CH. 20

ELECTROCHEMISTRY MAIN CONCEPTS

Galvanic Cell Electolytic Cell -electroplating Cell PotentialsNernst EquationBalance 1/2-Rxns -Acidic/Basic Solns

Electrochemical Cells: classified as Galvanic Cells *spontaneous chem rxn that produce E Electolytic Cell

Commercial Importance Galvanic Zn|MnO2 & Zn|Ag2O cells in watches Fuel Cells in space crafts H2 - O2

Electrolytic NaOH purify metals electroplating

Conversion Eelec to Echem

ELECTROLYTIC CELLS

conversion Echem to Eelec

GALVANIC - VOLTAIC CELLS

ELECTROLYTIC CELLS 1. process of electrolysis 2. pass electricity thru soln w/ E to cause nonspont redox 3. commercial importance-- NaOH purify ores electroplate

GALVANIC CELLS 1. provides source electricity thru spont redox 2. batteries

ZnC Alkaline Ag2O Pb Storage NiCd NiMH Fuel Cells

REVIEW REDOX RXN

REDUCEDH +1 ---> 0oxidizing agent

)g(02(s)

12

2 )aq(

11)s(

0 H ClFe ClH 2 Fe

OXIDIZEDFe 0 ---> +2reducing agent

OXIDATION REDUCTIONgain of O atoms loss of O atoms loss of H gain of H lose e- gain e-

LEO GER OIL RIG incr ox # decr ox #

SPONTANEOUS REDOX RXN

Two ½-Reactions:

)aq(2

)(0

)(2

)(0 Zn Cu Cu Zn saqs

Cu+2 Zn+2Cu0

Zn0

Oxidation: Zn0 (s) ---> Zn+2 (aq) + 2 e-

lose 2 e-, OX, ox # incr

Reduction: Cu+2 (aq) + 2 e- ---> Cu0 (s)

gain 2 e-, RED, ox # decr

No Electrical Work produced heat E released

BALANCING 1/2-REACTIONS1st Separate into 0.5-rxns2nd Balance all non-O & non-H atoms

IF 3rd Balance O

Balance Hw/ H2O

w/ H+

4th Add # e- to balance charge

5th

Combine 0.5- Multiply to balance

reactions

#e- in each 0.5-rxn6th Add 0.5-rxns together

ACIDIC BASIC

7th CheckAdd # OH- both sides

to balance H+ & H2O

8th XXXXXX Check

Write balanced molecular, add # moles of spectator ions to get neutral cmpds

)aq(-13)(

-24 )(

-1)(

-14 IO MnO I MnO aqaqaq

Potassium permanganate & potassium iodide in basic soln

Notice: K+1 spectator

)(-24)(

-14 MnO MnO :RXNS-0.5 aqaq )aq(

-13 )(

-1 IO I aq

)(-24)(

-14

- MnO MnO e 1 aqaq )aq(-13 )(

-12 IO I OH 3 aq

-)aq(

-13 )(

-12 e 6 H 6 IO I OH 3

aq

)MnO MnO e 6(1 )(-24)(

-14

-aqaq

-)aq(

1-3 )(

1-2

)(-24)(

-14

-

e 6 H 6 IO I OH 3

MnO 6 MnO 6 e 6

aq

aqaq

H 6IO I OH 3 )(-13 )(

-12 aqaq

H 6 IOMnO 6 MnO 6 I OH 3 )aq(-13)(

-24)(

-14 )(

-12 aqaqaq

Step 1- step 2 not needed as balanced

Step 3- step 4

Step 5

Step 6

H 6 IOMnO 6 MnO 6 I OH 3 )aq(-13)(

-24)(

-14 )(

-12 aqaqaq

OH 6 H 6 IOMnO 6

OH 6 MnO 6 I OH 3

)aq(1-

3)(2-

4

)(-14 )(

-12

aq

aqaq

)(2)aq(-13)(

-24)(

-14 )(

-1 OH 3 IOMnO 6 OH 6 MnO 6 I laqaqaq

molecular

)(2)aq(3)(42

)(4 )(

OH 3 IOKMnOK 6

KOH 6 KMnO 6 KI

laq

aqaq

Step 7

Step 8

ENERGY CAPTURE

E released in spont REDOX rxn is captured to perform electrical work

E of water wheel depends on 1) vol water 2) PE of H2O

Work from electrochem cell

w = (vol H2O)(E released/unit vol)

Right is Reduction - Cathode (+) ox- + ne- ----> Red e- flows into cathode reacts w/ oxidized species forms reduced species

Left is Oxidation - Anode (-) Red’ ---> ox-’ + ne-’

e- flow out of anode

GALVANIC CELL

Anode (-)

OXIDATION

Cathode (+)

REDUCTION

e-

e-

MAKE UP OF CELL consists of:

1. 2. 2 0.5-cells; anode(-) & cathode(+) Salt bridge, electrolyte (Na+NO3-)

allows slow mixing of ions

--

--

--

-

Anode(-)

Zn+2

Spont Rxn continuous e- flow thru external wire

Ions flow thru soln of redox rxn @ eletrodes

Zn0

Cu+2 + 2 e- ---> CuZn ---> Zn+2 + 2 e-

Cu+2(aq) + Zn (s) -----> Cu (s) + Zn-2(aq)

Cathode(+)

++

++

++

+

Cu+2Cu0

Volt Metere-

e-

Zn(s)|Zn+2, 1M(aq) || Cu+2, 1M (aq)|Cu(s)

NO3-1

NO3-1

“LINE” Notation of Electrochemical Cell

Zn(s)|Zn+2, 1M(aq) || Cu+2, 1M (aq)|Cu(s)

--

--

--

-

Anode(-)

Zn+2

Zn0

Cu+2 + 2 e- ---> CuZn ---> Zn+2 + 2 e-

Cathode(+)+

+

++

++

+

Cu+2Cu0

Volt Meter

AnodeOx electrode

CathodeRed electrode

Salt Bridge

Phase Boundary

Aqueous Solutions

Voltmeter shows diff in electrical potential bet 2-1/2 cells known as “cell potential”; electromotive force (emf) Eo

cell = Eoox+ Eo

red

Spont Rxn: Eocell> 0 Std H2 Electrode (SHE) reference electrode

Eo = 0 V 2 H+1 (aq) + 2 e- <---> H2

0 (g)What measures cell potential

Eoox??? Eo

red???

--

--

--

-

Volt Meter

E0cell = 0.76 V

Zn0

Zn+2

Cathode (+)

H2

1 atm

1 M H+

Pt

Anode(-)

Std Electrode Potential in Water @ 25 oC

Std Red. Pot. V Reduction 1/2-rxn 2.87 F2(g) + 2 e- -----> 2 F-1(aq)

0.80 Ag+1(aq) + 1 e- -----> Ag (s)

0.34 Cu+2(aq) + 2 e- -----> Cu (s)

0 2 H+1(aq) + 2 e- -----> H2 (g)

-0.28 Ni+2(aq) + 2 e- -----> Ni (s)

-0.76 Zn+2(aq) + 2 e- -----> Zn(s)

-3.05 Li+1(aq) + 1 e- -----> Li (s)

EX: Build galvanic cell --- Ag & Zn

What is reduced? Cathode?

What is oxidized? Anode?

What is the cell potential, emf?

Highest 0.080 V Ag+1(aq) + 1 e- -----> Ag (s)

Lowest -0.76 Zn(s) -----> Zn+2(aq) + 2 e-

Eocell = Eo

ox,Zn+ Eored,Ag

= +0.76 V + 0.80 V = +1.56 V

Write the “line” notationZn(s)|Zn+2, 1M(aq) || Ag+1, 1M (aq)|Ag(s)

Left is Oxidation - Anode (+) Red ----> ox- + ne-

Right is Reduction - Cathode (-) ox-’ + ne-’ = RED’

ELECTROLYTIC CELLdrive a nonspontaneous reaction Eo

cell < 0

Anode (+)

OXIDATION

Cathode (-)

REDUCTION

e-

e-

“e- pump”e- being push by external power source

+ -

ELECTROLYSISDriven by outside source E (nonspont)

MAKE UP OF CELLconsists of:1.2.

2 electrodes (in molten salt)

)g(2)( )( Cl Na 2 NaCl 2 ll

driven by DC source; e- pump DC source

++

++

++

+

--

--

--

-

Anode (+) Cathode (-)

Na+

Cl-

Na0

Na+ ions gain e- @ cathode; reduce

Cl- ions lose e- @ anode; oxidized

Cl20

Na+ + e- ---> Na2 Cl- ---> Cl2 + 2 e-

e-e-

Electrical E (V) needed to drive NaCl rxn

Oxidize Reduce

2 Cl- (aq) ---> Cl2 (g) + 2 e- Eox = -1.36 Na+ (aq) + e- ---> Na (s) Ered = -2.71 V

Ecell = -4.07 V

Is rxn spontaneous as written? Write balanced spont. rxn. Fe+2(aq) -----> Fe(s) + Fe+3(aq)

2 e- + Fe+2(aq) -----> Fe(s) Eo =Fe+2(aq) -----> Fe+3(aq) + 1 e- Eo =

-0.44 V-0.77 V2( )

Fe+2 + 2 Fe+2(aq) -----> Fe(s) + 2 Fe+3 (aq)

3 Fe+2(aq) -----> Fe(s) + 2 Fe+3 (aq) Eo = -1.21 V

Fe(s) + 2 Fe+3(aq) -----> 3 Fe+2 (aq) Eo = +1.21 V

SET #1 Using STD reduction potentials, which rxns are spontaneous?1) I2(s) + 5 Cu+2(aq) + 6 H2O(l) ------> 2 IO3

-(aq) + 5 Cu(s) + 12 H+(aq) 2) Hg2

+2(aq) + 2 I-(aq) ------> 2 Hg(l) + I2(s)

3) H2SO3(aq) + 2 Mn(s) + 4 H+(aq) ------> S(s) + 2 Mn+2(aq) + 3 H2O(l)

1) OX: I2(s) + 6 H2O(l) ------> 2 IO3-(aq) + 12 H+(aq) + 10 e-

RED: 5[Cu+2(aq) + 2 e- ------> Cu(s)]

I2(s) + 5 Cu+2(aq) + 6 H2O(l) ------> 2 IO3-(aq) + 5 Cu(s) + 12 H+(aq)

EOX = -1.195ERED = 0.337

Ecell = 0.337 + (-1.195) = -0.858 V NONSPONT

2) OX: 2 I-(aq) ------> I2(s) + 2 e-

RED: Hg2+2(aq) + 2 e- ------> 2 Hg(l)

2 I-(aq) + Hg+2(aq) ------> I2(aq) + Hg(l)

EOX = -0.536ERED = 0.789

Ecell = 0.789 + (-0.536) = 0.253 V SPONT

3) OX: 2 [Mn(s)------> Mn+2(aq) + 2 e-] RED: H2SO3(aq) + 4 H+(aq) + 4 e- ------> S(s) + 3 H2O(l)

H2SO3(aq) + 2 Mn(s) + 4 H+(aq) ------> S(s) + 2 Mn+2(aq) + 3 H2O(l)

EOX = 1.18ERED = 0.45

Ecell = 0.45 + 1.18 = 1.63 V SPONT

EMF & G

Change in Free E is measure of spontaneity @ constant T & P relationship bet emf & G is: Go = -nFEo

n = # e- pushed F: Faraday; 1 F = 96,500 C/mol = 96,500 J/V-mol

K relationship Go = -RT LnK R=8.314 J/mol-K

4 Ag(s) + O2(g) + 4 H+(aq) ------> 4 Ag+(aq) + 2 H2O(l) Need values for Eo, Go, & KUse eqn Go = -RT LnK

STEP1: RED: O2(g) + 4 H+(aq) + 4 e- ---> 2 H2O(l) OX: 4 Ag(s) ---> 4 Ag+(aq) + 4 e-

Go very large, so very favored,expect K to be very large also

STEP 2: Go = -nFE = -(4 e-)(96,500 J/V-mol)(+0.43 V) = -170,000 J/mol What can be deduced from this?

Ered = +1.23 VEox = -0.80 VEcell = +0.43 V

At 25oC, find standard Go and K for:

STEP 3: equilib constant K Go = -RT LnK ===> Ln K = Go/-(RT)

K = e69 = 9.3*102969 K) K)(298molJ (8.314-

molJ 170,000- K Ln

//

Notice example pg. 864 part brxn written 1/2 of original

All values remain the same, even though half of quantities K is 1/2, why????

ELECTROLYTIC CELL ELECTROPLATING

)(0-

)(1-

)(1

)(0 Ag e 1 Ag :Cathode e 1 Ag Ag :Anode saqaqs

++

++

++

+

Anode (+)platingmaterial

Cathode (-)object to be plated

Ag+Ag+ Ag0

Ag0

e-e-

DC source

NO3-

Coulumb (C) amt of charge that passes thru a pt w/ current 1 ampere @ 1 sec

1 C = 1 A•s

1 F = 96,500 C/mol e-

ELECTRICAL WORK

-G = wmax

wmax = nFE

wmax = n * F * E J = (mol) * (C/mol)*(J/C)

Steps1. Rxn Ag deposited, gain e-, red Ag+1 (aq) + 1 e- ---> Ag (s)

2. Relationship 1 mol Ag ~ 1 mol e-

= 96,500 C3. Current & time, find C 1.5 Ah*[2*(3600 s/h)] = 10,800 A•s = 10,800 C

10,800 C*(1 mol e-/96,500 C)*(1 mol Ag/1 mol e-) *(107.8 g Ag/1 mol Ag) = 12.1 g Ag

EX. How many grams Ag deposited from AgNO3 soln by current 1.5 Ah over 2 hrs.

QUANTITATIVE

Michael Faraday 1st to describe extent of current used to chem change @ electrodes

Faraday (F) amt of electricity supplied to deliver 1 mol of e-; “a mol of e-”

1 mol Ag = 107.9 g, know 1 mol e- passed

Amt Change - related to amt electricity passed

amt moles e- lost/gain in redox rxn

Ag+ (aq) + e- ---> Ag (s)

Coulumb (C) amt of charge that passes thru a pt w/ current 1 ampere @ 1 sec

Steps1. Rxn Cu deposited, gain e-, red Cu+2 (aq) + 2 e- ---> Cu (s)

2. Relationship 1 mol Cu ~ 2 mol e-

= 2 F

3. Current & time, find C 1.5 A*[2*(3600 s/hr)] = 11,000 A•s = 11,000 C 11,000 C*(2 mol Cu/2 mol e-)*(1 F/96,500 C)*(63.6 g/1 mol) = 3.5 g Cu

EX. How many grams Cu deposited from CuSO4 soln by current 1.5 A over 2hrs.

1 C = 1 A•s

1 F = 96,500 C = 1 mol e-

CELL POTENTIAL emf 1 V = 1 J/CEcell = Eox + Ered

Znox = -0.76 = +0.76 VCured = +0.34 VZn more diff to reduceEcell = 0.76 + 0.34 = 1.10 V

STEPS1. E1 = -0.74 V E2 = +1.28 V Red2 > Red1 #1 must be oxidized, reverse2. Rewrite, balance e-, & sum

EX. What is the rxn and Ecell from the following: 1) Cr+3 (aq) + 3 e- -----> Cr (s) 2) MnO2 (s) + 4 H+ (aq) + 2 e- ----> Mn+2 (aq) + 2 H2O (l)

2 [Cr(s) -----> Cr+3(aq) + 3 e-]3[MnO2(s) + 4 H+(aq) + 2 e- ----> Mn+2(aq) + 2 H2O(l)]

3 MnO2(s) + 12 H+(aq) + 2 Cr(s) ----> 3 Mn+2(aq) + 2 Cr+3(aq) + 6 H2O(l)

Ecell = 1.28 + 0.74 = 2.02 V

Zn|Cu Cell move 2 e-

@ 25oC 1 V = 1J/C (2.303RT/F) is const = 0.0592 J/CE = (0.0592/n) Log Kc

G = (2 mol e-)(96,500 C/mol)(1.10 J/C) =

G = -2.303RT Log Kc

Combine EqnsnFE = 2.303 Log Kc

E = (2.303RT/nF) Log Kc

spont very ,3710*2 cK

37.2 0592.0

V 1.102 cK Log

QUANTITATIVE, NERST EQN

aA + bB ----> cC + dD

@ 25oC (2.303RT/F) = 0.0592/n

Zn|Cu Cell Eo = 1.10 V n = 2

bBaA

dDcC Q Log

nF

2.303RT oE E

2Cu

2ZnLog

2

0.0592 1.10 E

Application Nernst 1) w/ varying concentrations 2) Ksp, solubility product constant 3) pH

R: gas constant, 8.314 T: temp, K n: # e- transferredF: Faraday constant, 96,500 Q: rxn quotient

Find the standard potential (Eo) for the rxn: Cd(s) + Cu+2(aq) <----> Cd+2(aq) + Cu(s) [Cu+2] = 0.80 [Cd+2] = 0.20

1/2-rxns Anode (ox) Cd(s) <----> Cd+2(aq) + 2 e- Cathode (red) Cu+2(aq) + 2 e- <----> Cu(s)

Eo

0.40 V0.34 V

Cd+2(aq) + Cu(s) <----> Cu+2(aq) + Cd(s) Eo = 0.40 + 0.34 = 0.74 V

Nernst Eqn: E = 0.74 - 0.0592/2*Log[0.20]/[0.80] = 0.74 - 0.0296*(-0.602) = 0.75 V

Nernst Eqn E = Eocell – (0.0592 V/n) Log([Fe+2]dil/[Fe+2]conc)

E = 0.0 – (0.0592 V/2) Log([0.003]/[1.5]) = -(0.0296 V) Log(0.002) = -(0.0296 V)(-2.70) = 0.080 V

Concen Cell: system consists of 2 half-cells; cell #1: strip Fe metal in 1.5 M Fe+3 soln cell #2: strip Fe metal in 0.003 M Fe+3 solnWhat is the emf?

Anode: Fe(s) --- Fe+2(aq) + 2 e- Eo = +0.44Cathode: Fe+2(aq) + 2 e- --- Fe(s) Eo = -0.44 Fe+2(aq, conc) ---- Fe+2(aq,dilute) Eo

cell = 0.44 + (-0.44) = 0.0 V

Ecell = 0.00 - (0.0592/2) Log([3.73*10-4]/[1.35]) = -(0.0296)Log(2.76*10-4) = -0.0296(-3.56)

Oxidize Zn (s) ---> Zn+2 (aq) + 2 e- Eox = 0.763 V

Reduced Zn+2 (aq) +2 e- ----> Zn(s) Ered = -0.763V

Concentration cell with 2 Zn(s)-Zn+2(aq) half-cells.1st half-cell [Zn+2]= 1.35 M 2nd cell [Zn+2] = 3.73*10-4 M a) which half-cell is anode? b) What is emf?

1st Cell;more concentrated

Eo = 0.00 V

Nernst Eqn E = Eocell – (0.0592 V/n) Log([Zn+2]dil/[Zn+2]conc)

Ecell = 0.105 V

emf - Electromotive Forces (V)

Ecell = Eox(Cl-) + Ered (H2O) = (-1.36) + (-.083) + -2.19 V

Soln: CuCl2 (s) ----> Cu (s) & Cl2 (g)

EX. Explain why CuCl2 produces Cu (s) & Cl2 (g).

Min emf

Ox: Anode 2 Cl- (l) ---> Cl2 (g) + 2 e-

Red: Cathode 2 Na+ (l) + 2 e- ---> Na (l)

Cell Rxn 2 Na+ (l) + 2 Cl- (l) ---> Cl2 (g) + Na (l)

Ecell = (0.34) + (-1.36) = -1.02 V

Oxidize 2 Cl- (aq) ---> Cl2 (g) + 2 e- Eox = -1.36 V anode 2 H2O (l) ---> O2 (g) + 4 H+

(aq) + 4 e- Eox = -1.23 V

Reduce 2 H2O (l) + 2 e- ---> H2 (g) + 2 OH- (aq) Ered = -0.83 V cathode Cu (aq) + 2 e- ---> Cu (s) Ered = +0.34 V H2 produced @ cathode

Half Reactions

Ox - Anode 2 Cl- (l) ---> Cl2 (g) + 2 e-

Red -Cathode 2 Na+ (l) + 2 e- ---> Na (l)

Electro- Cathode: e- forced unto (-) Anode: e- withdrawn (+)

more complex due to ability H2O to RED & OX

Possible red-ox solvent & ions solute; whether the solute anion or H2O, or the solute cation or H2O

Cell Rxn 2 Na+ (l) + 2 Cl- (l) ---> Cl2 (g) + Na (l)

AQUEOUS SOLN

Reduce 2 H2O (l) + 2 e- ---> H2 (g) + 2 OH- (aq) Ered = -0.83 V Na+ (aq) + e- ---> Na (s) Ered = -2.71 V H2 produced @ cathode

Acidic Soln 2 H+ (aq) + 2 e- ---> H2 (g)

not major dil. aq soln

Oxidize 2 Cl- (aq) ---> Cl2 (g) + 2 e- Eox = -1.36 V 2 H2O (l) ---> O2 (g) + 4 H+

(aq) + 4 e- Eox = -1.23 V

Overcharge??O2 formation is high to permit Cl- oxid than H2O“BRINES” produce H2 & Cl2

Anode: E = 0.356 V Cathode: E = 1.685 V Ecell = 0.356 + 1.685 = 2.041 V

Pb Storage Battery - car 12 V

Consists of: 6 cells (2 V) anode: Pb Cathode: PbO2

H2SO4 (aq)

DischargeAnode Pb(s) + SO4

-2 (aq) -----> PbSO4(s) + 2 e-

Cathode PbO2(s) + SO4-2(aq) + 4 H+ (aq) + 2 e- ----> PbSO4(s) + 2 H2O(l)]

Pb(s) + PbO2(s) + 4 H+(aq) + 2 SO4-2(aq) ----> 2 PbSO4(s) + 2 H2O(l)

Reactants: Pb; PbO2

DH2SO4 = 1.8 Charge D = 1.25 - 1.30 Recharge D < 1.20

- +

Pb PbO2

H2SO4

Find the equilibrium constant for the rxn @ 25oC of Ni(s) & Ag+1 (aq)

Anode: Ag 0.80Cathode: Ni 0.25n = 2

Log K = (2 * 1.05 V)/0.0592 = 35.473 K = 1035.473

= 3*1035

Ecell = 0.80 + 0.25 = 1.05 V

cathodeanodeocell

o

E E E 0592.0

nE K Log

Find Ecell and Go @ 25oC for n = 2 & K = 5.0*10-6

Go = -RT Ln K = -(8.314 J/mol•K)*(298.15 K) Ln(5.0*10-6) = -(2478.8)*(-12.2) = 30200 J/mol

= -0.16 V

Ecell = -(30200 J/mol)/[(2 mol e-)*(96,500 C/mol e-)]*(V/JC-1)

Ecell = -Go/nF

Ex. MnO4-2 is stable in strong basic soln. (frames 6 & 7 basic example)

In acidic soln, reacts to form permanganate and MnO2(s).Wrtie balanced overall rxn from the two half-rxns.

e 1 MnO MnO -)(

-14 )(

-24 aqaq

)(2)(2 )(-24 OH 2 MnO MnO lsaq

)(2)(2 )( )(2-

4

-)(

-14 )(

-24

OH 2 MnO e 2 H 4 MnO

e 2 MnO 2 MnO 2

lsaqaq

aqaq

)(2)(2)(-14 )( )(

-24 OH 2 MnO MnO 2 H 4 MnO 3 lsaqaqaq

]e 1 MnO 2[MnO -)(

-14 )(

-24 aqaq

-)(

-14 )(

-24 e 2 MnO 2 MnO 2 aqaq

)(2)(2 )( )(-24 OH 2 MnO H 4 MnO lsaqaq

)(2)(2 )( )(-24 OH 2 MnO e 2 H 4 MnO lsaqaq

Ex. Write balance eqn of CuS in 3M HNO3.Cu(s) + H+(aq) + NO3

-1(aq) -----> Cu+2(aq) + S(s) + NO(g) + H2O(l)

)()(-13

-)()(

2 )( NO NO e 2 S Cu CuS gaqsaqs

)(2)( )( )(1-

3

-)()(

2 )(

OH 4 NO 2 e 6 H 8 NO 2

e 6 S 3 Cu 3 CuS 3

lgaqaq

saqs

)(2)()()(2

)()(-13 )( OH 4 NO 2 S 3 Cu 3 H 8 NO 2 CuS 3 lgsaqaqaqs

)(2)( )(-13 OH 2 NO NO lgaq

)(2)(-

)()(-13 OH 2 NO e 3 H 4 NO lgaqaq

]OH 2 NO e 3 H 4 2[NO

]e 2 S Cu 3[CuS

)(2 )(-

)(1-

3

-)()(

2 )(

lgaq

saqs