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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
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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
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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 #
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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
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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
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)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
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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
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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)
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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-
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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
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“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
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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(-)
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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)
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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)
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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
+ -
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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-
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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
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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
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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
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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
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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:
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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????
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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-
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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)
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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.
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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)
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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-
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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)
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Reactants: Pb; PbO2
DH2SO4 = 1.8 Charge D = 1.25 - 1.30 Recharge D < 1.20
- +
Pb PbO2
H2SO4
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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
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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
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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
Page 43
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