Chapter 20 Electrochemistry
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Transcript of Chapter 20 Electrochemistry
Electrochemistry
Chapter 20Electrochemistry
Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
John D. BookstaverSt. Charles Community College
St. Peters, MO 2006, Prentice Hall, Inc.
Electrochemistry
Electrochemical Reactions
In electrochemical reactions, electrons are transferred from one species to another.
Metals tend to lose electrons and are oxidized, non metals tend to gain electrons and are reduced.
Electrochemistry
LEOLEO GERGER
LLosingosing EElectrons islectrons is OOxidation.xidation.
GGainingainingEElectrons islectrons isRReductioneduction
Electrochemistry
OIL RIGOIL RIG
• OOxidationxidation• IIss• LLoss.oss.
• RReduction eduction • IIss GGain.ain.
Electrochemistry
REDOX REACTIONSREDOX REACTIONS
• Are reduction – Are reduction – oxidation reactions.oxidation reactions.
• Electrons are Electrons are transferred.transferred.
• When an atom is When an atom is losing electrons its losing electrons its O.N. increases. It is O.N. increases. It is being oxidized.being oxidized.
• When an atom gains When an atom gains electrons its O.N. electrons its O.N. decreases. It is being decreases. It is being reducedreduced
Electrochemistry
Oxidation Numbers
In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.
Electrochemistry
Assigning Oxidation Numbers1. Elements in their elemental form have an ON= 0.2. The oxidation number of a monatomic ion is its
charge.3. Nonmetals tend to have negative oxidation
numbers, although some are positive in certain compounds or ions.
Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.
Hydrogen is +1 except in metal hydrides when is −1 .
Fluorine always has an oxidation number of −1.
Electrochemistry
Assigning Oxidation Numbers The other halogens have an oxidation
number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions
4. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
5. The sum of the oxidation numbers in a neutral compound is 0.
Electrochemistry
Oxidation and Reduction
• A species is oxidized when it loses electrons.Here, zinc loses two electrons to go from neutral
zinc metal to the Zn2+ ion.
Electrochemistry
Oxidation and Reduction
• A species is reduced when it gains electrons.Here, each of the H+ gains an electron and they
combine to form H2.
Electrochemistry
Oxidation and Reduction
• What is reduced is the oxidizing agent.H+ oxidizes Zn by taking electrons from it.
• What is oxidized is the reducing agent.Zn reduces H+ by giving it electrons.
Electrochemistry
• Zn added to HCl yields the spontaneous reaction
Zn(s) + 2H+(aq) Zn2+(aq) + H2(g).
• The oxidation number of Zn has increased from 0 to 2+.• The oxidation number of H has reduced from 1+ to 0.
• Zn is oxidized to Zn2+ while H+ is reduced to H2.
• H+ causes Zn to be oxidized and is the oxidizing agent.• Zn causes H+ to be reduced and is the reducing agent.• Note that the reducing agent is oxidized and the oxidizing
agent is reduced.
Oxidation-Reduction Oxidation-Reduction ReactionsReactions
Electrochemistry
• Law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end.
• Conservation of charge: electrons are not lost in a chemical reaction.
Half Reactions• Half-reactions are a convenient way of separating
oxidation and reduction reactions.
Balancing Oxidation-Balancing Oxidation-Reduction ReactionsReduction Reactions
Electrochemistry
Half Reactions• The half-reactions for
Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)
are
Sn2+(aq) Sn4+(aq) +2e-
2Fe3+(aq) + 2e- 2Fe2+(aq)• Oxidation: electrons are products.• Reduction: electrons are reactants.
Loss of Gain of
Electrons is Electrons is
Oxidation Reduction
Electrochemistry
Balancing Equations by the Method of Half Reactions
• Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple).
• MnO4- is reduced to Mn2+ (pale pink) while the C2O4
2- is oxidized to CO2.
• The equivalence point is given by the presence of a pale pink color.
• If more KMnO4 is added, the solution turns purple due to the excess KMnO4.
Electrochemistry
What is the balanced chemical equation?
1. Write down the two half reactions.
2. Balance each half reaction:a. First with elements other than H and O.
b. Then balance O by adding water.
c. Then balance H by adding H+.
d. If it is in basic solution, remove H+ by adding OH-
e. Finish by balancing charge by adding electrons.3. Multiply each half reaction to make the number of electrons
equal.4. Add the reactions and simplify.5. Check!
Electrochemistry
Half-Reaction Method
Consider the reaction between MnO4− and C2O4
2− :
MnO4−(aq) + C2O4
2−(aq) Mn2+(aq) + CO2(aq)
Electrochemistry
Balancing Equations by the Method of Half Reactions
• Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple).
• MnO4- is reduced to Mn2+ (pale pink) while the C2O4
2- is oxidized to CO2.
• The equivalence point is given by the presence of a pale pink color.
• If more KMnO4 is added, the solution turns purple due to the excess KMnO4.
Electrochemistry
Half-Reaction Method
First, we assign oxidation numbers.
MnO4− + C2O4
2- Mn2+ + CO2
+7 +3 +4+2
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.
Electrochemistry
For KMnO4 + Na2C2O4:
• The two incomplete half reactions are
MnO4-(aq) Mn2+(aq)
C2O42-(aq) 2CO2(g)
2. Adding water and H+ yields
8H+ + MnO4-(aq) Mn2+(aq) + 4H2O
• There is a charge of 7+ on the left and 2+ on the right. Therefore, 5 electrons need to be added to the left:
5e- + 8H+ + MnO4-(aq) Mn2+(aq) + 4H2O
Electrochemistry
• In the oxalate reaction, there is a 2- charge on the left and a 0 charge on the right, so we need to add two electrons:
C2O42-(aq) 2CO2(g) + 2e-
3. To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. Multiplying gives:
10e- + 16H+ + 2MnO4-(aq) 2Mn2+(aq) + 8H2O
5C2O42-(aq) 10CO2(g) + 10e-
Electrochemistry
4. Adding gives:
16H+(aq) + 2MnO4-(aq) + 5C2O4
2-(aq) 2Mn2+(aq) + 8H2O(l) + 10CO2(g)
5. Which is balanced!
Electrochemistry
Balancing in Basic Solution
• If a reaction occurs in basic solution, one can balance it as if it occurred in acid.
• Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place.
• If this produces water on both sides, you might have to subtract water from each side.
Electrochemistry
Examples – Balance the following oxidation-reduction reactions:
1. Cr (s) + NO3- (aq) Cr3+ (aq) + NO (g) (acidic)
2. Al (s) + MnO4- (aq) Al3+ (aq) + Mn2+ (aq) (acidic)
3. PO33- (aq) + Mn4
- (aq) PO43- (aq) + MnO2 (s)
(basic)
4. H2CO (aq) + Ag(NH3)2+ (aq) HCO3
- (aq) + Ag (s) + NH3 (aq) (basic)
Electrochemistry
Section 20-3 20-4
• Voltaic Cells – Spontaneous reactions
• ELECTRODES – POLARITIES
• SALT BRIDGE
• DRIVING FORCE- EMF
• STANDARD REDUCTION POTENTIALS
• HW Q 23, 25, 31, 33 (a, b ) , 35
Electrochemistry
Voltaic Cells
In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.
Electrochemistry
Voltaic Cells
• We can use that energy to do work if we make the electrons flow through an external device.
• We call such a setup a voltaic cell.
Electrochemistry
Voltaic Cells
• A typical cell looks like this.
• The oxidation occurs at the anode.
• The reduction occurs at the cathode.
Electrochemistry
Voltaic Cells
Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.
Electrochemistry
Voltaic Cells
• Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced.Cations move toward
the cathode.Anions move toward
the anode.
Electrochemistry
Voltaic Cells• In the cell, then,
electrons leave the anode and flow through the wire to the cathode.
• As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.
Electrochemistry
Voltaic Cells• As the electrons
reach the cathode, cations in the cathode are attracted to the now negative cathode.
• The electrons are taken by the cation, and the neutral metal is deposited on the cathode.
Electrochemistry
Electromotive Force (emf)• Water only
spontaneously flows one way in a
waterfall.• Likewise, electrons
only spontaneously flow one way in a
redox reaction—from higher to lower
potential energy.
Electrochemistry
Electromotive Force (emf)
• The potential difference between the anode and cathode in a cell is called the electromotive force (emf).
• It is also called the cell potential, and is designated Ecell.
Electrochemistry
Cell Potential
Cell potential is measured in volts (V).
One volt is the potential difference required to impart 1J of energy to a charge of 1 coulomb. (1 electron has a charge of 1.6 x 10-19 Coulombs). The potential difference between the 2 electrode provides the driving force that pushes the electron through the external circuit.
Electrochemistry
Cell Potential or Electromotive Force (emf)
• The “pull” or driving force on the electrons.
• ELECTROMOTIVE FORCE EMF
• CAUSES THE ELECTRON MOTION!
1 V = 1 JC
Electrochemistry
Standard Reduction Potentials
Reduction potentials for
many electrodes have been
measured and tabulated.
Electrochemistry
Standard Hydrogen Electrode
• Their values are referenced to a standard hydrogen electrode (SHE).
• By definition, the reduction potential for hydrogen is 0 V:
2 H+ (aq, 1M) + 2 e− H2 (g, 1 atm)
Electrochemistry
Standard Cell Potentials
The cell potential at standard conditions can be found through this equation:
Ecell = Ered (cathode) − Ered (anode)
Because cell potential is based on the potential energy per unit of charge, it is an intensive property.
Electrochemistry
For the reaction to be SPONTANEOUS
E has to be positive
Electrochemistry
Cell Potentials
• For the oxidation in this cell,
• For the reduction,
Ered = −0.76 V
Ered = +0.34 V
Electrochemistry
Cell Potentials
Ecell = Ered (cathode) − Ered (anode)
= +0.34 V − (−0.76 V)
= +1.10 V
Electrochemistry
• Since Ered = -0.76 V we conclude that the reduction of Zn2+ in the presence of the SHE is not spontaneous.
• The oxidation of Zn with the SHE is spontaneous.• Changing the stoichiometric coefficient does not affect
Ered.
• Therefore,
2Zn2+(aq) + 4e- 2Zn(s), Ered = -0.76 V.
• Reactions with Ered > 0 are spontaneous reductions relative to the SHE.
Electrochemistry
• Reactions with Ered < 0 are spontaneous oxidations relative to the SHE.
• The larger the difference between Ered values, the larger Ecell.
• In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is more positive than Ered(anode).
• Recall anodecathode oxredcell EEE
Electrochemistry
PNEUMONICS
• LEO GER
• OIL RIG
• RED CAT
• AND ALWAYS THE SOURCE OF ELECTRONS IS THE NEGATIVE ELECTRODE!!!
Electrochemistry
• In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is more positive than Ered(anode) since
Or
• More generally, for any electrochemical process
• A positive E indicates a spontaneous process (galvanic cell).
• A negative E indicates a nonspontaneous process.
Spontaneity of Redox Spontaneity of Redox ReactionsReactions
anodecathode oxredcell EEE
processoxidation processreduction oxredcell EEE
anodecathode redredcell EEE
Electrochemistry
Cell Potential Calculations
• To Calculate cell potential using Standard Reduction Potentials:
• 1. One reaction and its cell potential’s sign must be reversed--it must be chosen such that the overall cell potential is positive.
• 2. The half-reactions must often be multiplied by an integer to balance electrons--this is not done for the cell potentials.
Electrochemistry
Cell Potential Calculations Continued
• Fe3+(aq) + Cu(s) ----> Cu2+
(aq) + Fe2+(aq)
• Fe3+(aq) + e- ----> Fe2+
(aq) Eo = 0.77 V
• Cu2+(aq) + 2 e- ----> Cu(s) Eo = 0.34 V
• Reaction # 2 must be reversed.
Electrochemistry
Cell Potential Calculations Continued
• 2 (Fe3+(aq) + e- ----> Fe2+
(aq)) Eo = 0.77 V
• Cu(s) ----> Cu2+(aq) + 2 e- Eo = - 0.34 V
• 2Fe3+(aq) + Cu(s) ----> Cu2+
(aq) + 2Fe2+(aq)
• Eo = 0.43 V
Electrochemistry
Oxidizing and Reducing Agents
The greater the difference between the two, the greater the voltage of the cell.
REMEMBER TO REVERSE THE SIGN OF THE SPECIE THAT GETS OXIDIZED (THE ONE BELOW!!!!!)
Electrochemistry
• Section 4.5 – 4.6
• Free energy and redox rx
• Cell EMF under nonstandard conditions
• Nerst Equation- Concentration cells
• Redox Applications - Batteries and fuel cells – Corrosion prevention
• HW 47, 51 a, 59, 63
Electrochemistry
Oxidizing and Reducing Agents• The strongest
oxidizers have the most positive reduction potentials.
• O.A. PULL electrons• The strongest
reducers have the most negative reduction potentials.
• R.A. PUSH their electrons
Electrochemistry
Examples – Sketch the cells containing the following reactions. Include E°cell, the direction of electron flow, direction of ion migration through salt bridge, and identify the anode and cathode:
1. Cr3+ + Cl2 (g) Cr2O72- + Cl-
2. Cu2+ + Mg (s) Mg2+ + Cu (s)
3. IO3- + Fe2+ Fe3+ + I2
4. Zn (s) + Ag+ Zn2+ + Ag
Electrochemistry
Free Energy
G for a redox reaction can be found by using the equation
G = −nFE
where n is the number of moles of electrons transferred, and F is a constant, the Faraday.
1 F = 96,485 C/mol = 96,485 J/V-mol
Electrochemistry
Free Energy and Cell Potential
Under standard conditions G = nFE
• n = number of moles of electrons• F = Faraday = 96,485 coulombs per
mole of electrons = 96,485 J/Vmol• E = V
Electrochemistry
Find the free energy for the reaction of
Zn + Cu2+ ---> Zn2+ + Cu
Electrochemistry
The Nernst Equation• A voltaic cell is functional until E = 0 at which point
equilibrium has been reached.• The point at which E = 0 is determined by the
concentrations of the species involved in the redox reaction.
• The Nernst equation relates emf to concentration using
and noting that
Effect of Concentration Effect of Concentration on Cell EMFon Cell EMF
QRTGG ln
QRTnFEnFE ln
Electrochemistry
The Nernst Equation• This rearranges to give the Nernst equation:
• The Nernst equation can be simplified by collecting all the constants together using a temperature of 298 K:
• (Note that change from natural logarithm to base-10 log.)• Remember that n is number of moles of electrons.
QnFRT
EE ln
Qn
EE log0592.0
Electrochemistry
Cell EMF and Chemical Equilibrium• A system is at equilibrium when G = 0.• From the Nernst equation, at equilibrium and 298 K (E =
0 V and Q = Keq):
0592.0log
log0592.0
0
nEK
Kn
E
eq
eq
Electrochemistry
Examples
1. Calculate E°cell for the following reaction:
Zn (s) + Cu2+ Zn2+ + Cu (s)
2. Calculate Ecell if [Zn2+] = 1.88 M and [Cu2+] = 0.020 M
3. Calculate E°cell for the following reaction:
IO3- (aq) + Fe2+ (aq) Fe3+ (aq) + I2 (aq)
4. Calculate Ecell if [IO3-] = 0.20 M, [Fe2+] = 0.65
M, [Fe3+] = 1.0 M, and [I2] = 0.75 M
Electrochemistry
Concentration Cells
• Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.
• For such a cell, would be 0, but Q would not.Ecell
• Therefore, as long as the concentrations are different, E will not be 0.
Electrochemistry
Concentration Cells• We can use the Nernst equation to generate a cell that has
an emf based solely on difference in concentration.• One compartment will consist of a concentrated solution,
while the other has a dilute solution.• Example: 1.00 M Ni2+(aq) and 1.00 10-3 M Ni2+(aq).• The cell tends to equalize the concentrations of Ni2+(aq)
in each compartment.• The concentrated solution has to reduce the amount of
Ni2+(aq) (to Ni(s)), so must be the cathode.
Electrochemistry
Concentration Cells
Electrochemistry
PROBLEM
• Use the Nernst equation to determine the (emf) at 25oC of the cell. Zn(s)│Zn2+(0.00100M ║Cu2+(10.0M)│Cu(s).
• The standard emf of this cell is 1.10 V.
Electrochemistry
Applications of Oxidation-Reduction
Reactions
Electrochemistry
Batteries
•A battery is a galvanic cell or, more commonly, a group of galvanic cells connected in series.
Electrochemistry
Batteries
Electrochemistry
17_370
H2SO4
electrolytesolution
Anode (leadgrid filled withspongy lead) Cathode (lead
grid filled withspongy PbO2)
A lead storage battery consists of a lead anode, lead dioxide cathode, and an electrolyte of 38% sulfuric acid.
Electrochemistry
Lead Storage Battery
• Anode reaction:
Pb(s) +H2SO4(aq) ---> PbSO4(aq) + 2H+(aq) + 2e-
• Cathode reaction:
PbO2(s)+H2SO4(aq)+ 2e-+2H+(aq)->PbSO4(aq)+2H2O(l)
• Overall reaction:
Pb(s)+ PbO2(s) + 2H2SO4(aq)-->PbSO4(aq)+ 2H2O (l)
Electrochemistry
17_371
Cathode(graphite rod)
Anode(zinc inner case)
Paste of MnO2,NH4CL, andcarbon
Common dry cell and its components.
Electrochemistry
Alkaline Battery• Anode: Zn cap:
Zn(s) Zn2+(aq) + 2e-
• Cathode: MnO2, NH4Cl and C paste:
2NH4+(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) +
2H2O(l)
• The graphite rod in the center is an inert cathode.
• For an alkaline battery, NH4Cl is replaced with KOH.
Electrochemistry
• Anode: Zn powder mixed in a gel:
Zn(s) Zn2+(aq) + 2e-
• Cathode: reduction of MnO2.
Electrochemistry
Electrochemistry
Fuel Cells• Direct production of electricity from fuels occurs in a
fuel cell.
• On Apollo moon flights, the H2-O2 fuel cell was the primary source of electricity.
• Cathode: reduction of oxygen:
2H2O(l) + O2(g) + 4e- 4OH-(aq)
• Anode:
2H2(g) + 4OH-(aq) 4H2O(l) + 4e-
Electrochemistry
17_372
Cathode (steel)
Anode (zinc container)
Solution of HgO (oxidizingagent) in a basic medium (KOHand Zn(OH)2)
Insulation
Mercury battery used in calculators.
Electrochemistry
Fuel Cells
• . . . galvanic cells for which the reactants are continuously supplied.
• 2H2(g) + O2(g) 2H2O(l)
• anode: 2H2 + 4OH 4H2O + 4e
• cathode: 4e + O2 + 2H2O 4OH
Electrochemistry
Hydrogen Fuel Cells
Electrochemistry
17_369
Reference solution ofdilute hydrochloric acid
Silver wire coated withsilver chloride
Thin-walled membrane
Ion selective electrodes are glass electrodes that measures a change in potential when [H+] varies. Used to measure pH.
Electrochemistry
Corrosion of Iron
• Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen.
• Cathode: O2(g) + 4H+(aq) + 4e- 2H2O(l).
• Anode: Fe(s) Fe2+(aq) + 2e-.• Dissolved oxygen in water usually causes the oxidation
of iron.• Fe2+ initially formed can be further oxidized to Fe3+
which forms rust, Fe2O3.xH2O(s).
CorrosionCorrosion
Electrochemistry
• Oxidation occurs at the site with the greatest concentration of O2.
Preventing Corrosion of Iron• Corrosion can be prevented by coating the iron with paint
or another metal.• Galvanized iron is coated with a thin layer of zinc.
Electrochemistry
Electrochemistry
• Zinc protects the iron since Zn is the anode and Fe the cathode:
Zn2+(aq) +2e- Zn(s), Ered = -0.76 V
Fe2+(aq) + 2e- Fe(s), Ered = -0.44 V
• With the above standard reduction potentials, Zn is easier to oxidize than Fe.
Electrochemistry
Electrochemistry
Preventing Corrosion of Iron• To protect underground pipelines, a sacrificial anode is
added.• The water pipe is turned into the cathode and an active
metal is used as the anode.• Often, Mg is used as the sacrificial anode:
Mg2+(aq) +2e- Mg(s), Ered = -2.37 V
Fe2+(aq) + 2e- Fe(s), Ered = -0.44 V
Electrochemistry
Electrochemistry
Corrosion•Some metals, such as copper, gold, silver and platinum, are relatively difficult to oxidize. These are often called noble metals.
•About 1/5 of all iron and steel produced each year is used to replace rusted metal.
Electrochemistry
Self-protecting Metals
• Some metals such as aluminum, copper, and silver form a protective coating that keeps them from corroding further.
• The protective coating for iron and steel flakes away opening new layers of metal to corrosion.
Electrochemistry
Prevention of Corrosion• Coating--painting or applying oil to
keep out oxygen and moisture.• Galvanizing--dipping a metal in a more
active metal -- galvanized steel bucket.• Alloying -- mixing metals with iron to
prevent corrosion -- stainless steel.• Cathodic protection -- attaching a more
active metal. Serves as sacrificial metal--used to protect ships, gas lines, and gas tanks.
Electrochemistry
Redox Titrations
• Same as any other titration.• the permanganate ion is used often because
it is its own indicator. MnO4- is purple, Mn+2 is
colorless. When reaction solution remains clear, MnO4
- is gone.
• Chromate ion is also useful, but color change, orangish yellow to green, is harder to detect.
Electrochemistry
Example• The iron content of iron ore can be
determined by titration with standard KMnO4 solution. The iron ore is dissolved in excess HCl, and the iron reduced to Fe+2 ions. This solution is then titrated with KMnO4 solution, producing Fe+3 and Mn+2 ions in acidic solution. If it requires 41.95 mL of 0.205 M KMnO4 to titrate a solution made with 10.613 g of iron ore, what percent of the ore was iron?