SPRING REVIEW PART TWO. PRECIPITATION REACTIONS Chapter 19 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to

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


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


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


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??


2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)


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..


2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)


SolutionSolution

[Cl[Cl--] that can exist when [Hg] that can exist when [Hg222+2+] = 0.010 ] = 0.010

M,M,


2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)


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


2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)


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


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?



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?



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?



2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. PbCrO. PbCrO44 ppts. first. ppts. first.




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?



2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. .




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



2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. .




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


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


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


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.


CHCH44 248.2248.2

CC22HH66 336.1 336.1

CC33HH88 419.4419.4


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


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+ 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+



- +

Volts

ZnH2

Salt Bridge

Zn2+ H+



- +

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-?







But how to measure moles of e-?But how to measure moles of e-?Current =

charge passingtime

Current = charge passing

time






But how to measure moles of e-?But how to measure moles of e-?Current = charge passing

timeCurrent =

charge passingtime

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




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




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





SolutionSolution





1350 C • 1 mol e -96, 500 C

0.0140 mol e -1350 C • 1 mol e -96, 500 C

0.0140 mol e -



SolutionSolution



(c)(c) Calc. quantity of AgCalc. quantity of Ag



1350 C • 1 mol e -96, 500 C

0.0140 mol e -1350 C • 1 mol e -96, 500 C

0.0140 mol e -



SolutionSolution



(c)(c) Calc. quantity of AgCalc. quantity of Ag



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



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 -



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



c)c) Calculate chargeCalculate charge



= 4.38 mol e -



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



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



= 4.38 mol e -




SolutionSolution


b)b) Mol of e- = 4.38 molMol of e- = 4.38 mol

c)c) Charge = 423,000 CCharge = 423,000 C




SolutionSolution




d)d) Calculate time Calculate time




SolutionSolution




d)d) Calculate time Calculate time Time (s) = Charge (C)

I (amps)Time (s) =

Charge (C)I (amps)




SolutionSolution




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

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SPRING REVIEW PART TWO. PRECIPITATION REACTIONS Chapter 19 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to