2/18/2015 1 George Mason University General Chemistry 212 Chapter 21 Electrochemistry Acknowledgements Course Text: Chemistry: the Molecular Nature of Matter and Course Text: Chemistry: the Molecular Nature of Matter and Change, 7 Change, 7 th th edition, 2011, McGraw-Hill edition, 2011, McGraw-Hill Martin S. Silberberg & Patricia Amateis Martin S. Silberberg & Patricia Amateis The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material. Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor.
Transcript
2/18/2015 1
George Mason UniversityGeneral Chemistry 212
Chapter 21Electrochemistry
Acknowledgements
Course Text: Chemistry: the Molecular Nature of Matter and Course Text: Chemistry: the Molecular Nature of Matter and Change, 7Change, 7thth edition, 2011, McGraw-Hill edition, 2011, McGraw-HillMartin S. Silberberg & Patricia AmateisMartin S. Silberberg & Patricia Amateis
The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material.Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor.
2/18/2015 2
Electrochemistry Redox Reactions and Electrochemical Cells
Review of Oxidation Reduction Concepts Half-Reaction Method for Balancing Redox
Reactions Electrochemical Cells
Voltaic Cell: Using Spontaneous Reactions to Generate Electrical Energy Construction and Operation Cell Notation Why Does the Cell Work
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Electrochemistry Cell Potential: Output of a Voltaic Cell
Standard Cell Potentials Strengths of Oxidizing and Reducing
Agents Free Energy and Electrical Work
Standard Cell Potential Effect of Concentration of Ecell Changes in Ecell During Cell Operation Concentration Cells
Electrochemical Processes in Batteries Primary (Nonrechargeable Batteries) Fuel Cells
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Electrochemistry Corrosion: A case of Environmental
Electrochemistry
Corrosion of Iron
Protecting Against Corrosion
Electrolytic Cells: Using electrical energy to drive Nonspontaneous Reactions
Construction and Operation
Predicting Electrolysis Products
Stoichiometry of Electrolytes
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Electrochemistry Electrochemistry
The study of the relationship between chemical change (reactions) and the flow of electrons (electrical work)
Electrochemical Systems
Electrolytic – Work done by absorbing free energy from a source (passage of an electrical current through a solution) to drive a nonspontaneous reaction
Voltaic/Galvanic – Release of free energy from a spontaneous reaction to produce electricity (Batteries)
Reduction (e- gain) Total number of electrons gained by the
atoms/ions of the oxidizing agent always equals the total number of electrons lost by the reducing agent
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Electrochemistry
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Electrochemistry Oxidation Number
A number equal to the magnitude of the charge an atom would have if its shared electrons were held completely by the atom that attracts them more strongly
The oxidation number in a binary ionic compound equals the ionic charge
The oxidation number for each element in a covalent compound (or polyatomic ion) are assigned according to the relative attraction of an atom for electrons
See next slide for a summary of the rules for assigning oxidation numbers
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Electrochemistry
Rules for Assigning an oxidation Number (O.N.)
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Electrochemistry Balancing Redox Reactions
Oxidation Number Method
Half-Reaction Method
The balancing process must insure that:
The number of electrons lost by the
reducing agent equals the number of
electrons gained by the oxidizing agent
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Electrochemistry Oxidation Number Method
Assign oxidation numbers to all elements in the reaction
From changes in oxidation number of given elements, identify oxidized and reduced species
For each element that undergoes a change of oxidation number, compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number change (Draw tie-lines between these atoms)
Multiply one or both these number by appropriate factors to make the electrons lost equal to the electrons gained
Use factors as coefficients in reaction equation
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Practice ProblemBalance equation with Oxidation Number method:
3 3 2 2 2Cu(s) + HNO (aq) Cu(NO ) + NO (g) + H O(l)
3 3 2 2 2Cu(s) + HNO (aq) Cu(NO ) + NO (g) + H O(l)
-2+5 +50 +1 -2 -2-2+2 +4 +1
3 3 2 2 2Cu(s) + HNO (aq) Cu(NO ) + NO (g) + H O(l)
Add OH- to neutralize H+ , balance H2O, and form “basic” solution
- + -4 2 26e + 8H + 2MnO 2MnO + 4H O
2- 2- + -2 2 4 36H O + 3 C O 6CO + 12H + 6e
- 2- 2- +4 2 2 4 2 32MnO + 2H O + 3 C O 2MnO + 6CO + 4H
- 2- - 2- + -4 2 2 4 2 32MnO + 2H O + 3 C O + 4OH 2MnO + 6CO + 4H + 4OH
- 2- - 2-4 2 2 4 2 3 22MnO + 2H O + 3 C O + 4OH 2MnO + 6CO + 4H O
- 2- - 2-4 2 4 2 3 22MnO + 3 C O + 4OH 2MnO + 6CO + 2H O
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Electrochemistry Electrochemical Cells
Voltaic (Galvanic) Cells
Use spontaneous reaction (G < 0) to generate electrical energy
Difference in Chemical Potential energy between higher energy reactants and lower energy products is converted to electrical energy to power electrical devices
Thermodynamically - The system does work on the surroundings
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Electrochemistry Electrochemical Cells
Electrolytic Cells
Uses electrical energy to drive nonspontaneous reaction (G > 0)
Electrical energy from an external power supply converts lower energy reactants to higher energy products
Thermodynamically – The surroundings do work on the system
Examples – Electroplating and recovering metals from ores
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Electrochemistry
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Electrochemistry Electrochemical Cells
Cell notation is used to describe the structure of a voltaic (galvanic) cell
For the Zn/Cu cell, the cell notation is:Zn(s) Zn2+(aq) Cu2+(aq) Cu(s)
= phase boundary (solid Zn vs. Aqueous Zn2+) = salt bridge Anode reaction (oxidation) is left of the salt bridge Cathode reaction (reduction) is right of the salt bridge Half-cell components usually appear in the same order
as in the half-reactions (Zn(s) + 2e- Zn2+). Zinc solid loses 2 e- (oxidized) to produce zinc(II) at
the negative ANODE Copper(II) gains 2e- (reduced) to form copper metal at
positive CATHODE
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Electrochemistry Voltaic (Galvanic) Cells
Zinc metal (Zn) in solution of Cu++ ions
Construction of a Voltaic Cell
The oxidizing agent (Zn) and reducing agent (Cu2+) in the same beaker will not generate electrical energy
Separate the half-reactions by a barrier and connect them via an external circuit (wire)
Set up salt bridge between chambers to maintain neutral charge in electrolyte solutions
2+ -Cu (aq) + 2e Cu(s) [reduction] 2+ -Zn(s) Zn (aq) + 2e [oxidation]
2+ 2+Zn(s) + Cu (aq) Zn + Cu(s)
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Electrochemistry Oxidation Half-Cell
Anode Compartment – Oxidation of Zinc (An Ox) Zinc metal in solution of Zn2+ electrolyte (ZnSO4)
Zn is reactant in oxidation half-reaction
Conducts released electrons (e-) out of its half-cell
Reduction Half-Cell
Cathode Compartment – Reduction of Copper (Red Cat)
Copper bar in solution of Cu2+ electrolyte (CuSO4)
Copper metal is product in reduction half-cell reaction
Conducts electrons into its half-cell
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ElectrochemistryZinc-Copper Voltaic Cell
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Electrochemistry Relative Charges on the Anode/Cathode
electrodes
Electrode charges are determined by the source of the electrons and the direction of electron flow
Zinc atoms are oxidized (lose 2 e-) to form Zn2+ at the anode
Anode – negative charge (e- rich)
Released electrons flow to right toward cathode to be accepted by Cu2+ to form Cu(s)
Cathode – positive charge (e- deficient)
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Electrochemistry Purpose of Salt Bridge
Electrons from oxidation of Zn leave neutral ZnSO4 solution producing net positive charge
Incoming electrons to CuSO4 solution would produce net negative charge in solution as copper ions are reduced to copper metal
Resulting charge imbalance would stop reaction
Salt bridge provides “liquid wire” allowing ions to flow through both compartments completing circuit
Salt bridge constructed of an inverted “U-tube” containing a solution of non-reactingNa+ & SO4
2- ions in a gel
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Electrochemistry Active vs Inactive Electrodes
Active Electrodes Electrodes in Zn/Cu2+ cell are active Zinc & Copper bars are components of
the cell reactions Mass of Zn bar decreases as Zn2+ ions in
cell solution increase Mass of Copper bar increases as Cu2+
ions accept electron to form more copper metal
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Electrochemistry Active vs Inactive Electrodes
Inactive Electrodes In many Redox reactions, one or the
other reactant/product is not capable of serving as an electrode
Inactive electrodes - Graphite or Platinum Can conduct electrons into and out of
half-cells Cannot take part in the half-reactions
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Electrochemistry
Voltaic Cell
with
Inactive Graphite Electrodes
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Practice ProblemA mercury battery, used for hearing aids and electric watches, delivers a constant voltage (1.35 V) for long periods. The half reactions are given below. Which half reaction occurs at the Anode and which occurs at the Cathode? What is the overall cell reaction?
HgO(s) + H2O(l) + 2e- Hg(l) + 2 OH-(aq)
Zn(s) + 2 OH-(aq) Zn(OH)2(s) + 2e-
Ans: Reduction occurs at Cathode (Red Cat)
Hg2+ gains 2 e- (reduced) to form Hg
Oxidation occurs at the Anode (An Ox)
Zn loses 2 e- (oxidized) to form Zn2+
2 2HgO(s) + Zn(s) + H O Zn(OH) + Hg(l)
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Practice ProblemWrite the cell notation for a voltaic cell with the following cell reaction
Anode (oxidation) isrepresented on left sideof Cell notation
Cathode (reduction) isrepresented on right sideof cell notation
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Practice ProblemWrite the cell reaction for the following voltaic cell
Pt|H2(g) | H+(aq) ║ Br2(l) | Br-(aq)|Pt
Note: Platinum (Pt) serves as a reaction site at the anode, but does not participate in the reaction
Ans:
+ - -2H (g) 2H (aq) + 2e lose 2 e Oxidation @ Anode (An Ox)
- - -2Br (l) + 2e 2Br (aq) gain 2 e Reduction @ Cathode (Red Cat)
+ -2 2Pt H (g) + Br (l) 2H (aq) + 2Br (aq) PtI I
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Electrochemistry Cell Potential
The movement of electrons is analogous to the pumping of water from one point to another
Water moves from a point of high pressure to a point of lower pressure. Thus, a pressure difference is required
The work expended in moving the water through a pipe depends on the volume of water and the pressure difference
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Electrochemistry Cell Potential
Movement of Electrons
An electric charge moves from a point of high electrical potential (high electrical pressure) to one of lower electrical potential
The work expended in moving the electrical charge through a conductor depends on the potential difference and the amount of charge
cell
Work(w) = Potential difference (E) Charge
w = E Charge
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Electrochemistry Cell Potential
Purpose of a voltaic cell is to convert the free energy of a spontaneous reaction into the kinetic energy of electrons moving through an external circuit (electrical energy)
Electrical energy is proportional to the difference in the electrical potential between the two cell electrodes
Cell Potential
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Electrochemistry Cell Potential
Positive Cell Potential – Electrons flow spontaneously from the negative electrode (Anode) to the positive electrode (Cathode)
Negative cell potential is associated with a “nonspontaneous” cell reaction
Cell potential for a cell reaction at equilibrium would be “0”
As with Entropy, there is a clear relationship between Ecell , K, and G
cellE < 0 for a nonspontaneous process
cellE = 0 for an equilibrium process
cellE > 0 for a spontaneous process
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Electrochemistry Units of Cell Potential
The SI (metric) unit of electrical charge is the:
Coulomb (C) The SI (metric) unit of current is the:
Ampere (A)
The SI (metric) unit of electrical potential is the:
“Volt (V)” By definition, the energy released by a potential
difference of one volt moving between the anode and cathode of a voltaic cell releases 1 joule of work per coulomb of charge
1 coulomb1 ampere = 1A = 1 C / s
second
1 J 1 J1 volt = 1 C =
C V
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Electrochemistry The charge (F) that flows through a cell equals
the number of moles of electrons (n) transferred times the charge of 1 mol of electrons
4
- -
J JF = 96,485 × = 9.6×10 ×
V •mol e V •mol e
-
-
chargeCharge = moles of e
mol e
Charge = nF
-
96,485 C JF = C - Coulomb , SI unit of charge
mol e V
-Moles e = n
- -
Charge 96,485 C = F = Faraday Constant =
Mol e mol e
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Electrochemistry The total amount of a substance undergoing a
ReDox reaction in a cell depends on the total charge flowing through the cell, which is a function of the current (amperes) and the time (seconds)
The ratio of the total charge and the charge per mole of electrons determines the number moles of substance that will react
Thus,
- -
-
-
Charge (C)Moles of substance =
moles e (n) Charge / moles e (F)
Current (i)× Time (t) amperes seconds Moles of substance = =
n × F moles e 96,485 Coulombs×
moles substance mole e
Moles of substai × t
nce = n = total moles electronsn × F
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ElectrochemistryHow long in seconds would it take to deposit 38.93 g of Pb from an aqueous solution of Pb2+ with a current of 1.679 A?
- -
-
-
i t Current (Amperes) × Time (seconds)Moles = =
n F (moles e ) × (Faradays (Coulombs / mole e ))
Moles mole e Faradays Time =
Current
1 mol Pb 2 mol e39.93 g Pb
207.2 g mo Time =
-
4
96,485 C
l Pb mol e
1.679 A
Time = 2.215×10 sec
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Electrochemistry Standard Cell Potential
Eocell – The potential measured at a
specific temperature (298 K) with no current flowing and all concentrations in their “Standard States”
1 atm for gases
1 M for solutions
Pure solids for electrodes
2+ 2+ ocellZn(s) + Cu (aq; 1M) Zn (aq; 1M) + Cu(s) E = 1.10V
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Electrochemistry Standard Electrode Half-Cell Potentials
Eohalf-cell – Potential associated with a given half-
cell reaction (electrode compartment) when all components are in “Standard States”
Standard Electrode Potential for a half-cell reaction, whether anode (oxidation) or cathode (reduction) is written and presented in Appendix D as a “reduction”
Ex.
would be written in the table as:
2+ -Cu (aq) + 2e Cu(s) [reduction - cathode] 2+ -Zn(s) Zn (aq) + 2e [oxidation - anode]
2+ - o oCopper cathodeCu (aq) + 2e Cu(s) E (E ) [reduction]
2+ - o oZinc anodeZn (aq) + 2e Zn(s) E (E ) [reduction]
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Electrochemistry Standard Electrode Half-Cell Potentials
Electrons flow spontaneously from Anode (negative) to Cathode (positive)
Cathode must have a more “Positive” Eohalf-cell than
the Anode For a “positive” Eo
cell ; i.e., a spontaneous reaction
The standard cell potential is the difference between the standard electrode potential of the “Cathode” (reduction) half-cell and the standard electrode potential of the “Anode” (oxidation) half-cell
Standard half-cell potentials are “intensive” properties, thus their values do NOT have to be adjusted for stoichiometry (# of moles)
o o ocell cathode (reduction) anode (oxidation)E = (E - E ) > 0
o o ocell copper ZincE = E - E
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Practice ProblemWrite out the overall equation for the cell reaction and determine the standard cell potential for the following galvanic cell
[Eo (Ag+/Ag) = 0.80 V; Eo (Ni2+/Ni) = - 0.26 V]
+ -2Ag (aq) + 2e 2Ag(s) [reduction (cathode)] 2+ -Ni(s) Ni (aq) + 2e [oxidation (anode)]
+ 2+Ni(s) + 2Ag (aq) Ni + 2Ag(s)
2+ +Ni(s) Ni (aq) Ag (aq) Ag(s)I I
+ -2Ag (aq) + 2e 2Ag(s) [reduction (cathode)] 2+ -Ni (aq) + 2e Ni(s) [reduction (anode)]
o o ocell Silver(Red) Nickel(Ox)E = E - E
ocellE = 0.80 - (-0.26) = 1.06 V
Write both reactions in "reduction" form
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Electrochemistry The Standard Hydrogen Electrode
Half-cell potentials are not absolute quantities
The values found in tables are determined relative to a “Standard”
The Standard Electrode potential is defined as zero
(Eoreference) = 0.00
The “standard reference half-cell” is a standard
“Hydrogen” electrode
Specially prepared Platinum electrode immersed in a1 M aqueous solution of a strong acid through which H2 gas at 1 atm is bubbled + - o
2 reference2H (aq; 1 M) + 2e H (g); 1 atm) E = 0.00V
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Electrochemistry Reference Half-Cell and Unknown Half-Cell
The “Standard” electrode can act as either the “Anode” or the “Cathode”
Oxidation of H2 (lose e-) at anode half-cell and reduction of unknown at cathode half-cell
Reduction of H+ (gain e-) at cathode half-cell and oxidation of unknown at anode half-cell
o o o o ocell cathode anode unknown referenceE = E - E E - E
o o ocell unknown unknownE = E - 0.00 V = E
o o o o ocell cathode anode reference unknownE = E - E E - E
o o ocell unknown unknownE = 0.00 V - E = - E
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Practice ProblemDetermine the standard electrode potential, Eo
zinc, using a voltaic cell consisting of the Zn/Zn2+ half-reaction and the H+/H2 half-reaction.
Eocell = + 0.76 V
Ans: Zinc is being oxidized (loses 2e-) producing electrons at the negative anode; H+ gains e- at
positive cathode+ - o
2 reference2H (aq) + 2e H (g) E = 0.00 V 2+ - o
zincZn(s) Zn (aq) + 2e E = ? V + 2+ o
2Zn(s) + 2H (aq) Zn (aq) + H (g) E cell = 0.76V
o o o o ocell cathode anode reference ZincE = E - E E - E
o o oZinc reference cellE = E - E = 0.00 V - 0.76 V = - 0.76 V
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Electrochemistry Relative Strength of Oxidizing and Reducing Agents
The more positive the Eo value, the more readily the reaction occurs Strength of Oxidizing Agents – Cu2+ > H+ > Zn2+
Strength of Reducing Agents – Zn > H2 > Cu
Oxidizing agents decrease in strength as the value of Eo decreases, while the strength of the reducing agents increases as the value of Eo decreases
Cu2+ is the stronger Oxidizing agent
Zn metal (not the ion) is the stronger Reducing agent
2+ - oCu (aq) + 2e = Cu(s) E = 0.34 V
+ - o22H (aq) + 2e = H (g) E = 0.00 V
2+ - oZn (aq) + 2e = Zn(s) E = - 0.76 V
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Electrochemistry Table of Standard Electrode Potentials
(The emf Series)
All Values are relative to the “standard hydrogen (reference) electrode
All reactions are written as “reductions”
Selected Standard Electrode Potentials (298oC)
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Electrochemistry EMF Series
All Values are relative to the “standard hydrogen (reference) electrode
All reactions are written as “reductions” F2 is the strongest oxidizing agent (high, positive Eo)
Fluorine is very electronegative with a high ionization potential (easily reduced by gaining electrons)
By gaining e- it forms the weakest reducing agent, F- , which is very reluctant to lose electrons)
Li metal is strongest reducing agent (low, more negative Eo) Lithium has a Low ionization energy and is easily
oxidized by losing electrons By losing e- , Lithium forms the weakest oxidizing
agent, Li+
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Electrochemistry Similarities – Acid/Base vs Redox
Acid Strength vs Base Strength using Ka & Kb values Redox (Oxidizing agent vs Reducing agent) using
Standard Electrode Potential (Eo) values Appendix D (Table of Standard Electrode Potentials)
The stronger oxidizing agent (species on left side of table) has a half-reaction with a larger more positive (less negative) Eo than a species lower in the list
The stronger reducing agent (species on the right side of table) has a half-reaction with a smaller (less positive) Eo value than a species higher in the list
A spontaneous reaction between a metal, acting as a reducing agent by losing electrons (oxidized) to be gained by the ion of another element (reduced) would occur if the Eo potential of the metal is less than that of the metal ion.
between an oxidizing agent and any reducing agent that lies below it in the table
Zn(s) (reducing agent (Eo = -0.76 V) will react spontaneously with Cu2+ (oxidizing agent) (Eo = +0.34 V), which lies above Zn in the table
The Reduction of Cu2+ to Cu will occur at the Cathode
The Oxidation of Zn to Zn2+ will occur at the Anode
A spontaneous reaction will occur when the Eo of the reaction at the Cathode (reduction) minus the Eo of the reaction at the Anode (oxidation) is > 0 (Positive)
o o ocell cathode (reduction) anode (oxidation)E = (E - E ) > 0
Combing half-cell reactions to form a “Net” reaction In the Standard Electrode Potential table
(Appendix D), both reactions are written as “reductions” (e- gain)
One of the reactions will have to be reversed so that the applicable reactants are on the left side of the net equation and the products on the right side
The sign of Eo for the reversed reaction need not be reversed
Which reaction to reverse?Scenario #1 - Net reaction is supplied
Balanced equation clearly indicates which species is oxidized and which species is reduced
Consult table to see which half-cell reaction was reversed (Con’t)
Combing half-cell reactions to form a “Net” reactionScenario #2 - Net reaction to be determined
from half-cell reactions (con’t) Since -0.14 < +0.80, the Sn2+ + 2 e-
reaction must be reversed (converted to the Anode Oxidation term)
Balance equations to account for reaction coefficients (e- lost = e- gained)
+ - o2Ag (aq) + 2e = 2Ag(s) E = + 0.80 V (Cathode Reduction)2+ - oSn(s) = Sn (aq) + 2e E = - 0.14 V (Anode Oxidaton)
+ 2+Sn(s) + 2Ag (aq) Sn (aq) + 2Ag(s) (overall reaction) o o o
cell cathode reduction(Ag) anode oxidation(Sn)E = E - E
ocellE = 0.80 - (-0.14) = + 0.94 V (Spontaneous)
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Practice ProblemConsult the table of standard electrode potentials in your textbook in order to decide which one of the following reagents is capable of reducing I2(s) to I-
(aq, 1 M)
a. Br-(aq)
b. Ag(s)
c. Sn(s)
d. Zn2+ (aq, 1 M)
e. Sn4+ (aq,1 M)
Br2(l) + 2e- 2Br-(aq) +1.07V
Ag+(aq) + e- Ag(s) +0.80V
Sn2+(aq) + 2e- Sn(s) -0.14V
Zn2+ (aq) + 2e- Zn(s) -0.76V
Sn4+ (aq) + 2e- Sn2+ (aq) +0.13V
I2(s) + 2e- 2I-(aq) +0.53VReduction Form
(con’t)
2/18/2015 59
Practice Problem (con’t)The I2 is being reduced – gaining electrons – at Cathode (Red Cat)
The other solution must be capable of being oxidized at Anode (AnOx)Only answers a, b, c reflect oxidizable elements, i.e., can lose electrons
a. The Br- solution undergoes oxidation (lose e-) at the Anode Thus
b. The Silver (Ag(s) undergoes oxidation (loses e-) to form Ag+
Thus:
o o o ocell cathode anodeE = E - E + 0.53 - (+1.07) = - 0.54E
o -cell 2Neg E Bromide (Br ) will not reduce I
o o o ocell cathode anodeE = E - E + 0.53 - (+0.80) = - 0.27
ocell 2Neg E Ag(s) will not reduce I
(con’t)
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Practice Problem (con’t)c. The Sn(s) will undergo oxidation to form Sn2+
d. The Zn2+ solution is already oxidized (will not lose any more e-)Thus:
e. The Sn4+ solution is already oxidized Thus:
2+2Zn will not reduce I
42Sn will not reduce I
o o o ocell cathode anodeE = E - E + 0.53 - (-0.14) = + 0.67
ocell 2Pos E Sn(s) will reduce I
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Practice Problem Combine the following 3 half-cell reactions into 3
balanced spontaneous reactions; Calculate Eocell for
each; rank the relative strengths of the oxidizing & reducing agents
(A)
(B)(C)
1. Combine & Balance half-reactions A & B
Reverse B Eo in (A) is more positive than Eo in (B)
(A – cathode reduction; B – anode oxidation)
(A)
(B)
- + - o3 2NO (aq) + 4H (aq) + 3 e NO(g) + 2H O(l) E = 0.96 V
+ - + o2 2 5N (g) + 5H + 4e N H E = - 0.23 V
+ - 2+ o o2 2MnO (s) + 4H (aq) + 2e Mn (aq) + 2H (l) E = 1.23V
- + - o3 24NO (aq) + 16H (aq) + 12 e 4NO(g) + 8H O(l) E = 0.96 V
+ - 2+ o2 22MnO (s) + 8H (aq) + 4e 2Mn (aq) + 4H o(l) E = 1.23 V
+ + - o2 5 2N H N (g) + 5H + 4e E = -0.23 V
+ 2+2 5 2 2 2N H (aq) + 2MnO (s) + 3H (aq) N (g) + 2Mn (aq) + 4H O(l)
ocellE (B + C) = 1.23 V - (-0.23) = 1.46 V Con’t
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Practice Problem (con’t)Overall Ranking of Oxidizing & Reducing Agents
Rank Oxidizing & Reducing Agents Within Each Equation
Equation 1 (A+B): Oxidizing Agent – NO3- > N2
Reducing Agent – N2H5
+ > NO
Equation 2 (C+A): Oxidizing Agent – MnO2 > NO3-
Reducing Agent – NO > Mn+2
Equation 3 (C+B): Oxidizing Agent – MnO2 > N2
Reducing Agent – N2H5
+ > Mn+2
- + - o3 2(A) NO (aq) + 4H (aq) + 3 e NO(g) + 2H O(l) E = 0.96 V
+ - o2 2 5(B) N (g) + 5H + 4e N H E = -0.23 V
+ - 2+ o o2 2(C) MnO (s) + 4H (aq) + 2e Mn (aq) + 2H (l) E = 1.23 V
-2 3 2Oxidizing Agents : MnO > NO > N+ 2+
2 5Reducing Agents : N H > NO > Mn
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Electrochemistry Relative Reactivities of Metals
Metals that displace H2 from acid If the Eo
cell for the reaction of H+ is more positive for metal A than it is for metal B, metal A is a stronger reducing agent than metal B and a more active metal
Metals Li through Pb (includes Fe) in the standard electrode potential list (appendix D) lie below H+ and give positive Eo
cell when reducing H+ to H2, i.e., Hydrogen gas is released
+2 - oFe(s) Fe (aq) + 2e E = - 0.44 V (anode - oxidation)
+ - o
22H (aq) + 2e H (g) E = 0.00 V (cathode - reduction)
+ 2+
2Fe(s) + 2H (aq) H (g) + Fe (aq)o
cellE = 0.00 - (-0.44) = 0.44 V
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Electrochemistry Relative Reactivities of Metals
Metals that cannot displace H2 from acid Metals that lie above the standard hydrogen
reference half-reaction cannot reduce H+ from acids The Eo
cell for the reversed metal half-reaction is negative and the reaction does not occur
The higher the metal in the list, the more negative is its Eo
cell for the reduction of H+ to H2, thus its reducing strength (and reactivity) is less
Thus, Gold (Au3+, Eo = +1.5V) is less active than Silver (Ag+, Eo = +0.8V) and does not release H2 gas
+ - o2 Ag(s) 2Ag (aq) + 2e E = 0.80 V (anode - oxidation) + - o
22H (aq) + 2e H (g) E = 0.00 V (cathode - reduction)
+ +
22Ag(s) + 2H (aq) H (g) + 2Ag (aq)o
cellE = 0.00 V - 0.80 V = - 0.80 V (not spontaneous)
2/18/2015 66
Electrochemistry Relative Reactivities of Metals
Metals that displace H2 from water Metals that lie below the half-cell reaction
potential for water can displace H2 from water In the reaction below the E value for water is not
the standard state value listed in the table because in pure water, [OH-] is 1.0 x 10-7 M, not the standard state value of 1 M (-0.83 V)
Ecell > 0 Sodium displaces Hydrogen from water
- -
2 22H O(l) + 2e H (g) + 2OH E = -0.42V
+ - o2Na(s) 2Na (aq) + 2e E = -2.71 V+ -
2 22Na(s) + 2H O(l) 2Na (aq) + H + 2OHo
cellE = - 0.42 V - (-2.71 V) = + 2.29 V
2/18/2015 67
Electrochemistry Relative Reactivities of Metals
Metals that can displace other metals from solution Any metal that is lower in the standard electrode
half-cell list can reduce the ion of a metal that is higher in the list, thus displacing that metal from solution (See next slide and slide #48)
Ecell > 0 Zinc is the stronger reducing agent reducing
Fe2+ to Fe and displacing it from solution
2+ - oZn(s) Zn (aq) + 2e E = -0.76V (anode; oxidation)
2+ - oFe (aq) + 2e Fe(s) E = -0.44V (cathod; reduction)
o
cellE = - 0.44 V - (-0.76 V) = 0.32 V
2+ 2+Zn(s) + Fe Zn (aq) + Fe(s)
2/18/2015 68
Electrochemistry
2/18/2015 69
Electrochemistry Free Energy and Electrical Work
Electrical Work Potential (Ecell, in volts) times the charge
Ecell measured with no current flowing No energy lost to heating Ecell voltage is maximum possible for cell Work is maximum possible Only reversible process can do maximum work Reversible process with no current flow:
The voltaic cell loses energy as it does work on the surroundings; thus the work term (wmax) is negativemax cellw = - E charge
max cellw = - E charge = G
max cellG = w = - E charge
o o
cellΔG = - E n F (components in standard states)
max cellG = w = - E nF
4 4
-
Coulombs JF = 9.65 10 = 9.65 10
mole V •mol e
n = moles
maxΔG = w
Recall :
cellΔG - E
Recall Slide 70 from Chapter 20
(Recall slide # 39)
2/18/2015 71
Electrochemistry
oΔG = - RT lnK
o
cell-E n F = - RT lnK
)
o
cell -4
-
J8.314 298.15 K RT mol rxn K E = lnK = 2.303(log K)n mol e Jn F (9.65 10mol rxn V •mol e
o
cell
0.0592 VE = log K
no
cellnElog K = (at 298.15K)
0.0592 V
(Slide 82 from Chapter 20)
o o
cellΔG = - E n F
2/18/2015 72
Electrochemistry
Summary Relationship between
Go Eocell K
2/18/2015 73
Electrochemistry Effect of Concentration on Cell Potential
Most cells do not start with concentrations in their “standard” states
Recall: oΔG = ΔG + RTlnQ
o o
cellΔG = - nF × E (Standard State)
cellG = - nF E
o ocell cellΔG = - nF × E = ΔG + RTlnQ = - n F × E + RTlnQ
ocell cell
nF RTlnQE = E - (Nernst Equation)
nF nF
)
o
cell cell -4
-
J8.314 298.15 K
mol rxn KE = E - 2.303 log Qn mol e J
(9.65 10mol rxn V •mol e
o
cell cell
0.0592 VE = E - × log Q (at 298.15 K)
n
(Slide 86 from Chapter 20)
ocell cell-nF × E = - n F × E + RT lnQ
2/18/2015 74
Electrochemistry Changes in Potential During Cell Operation
The potential of a cell changes as the concentration of the cell components change
o
2+ 2+
2+
cell 2+
Zn(s) + Cu (aq) Zn (aq) + Cu(s)[Zn ]
E = 1.10 V Q = [Cu ]
o
cell cell
0.0592 VE = E - × log Q (at 298.15 K)
no
cell cell
ocell cell
ocell cell
cell
Stage 1 When Q < 1 E > E
Stage 2 When Q = 1 E = E
Stage 3 When Q > 1 E < E
Stage 4 When Q = K E = 0 (Equilibrium)
cell
cell
cell
If Q / K < 1, E is positive (+) (cell does work)
If Q / K = 1, E = 0 (equilibrium - cell no longer does work)
If Q / K > 1, E is negative (-) (reaction reverses until Q / K = 1)
2/18/2015 75
Electrochemistry
cell
cell
cell
If Q / K < 1, E is positive (+)(cell does work)
If Q / K = 1, E = 0(cell no longer does work)
If Q / K > 1, E is negative (-)(reaction reverses until Q / K = 1)
o
cell cell
0.0592 VE = E - × log Q
n
2/18/2015 76
Electrochemistry Concentration Cells
In a cell composed of the same substance, but differing concentrations in the two half-cells, the two concentrations move to equilibrate producing electrical energy
The cell reaction is the “sum” of identical half-cell reactions written in opposite directions
The Standard Electrode Potentials (Eocell) are both
based on a 1 M solution (standard conditions), so they “cancel” each other, i. e., Eo
cell = 0
The non-standard cell potential, Ecell, depends on the ratio of the two concentrations [A]dil / [A]conc = Q
2/18/2015 77
Electrochemistry How the Concentration Cell Works The dilute solution is in the Anode compartment
(oxidation) and the concentrated solution is in the Cathode compartment (Reduction)
In the Anode (dilute) half-cell, Cu atoms give up 2 electrons and the resulting Cu2+ ions enter the solution and make it more concentrated
In the Cathode (conc) half-cell, Cu2+ ions gain 2 electrons and the resulting Cu atoms plate out on the electrode, making the solution less concentrated
In this type of Voltaic cell, the dilution continues until equilibrium is attained, i.e.,
Ecell decreases until Ecell = 0 (Q = K)
2+ -
2+( -
Cu(s) Cu (aq;0.10M) + 2e (Oxidation)
Cu aq;1.0M) + 2e Cu(s) Reduction
2/18/2015 78
Electrochemistry Alkaline Battery
Nickel-Metal Hydride (Ni-MH) Battery
Lithium Ion Battery
- -2Zn(s) + 2OH (aq) ZnO(s) + H O(l) + 2e
- -2 2 2MnO (s) + 2H O(l) + 2e Mn(OH) (s) + 2OH (aq)
[Anode (Oxidation]
[Cathode (Reduction]
2 2 2 cellZn(s) + MnO (s) + H O(l) ZnO(s) + Mn(OH) (s) E = 1.5 V Overall Cell Reaction
- -2MH(s) + OH (aq) M(s) + H O(l) + e
- -2 2NiO(OH)(s) + H O(l) + e Ni(OH) (s) + OH (aq)
[Anode (Oxidation]
[Cathode (Reduction]
Overall Cell Reaction2 cellMH(s) + NIO(OH)(s) M(s) + Ni(OH) (s) E = 1.4 V
+ -x 6 6Li C xLi + xe + C (s)
+ - -1-x 2 4 2 4Li Mn O (s)(s) + xLi + xe LiMn O (s) + OH (aq)
x 6 1-x 2 4 2 4 cellLi C + Li Mn O (s) LiMn O (s) + C6(s) E = 3.7 V
[Anode (Oxidation]
[Cathode (Reduction]
Overall Cell Reaction
2/18/2015 79
Practice ProblemCalculate Ecell for a voltaic cell containing the following half-cells:
[Zn2+] = 0.010 M [H+] = 2.5 M PH2 = 0.30 atm
Construct Eocell
Calculate Q
2+ +2Zn(s) Zn (aq) H H (g)I II I
+ - o2 2H (aq) + 2e H (g) (E = 0.00 V) (red; cathode)
ocellE = 0.00 V - (-0.76 V) = 0.76 V
2+-4H2
+ 2 2
P [Zn ] 0.30 0.010Q = = = 4.8 10
[H ] 2.5
o ocell cell cell
RT 0.0592 VE = E - 2.303× log Q = E - log Q
nF n
-4cell
0.0592 VE = 0.76V - log(4.8 10 ) = 0.76V - (-0.0982 V = 0.86 V
2
2+ o Zn(s) Zn + 2e - (E = - 0.76 V) (ox; anode)+ 2+
22H (aq) + Zn(s) H (g) + Zn (aq)
2/18/2015 80
Practice ProblemWhat is the equilibrium constant for the following reaction
[Eo(Ce4+/Ce3+) = 1.72 V Eo(Cl2/Cl-) = 1.36 V
2 Cl-(aq) + 2 Ce4+(aq) Cl2(g) + 2 Ce3+(aq)- 4+ 3+
2Cl (aq) Cl (g) Ce (aq) Ce (aq)I II I
4+ - 3+ o2Ce (aq) + 2e 2Ce (aq) (E = 1.72 V) (red; cathode)
ocellE = 1.72V - 1.36 V = 0.36 V
o
cellnE 2 0.36 Vlog K = = = 12.2 (at 298.15 K)
0.0592 V 0.0592 V
- o22Cl (aq) Cl + 2e - (E = 1.36 V) (ox; anode)
4+ - 3+22Ce (aq) + 2Cl Cl (g) + 2Ce (aq)
o
cell
0.0592 VE = log K
n
2/18/2015 81
Practice ProblemWrite out the overall equation for the cell reaction and determine the standard cell potential for the following galvanic cell.
[Eo (Ag+/Ag) = 0.80 Eo (Ni2+/Ni) = -0.26]
Ni(s)|Ni2+(aq)║Ag+(aq)|Ag(s)+ - + o 2Ag (aq) + 2e 2Ag (aq) (E = 0.80 V) (red; cathode)
ocellE = 0.80V - (- 0.26) = 1.06 V
+ 2+ +2Ag (aq) + Ni(aq) Ni (aq) + 2Ag (aq)
2+ oNi(aq) Ni + 2e - (E = - 0.26 V) (ox; anode)
2/18/2015 82
Practice ProblemWhat is the maximum work you can obtain from 15.0 g of Ni in the galvanic cell shown in the previous problem when the Ecell is 0.97 V?
