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Transcript of SPRING REVIEW PART TWO. PRECIPITATION REACTIONS Chapter 19 Copyright © 1999 by Harcourt Brace &...
SPRING REVIEWPART TWO
PRECIPITATION REACTIONSPRECIPITATION REACTIONSChapter 19Chapter 19PRECIPITATION REACTIONSPRECIPITATION REACTIONSChapter 19Chapter 19
Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
Analysis of Silver GroupAnalysis of Silver GroupAll salts formed in All salts formed in
this experiment are this experiment are said to be said to be INSOLUBLEINSOLUBLE and and form when mixing form when mixing moderately moderately concentrated concentrated solutions of the solutions of the metal ion with metal ion with chloride ions.chloride ions.
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Analysis of Silver GroupAnalysis of Silver Group
Although all salts formed in this Although all salts formed in this experiment are said to be insoluble, experiment are said to be insoluble, they do dissolve to some SLIGHT they do dissolve to some SLIGHT extent.extent.
AgCl(s) AgCl(s) AgAg++(aq) + Cl(aq) + Cl--(aq)(aq)
When equilibrium has been established, When equilibrium has been established, no more AgCl dissolves and the solution no more AgCl dissolves and the solution is is SATURATEDSATURATED..
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Analysis of Silver GroupAnalysis of Silver Group
AgCl(s) AgAgCl(s) Ag++(aq) + Cl(aq) + Cl--(aq)(aq)
When solution is When solution is SATURATEDSATURATED, expt. , expt. shows that [Agshows that [Ag++] = 1.67 x 10] = 1.67 x 10-5-5 M. M.
This is equivalent to the This is equivalent to the SOLUBILITYSOLUBILITY of of AgCl.AgCl.
What is [ClWhat is [Cl--]? ]?
This is also equivalent to the AgCl solubility.This is also equivalent to the AgCl solubility.
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Analysis of Silver GroupAnalysis of Silver Group
AgCl(s) AgCl(s) Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)
Saturated solution has Saturated solution has [Ag[Ag++] = [Cl] = [Cl--] = 1.67 x 10] = 1.67 x 10-5-5 M M
Use this to calculate KUse this to calculate Kcc
KKcc = [Ag = [Ag++] [Cl] [Cl--]]
= (1.67 x 10= (1.67 x 10-5-5)(1.67 x 10)(1.67 x 10-5-5) )
= 2.79 x 10= 2.79 x 10-10-10
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Analysis of Silver GroupAnalysis of Silver Group
AgCl(s) AgCl(s) Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)
KKcc = [Ag = [Ag++] [Cl] [Cl--] = 2.79 x 10] = 2.79 x 10-10-10
Because this is the product of Because this is the product of “solubilities”, we call it “solubilities”, we call it
KKspsp = solubility product constant = solubility product constant
See Table 19.2 and Appendix JSee Table 19.2 and Appendix J
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Precipitating an Insoluble SaltPrecipitating an Insoluble SaltHgHg22ClCl22(s) (s) Hg Hg22
2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)
KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]]22
If [HgIf [Hg222+2+] = 0.010 M, what [Cl] = 0.010 M, what [Cl--] is req’d to ] is req’d to
just begin the precipitation of Hgjust begin the precipitation of Hg22ClCl22??
That is, what is the maximum [ClThat is, what is the maximum [Cl--] that can ] that can
be in solution with 0.010 M Hgbe in solution with 0.010 M Hg222+2+ without without
forming Hgforming Hg22ClCl22??
Precipitating an Insoluble SaltPrecipitating an Insoluble SaltHgHg22ClCl22(s) (s) Hg Hg22
2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)
KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]]22
Recognize thatRecognize that
KKspsp = product of = product of
maximum ion concs.maximum ion concs.
Precip. begins when product of Precip. begins when product of
ion concs. EXCEEDS the Kion concs. EXCEEDS the Kspsp..
Precipitating an Insoluble SaltPrecipitating an Insoluble SaltHgHg22ClCl22(s) (s) Hg Hg22
2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)
KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]]22
SolutionSolution
[Cl[Cl--] that can exist when [Hg] that can exist when [Hg222+2+] = 0.010 ] = 0.010
M,M,
Precipitating an Insoluble SaltPrecipitating an Insoluble SaltHgHg22ClCl22(s) (s) Hg Hg22
2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)
KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]]22
SolutionSolution
[Cl[Cl--] that can exist when [Hg] that can exist when [Hg222+2+] = 0.010 ] = 0.010
M,M,[Cl ] =
Ksp
0.010 = 1.1 x 10-8 M[Cl ] =
Ksp
0.010 = 1.1 x 10-8 M
Precipitating an Insoluble SaltPrecipitating an Insoluble SaltHgHg22ClCl22(s) (s) Hg Hg22
2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)
KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]]22
SolutionSolution
[Cl[Cl--] that can exist when [Hg] that can exist when [Hg222+2+] = 0.010 M,] = 0.010 M,
If this conc. of ClIf this conc. of Cl-- is just exceeded, Hg is just exceeded, Hg22ClCl22
begins to precipitate.begins to precipitate.
[Cl ] = Ksp
0.010 = 1.1 x 10-8 M[Cl ] =
Ksp
0.010 = 1.1 x 10-8 M
Precipitating an Insoluble SaltPrecipitating an Insoluble SaltHgHg22ClCl22(s) (s) Hg Hg22
2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)
KKspsp = 1.1 x 10 = 1.1 x 10-18-18
Now raise [ClNow raise [Cl--] to 1.0 M. What is the value ] to 1.0 M. What is the value of [Hgof [Hg22
2+2+] at this point?] at this point?
SolutionSolution
[Hg[Hg222+2+] = K] = Ksp sp / [Cl/ [Cl--]]22
= K= Kspsp / (1.0) / (1.0)22 = 1.1 x 10 = 1.1 x 10-18-18 M M
The concentration of HgThe concentration of Hg222+2+ has been has been
reduced by 10reduced by 101616 ! !
Separating Metal Ions Separating Metal Ions CuCu2+2+, Ag, Ag++, Pb, Pb2+2+
Separating Metal Ions Separating Metal Ions CuCu2+2+, Ag, Ag++, Pb, Pb2+2+
Ksp ValuesAgCl 1.8 x 10-10
PbCl2 1.7 x 10-5
PbCrOPbCrO4 4 1.8 x 101.8 x 10-14-14
Ksp ValuesAgCl 1.8 x 10-10
PbCl2 1.7 x 10-5
PbCrOPbCrO4 4 1.8 x 101.8 x 10-14-14
Separating Salts by Differences in KspSeparating Salts by Differences in KspA solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add . Add
CrOCrO442-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow
PbCrOPbCrO44. Which precipitates first?. Which precipitates first?
KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12
KKsp sp for PbCrOfor PbCrO4 4 = 1.8 x 10= 1.8 x 10-14-14
SolutionSolution
The substance whose KThe substance whose Kspsp is first is first exceeded precipitates first. exceeded precipitates first.
The ion requiring the lesser amount The ion requiring the lesser amount of CrOof CrO44
2-2- ppts. first. ppts. first.
Separating Salts by Differences in KspSeparating Salts by Differences in KspA solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. .
Add CrOAdd CrO442-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and and
yellow PbCrOyellow PbCrO44. Which precipitates first?. Which precipitates first?
KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12
KKsp sp for PbCrOfor PbCrO4 4 = 1.8 x 10= 1.8 x 10-14-14
SolutionSolution
Calculate [CrOCalculate [CrO442-2-] required by each ion. ] required by each ion.
Separating Salts by Differences in KspSeparating Salts by Differences in Ksp
[CrO[CrO442-2-] to ppt. PbCrO] to ppt. PbCrO4 4 = K= Ksp sp / [Pb/ [Pb2+2+] ]
= 1.8 x 10= 1.8 x 10-14-14 / 0.020 = 9.0 x 10 / 0.020 = 9.0 x 10-13-13 M M
[CrO[CrO442-2-] to ppt. Ag] to ppt. Ag22CrOCrO4 4 = K= Ksp sp / [Ag/ [Ag++]]22
= 9.0 x 10= 9.0 x 10-12-12 / (0.020) / (0.020)22 = 2.3 x 10 = 2.3 x 10-8-8 M M
PbCrOPbCrO44 precipitates first. precipitates first.
A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add . Add CrOCrO44
2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. Which precipitates first?. Which precipitates first?
KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12
KKsp sp for PbCrOfor PbCrO4 4 = 1.8 x 10= 1.8 x 10-14-14
SolutionSolution
A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. . Add CrOAdd CrO44
2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and and yellow PbCrOyellow PbCrO44. PbCrO. PbCrO44 ppts. first. ppts. first.
KKspsp (Ag (Ag22CrOCrO44)= 9.0 x 10)= 9.0 x 10-12-12
KKsp sp (PbCrO(PbCrO44) = 1.8 x 10) = 1.8 x 10-14-14
How much PbHow much Pb2+2+ remains in solution when Ag remains in solution when Ag++ begins to precipitate?begins to precipitate?
Separating Salts by Differences in KspSeparating Salts by Differences in Ksp
A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add . Add CrOCrO44
2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. PbCrO. PbCrO44 ppts. first. ppts. first.
KKspsp (Ag (Ag22CrOCrO44)= 9.0 x 10)= 9.0 x 10-12-12
KKsp sp (PbCrO(PbCrO44) = 1.8 x 10) = 1.8 x 10-14-14
How much PbHow much Pb2+2+ remains in solution when Ag remains in solution when Ag++ begins to precipitate?begins to precipitate?
SolutionSolutionWe know that [CrOWe know that [CrO44
2-2-] = 2.3 x 10] = 2.3 x 10-8-8 M to begin to M to begin to ppt. Agppt. Ag22CrOCrO44. .
What is the PbWhat is the Pb2+2+ conc. at this point? conc. at this point?
Separating Salts by Differences in KspSeparating Salts by Differences in Ksp
A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add . Add CrOCrO44
2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. .
KKspsp (Ag (Ag22CrOCrO44)= 9.0 x 10)= 9.0 x 10-12-12
KKsp sp (PbCrO(PbCrO44) = 1.8 x 10) = 1.8 x 10-14-14
How much PbHow much Pb2+2+ remains in solution when Ag remains in solution when Ag++ begins to precipitate?begins to precipitate?
SolutionSolution
[Pb[Pb2+2+] = K] = Kspsp / [CrO / [CrO442-2-] = 1.8 x 10] = 1.8 x 10-14-14 / 2.3 x 10 / 2.3 x 10-8-8 M M
= 7.8 x 10= 7.8 x 10-7-7 M M
Separating Salts by Differences in KspSeparating Salts by Differences in Ksp
A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add . Add CrOCrO44
2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. .
KKspsp (Ag (Ag22CrOCrO44)= 9.0 x 10)= 9.0 x 10-12-12
KKsp sp (PbCrO(PbCrO44) = 1.8 x 10) = 1.8 x 10-14-14
How much PbHow much Pb2+2+ remains in solution when Ag remains in solution when Ag++ begins to precipitate?begins to precipitate?
SolutionSolution
[Pb[Pb2+2+] = K] = Kspsp / [CrO / [CrO442-2-] = 1.8 x 10] = 1.8 x 10-14-14 / 2.3 x 10 / 2.3 x 10-8-8 M M
= 7.8 x 10= 7.8 x 10-7-7 M M
Lead ion has dropped from 0.020 M to < 10Lead ion has dropped from 0.020 M to < 10-6-6 M M
Separating Salts by Differences in KspSeparating Salts by Differences in Ksp
Entropy and Free EnergyEntropy and Free EnergyEntropy and Free EnergyEntropy and Free EnergyHow to predict if a How to predict if a
reaction can occur, reaction can occur, given enough time?given enough time?
THERMODYNAMICSTHERMODYNAMICS
How to predict if a How to predict if a reaction can occur at a reaction can occur at a reasonable rate?reasonable rate?
KINETICSKINETICS
Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
ThermodynamicsThermodynamicsThermodynamicsThermodynamicsIs the state of a chemical system such that Is the state of a chemical system such that
a rearrangement of its atoms and molecules a rearrangement of its atoms and molecules would decrease the energy of the system? would decrease the energy of the system?
If yes, system is favored to react — a If yes, system is favored to react — a product-favoredproduct-favored system.system.
Most product-favored reactions are Most product-favored reactions are exothermic.exothermic.
Often referred to as Often referred to as spontaneousspontaneous reactions.reactions.
Spontaneous does not imply anything about Spontaneous does not imply anything about time for reaction to occur.time for reaction to occur.
Product-Favored ReactionsProduct-Favored ReactionsIn general, product-In general, product-
favored reactions are favored reactions are exothermicexothermic..
FeFe22OO33(s) + 2 Al(s) (s) + 2 Al(s)
---> 2 Fe(s) + ---> 2 Fe(s) + AlAl22OO33(s)(s)
H = - 848 kJH = - 848 kJ
Entropy, SEntropy, SEntropy, SEntropy, SOne property common to One property common to
product-favored processes is product-favored processes is that the final state is more that the final state is more DISORDEREDDISORDERED or or RANDOMRANDOM than the original.than the original.
Spontaneity is related to an Spontaneity is related to an increase in randomness.increase in randomness.
The thermodynamic property The thermodynamic property related to randomness is related to randomness is ENTROPY, SENTROPY, S..
Reaction of Reaction of K with K with waterwater
How probable is it that reactant How probable is it that reactant molecules will react? molecules will react?
PROBABILITYPROBABILITY suggests that a suggests that a product-favored reaction will product-favored reaction will result in the result in the dispersal of dispersal of energy or of matter energy or of matter or both.or both.
Directionality of ReactionsDirectionality of Reactions
S (gases) > S (liquids) > S S (gases) > S (liquids) > S (solids)(solids)
SSoo (J/K•mol) (J/K•mol)
HH22O(liq)O(liq) 69.9169.91
HH22O(gas)O(gas)188.8 188.8
SSoo (J/K•mol) (J/K•mol)
HH22O(liq)O(liq) 69.9169.91
HH22O(gas)O(gas)188.8 188.8
Entropy, SEntropy, SEntropy, SEntropy, S
Increase in molecular complexity Increase in molecular complexity generally leads to increase in S.generally leads to increase in S.
SSoo (J/K•mol) (J/K•mol)
CHCH44 248.2248.2
CC22HH66 336.1 336.1
CC33HH88 419.4419.4
SSoo (J/K•mol) (J/K•mol)
CHCH44 248.2248.2
CC22HH66 336.1 336.1
CC33HH88 419.4419.4
Entropy, SEntropy, SEntropy, SEntropy, S
Entropy Changes for Phase Entropy Changes for Phase ChangesChanges
For a phase change, For a phase change,
S = q/TS = q/Twhere q = heat transferred in where q = heat transferred in
phase changephase change
For HFor H22O (liq) ---> HO (liq) ---> H22O(g)O(g)
H = q = +40,700 J/molH = q = +40,700 J/mol
Consider 2 HConsider 2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)
SSoo = 2 S = 2 Soo (H (H22O) - [2 SO) - [2 Soo (H (H22) + S) + Soo (O (O22)])]
SSoo = 2 mol (69.9 J/K•mol) - = 2 mol (69.9 J/K•mol) - [2 mol (130.7 J/K•mol) + [2 mol (130.7 J/K•mol) +
1 mol (205.3 1 mol (205.3 J/K•mol)]J/K•mol)]
SSoo = -326.9 J/K = -326.9 J/K
Note that there is a Note that there is a decrease in S decrease in S because because 3 mol of gas give 2 mol of liquid.3 mol of gas give 2 mol of liquid.
Calculating Calculating S for a ReactionS for a Reaction
SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)
2nd Law of Thermodynamics2nd Law of ThermodynamicsA reaction is spontaneous (product-favored) if A reaction is spontaneous (product-favored) if
²S for the universe is positive.²S for the universe is positive.
SSuniverseuniverse = = SSsystemsystem + + SSsurroundingssurroundings
SSuniverseuniverse > 0 for product-favored > 0 for product-favored processprocess
First calc. entropy created by matter dispersal First calc. entropy created by matter dispersal ((SSsystemsystem))
Next, calc. entropy created by energy dispersal Next, calc. entropy created by energy dispersal ((SSsurroundsurround))
CELL POTENTIAL, ECELL POTENTIAL, ECELL POTENTIAL, ECELL POTENTIAL, E
Electrons are “driven” from anode to cathode Electrons are “driven” from anode to cathode by an electromotive force or by an electromotive force or emfemf..
For Zn/Cu cell, this is indicated by a voltage of For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 1.10 V at 25 C and when [ZnC and when [Zn2+2+] and [Cu] and [Cu2+2+] = ] = 1.0 M.1.0 M.
Zn and ZnZn and Zn2+2+,,anodeanode
Cu and CuCu and Cu2+2+,,cathodecathode
Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
CELL POTENTIAL, ECELL POTENTIAL, ECELL POTENTIAL, ECELL POTENTIAL, EFor Zn/Cu cell, voltage is 1.10 V at 25 For Zn/Cu cell, voltage is 1.10 V at 25 C C
and when [Znand when [Zn2+2+] and [Cu] and [Cu2+2+] = 1.0 M.] = 1.0 M.This is the This is the STANDARD CELL STANDARD CELL
POTENTIAL, EPOTENTIAL, Eoo
——a quantitative measure of the tendency a quantitative measure of the tendency of reactants to proceed to products when of reactants to proceed to products when all are in their standard states at 25 all are in their standard states at 25 C. C.
Calculating Cell VoltageCalculating Cell VoltageCalculating Cell VoltageCalculating Cell VoltageBalanced half-reactions can be added Balanced half-reactions can be added
together to get overall, balanced equation. together to get overall, balanced equation.
If we know EIf we know Eoo for each half-reaction, we for each half-reaction, we could get Ecould get Eoo for net reaction. for net reaction.
2 I- ---> I2 + 2e-
2 H2O + 2e- ---> 2 OH- + H2
-------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
2 I- ---> I2 + 2e-
2 H2O + 2e- ---> 2 OH- + H2
-------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
CELL POTENTIALS, ECELL POTENTIALS, EooCELL POTENTIALS, ECELL POTENTIALS, Eoo
Can’t measure 1/2 reaction Eo directly. Therefore, measure it relative to a STANDARD HALF CELL, SHE.
2 H2 H++(aq, 1 M) + 2e- (aq, 1 M) + 2e- -->--> H H22(g, 1 (g, 1 atm)atm)
2 H2 H++(aq, 1 M) + 2e- (aq, 1 M) + 2e- -->--> H H22(g, 1 (g, 1 atm)atm)
EEoo = 0.0 V = 0.0 V
Volts
ZnH2
Salt Bridge
Zn2+ H+
Zn Zn2+ + 2e- OXIDATION ANODE
2 H+ + 2e- H2REDUCTIONCATHODE
- +
Volts
ZnH2
Salt Bridge
Zn2+ H+
Zn Zn2+ + 2e- OXIDATION ANODE
2 H+ + 2e- H2REDUCTIONCATHODE
- +
Zn/Zn2+ half-cell hooked to a SHE.Eo for the cell = +0.76 V
Zn/Zn2+ half-cell hooked to a SHE.Eo for the cell = +0.76 V
Volts
ZnH2
Salt Bridge
Zn2+ H+
Zn Zn2+ + 2e- OXIDATION ANODE
2 H+ + 2e- H2REDUCTIONCATHODE
- +
Volts
ZnH2
Salt Bridge
Zn2+ H+
Zn Zn2+ + 2e- OXIDATION ANODE
2 H+ + 2e- H2REDUCTIONCATHODE
- +
Overall reaction is reduction of HOverall reaction is reduction of H++ by Zn by Zn
metal.metal.
Zn(s) + 2 HZn(s) + 2 H++ (aq) --> Zn (aq) --> Zn2+2+ + H + H22(g)(g)
EEoo = +0.76 V = +0.76 V
Therefore, ETherefore, Eoo for Zn ---> Zn for Zn ---> Zn2+2+ (aq) + (aq) +
2e- is2e- is
+0.76 V+0.76 V..
Zn is a Zn is a betterbetter reducing agent than H reducing agent than H22..
TABLE OF STANDARD POTENTIALSTABLE OF STANDARD POTENTIALSTABLE OF STANDARD POTENTIALSTABLE OF STANDARD POTENTIALS
Eo (V)
Cu2+ + 2e- Cu +0.34
2 H+ + 2e- H2 0.00
Zn2+ + 2e- Zn -0.76
oxidizingability of ion
reducing abilityof element
Eo (V)
Cu2+ + 2e- Cu +0.34
2 H+ + 2e- H2 0.00
Zn2+ + 2e- Zn -0.76
oxidizingability of ion
reducing abilityof element
EEoo and and GGooEEoo and and GGoo
EEoo is related to is related to GGoo, the free , the free energy change for the reaction.energy change for the reaction.
GGoo = - n F E = - n F Eoo where F = Faraday constant where F = Faraday constant
= 9.6485 x 10= 9.6485 x 1044 J/V•molJ/V•mol
and n is the number of moles of and n is the number of moles of electrons transferredelectrons transferred
Michael FaradayMichael Faraday1791-18671791-1867
Michael FaradayMichael Faraday1791-18671791-1867Michael FaradayMichael Faraday1791-18671791-1867
Originated the terms anode, Originated the terms anode, cathode, anion, cation, cathode, anion, cation, electrode.electrode.
Discoverer of Discoverer of electrolysiselectrolysismagnetic props. of mattermagnetic props. of matterpretty cool guypretty cool guyelectromagnetic inductionelectromagnetic inductionbenzene and other organic benzene and other organic
chemicalschemicalsWas a popular lecturer.Was a popular lecturer.
EEoo and and GGooEEoo and and GGoo
GGoo = - n F E = - n F Eoo For a For a product-favoredproduct-favored reaction reaction Reactants ----> ProductsReactants ----> Products
GGo o < 0 and so E < 0 and so Eo o > 0 > 0EEoo is positive is positive
For a For a reactant-favoredreactant-favored reaction reaction Reactants <---- ProductsReactants <---- Products
GGo o > 0 and so E > 0 and so Eo o < 0 < 0EEoo is negative is negative
Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistryQuantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
Consider electrolysis of aqueous silver Consider electrolysis of aqueous silver ion.ion.
AgAg++ (aq) + e- ---> Ag(s) (aq) + e- ---> Ag(s)
1 mol e-1 mol e- ---> 1 mol Ag---> 1 mol Ag
If we could measure the moles of e-, we If we could measure the moles of e-, we could know the quantity of Ag formed.could know the quantity of Ag formed.
But how to measure moles of e-?But how to measure moles of e-?
Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistryQuantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
Consider electrolysis of aqueous silver Consider electrolysis of aqueous silver ion.ion.
AgAg++ (aq) + e- ---> Ag(s) (aq) + e- ---> Ag(s)
1 mol e-1 mol e- ---> 1 mol Ag---> 1 mol Ag
If we could measure the moles of e-, we If we could measure the moles of e-, we could know the quantity of Ag formed.could know the quantity of Ag formed.
But how to measure moles of e-?But how to measure moles of e-?Current =
charge passingtime
Current = charge passing
time
Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistryQuantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
Consider electrolysis of aqueous silver Consider electrolysis of aqueous silver ion.ion.
AgAg++ (aq) + e- ---> Ag(s) (aq) + e- ---> Ag(s)
1 mol e-1 mol e- ---> 1 mol Ag---> 1 mol Ag
If we could measure the moles of e-, we If we could measure the moles of e-, we could know the quantity of Ag formed.could know the quantity of Ag formed.
But how to measure moles of e-?But how to measure moles of e-?Current = charge passing
timeCurrent =
charge passingtime
I (amps) = coulombsseconds
I (amps) = coulombsseconds
But how is charge related to moles of But how is charge related to moles of electrons?electrons?
Charge on 1 mol of e- = Charge on 1 mol of e- =
(1.60 x 10(1.60 x 10-19 -19 C/e-)(6.02 x 10C/e-)(6.02 x 102323 e-/mol) e-/mol)
= = 96,500 C/mol e- 96,500 C/mol e- = = 1 Faraday1 Faraday
Current = charge passing
timeCurrent =
charge passingtime
I (amps) = coulombsseconds
I (amps) = coulombsseconds
Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistryQuantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for (aq) solution for 15.0 min. What mass of Ag metal is 15.0 min. What mass of Ag metal is deposited?deposited?
SolutionSolution
(a)(a) Calc. chargeCalc. charge
Coulombs = amps x timeCoulombs = amps x time
= (1.5 amps)(15.0 min)(60 s/min) = = (1.5 amps)(15.0 min)(60 s/min) = 1350 C1350 C
I (amps) = coulombsseconds
I (amps) = coulombsseconds
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 min. What (aq) solution for 15.0 min. What mass of Ag metal is deposited?mass of Ag metal is deposited?
SolutionSolution
(a)(a) Charge = 1350 CCharge = 1350 C
(b)(b) Calculate moles of e- usedCalculate moles of e- used
I (amps) = coulombsseconds
I (amps) = coulombsseconds
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 min. What (aq) solution for 15.0 min. What mass of Ag metal is deposited?mass of Ag metal is deposited?
SolutionSolution
(a)(a) Charge = 1350 CCharge = 1350 C
(b)(b) Calculate moles of e- usedCalculate moles of e- used
I (amps) = coulombsseconds
I (amps) = coulombsseconds
1350 C • 1 mol e -96, 500 C
0.0140 mol e -1350 C • 1 mol e -96, 500 C
0.0140 mol e -
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 min. What (aq) solution for 15.0 min. What mass of Ag metal is deposited?mass of Ag metal is deposited?
SolutionSolution
(a)(a) Charge = 1350 CCharge = 1350 C
(b)(b) Calculate moles of e- usedCalculate moles of e- used
(c)(c) Calc. quantity of AgCalc. quantity of Ag
I (amps) = coulombsseconds
I (amps) = coulombsseconds
1350 C • 1 mol e -96, 500 C
0.0140 mol e -1350 C • 1 mol e -96, 500 C
0.0140 mol e -
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 min. What (aq) solution for 15.0 min. What mass of Ag metal is deposited?mass of Ag metal is deposited?
SolutionSolution
(a)(a) Charge = 1350 CCharge = 1350 C
(b)(b) Calculate moles of e- usedCalculate moles of e- used
(c)(c) Calc. quantity of AgCalc. quantity of Ag
I (amps) = coulombsseconds
I (amps) = coulombsseconds
1350 C • 1 mol e -96, 500 C
0.0140 mol e -1350 C • 1 mol e -96, 500 C
0.0140 mol e -
0.0140 mol e - • 1 mol Ag1 mol e -
0.0140 mol Ag or 1.51 g Ag0.0140 mol e - • 1 mol Ag1 mol e -
0.0140 mol Ag or 1.51 g Ag
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryThe anode reaction in a lead storage battery The anode reaction in a lead storage battery
isis
Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + (aq) +
2e-2e-
If a battery delivers 1.50 amp, and you have If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?454 g of Pb, how long will the battery last?
SolutionSolution
a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb
b)b) Calculate moles of e-Calculate moles of e-
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryThe anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is
Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-
If a battery delivers 1.50 amp, and you have 454 g of Pb, If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?how long will the battery last?
SolutionSolution
a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb
b)b) Calculate moles of e-Calculate moles of e-
2.19 mol Pb • 2 mol e -1 mol Pb
= 4.38 mol e -2.19 mol Pb • 2 mol e -1 mol Pb
= 4.38 mol e -
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryThe anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is
Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-
If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?the battery last?
SolutionSolution
a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb
b)b) Calculate moles of e-Calculate moles of e-
c)c) Calculate chargeCalculate charge
2.19 mol Pb • 2 mol e -1 mol Pb
= 4.38 mol e -2.19 mol Pb • 2 mol e -1 mol Pb
= 4.38 mol e -
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryThe anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is
Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-
If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?the battery last?
SolutionSolution
a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb
b)b) Calculate moles of e-Calculate moles of e-
c)c) Calculate chargeCalculate charge
4.38 mol e- • 96,500 C/mol e- = 423,000 C4.38 mol e- • 96,500 C/mol e- = 423,000 C
2.19 mol Pb • 2 mol e -1 mol Pb
= 4.38 mol e -2.19 mol Pb • 2 mol e -1 mol Pb
= 4.38 mol e -
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryThe anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is
Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-
If a battery delivers 1.50 amp, and you have 454 g of Pb, If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?how long will the battery last?
SolutionSolution
a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb
b)b) Mol of e- = 4.38 molMol of e- = 4.38 mol
c)c) Charge = 423,000 CCharge = 423,000 C
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryThe anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is
Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-
If a battery delivers 1.50 amp, and you have 454 g of Pb, If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?how long will the battery last?
SolutionSolution
a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb
b)b) Mol of e- = 4.38 molMol of e- = 4.38 mol
c)c) Charge = 423,000 CCharge = 423,000 C
d)d) Calculate time Calculate time
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryThe anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is
Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-
If a battery delivers 1.50 amp, and you have 454 g of Pb, If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?how long will the battery last?
SolutionSolution
a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb
b)b) Mol of e- = 4.38 molMol of e- = 4.38 mol
c)c) Charge = 423,000 CCharge = 423,000 C
d)d) Calculate time Calculate time Time (s) = Charge (C)
I (amps)Time (s) =
Charge (C)I (amps)
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryThe anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is
Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-
If a battery delivers 1.50 amp, and you have 454 g of Pb, If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?how long will the battery last?
SolutionSolution
a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb
b)b) Mol of e- = 4.38 molMol of e- = 4.38 mol
c)c) Charge = 423,000 CCharge = 423,000 C
d)d) Calculate time Calculate time Time (s) = Charge (C)
I (amps)Time (s) =
Charge (C)I (amps)
Time (s) = 423, 000 C1.50 amp
= 282,000 sTime (s) = 423, 000 C1.50 amp
= 282,000 s About 78 hoursAbout 78 hours