1 Electrochemistry Chapter 20. 2 Half-reaction method…remember? Example Example Al (s) + Cu 2+...

11

ElectrochemistryElectrochemistry

Chapter 20Chapter 20

22

Half-reaction method…Half-reaction method…remember?remember?• ExampleExample

AlAl(s)(s) + Cu + Cu2+2+(aq)(aq) Al Al3+3+

(aq)(aq) + Cu + Cu(s)(s)

• Oxidation: AlOxidation: Al(s) (s) Al Al3+3+(aq)(aq) + 3e + 3e--

• Reduction: 2eReduction: 2e- - + Cu+ Cu2+2+(aq)(aq) Cu Cu(s)(s)

– Use lowest common multiple to make both equivalent in number of Use lowest common multiple to make both equivalent in number of electronselectrons

• Oxidation Oxidation multiply by 2 multiply by 2– 2Al2Al(s) (s) 2Al 2Al3+3+

(aq)(aq) + 6e + 6e--

• Reduction Reduction multiply by 3 multiply by 3– 6e6e- - + 3Cu+ 3Cu2+2+

(aq)(aq) 3Cu 3Cu(s)(s)

– Collate (electrons cross out)Collate (electrons cross out)• Net reaction:Net reaction:

2Al2Al(s)(s) + 3Cu + 3Cu2+2+(aq)(aq) 2Al 2Al3+3+

(aq)(aq) + 3Cu + 3Cu(s)(s)

DOES EVERYTHING BALANCE?DOES EVERYTHING BALANCE?(Make sure to balance after every step!)(Make sure to balance after every step!)

33

In acidic milieuIn acidic milieu

FeFe2+2+(aq)(aq) + MnO + MnO44

--(aq)(aq) Fe Fe3+3+

(aq)(aq) + Mn + Mn2+2+(aq)(aq)

• Oxidation:Oxidation:FeFe2+2+

(aq)(aq) Fe Fe3+3+(aq)(aq) + e + e--

• Reduction: Reduction: 8H8H++

(aq)(aq) + MnO + MnO44--(aq)(aq) 4H 4H22OO(l) (l) + Mn+ Mn2+2+

(aq)(aq)– What did I do in the above half-rxn?What did I do in the above half-rxn?– Is it fully balanced? Is it fully balanced?

5e5e-- + 8H + 8H++(aq)(aq) + MnO + MnO44

--(aq)(aq) 4H 4H22OO(l) (l) + Mn+ Mn2+2+

(aq)(aq)

• Balance both half-reactions:Balance both half-reactions:5Fe5Fe2+2+

(aq)(aq) 5Fe 5Fe3+3+(aq)(aq) + 5e + 5e- - (multiply by 5; why?)(multiply by 5; why?)

5e5e-- + 8H + 8H++(aq)(aq) + MnO + MnO44

--(aq)(aq) 4H 4H22OO(l) (l) + Mn+ Mn2+2+

(aq)(aq)

• Collate Collate • Net rxn:Net rxn:

5Fe5Fe2+2+(aq)(aq) + 8H + 8H++

(aq)(aq) + MnO + MnO44--(aq)(aq) 5Fe 5Fe3+3+

(aq)(aq) +4H +4H22OO(l) (l) + Mn+ Mn2+2+(aq)(aq)

44

SolveSolve

VOVO22++

(aq) (aq) + Zn+ Zn(s)(s) VO VO2+2+(aq)(aq) + Zn + Zn2+2+

(aq)(aq)

55

Answer Answer

VOVO22++

(aq) (aq) + Zn+ Zn(s)(s) VO VO2+2+(aq)(aq) + Zn + Zn2+2+

(aq)(aq)

• Oxidation:Oxidation:ZnZn(s)(s) Zn Zn2+2+

(aq) (aq) + 2e+ 2e--

• Reduction: Reduction: ee- - ++ 2H2H++

(aq)(aq) + VO + VO22++

(aq) (aq) VO VO2+2+(aq)(aq) + H + H22OO(l)(l)

• Balancing both half-reactions:Balancing both half-reactions:ZnZn(s)(s) Zn Zn2+2+

(aq) (aq) + 2e+ 2e--

2e2e- - ++ 4H4H++(aq)(aq) + 2VO + 2VO22

++(aq) (aq) 2VO 2VO2+2+

(aq)(aq) + 2H + 2H22OO(l)(l)

• CollateCollate• Net reaction:Net reaction:ZnZn(s) (s) ++ 4H4H++

(aq)(aq) + 2VO + 2VO22++

(aq) (aq) 2VO 2VO2+2+(aq)(aq) + 2H + 2H22OO(l) (l) ++ ZnZn2+2+

(aq) (aq)

66

In basic milieuIn basic milieu

II--(aq)(aq) + MnO + MnO44

--(aq)(aq) I I2(aq)2(aq) + MnO + MnO2(s)2(s)

• Oxidation:Oxidation:II--

(aq)(aq) I I2(aq)2(aq) + e + e--

2I2I--(aq)(aq) I I2(aq)2(aq) + 2e + 2e--

• Reduction: Reduction: MnOMnO44

--(aq)(aq) 2OH 2OH--

(aq) (aq) + MnO+ MnO2(s)2(s)

2H2H++(aq)(aq) + 2OH + 2OH--

(aq) (aq) + MnO+ MnO44--(aq)(aq) 2OH 2OH--

(aq)(aq)+ MnO+ MnO2(s)2(s)

2H2H22OO(l) (l) + MnO+ MnO44--(aq)(aq) 2OH 2OH--


2H2H22OO(l) (l) + MnO+ MnO44--(aq)(aq) 4OH 4OH--


3e3e-- + 2H + 2H22OO(l) (l) + MnO+ MnO44--(aq)(aq) 4OH 4OH--


• Balance both half-reactions:Balance both half-reactions:6I6I--

(aq)(aq) 3I 3I2(aq)2(aq) + 6e + 6e--

6e6e-- + 4H + 4H22OO(l) (l) + 2MnO+ 2MnO44--(aq)(aq) 8OH 8OH--

(aq)(aq)+ 2MnO+ 2MnO2(s)2(s)

• Collate Collate • Net rxn:Net rxn:

6I6I--(aq) (aq) + 4H+ 4H22OO(l) (l) + 2MnO+ 2MnO44

--(aq)(aq) 3I 3I2(aq)2(aq) + 8OH + 8OH--

(aq)(aq)+ 2MnO+ 2MnO2(s)2(s)

77

Solve Solve

• AlAl(s)(s) + H + H22OO(l)(l) Al(OH) Al(OH)44--(aq)(aq) + H + H2(g)2(g)

88

Answer Answer

• AlAl(s)(s) + H + H22OO(l)(l) Al(OH) Al(OH)44--(aq)(aq) + H + H2(g)2(g)

• Oxidation:Oxidation:AlAl(s)(s) + 4OH + 4OH--

(aq) (aq) Al(OH) Al(OH)44--(aq) (aq) + 3e+ 3e--

• Reduction:Reduction:2e2e-- + 2H + 2H22OO(l)(l) 2OH 2OH--

(aq)(aq) + H + H2(g)2(g)

• Balance each half-reaction:Balance each half-reaction:2Al2Al(s)(s) + 8OH + 8OH--

(aq) (aq) 2Al(OH) 2Al(OH)44--(aq) (aq) + 6e+ 6e--

6e6e-- + 6H + 6H22OO(l)(l) 6OH 6OH--(aq)(aq) + 3H + 3H2(g)2(g)

• Collate Collate • Net-reaction:Net-reaction:

2Al2Al(s)(s) + 2OH + 2OH--(aq) (aq) + 6H+ 6H22OO(l) (l) 2Al(OH) 2Al(OH)44

--(aq) (aq) +3H+3H2(g)2(g)

99

Electricity Electricity

• Movt of electronsMovt of electrons• Movt of electrons through wire connecting 2 Movt of electrons through wire connecting 2

half-reactions half-reactions electrochemical cellelectrochemical cell• Also called Also called voltaicvoltaic or or galvanicgalvanic cell cell• Cell produces current from spontaneous rxnCell produces current from spontaneous rxn

– Example: copper in solution of AgNOExample: copper in solution of AgNO3 3 is is spontaneousspontaneous

• On the other hand, an On the other hand, an electrolytic cellelectrolytic cell uses uses electrical current to drive a electrical current to drive a non-spontaneousnon-spontaneous chemical rxnchemical rxn

1010

Voltaic cellVoltaic cell

• Solid Zn in zinc ion solution Solid Zn in zinc ion solution = = half-cellhalf-cell

• Likewise, Cu/Cu-ion solutionLikewise, Cu/Cu-ion solution• Wire attached to each solidWire attached to each solid• Salt bridgeSalt bridge = = • 1. contains electrolytes, 1. contains electrolytes, • 2. connects 2 half-cells, 2. connects 2 half-cells, • 3. anions flow to neutralize 3. anions flow to neutralize

accumulated cations at accumulated cations at anode and cations flow to anode and cations flow to neutralize accumulated neutralize accumulated anions at cathode anions at cathode (completes circuit)(completes circuit)

• ““An Ox” = anode An Ox” = anode oxidationoxidation

• Has negative charge Has negative charge because releases electronsbecause releases electrons

• ““Red Cat” = reduction Red Cat” = reduction cathodecathode

• Has positive charge Has positive charge because takes up electronsbecause takes up electrons

1111

Electrical currentElectrical current

• Measured in amperes (A)Measured in amperes (A)• 1 A = 1 C/s1 A = 1 C/s• Coulomb = unit of electric chargeCoulomb = unit of electric charge• ee-- = 1.602 x 10 = 1.602 x 10-19-19 C C• 1 A = 6.242 x 101 A = 6.242 x 101818 e e--/s/s• Electric current driven by difference in Electric current driven by difference in

potential energy per unit of charge: J/Cpotential energy per unit of charge: J/C• Potential difference (electromotive force or Potential difference (electromotive force or

emf) = volt (V)emf) = volt (V)• Where 1 V = 1 J/CWhere 1 V = 1 J/C

1212

More…More…

• In the voltaic cell, potential difference (emf) In the voltaic cell, potential difference (emf) between cathode and anode is referred to asbetween cathode and anode is referred to as– Cell potential (ECell potential (Ecellcell))

• Under standard conditions (1 M, 1 atm, 25°C), Under standard conditions (1 M, 1 atm, 25°C), cell potential iscell potential is

• Standard cell potential = EStandard cell potential = E°°cellcell

• Cell potential = measure of overall tendency Cell potential = measure of overall tendency of redox rxn to occur spontaneouslyof redox rxn to occur spontaneously

• Thus, the higher the EThus, the higher the E°°cellcell, the greater the , the greater the

spontaneityspontaneity

1313

Electrochemical notationElectrochemical notation

• CuCu(s)(s)|Cu|Cu2+2+(aq)(aq)||Zn||Zn2+2+

(aq)(aq)|Zn|Zn(s)(s)

• Notation describes voltaic cellNotation describes voltaic cell

• An ox on leftAn ox on left

• Red cat on rightRed cat on right

• Separated by double vertical line (salt Separated by double vertical line (salt bridge)bridge)

• Single vertical line separates diff phasesSingle vertical line separates diff phases

1414

Electrochemical notationElectrochemical notation

• Some redox rxns reactants & products Some redox rxns reactants & products in same phasein same phase

• Mn doesn’t precipitate out Mn doesn’t precipitate out uses Pt at uses Pt at cathodecathode– Pt is inert, but provides area for electron Pt is inert, but provides area for electron

gain/lossgain/loss

• FeFe(s)(s)|Fe|Fe2+2+(aq)(aq)||MnO||MnO44

--(aq)(aq), H, H++

(aq)(aq), Mn, Mn2+2+(aq)(aq)|Pt|Pt(s)(s)

• Write out net reactionWrite out net reaction

1515

Answer Answer

• FeFe(s)(s)|Fe|Fe2+2+(aq)(aq)||MnO||MnO44

--(aq)(aq), H, H++

(aq)(aq), Mn, Mn2+2+(aq)(aq)|Pt|Pt(s)(s)

• Oxidation: Oxidation:

FeFe(s)(s) Fe Fe2+2+(aq)(aq) + 2e + 2e--

• Reduction:Reduction:5e5e-- + MnO + MnO44

--(aq)(aq) + 8H + 8H++

(aq)(aq) Mn Mn2+2+(aq) (aq) + 4H+ 4H22OO(l)(l)

• Net-reaction:Net-reaction:5Fe5Fe(s) (s) + 2MnO+ 2MnO44

--(aq)(aq) + 16H + 16H++

(aq)(aq) 5Fe 5Fe2+2+(aq)(aq) + 2Mn + 2Mn2+2+

(aq) (aq) + 8H+ 8H22OO(l)(l)

1616

Standard reduction Standard reduction potentialspotentials

• One half-cell must have a potential of One half-cell must have a potential of zero to serve as referencezero to serve as reference– Standard hydrogen electrode (SHE) Standard hydrogen electrode (SHE)

half-cellhalf-cell

• Comprises Pt electrode in 1 M HCl w/ Comprises Pt electrode in 1 M HCl w/ HH22 bubbling at 1 atm: bubbling at 1 atm:

• 2H2H++(aq)(aq) + 2e + 2e-- H H2(g)2(g); E; E°°

red red = 0.00 V= 0.00 V

1717

Example Example

• Throw zinc into 1M HClThrow zinc into 1M HCl

• ZnZn(s)(s)|Zn|Zn2+2+(aq)(aq)||2H||2H++

(aq)(aq)|H|H2(g)2(g)

• EE°°cell cell = E= E°°

ox ox + E+ E°°redred = 0.76 V= 0.76 V

• If EIf E°°red red = 0.00 V (as the reference)= 0.00 V (as the reference)

• Then EThen E°°ox ox = 0.76 V (= oxid of Zn half-rxn)= 0.76 V (= oxid of Zn half-rxn)

• Reduction of Zn-ion Reduction of Zn-ion – Is = -0.76 V (non-spontaneous)Is = -0.76 V (non-spontaneous)

1818

Problem Problem

• CrCr(s)(s)|Cr|Cr3+3+(aq)(aq)||Cl||Cl--(aq)(aq)|Cl|Cl2(g)2(g)

• What is the std cell pot (EWhat is the std cell pot (E°°cellcell) given ) given

oxid of Cr = 0.73 V and Cl red = oxid of Cr = 0.73 V and Cl red = 1.36V?1.36V?

• Hint: standard electrode potentials are Hint: standard electrode potentials are intensive properties; e.g., like densityintensive properties; e.g., like density– Stoichiometry irrelevant!Stoichiometry irrelevant!

1919

Solution Solution


ox ox + E+ E°°red red = 0.73V + 1.36V = 0.73V + 1.36V

= = 2.09V 2.09V

2020

Appendix M, pages A-33-35Appendix M, pages A-33-35

• Standard reduction potentials in Standard reduction potentials in aqueous solution @ 25°Caqueous solution @ 25°C

• Also, pg. 967, Table 20.1 (gives Also, pg. 967, Table 20.1 (gives increasing strengths of ox/red agents)increasing strengths of ox/red agents)– Let’s take a look at itLet’s take a look at it

•Does increasing strengths of ox/red agents Does increasing strengths of ox/red agents make sense?make sense?

•What happens to oxidizing agent, reducing What happens to oxidizing agent, reducing agent?agent?

2121

Problem Problem

• Calculate the standard cell potential Calculate the standard cell potential for the following:for the following:AlAl(s)(s) + NO + NO33

--(aq)(aq) + 4H + 4H++

(aq)(aq) Al Al3+3+(aq)(aq) + NO + NO(g)(g) + +

2H2H22OO(l)(l)

2222

Answer Answer

• OxidationOxidationAlAl(s)(s) Al Al3+3+

(aq)(aq) + 3e + 3e--; E; E°°ox ox = 1.66V= 1.66V

• ReductionReduction

NONO33--(aq)(aq) + 4H + 4H++

(aq)(aq) + 3e + 3e-- NO NO(g)(g) + 2H + 2H22OO(l)(l); ;

EE°°red red = 0.96V= 0.96V


ox ox + E+ E°°red red =1.66V + 0.96V = =1.66V + 0.96V =

2.62V2.62V

2323

Predicting the spontaneous Predicting the spontaneous direction of a redox rxndirection of a redox rxn

• Generally, any reduction half-rxn is Generally, any reduction half-rxn is spontaneous when paired w/reverse spontaneous when paired w/reverse of half-rxn below it in table of of half-rxn below it in table of standard reduction potentialsstandard reduction potentials

• Let’s look at tableLet’s look at table• Predict the exact value and Predict the exact value and

spontaneity for the following:spontaneity for the following:

FeFe(s)(s) + Mg + Mg2+2+(aq)(aq) Fe Fe2+2+

(aq)(aq) + Mg + Mg(s)(s)

2424

Answers Answers

FeFe(s)(s) + Mg + Mg2+2+(aq)(aq) Fe Fe2+2+

(aq)(aq) + Mg + Mg(s)(s)

• OxidationOxidationFeFe(s)(s) Fe Fe2+2+

(aq)(aq) + 2e + 2e--; E; E°°ox ox = 0.45V= 0.45V

• ReductionReductionMgMg2+2+

(aq)(aq) + 2e + 2e-- Mg Mg(s)(s); E; E°°red red = -2.37V= -2.37V


ox ox + E+ E°°red red =0.45V + -2.37V = =0.45V + -2.37V =

-1.92V-1.92V

nonspontaneousnonspontaneous

2525

Will metal X dissolve in Will metal X dissolve in acid?acid?

• Metals whose reduction half-rxns lie Metals whose reduction half-rxns lie below reduction of proton to below reduction of proton to hydrogen gas will dissolve in acidshydrogen gas will dissolve in acids

• Why?Why?– Just look at the table!Just look at the table!

• Nitric acid is exceptionNitric acid is exception– Let’s take a lookLet’s take a look

2626

EE°°cellcell, , GG°°, K, K

• What must the values for EWhat must the values for E°°cellcell, , GG°°, & , &

K be in order to have a spontaneous K be in order to have a spontaneous rxn?rxn?

GG°°<0 <0

• EE°°cellcell>0 >0

• K>1K>1– Product-favoredProduct-favored

2727

Relationship between Relationship between GG°° & & EE°°

cellcell

• Faraday’s Constant (F) = 96,485 Faraday’s Constant (F) = 96,485 C/mol eC/mol e--

GG° ° =-n=-nee--FEFE°°cellcell

• Problem:Problem:

• CalculateCalculate GG° ° for for

II2(s)2(s) + 2Br + 2Br--(aq)(aq) 2I 2I--

(aq)(aq) + Br + Br2(l)2(l)

Is it spontaneous?Is it spontaneous?

2828

Solution: it’s Solution: it’s nonspontaneous!nonspontaneous!

II2(s)2(s) + 2Br + 2Br--(aq)(aq) 2I 2I--

(aq)(aq) + Br + Br2(l)2(l)

• OxidationOxidation2Br2Br--

(aq)(aq) Br Br2(l) 2(l) + 2e+ 2e--; E; E°°ox ox = -1.09V = -1.09V

• ReductionReductionII2(s)2(s) + 2e + 2e-- 2I 2I--

(aq)(aq); E; E°°red red = 0.54V = 0.54V

• EE°°cell cell = -1.09V + 0.54V = -0.55V= -1.09V + 0.54V = -0.55V

- 5e- cell -

96,485 JG° =-n FE° =-(2mol e ) ( ) ( 0.55V) 1.1 10 J (where V= )

mol e C

C

2929

Problem Problem

2Na2Na(s)(s) + 2H + 2H22OO(l)(l) H H2(g)2(g) + 2OH + 2OH--(aq)(aq) + 2Na + 2Na++

(aq)(aq)

Is it spontaneous?Is it spontaneous?

3030

Solution: it’s spontaneous! Solution: it’s spontaneous!

• OxidationOxidation

2Na2Na(s)(s) 2Na 2Na++(aq) (aq) + 2e+ 2e--; ; EE°°

ox ox = 2.71V = 2.71V


2H2H22OO(l)(l) + 2e + 2e- - H H2(g)2(g) + 2OH + 2OH--(aq) (aq) EE°°

red red = -= -0.83V 0.83V

• EE°°cell cell = 2.71V + -0.83V = 1.88V= 2.71V + -0.83V = 1.88V

- 5e- cell -

96,485 JG° =-n FE° =-(2mol e ) ( ) (1.88V) -3.63 10 J (where V= )

mol e C

C

3131

Relationship between ERelationship between E°°cell cell & &

K K

-

cell

-

cell

e

RTE° = lnK

nFJ CWhere R = 8.314 , T = 298.15 K, F = 96,485 , &mol K mol e

lnK = 2.303logK

0.0592VE° = log

nK

3232

ProblemProblem

• Calculate K forCalculate K for

2Cu2Cu(s)(s) + 2H + 2H++(aq)(aq) Cu Cu2+2+

(aq)(aq) + H + H2(g)2(g)

3333

Solution: is it product-Solution: is it product-favored? favored?


2Cu2Cu(s)(s) Cu Cu2+2+(aq)(aq) + 2e + 2e--; E; E°°

ox ox = -0.34V= -0.34V


2H2H++(aq)(aq) + 2e + 2e-- H H2(g)2(g); E; E°°

red red = 0.00V= 0.00V

EE°°cell cell = -0.34V= -0.34V

-

cell

e

12

0.0592VE° = log

n

0.0592V0.34 log

2

3.3 10

K

V K

K

3434

Cell potential & concentration: Cell potential & concentration: Nernst EquationNernst Equation

• Concentration ≠ Concentration ≠ 1M1M– Non-standard Non-standard

conditionsconditions

• Under standard Under standard conditions, Q = 1conditions, Q = 1

EEcellcell = E = E°°cell cell

-

cell cell

e

0.0592 VE = E° - logQ

n

3535

ProblemProblem

• Compute the cell potential, givenCompute the cell potential, givenCuCu(s)(s) Cu Cu2+2+(aq, 0.010 M) + 2e(aq, 0.010 M) + 2e- -

MnOMnO44--(aq, 2.0 M) + 4H(aq, 2.0 M) + 4H++(aq, 1.0M) + 3e(aq, 1.0M) + 3e-- MnO MnO2(s)2(s) + +

2H2H22OO(l)(l)

3636

Solution Solution

• Balance the equation!Balance the equation!

• OxidationOxidation3Cu3Cu(s)(s) 3Cu 3Cu2+2+

(aq)(aq) + 6e+ 6e--; E; E°°

ox ox = -0.34V= -0.34V

• Reduction Reduction 2MnO2MnO44

--(aq)(aq)+ 8H+ 8H++

(aq)(aq) + 6e + 6e-- 2MnO 2MnO2(s)2(s) + 4H + 4H22OO(l) (l) ; E; E°°red red = 1.68V= 1.68V

-

cell cell

e

2+ 3

cell - 2 + 84

3

cell 2 8

cell

0.0592 VE = E° - logQ

n

0.0592 V [Cu ]E = 1.34V- log

6 [MnO ] [H ]

0.0592 V [0.010]E = 1.34V- log

6 [2.0] [1.0]

E = 1.34V-(-0.065V) = 1.41V

3737

To summarizeTo summarize

• If Q<1, rxn goes to products If Q<1, rxn goes to products – EEcell cell > E> E°°

cellcell

• If Q>1, rxn goes to reactantsIf Q>1, rxn goes to reactants– EEcell cell < E< E°°

cellcell

• If Q = K, @ eq.,If Q = K, @ eq.,– EE°°

cellcell = 0 (& E = 0 (& Ecellcell = 0) = 0)•Explains why all batteries dieExplains why all batteries die

3838

Concentration cellsConcentration cells

• Voltaic cells can be constructed from two similar half-rxns Voltaic cells can be constructed from two similar half-rxns where where difference in concentration drives current flowdifference in concentration drives current flow

CuCu(s)(s) + Cu + Cu2+2+(aq, 2.0M) (aq, 2.0M) Cu Cu2+2+(aq, 0.010M) + Cu(aq, 0.010M) + Cu(s)(s)

– EE°°cell cell = 0 since both half-rxns are the same= 0 since both half-rxns are the same

• However, using Nernst equation, different concentrations However, using Nernst equation, different concentrations yield 0.068Vyield 0.068V– Let’s take a lookLet’s take a look

• Flow is from lower Cu-ion concentration half-cell to Flow is from lower Cu-ion concentration half-cell to higher onehigher one– Down the concentration gradientDown the concentration gradient

• The electrons will flow to the concentrated cell where they dilute The electrons will flow to the concentrated cell where they dilute the Cu-ion concentrationthe Cu-ion concentration

• Results in Results in Cu-ion concentration in dilute cell & Cu-ion concentration in dilute cell & Cu-ion concentration in concentrated cellCu-ion concentration in concentrated cell

3939

Batteries Batteries

• Dry-cell batteriesDry-cell batteries– Don’t contain large amounts of waterDon’t contain large amounts of water

• AnodeAnodeZnZn(s)(s) Zn Zn2+2+

(aq)(aq) + 2e + 2e--

• CathodeCathode2MnO2MnO2(s)2(s) + 2NH + 2NH44

++(aq)(aq) + 2e + 2e-- Mn Mn22OO3(s)3(s) + 2NH + 2NH3(g)3(g) + H + H22OO(l)(l)

– Cathode is carbon-rod immersed in moist (acidic) Cathode is carbon-rod immersed in moist (acidic) paste of MnOpaste of MnO22 that houses NH that houses NH44ClCl

• 1.5 V 1.5 V

4040

Batteries Batteries

• More common dry-cell typeMore common dry-cell type– Alkaline batteryAlkaline battery

• AnodeAnode

ZnZn(s)(s) + 2OH + 2OH--(aq)(aq) Zn(OH) Zn(OH)2(s)2(s) + 2e + 2e--

• CathodeCathode 2MnO2MnO2(s)2(s) + 2H + 2H22OO(l)(l) + 2e + 2e-- 2MnO(OH) 2MnO(OH)(s)(s) + 2OH + 2OH--

(aq)(aq)

• Longer shelf-life, “live” longerLonger shelf-life, “live” longer• Cathode in basic pasteCathode in basic paste

4141

Car Batteries Car Batteries

• Lead-acid storage batteriesLead-acid storage batteries• 6 electrochemical cells (2V) in series6 electrochemical cells (2V) in series• AnodeAnode

PbPb(s)(s) + HSO + HSO44--(aq)(aq) PbSO PbSO4(s)4(s) + H + H++

(aq)(aq) + 2e + 2e--

• CathodeCathodePbOPbO2(s)2(s) + HSO + HSO44

--(aq)(aq) + 3H + 3H++

(aq)(aq) + 2e + 2e-- PbSO PbSO4(s)4(s) + 2H + 2H22OO(l)(l)

• In 30% soln of sulfuric acidIn 30% soln of sulfuric acid• If dead due to excess PbSOIf dead due to excess PbSO44 covering electrode covering electrode

surfacessurfaces• Re-charge (reverse rxn) Re-charge (reverse rxn) converts PbSO converts PbSO44 to Pb and to Pb and

PbOPbO22

4242

Rechargeable batteries Rechargeable batteries

• Ni-CdNi-Cd• AnodeAnode

CdCd(s)(s) + 2OH + 2OH--(aq)(aq) Cd(OH) Cd(OH)2(s)2(s) + 2e + 2e--

• Cathode Cathode 2NiO(OH)2NiO(OH)(s)(s) + 2H + 2H22OO(l)(l) + 2e + 2e-- 2Ni(OH) 2Ni(OH)2(s)2(s) + 2OH + 2OH--

(aq)(aq)

• KOH, usually, usedKOH, usually, used• 1.30 V1.30 V• Reverse rxn recharges batteryReverse rxn recharges battery• Excess recharging Excess recharging electrolysis of water electrolysis of water• EXPLOSION!!!EXPLOSION!!!• Muhahahaha!Muhahahaha!

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Rechargeable batteriesRechargeable batteries

• Since Cd is toxicSince Cd is toxic– Developed safer alternativeDeveloped safer alternative

• Ni-MH Ni-MH

• Hybrid car batteries: high energy densityHybrid car batteries: high energy density• Same cathode rxn as previousSame cathode rxn as previous• AnodeAnode

MHMH(s)(s) + OH + OH--(aq)(aq) M M(s)(s) + H + H22OO(l)(l) + e + e--

• Commonly, M = ABCommonly, M = AB55, where A is rare earth mixture , where A is rare earth mixture of La, Ce, Nd, Pr, and B is Ni, Co, Mn, and/or Mnof La, Ce, Nd, Pr, and B is Ni, Co, Mn, and/or Mn

• Very few use ABVery few use AB22, where A = Ti and/or V, where A = Ti and/or V

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Rechargeable batteriesRechargeable batteries

• Anode made of graphite w/incorporated Anode made of graphite w/incorporated Li-ions between carbon layersLi-ions between carbon layers

• Ions spontaneously migrate to cathodeIons spontaneously migrate to cathode• Cathode = LiCoOCathode = LiCoO22 or LiMn or LiMn22OO44

• Transition metal reducedTransition metal reduced• Used in laptop computers, cell phones, Used in laptop computers, cell phones,

digital camerasdigital cameras• Light weight and high E densityLight weight and high E density

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Fuel cellFuel cell

• Reactants flow through batteryReactants flow through battery– Undergo redox rxnUndergo redox rxn

• Generate electricityGenerate electricity

• Hydrogen-oxygen fuel cellHydrogen-oxygen fuel cell• AnodeAnode

2H2H2(g)2(g) + 4OH + 4OH--(aq)(aq) 4H 4H22OO(l)(l) + 4e + 4e--

• CathodeCathodeOO2(g)2(g) + 2H + 2H22OO(l)(l) + 4e + 4e-- 4OH 4OH--

(aq)(aq)

• Used in space-shuttle programUsed in space-shuttle program– And Arnold’s HummahAnd Arnold’s Hummah

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Electrolysis Electrolysis

• Electrical current Electrical current used to drive used to drive nonspontaneous nonspontaneous redox rxnredox rxn

• In electrolytic cellsIn electrolytic cells• Used inUsed in

– Electrolysis of waterElectrolysis of water– Metal plating: silver Metal plating: silver

coated on metal, coated on metal, jewelry, etc.jewelry, etc.

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Electrolytic cells: using Electrolytic cells: using electricity to run a rxnelectricity to run a rxn

• Anode is “+” Anode is “+” gives electrons, gives electrons, connected to positive terminal of connected to positive terminal of power sourcepower source

• Cathode is “-” Cathode is “-” takes electrons, takes electrons, connected to negative terminal of connected to negative terminal of power sourcepower source

• Opposite scheme of voltaic cell!Opposite scheme of voltaic cell!

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Predicting the products of Predicting the products of electrolysiselectrolysis

1.1. Pure molten saltsPure molten salts– Anion oxidized/cation reducedAnion oxidized/cation reduced

• Obtain 2NaObtain 2Na(s)(s) and Cl and Cl2(g)2(g) from electrolysis of NaCl from electrolysis of NaCl

2.2. Mixture of cations or anionsMixture of cations or anions– KK++/Na/Na++ and Cl and Cl--/Br/Br-- present present

• Look at page 967 & compare half-cell Look at page 967 & compare half-cell potentialspotentials

– Cation/anion preferably reduced that has least Cation/anion preferably reduced that has least negative, or most positive, half-cell potentialnegative, or most positive, half-cell potential

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Example Example

• Predict the half-rxn occurring at the anode and the cathode Predict the half-rxn occurring at the anode and the cathode for electrolysis of for electrolysis of

AlBrAlBr33 & MgBr & MgBr22

• OxidationOxidationBrBr--

(l)(l) Br Br2(g)2(g) + 2e + 2e--;; EE°°ox ox = = --1.09V1.09V

Bromide will be oxidized at the anodeBromide will be oxidized at the anode• Reduction Reduction

AlAl3+3+(l)(l) + 3e + 3e-- Al Al(s)(s);; EE°°

red red = = --1.66V1.66VMgMg2+2+

(l)(l) + 2e + 2e-- Mg Mg(s) (s) ;; EE°°red red = -2.37V= -2.37V

Reduction of Al will occur at the cathode since its potential is Reduction of Al will occur at the cathode since its potential is greater than Mg’sgreater than Mg’s

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Predicting the products of Predicting the products of electrolysiselectrolysis

3.3. aqueous solns: same as #2aqueous solns: same as #2– Water redox might occur Water redox might occur

simultaneouslysimultaneously

2H2H22OO(l)(l) O O2(g)2(g) + 4H + 4H++(aq)(aq) + 4e + 4e--

2H2H22OO(l)(l) + 2e- + 2e- H H2(g)2(g) + 2OH + 2OH--(aq)(aq)

EE°°ox ox = -0.82 V & E= -0.82 V & E°°

red red = -0.41 V = -0.41 V

EE°°cell cell = -1.23 V= -1.23 V

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Problem Problem

• Given oxidation of IGiven oxidation of I-- = -0.54 V & = -0.54 V & reduction of Lireduction of Li++ = -3.04 V, which, if = -3.04 V, which, if any, gases would be formed and any, gases would be formed and where; i.e., at cathode/anode?where; i.e., at cathode/anode?

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Solution Solution


2I2I--(aq)(aq) 2I 2I2(aq)2(aq) + 2e + 2e--; E; E°°

ox ox = -0.54V = -0.54V

2H2H22OO(l)(l) O O2(g)2(g) + 4H + 4H++(aq)(aq) + 4e + 4e--; E; E°°

ox ox = -0.82V = -0.82V

II-- will be oxidized at the anode will be oxidized at the anode


2Li2Li++(aq)(aq) + 2e + 2e-- 2Li 2Li(s)(s); E; E°°red red = -3.04 V = -3.04 V

2H2H22OO(l)(l) + 2e + 2e-- H H2(g)2(g) + 2OH + 2OH--(aq)(aq); E; E°°

red red = -0.41 V = -0.41 V

Water will be reduced at the cathodeWater will be reduced at the cathode

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Stoichiometry of Stoichiometry of electryolysiselectryolysis

• Can use eCan use e-- stoichiometric relations to stoichiometric relations to predict moles and/or grams of predict moles and/or grams of substancessubstances

• Remember, unit of current = ampere Remember, unit of current = ampere = A = 1 C (magnitude of current)/s = A = 1 C (magnitude of current)/s (time of current flow)(time of current flow)

• Also, F = 96,485 C/mole eAlso, F = 96,485 C/mole e--

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Problem Problem

• Gold can be plated out of a soln Gold can be plated out of a soln containing the Aucontaining the Au3+3+ according to according to

AuAu3+3+(aq)(aq)

+ 3e+ 3e-- Au Au(s)(s)

• What mass of gold (in grams) will be What mass of gold (in grams) will be plated by the flow of 5.5 A of current plated by the flow of 5.5 A of current for 25 mins?for 25 mins?

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Solution Solution

60 5.5 1mol Au 196.9725min 5.6 Au

1min 1 96,485 3 1

s C mol e gg

s C mol e molAu

1 Electrochemistry Chapter 20. 2 Half-reaction method…remember? Example Example Al (s) + Cu 2+...

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Transcript of 1 Electrochemistry Chapter 20. 2 Half-reaction method…remember? Example Example Al (s) + Cu 2+...