Dry Cell Battery Anode (-) Zn ---> Zn 2+ + 2e- Cathode (+) 2 NH 4 + + 2e- ---> 2 NH 3 + H 2 Common...

Dry Cell BatteryDry Cell BatteryDry Cell BatteryDry Cell BatteryAnode (-)Anode (-)

Zn ---> ZnZn ---> Zn2+2+ + 2e- + 2e-

Cathode (+)Cathode (+)

2 NH2 NH44++ + 2e- ---> + 2e- --->

2 2 NHNH33 + H + H22

Common Common dry celldry cell

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

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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

Nearly same reactions as in common dry Nearly same reactions as in common dry cell, but under basic conditions.cell, but under basic conditions.

Alkaline BatteryAlkaline BatteryAlkaline BatteryAlkaline Battery

33


Anode:Anode:

Zn is reducing agent under basic Zn is reducing agent under basic conditionsconditions

Cathode:Cathode:

HgO + HHgO + H22O + 2e- ---> Hg + 2 OHO + 2e- ---> Hg + 2 OH--

Mercury BatteryMercury BatteryMercury BatteryMercury Battery

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Lead Storage BatteryLead Storage BatteryLead Storage BatteryLead Storage BatteryAnode (-) Anode (-) EEoo = +0.36 V = +0.36 V

Pb + HSOPb + HSO44-- ---> PbSO ---> PbSO44 + H + H++ + 2e- + 2e-

Cathode (+) Cathode (+) EEoo = +1.68 V = +1.68 V

PbOPbO22 + HSO + HSO44-- + 3 H + 3 H++ + 2e- + 2e-

---> PbSO---> PbSO44 + 2 H + 2 H22OO

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Ni-Cad BatteryNi-Cad BatteryNi-Cad BatteryNi-Cad BatteryAnode (-)Anode (-)

Cd + 2 OHCd + 2 OH-- ---> Cd(OH) ---> Cd(OH)22 + 2e- + 2e-

Cathode (+) Cathode (+)

NiO(OH) + HNiO(OH) + H22O + e- ---> Ni(OH)O + e- ---> Ni(OH)22 + OH + OH--

Electrolysis of Aqueous NaOHElectrolysis of Aqueous NaOHElectrolysis of Aqueous NaOHElectrolysis of Aqueous NaOH

Anode (+) Anode (+) EEoo = -0.40 V = -0.40 V

4 OH4 OH-- ---> O ---> O22(g) + 2 H(g) + 2 H22O + 2e-O + 2e-

Cathode (-) Cathode (-) EEoo = -0.83 V = -0.83 V

4 H4 H22O + 4e- ---> 2 HO + 4e- ---> 2 H22 + 4 OH + 4 OH--

EEoo for cell = -1.23 V for cell = -1.23 V

Electric Energy ----> Chemical ChangeElectric Energy ----> Chemical Change


77


ElectrolysisElectrolysisElectric Energy ---> Chemical ChangeElectric Energy ---> Chemical Change

ElectrolysisElectrolysisElectric Energy ---> Chemical ChangeElectric Energy ---> Chemical Change

BATTERY

+

Na+Cl-

Anode Cathode

electrons

BATTERY

+

Na+Cl-

Anode Cathode

electrons

Electrolysis of Electrolysis of molten NaCl.molten NaCl.Here a battery Here a battery “pumps” “pumps” electrons from electrons from ClCl-- to Na to Na++..

88


Electrolysis of Molten NaClElectrolysis of Molten NaClElectrolysis of Molten NaClElectrolysis of Molten NaCl


2 Cl2 Cl-- ---> Cl ---> Cl22(g) + 2e-(g) + 2e-


NaNa++ + e- ---> Na + e- ---> Na


External energy needed External energy needed because Ebecause Eoo is (-). is (-).

Note that signs of electrodes Note that signs of electrodes are reversed from batteries.are reversed from batteries.

BATTERY

+

Na+Cl-

Anode Cathode

electrons

BATTERY

+

Na+Cl-

Anode Cathode

electrons

99


Electrolysis of Aqueous NaClElectrolysis of Aqueous NaClElectrolysis of Aqueous NaClElectrolysis of Aqueous NaClAnode (+) Anode (+) EEoo = -1.36 V = -1.36 V

2 Cl2 Cl-- ---> Cl ---> Cl22(g) + 2e-(g) + 2e-


2 H2 H22O + 2e- ---> HO + 2e- ---> H22 + 2 OH + 2 OH--


Note that HNote that H22O is more O is more easily reduced easily reduced than Nathan Na++. .

BATTERY

+

Na+Cl-

Anode Cathode

H2O

electrons

BATTERY

+

Na+Cl-

Anode Cathode

H2O

electrons

1010


Electrolysis of Aqueous NaClElectrolysis of Aqueous NaClElectrolysis of Aqueous NaClElectrolysis of Aqueous NaClCells like these are the source of NaOH Cells like these are the source of NaOH

and Cland Cl22..

In 1995In 1995

25.1 x 1025.1 x 1099 lb Cl lb Cl22

26.1 x 1026.1 x 1099 lb NaOH lb NaOH

Also the source Also the source of NaOCl for of NaOCl for use in bleach.use in bleach.

1111


Electrolysis of Aqueous CuClElectrolysis of Aqueous CuCl22Electrolysis of Aqueous CuClElectrolysis of Aqueous CuCl22


2 Cl2 Cl-- ---> Cl ---> Cl22(g) + 2e-(g) + 2e-

Cathode (-) Cathode (-) EEoo = +0.34 V = +0.34 V

CuCu2+2+ + 2e- ---> Cu + 2e- ---> Cu


Note that Cu is more Note that Cu is more easily reduced than easily reduced than either Heither H22O or NaO or Na++. .

BATTERY

+

Cu2+Cl-

Anode Cathode

electrons

H2O

BATTERY

+

Cu2+Cl-

Anode Cathode

electrons

H2O

1212


Producing AluminumProducing AluminumProducing AluminumProducing Aluminum2 Al2 Al22OO33 + 3 C ---> 4 Al + 3 CO + 3 C ---> 4 Al + 3 CO22

Charles Hall (1863-1914) developed Charles Hall (1863-1914) developed electrolysis process. Founded Alcoa.electrolysis process. Founded Alcoa.

Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry


Consider electrolysis of aqueous silver ion.Consider electrolysis of aqueous silver 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-?


1414









Current = charge passing

timeCurrent =

charge passingtime

1515










timeCurrent =

charge passingtime

I (amps) = coulombsseconds


1616


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


timeCurrent =

charge passingtime





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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 (aq) solution for 15.0 min. What mass of Ag metal is deposited?min. What mass of Ag metal is deposited?

SolutionSolution

(a)(a) Calc. chargeCalc. charge

Coulombs = amps x timeCoulombs = amps x time

= (1.5 amps)(15.0 min)(60 s/min) = 1350 C= (1.5 amps)(15.0 min)(60 s/min) = 1350 C



1818



1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 min. What mass (aq) solution for 15.0 min. What mass of Ag metal is deposited?of Ag metal is deposited?

SolutionSolution

(a)(a) Charge = 1350 CCharge = 1350 C

(b)(b) Calculate moles of e- usedCalculate moles of e- used



1919




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 -

2020




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 -

2121




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

2222



The 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) + 2e-(aq) + 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-

2323



The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is


If a battery delivers 1.50 amp, and you have 454 g of Pb, how If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?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 -

2424


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


If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?battery last?

SolutionSolution



c)c) Calculate chargeCalculate charge



= 4.38 mol e -

2525


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


If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?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 -

2626






SolutionSolution


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

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

2727






SolutionSolution




d)d) Calculate time Calculate time

2828






SolutionSolution




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

I (amps)Time (s) =

Charge (C)I (amps)

2929






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

Dry Cell Battery Anode (-) Zn ---> Zn 2+ + 2e- Cathode (+) 2 NH 4 + + 2e- ---> 2 NH 3 + H 2 Common...

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Transcript of Dry Cell Battery Anode (-) Zn ---> Zn 2+ + 2e- Cathode (+) 2 NH 4 + + 2e- ---> 2 NH 3 + H 2 Common...