chapter 19 s16 - uri department of chemistry chapter 19.pdf · composition (equilibrium constants)...
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Electrochemistry
The study of chemical reactions through electrical circuits. Monitor redox reactions by controlling electron transfer
REDOX: Shorthand for “REDuction-OXidation”
Use redox reactions to experimentally measure:Reaction progress (kinetics)Composition (equilibrium constants)Energy changes (thermodynamics)
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Oxidation-Reduction ReactionsOxidation-reduction reactions (REDOX reaction) occur when
electrons are transferred from one reactant to another during a chemical reaction. There is a change in oxidation number for
both substancesOxidation Number: Theoretical charge on an ion
Oxidation is the process where the oxidation number increases.Electrons are lost from the substance
Reduction is the process where the oxidation number decreases. Electrons are gained by the substance
Oxidation and reduction always accompany each other;Neither can occur alone
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Redox Reaction: Magnesium Burning
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Oxidation: Mg (s) Mg2+ + 2e-Reduction: 1/2O2(g) + 2e- O2-
Reaction: Mg (s) + 1/2O2(g) MgO(s)
Oxidation Number Rules: See Chapter 4The rule earlier in the list always takes
precedence.1) ON = 0 for a compound or ionic charge for an ion2)ON = +1 for IA elements and
ON = +2 2A elements3) ON= -2 for oxygen
4)ON= -1 for 7A elementsIf both elements in 7A, then the one higher in the list is -1
5)ON = -2 for 6A elements other than oxygen6) ON = -3 for 5A elements (very shaky!!!)
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Balancing Redox Reactions: Balance Elements
1. Write equation for the ions in acid solution: Fe2+ + Cr2O7
2- Fe3+ + Cr3+
2. Divide equation into 2 half reactionsFe2+ Fe3+ Cr2O7
2- 2Cr3+
3. Balance all elements except H and OFe2+ Fe3+ Cr2O7
2- 2Cr3+
4. Balance O with H2OFe2+ Fe3+ Cr2O7
2- 2Cr3+ + 7H2O5. Balance H with H+
Fe2+ Fe3+ Cr2O72- + 14H+ 2Cr3+ + 7H2O
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Balancing Redox Reactions: Charge6. Balance chargeFe2+ Fe3+ + 1e- 6e- + Cr2O7
2- + 14H+ 2Cr3+ + 7H2OOxidation: lose electron Reduction: gain electrons7. Multiply by an integer to equalize # electrons 6Fe2+ 6Fe3+ + 6e- 6e- + Cr2O7
2- + 14H+ 2Cr3+ + 7H2O
8. Add reactions together and cancel like species6Fe2+ + 6e- + Cr2O7
2- + 14H+ 2Cr3+ + 7H2O+ 6Fe3+ + 6e-
9. Check balance of final # of atoms and charge6Fe2+ + Cr2O7
2- + 14H+ 2Cr3+ + 7H2O+ 6Fe3+
(12+) + (2-) + (14+) = +24 (6+) + (0) + (18+) = +24
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Balancing Redox Reactions: Basic Reactions
Need OH- instead of H+ in final equation
1. Balance as if the reaction is in acidic solution.6Fe2+ + Cr2O7
2- + 14H+ 2Cr3+ + 7H2O+ 6Fe3+
2. Add OH- to both sides to match H+
14OH- + 6Fe2+ + Cr2O72- + 14H+ 2Cr3+ + 7H2O+ 6Fe3++14OH-
3. Combine OH- and H+ to make water14H2O + 6Fe2+ + Cr2O7
2- 2Cr3+ + 7H2O+ 6Fe3++14OH-
4. Cancel water from both sides7H2O + 6Fe2+ + Cr2O7
2- 2Cr3+ + 6Fe3++14OH-
5. Check to see if balanced
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Parts of an Electrochemical CellIonic SolutionsProvide ions to transfer chargeSolution + Electrode= Half-cell
ElectrodesAnode: oxidation occurs Cathode: reduction occurs
Salt bridge.Keeps 2 half-cells connectedIons flow, but solution doesn’t
Metal wires Connect the electrodes to the terminals of the voltmeter.Provide means of transporting electrons between electrodes
Voltmeter Measures the electron flow in the system
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Types of Cells
Voltaic or Galvanic CellNet oxidation/reduction reaction is spontaneousConvert energy to useful work Batteries
Electrolytic CellNet redox reaction is non-spontaneousWork is done on the cell, energy must be supplied: Recharge a battery, electroplating
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Cell DiagramReaction: Zn(s) + Cu2+ Zn2+ + Cu
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)anode solution salt bridge solution cathode
Single bar (|)Divides reduced from oxidizedShow phase differences
Double bars (||) Represent salt bridgeDivides redox half reactions
Goes left to right Anode written firstRemainder in order of actual cell
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Cell DiagramZn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)anode solution salt bridge solution cathode
Anode (oxidation) reaction: Zn(s)Zn2+(aq) + 2e–
Solid zinc (Zn(s)) oxidized to Zn2+(aq)Solid zinc is the physical electrode
Cathode (reduction) reaction: Cu2+(aq) + 2e–Cu(s) Cu2+(aq) is reduced to solid copper (Cu(s)) Solid copper is the physical electrode
Net Reaction: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
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Standard Electrode PotentialsStandard Cell Potentials
Eocell = Eo
reduction - Eooxidation
Standard Hydrogen Electrode (SHE)2H+(aq) + 2e– H2(g)
Standard Electrode Potential, Eored
[H+] = 1.00 MPH2= 1.00atmT = 298KEo
red= 0.000VCalculate potential of oxidation half reactions(Eo
oxid)Eo
oxid= Eored - Eo
cell = 0.000V - Eocell = -Eo
cell
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Spontaneous: Galvanic CellPt | H2(g) | H+(aq) || Cu2+(aq) | Cu(s)H2(g) + Cu2+(aq) 2H+(aq) + Cu(s)
Cell Potential:Eo
cell = Eored - Eo
oxidEo
cell = +.340VAnode Reaction:
H2(g) 2H+(aq) + 2e–
Eooxid = 0.000 V
Cathode Reaction: Cu2+(aq) + 2e– Cu(s)Eo
red = +.340V
+ Eocell; Reaction is spontaneous as written
(Galvanic)
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Nonspontaneous: Electrolytic CellPt | H2(g) | H+(aq) || Zn2+(aq) | Zn(s)H2(g)+ Zn2+(aq)2H+(aq) + Zn(s)
Cell Potential:Eo
cell = Eored - Eo
oxid Eocell = -0.763V
Anode Reaction: H2(g) 2H+(aq) + 2e– Eo
oxid = 0.00 V
Cathode Reaction: Zn2+(aq) + 2e– Zn(s) Eo
red = -.763V
- Eocell: Reaction nonspontaneous (electrolytic)
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Daniell cell: Cu2+(aq) + Zn(s)Cu(s) + Zn2+
(aq)
Cell ReactionsReduction: Cu2+(aq) + 2e–Cu(s) Oxidation: Zn(s) Zn2+(aq) + 2e–
From Reduction TablesCu2+ Eo
red = + 0.34 VZn2+ Eo
ox = – 0.76V
Cell potential, Eocell
Eocell= Eo
red - Eooxid = +0.34V - (-0.76 V)
Eocell= +1.10V
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Calculating Cell Potentials
25 oC
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For a Spontaneous ReactionReduction reaction is the more positive, higher in table
Oxidation reaction is lower in table and needs to be reversed
Redox Rules
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1. E0 is for the reaction as written
2. The more positive E0 the greater the tendency for the substance to be reduced
3. Half-cell reactions are reversibleThe sign of E0 changes when the reaction is reversed
4. Changing stoichiometric coefficients of a half-cell reaction does not change E0
Intensive Property: Amount doesn’t matter
Will Br2(l) spontaneously oxidize Fe2+(aq)? If so, what are the net cell reaction and the
Eocell?
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Figure out what is being asked: Oxidize: Cause to lose e-: Fe2+(aq) Fe3+(aq)
Write and balance chemical equation:2Fe2+(aq) + Br2(l) 2Fe3+(aq) + 2 Br– (aq)
Determine half-reactions and get Eo from table:Br2(l) + 2e– 2Br–(aq) reduction Eo = +1.07 VFe2+(aq) Fe3+(aq) + e– oxidation Fe3+(aq) + e– Fe2+(aq Eo = + 0.77 V from tableCell Potential
Eocell= Eo
reduction- Eooxidation= +1.07V - 0.77V = +0.30V
Eocell>0 Spontaneous reaction
Will I–(aq) reduce Cr3+(aq) to the free metal. If so, what are the net reaction and the Eo
cell?
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Figure out what is being asked: Reduce: Cause something to gain e-: Cr3+(aq) + 3e- Cr(s)
Write and balance chemical equation:2Cr3+(aq) + 6I–(aq) 2Cr(s) + 3I2(s)
Cell ReactionsOxidation: 2I–(aq) I2(s) + 2e–
Table: I2(s) + 2e– 2I–(aq) Eo = +0.53 VReduction: Cr3+(aq) + 3e– Cr(s) Eo = -0.74 V
Cell PotentialEo
cell= Eoreduction - Eo
oxidation = (-0.74) - (+.53V) = -1.27VEo
cell <0 Non-spontaneous Reaction NR
Gibb’s Energy and Cell PotentialsElectrical work, w
w (joules)= Total cell charge (coulombs) x cell potential (V)Total cell charge = # mol electrons(n) x (coulombs/mol e-s)Faraday’s constant, F= 96485 Coulombs/1 mol e-s
Gibb's Energy:GG =wmax = –nFEcellAt standard conditions: Go = –nFEo
cell = -RTlnK(Voltaic cells are spontaneous, Ecell must be positive)
Cell potential, Ecell–nFEo
cell = -RTlnK so Eocell=( RT/nF)lnK
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Summarizing the Important Relationships
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ΔG° K E°cell Spontaneity Direction(-) >1 (+) Spontaneous More products
0 1 0 At equilibrium Products= reactants
(+) <1 (-) Nonspontaneous More reactants
Find Keq at 25 oC for the reaction; Sn(s) + 2Cu2+(aq) Sn2+(aq) + 2Cu+(aq)
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Balance net reaction:Sn(s) + 2Cu2+(aq) Sn2+(aq) + 2Cu+(aq)
Cell ReactionsOxidation: Sn(s) Sn2+(aq) + 2e– Eo
ox= - 0.14VReduction : 2Cu2+(aq) + 2e- 2Cu+(aq) Eo
red= + 0.15 VLink Eo
cell and K and fill in constantsEo
cell = (RT/nF) lnKeq lnKeq= Eocell(nF/RT)
R = 8.314 J/mol K n = 2 Eocell= +0.29 V
T = 25°C = 298K F = 96485 coul/molSolve for Keq
ln Keq = 22.6 Keq = e22.6 = 7x109
The Nernst Equation:Cell Potentials at Nonstandard Conditions
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From Thermodynamics:G = Go + RT ln Q
Linking Thermodynamics & ElectrochemistryG = –nFE and Go = –nFEo
Substitute into first equation to remove G :(–nF)E = (–nF)Eo + RT ln Q
Divide by –nF to get the Nernst Equation:E = Eo
cell - (RT/nF) ln QCan now change temperature and concentrations
Concentration Cells:Daniell Cell at Nonstandard Conditins
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Electrochemical cell that does not have 1M solutions Find the EMF at the following conditions:
1. [Cu2+] = 1.00 M, [Zn2+] = 1.0×10–9 M 2. [Cu2+] = 0.10 M, [Zn2+] = 0.90 M
Daniell CellZn(s) + Cu2+(aq) Cu(s) + Zn2+(aq) Eocell = +1.10 V
Nernst EquationE = Eo
cell - (RT / nF) ln QQ = [Cu(s)] [Zn2+(aq)] / [Zn(s)] [Cu2+(aq)] = [Zn2+/Cu2+]
Plugging in from equationE = Eo
cell - (RT / nF) ln [Zn2+/Cu2+]
Concentration Cells: Daniell Cell at Nonstandard Conditions
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Solve for E
Conditions of First Cell:[Cu2+] = 1.00 M, [Zn2+] =1.0×10–9 M
E = 1.37 VConditions of Second Cell
[Cu2+] = 0.10 M [Zn2+] = 0.90 ME = 1.07 V
]][ln
/96485*2298*/314.810.1 2
2
CuZn
molcoulmolKmolKJVE
Changing Zinc Concentration
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After the initial drop, concentration has a minimal effect on voltage
The Dry Cell (Disposable Batteries)Portable electronic devices
Zn(s) | ZnCl2(aq), NH4Cl(aq) || Mn2O3(aq) | MnO2(s)|C(s)Oxidation: Zn Zn2+
(aq)+ 2e-Reduction: 2MnO2 + 2NH4
+ + 2e- Mn2O3 (s) + 2NH3 + H2OCell Reaction: Zn+2MnO2+ 2NH4
+ Zn2++Mn2O3+2NH3+ H2O
Cell VoltageCalculated:+ 1.36V Produced: + 1.5V
Irreversible ReactionCell is “dead” when reactants are used up
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Button Batteries: Mercury BatteryPacemakers, hearing aids, watches
Zn | ZnO, OH- || HgO, OH- | Hg(s) | Fe(s)Oxidation: Zn + 2OH- ZnO + H2O +2e-Reduction: HgO + H2O + 2e- Hg + 2OH-Cell Reaction: Zn+ HgO + H2O ZnO + Hg
Cell VoltageConstant OH- composition More constant voltageReported voltage: + 1.5V
Irreversible ReactionCell is “dead” when reactants are used up
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Lead Storage BatteryCar and boat batteries
Pb | H2SO4 (38%) || H2SO4 (38%) | PbO2Oxidation: Pb + SO4
2- PbSO4 +2e-Reduction: PbO2 + 4H++ SO4
2- + 2e- PbSO4 + 2H2OCell Reaction: Pb + PbO2 + 4H++ SO4
2- 2PbSO4 + 2H2OCell Voltage
Voltage: + 2.0VUsually 6 cells=12V
Reversible ReactionReaction reverses whenengine is running
Battery rechargesElectrolytic cell reaction
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Fuel CellsGalvanic cells: Continually renew reactants
C-Ni (catalyst) | H2 | OH- (aq)|| OH- (aq)| O2 | C-Ni (catalyst)
Oxidation: 2H2 + 4OH- 4H2O +4e-Reduction: O2 + 2H2O + 4e- 4OH-
Cell Reaction: 2H2 + O2 2H2OCell Voltage
Voltage: + 1.23VNonreversible Reaction
Need fresh reactantsRemoval of productsNeed electrocatalysts
Corrosion: Deterioration of metal through an electrochemical process
Oxidation of metals: Rust, tarnish, “patina”Rust: Fe2O3•xH2O
2Fe + O2 + 4H+ 2Fe2+ + 2H2O E=1.67V4Fe2+ + O2+ (4+2x)H2O 2Fe2O3•xH2O + 8H+ E>0
Cell ReactionSpontaneous reactionsSalts increase rate
Cathodic protectionMore reactive metal protects (Zn)Reacts in place of protected metal (Fe)
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Electrolytic CellsNon-spontaneous processes driven by the application of an external power supply
Side reactions from solvent and dissolved ions
Determine reactions that are most spontaneous
Least Negative Ecell
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A 1 M solution of potassium iodide is electrolyzed under acidic conditions.What are the products?
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Half-Cell Potentials from tableK+(aq) + e–K(s) Eo = –2.92 V Reduction2I–(aq) I2(s) + 2e– Eo = - 0.54 V Oxidation 2H+(aq) + 2e– H2(g) Eo = 0.00 V Reduction2H2O(l) O2(g) + 4H+(aq) + 4e– Eo = -1.23 V Oxidation
Dissociation Reaction: KI(aq) K+(aq) + I–(aq)Resulting Products: K+, I-, H+, H2O
Cation Reduction: K+(aq) + e–K(s)Acid Reduction to H2 : 2H+(aq) + 2e– H2(g)Anion Oxidation: 2I–(aq) I2(s) + 2e–
H2O Oxidation to O2: 2H2O(l) O2(g) + 4H+(aq) + 4e–
A 1 M solution of potassium iodide is electrolyzed under acidic conditions.
What are the products?
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Less Negative ReactionsReduction: 2H+(aq) + 2e– H2(g) Eo = 0.00 VOxidation: 2I–(aq) I2(s) + 2e– Eo = + 0.54 V
Standard PotentialEo = 0.00 - (+0.54) = – 0.54V
need at least .54V to start reaction
Net Reaction2H+(aq) + 2I–(aq) I2(s) + H2(g)
ProductsSolid Iodine:I2(s) and Hydrogen gas: H2(g)
Producing Products by ElectrolysisElectroplating
Coating one metal onto anotherSilver or gold over iron or steelCheaperProduct often more durable
Purification of copperImpure copper anodeMore reactive impurities oxidized (ions)Less reactive impurities drop to bottom
Gold, silver, etc. now separatedCopper built up on cathode 99.5% pure
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Quantitative ElectrolysisUse Electrolytic Cells to find
stoichiometry Measure the charge passed through the cell
The current (amps, A) is the rate of charge flow1amp = 1 coulomb per second.
nF = Atn = number of moles of electrons F = Faraday's constant = 96485 C/molA = current in amps = 1 coulomb/sect = time in seconds
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AxmA
AmAxA 2105.21000125 sec100.9
min1sec60min15 2xxt
Quantitative Electrolysis of Water: 2H2OO2 + 2H2
CxsCxAtnF 5.22sec)100.9)(/105.2( 22
2542 108.5103.2
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2 Oelectronselectrons
OO molxmolxx
molmolemol
mLLKmolKLatmmolx
PnRTV O 4.10014.0)298)(0821.0)(108.5( 2
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Water is electrolyzed in a cell at 25 mA for 15 minutes. How many mL of oxygen gas are produced at 1 atm & 25 °C?
electronsmolxmolC
CFCn 4103.2
/964855.22