
reaction to occur
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ACID DRY CELL
• Anode: zinc Zn Zn 2+ + 2 e -
• Cathode: carbon rod
• Total potential: 1.5 V
2 NH 4+ + 2 MnO 2 + 2 e - Mn 2 O 3 + 2 NH 3 + H 2 O
ALKALINE DRY CELL
• Same as the acid dry cell, but with KOH or NaOH instead of
NH 4 Cl in the cathode
– Alkaline batteries last longer because the zinc anode corrodes
more slowly under basic conditions
• Anode: Zn Zn 2+ + 2 e -
• Cathode: 2 MnO 2 + H 2 O + 2 e - Mn 2 O 3 + 2 OH -
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NICKEL-CADMIUM BATTERY
• Can be recharged (like lead-acid battery) because
the products stick to the electrodes
–Anode: Cd + 2 OH - Cd(OH) 2 + 2 e -
–Cathode: NiO 2 + 2 H 2 O + 2 e - Ni(OH) 2 + 2 OH -
OTHER BATTERIES
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FUEL CELLS
• A fuel cell is a galvanic cell for which the reactants are
continuously supplied (to “fuel” the spontaneous
electrochemical reaction)
– Hydrogen fuel cells
• Used in the space shuttle to produce water
• Automobiles
• Anode: 2 H 2 + 4 OH - 4 H 2 O + 4 e -
• Cathode: 4 e - + O 2 + 2 H 2 O 4 OH -
• Overall: 2 H 2 + O 2 2 H 2 O
CORROSION
Section 17.6
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CORROSION
• Corrosion occurs when metals oxidize to form the ores in
which they are naturally found
– Because common metals (except Au) all have standard reduction
potentials less positive than O 2 (g), oxygen can oxidize them.
– Recall that any oxidizing agent (reactant) higher on the list can
oxidize and reducing agent (product) lower on the list
• Cu, Au, Ag, and Pt are relatively difficult to oxidize and are sometimes
called the “noble metals”
• Be sure to read section 17.6 to learn about corrosion and
corrosion prevention
ELECTROLYSIS
Section 17.7
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ELECTROLYSIS
• A galvanic (aka voltaic) cell produces current when a
spontaneous redox reaction occurs
• An electrolytic cell uses electrical energy to produce
a non-spontaneous chemical change
– Electrolysis occurs by forcing a current through a cell to
produce a chemical change for which the cell potential (E)
is negative
– Applications: charging batteries, producing Al metal,
chrome plating
Ch 17 - Electrochemistry 55
GALVANIC VS. ELECTROLYTIC CELLS
Galvanic Cell
• Anode: Zn Zn 2+ + 2 e - (-E = +0.76 V)
• Cathode: Cu 2+ + 2 e - Cu (E = 0.34 V)
Zn + Cu 2+ Zn 2+ + Cu (E cell = 1.10 V)
– (E is positive, so ∆G is negative,
spontaneous)
– Electrons flow from Zn(s) to Cu(s)
Electrolytic Cell
• If an external power source
provides a potential greater than
1.10 V in the opposite direction,
then the reverse reaction occurs
Zn 2+ + Cu Zn + Cu 2+
– Electrons flow from Cu(s) to Zn(s)
• Cu/Cu 2+ half-cell becomes the
anode
• Zn/Zn 2+ half-cell becomes the
cathode
Ch 17 - Electrochemistry 56
GALVANIC VS. ELECTROLYTIC CELLS
ELECTROCHEMICAL STOICHIOMETRY
• How much chemical change occurs with the flow a given current for a
specified time
• Suppose we need to determine the mass of copper than is plated out
(deposited on the electrode by the reduction of metal ions) when a current
of 10.0 amps (an amp, or ampere, A, is 1 C/s) is passed for 30.0 min through
a solution containing Cu 2+ Cu 2+ + 2 e - Cu
• Problem Solving Steps
Current & time quantity of charge (C) moles of electrons moles of copper grams of copper
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ELECTROPLATING
• How long must a current of 5.00 A be applied to a
solution of Ag + to produce 10.5 g silver metal
ELECTROLYSIS OF WATER
– 31.3 min
Ch 17 - Electrochemistry 59
the reverse of the Hydrogen Fuel Cell reaction
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ELECTROLYSIS OF MIXTURES OF IONS
• By gradually increasing the applied voltage to a mixture of
ions, one can do a sort of “selective-electroplating” on to the
cathode
– Consider a mixture of Cu 2+ , Ag + , and Zn 2+
Ag + + e - Ag
Cu 2+ + 2 e - Cu
Zn 2+ + 2 e - Zn
E = 0.80 V
E = 0.34 V
E = -0.76 V
• The reduction of Ag + occurs most easily, so Ag(s) forms first,
then Cu(s), and finally Zn(s)
Ch 17 - Electrochemistry 61
RELATIVE OXIDIZING ABILITIES
• An acidic solution contains the ions Ce 4+ , VO 2+ , and Fe 3+ .
Using the standard reduction potential table, give the order
of oxidizing ability of these species and predict which one
will be reduced at the cathode of an electrolytic cell at the
lowest voltage.
– Ce 4+ > VO 2+ > Fe 3+
– Ce 4+ is reduced first
Ch 17 - Electrochemistry 62
COMMERCIAL ELECTROLYTIC PROCESSES
Section 17.8
COMMERCIAL ELECTROLYTIC PROCESSES
• Forcing a non-spontaneous redox reaction to occur has
many useful applications
– Production of aluminum – getting Al from ores
– Electrorefining of other metals – getting Cu, Zn, & Fe from ores
– Metal plating – silver plating, chrome plating
– Electrolysis of sodium chloride – producing Cl 2 and NaOH
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