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Annjaneth Briones and Nathalie Dagmang
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an electron is transferred from one
reactant to another
Zn(s)
+ Cu2+ Cu(s)
+ Zn2+
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An electron-exchange
reaction/oxidation or reduction
at an electrodeqa spontaneous redox reaction
produces energy or
a non-spontaneous redox
reaction uses up energy
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Direct Electrochemical Cell
No work done Work done/needed
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Spontaneous
reactionGenerates or
stores energy
ex. batteries
Non-spontaneous
reactionUses up electrical
ex. Metal plating,
recovery of metal
from ores
Voltaic/GalvanicCell ElectrolyticCell
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Voltaic/GalvanicCell ElectrolyticCell
load Power
supply
energy
energy
+- + -
Work done by
system to
surroundings
Work done on
system by
surroundings
cathodecathode anodeanode
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ANODE CATHODE
Rxn Ion movingtowards it
Sign Rxn Ion movingtowards it
Sign
Galvanic
Cell
oxidation anions _ reduction cations +
Electrolytic
celloxidation anions + reduction cations _
difference b/w the anode and cathode of galvanic and
electrolytic cells is their polarity
Flow of electrons: ANODE CATHODE (ALWAYS)
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Voltaic/Galvanic Cell
Volt-
meter
+
Cu
CuSO4
-
Zn
ZnSO4
Salt bridge
e- e-
2K+
SO42-
Zn(s)Zn2+
(aq) + 2e- Cu2+ (aq) + 2e
- Cu(s)
Cathode = ReductionAnode = Oxidation
Zn|Zn2+
||Cu2+
|Cu
1
2
3
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Has infinite resistance = wont consumeelectricity
Measures the potential difference of the 2half cells (Voltage)
parameter for the tendency of the reaction toproceed to an equilibrium state (same Vs)
potential continues to decrease until it reaches0.00 V.
Measures the Electromotive Force (EMF)
Electric work it can do/ can be extracted
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Prevents the direct reaction but stillmaintains electrical contact b/w the halfcells.
Provides ions to neutralize its charges
Negative ions positive half-cell (Zinc)
Positive ions negative half-cell (Copper)
Some SO42-ions (from the cathode vessel) and some
Zn2+ ions (from the anode vessel) also migrate intothe salt bridge.
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Solid conductor through which an electric
current enters or leaves
Collector or emitter of electric charge
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-
Oxidation takes place
Zn(s)Zn2+
(aq) + 2e-
ZnSO4electrolyte
Zn2+
SO42-
Zn(s)
e-
e-
Reducing agent causes the cathode
to reduce
less e-
affinity
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+
Reduction takes place
Cu2+ (aq) + 2e- Cu(s)
CuSO4electrolyte
Cu2+
SO42- Cu(s)
e-
e-
Oxidizing agent causes the anode
to oxidize
more e-
affinity
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Cu2+ (aq) + 2e- Cu(s)
Zn(s) Zn2+
(aq) + 2e-
Zn(s) + Cu2+ Cu(s) + Zn
2+
(reduction)
(oxidation)
(Over-all reaction)
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Electric current - transfer of charge throughelectrolytic and metallic conduction
Electrolytic or ionic conduction- charge is carried
by ions: cations move toward the cathode and anionsmove toward the anode
Metallic conduction- charge is carried by flow ofelectrons from the negative pole to the positive pole
the electrical conductors in an electrochemical cellare the electrodes (via metallic conduction), and theelectrolyte solution (via electrolytic conduction)
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Potential at standard conditions: 1.00 M
1 atm
25 C
Independent on number of moles Cu2+
(aq)+ 2e- Cu
(s)E = 0.34 V
2Cu2+ (aq) + 4e- 2Cu(s) E = 0.34 V
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Measure of the tendency for a reduction
to occur (when SHE is the anode) (+) V = spontaneous
(-) V = non-spontaneous
When reaction is reversed (turned into
oxidation), V changes sign
Standard reduction potential Potential when the SHE (Standard Hydrogen
Electrode) is the anode
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Potential assigned as 0.00 V
2H+ + 2e- H2
H2
2H+ + 2e-
E = 0.00 V
E = 0.00 V
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Cu2+ (aq) + 2e- Cu(s)
Zn(s) Zn2+
(aq) + 2e-
Zn(s) + Cu2+ Cu(s) + Zn
2+ Ecell = 1.103 V
E cathode = 0.340 V
E anode = -0.763 V
E cathode - E anode = E cell*reduction potentials are used
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Factors affecting cell potential: Concentration
Temperature
Nature of reactant
E = E - RT ln QnF
where:
R (gas constant) = 8.314 J/molK,
T = temperature in Kelvin,
n= number of moles e- in rxn
F (Faraday ) = 96, 485 Coulombs
Q = reaction quotient
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(Substituting the constants and assuming 25C
temperature)
E = E - 0.0592 log Qn
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At equilibrium, G = 0, Ecell = 0, and Q = Keq
thus,
becomes
Solving for Keq, we get:
Note that Keq can be Ksp, Ka, Kb, Kf, etc.
Qlog
n
0592.0EE cell
ocell !
eqcello
Klogn
0592.0E !
!
0592.0
)E)(n(logantiK
cello
eq
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Thus, the reaction is spontaneous only ifG isnegative and Ecell is positive
Example: Calculate the Go in J/mol for
3 Sn4+
+ 2 Cr
3 Sn2+
+ 2 Cr3+
Cathode: 3(Sn4+ + 2e- Sn2+) Eo = + 0.15 VAnode: 2(Cr Cr3+ +3e-) Eo = - 0.74 V
Eocell = (+0.15 V) (-0.74 V) = + 0.89 V
The very negative value ofG indicates that thereaction is product-favored. This is consistent with the
positive value of Ecell
cellnFEG !(
where F = 96485
C/mol e-, n = no. of
e
-
transferred
mol/J230,515V89.0emol
C485,96
rxnmol
emol6nFEG cell
oo !
!!(
Note: 1 J = 1 CV
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Let us now calculate for the Keq of theprevious example
Recall: E
o
cell = (+0.15 V) (-0.74 V) = + 0.89 VThus,
the very large value of Keq reinforces ourprevious conclusion that the reaction isproduct-favored
90eq 10x59.1
0592.0
)89.0)(6(loganti !
!
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C = AtWhereC = amount of electricity passing
A (ampere) = current
T= seconds
1 F = 96,500 Coulomb/mole e-
the mass of the product formed or the reactant
used is proportional to the amount of electricity
that passes through the system
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1 Set-up the following half-cells:
Zn
1 M ZnSO4
Zn|Zn2+(1 M)
C
2 M FeSO4+2 M FeCl3
Fe3+(1 M), Fe2+|Fe
Cu
1M CuSO4
Cu|Cu2+(1 M)
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C
2 M FeSO4+2 M FeCl2
Fe3+(1 M), Fe2+|Fe
* Why graphite and not Iron nail?
Desired reaction is
Fe3+ Fe2+ not Fe3+ Fe
When iron nail is used:
Fe3+ Fe2+ FeOFe3+ is oxidized to Fe2+ which will cause rusting
Graphite = semi-conductor
- inert; wont participate, will only conduct
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2Connect the Cu2+/Cu half-cell with avoltmeter and a KNO3 salt bridge
Volt-meter
X
CuSO4
Salt bridge
Cu
X-
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3Record Voltmeter reading
Volt-meter
X
CuSO4
Salt bridge
Cu
X-
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1 Set-up the following half-cells:
C
1 M KCl
Cl- (1 M), Cl2|C
1 M KI1M KBr
Br-(1 M), Br2|C I-(1 M), I2|C
C C C C C
Carbon = inert wont participate in the rxn
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2 Connect to 1.5 V dry cells for 1 minto generate X2
C
1 M KCl
Cl-(1 M), Cl2|C
1 M KX1M KBr
Br-(1 M), Br2|C I-(1 M), I2|C
C C C C C
batt batt batt
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* What happened:
C
KI + I2
I-(1 M), I2|C
C
batt
Bubbles
formed from
H20
Formed brownish
substance
(Iodine / I2)
2I- I2 + 2e-
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* What happened:
C
KCl + Cl2
Cl-(1 M), Cl2|C
KBr + Br2
Br-(1 M), Br2|C
C C C
batt batt
Bubbles Bubbles
Yell ish
2Cl- Cl2 + 2e- 2Br- Br2 + 2e
-
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3 Connect the electrolyzed solutions tothe Cu half cell
Volt-
meter
CCu
KX, X2
Salt bridge
CuSO4
X2 + 2e- 2X-
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3 Record voltmeter reading
Volt-meter
CCu
KX, X2
Salt bridge
CuSO4
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Cell Notation Volts
Cu|Cu2+||Cu2+|Cu 0.0896
Zn|Zn2+||Cu2+|Cu 1.08
Cu|Cu2+||Fe3+|Fe 0.414
C|Cl-,Cl2||Cu2+|Cu 0.468
C|Br-,Br2||Cu2+|Cu 0.269
C|I-,I2||Cu2+|Cu 0.121
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Finding for Eanode:
Take second set-up:
Zn|Zn2+||Cu2+|Cu
Given:
Cu2+ (aq) + 2e- Cu(s) , E = 0.34
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Finding for Eanode:
Take second set-up:
Zn|Zn2+||Cu2+|Cu
Manipulating the equation:
E cathode - E anode = E cell0.34 - E anode = E cell
E anode = 0.34 - E cell
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Zn|Zn2+
||Cu2+
|CuExperimental E cell = 1.08 V
E anode = 0.34 - 1.08
E anode
= - 0.74Zn2+ (aq) + 2e
- Zn(s) , E = 0.763
% error = [-0.763 (- 0.74)]/-0.763
= 3.01%
Finding for Eanode:
Take second set-up:
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Finding for Ecathode:
Take third set-up:
Cu|Cu2+||Fe3+, Fe2+|C
Given:
Cu2+ (aq) + 2e- Cu(s) , E = 0.34
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Finding for Ecathode:
Take third set-up:
Cu|Cu2+||Fe3+, Fe2+|C
Manipulating the equation:
E cathode - E anode = E cell
E cathode = E cell + 0.34
E cathode = 0.34 + E cell
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Finding for Ecathode:
Take third set-up:
Cu|Cu2+
||Fe3+
, Fe2+
|CExperimental E cell = 0.414 V
E cathode = 0.414 + 0.34
E cathode = 0.754 V
Fe3+ + e- Fe2+ E= 0.771
% error = [0.771- 0.754)]/0.771= 2.2 %
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Voltage reading: E = 0.468 V
Ecathode = Ecell + 0.34
Ecathode = 0.468 + 0.34 =0.808V
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Take 4th set-up:
C|Cl-, Cl2||Cu2+|Cu
Data:
Ampere = 0.065 A
Time = 270 seconds
Substituting:
C = 0.065 x 270 = 17.55 Coulombs
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Take 4th set-up:
C|Cl-, Cl2||Cu2+|Cu
From the equation,
2 mol e- = 2 mol Cl- transformed to Cl2Hence,
1.818 x 10-4 mol e- = 1.818 x 10-4 mol Cl-
2Cl- (aq) Cl2+ 2e-
17.55 C x(1 mole e-/96,500 C) = 1.818 x 10-4
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Take 4th set-up:
C|Cl-, Cl2||Cu2+|Cu
0.025 mol Cl- - 1.818 x 10-4 mol Cl- = 0.0248182 mol Cl- left0.0248182 mol Cl-/0.025 L = 0.992728 M
[Cl-] = 0.992728 M
Cl2 is a gas, so the activity of Cl2 is taken as 1,
so Q = __1__[Cl-]2
2Cl- (aq) Cl2+ 2e-
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E = E - 0.0592 log Q
n
Substituting E, n and Q in the equation
0.808 = E - 0.0592 log (1.014704)2
E = 0.808 + 0.0001876479346
= 0.8081876479V
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Cell Notation Cathode Anode
Cu|Cu2+||Cu2+|Cu Copper Copper
Zn|Zn2+||Cu2+|Cu Copper Zinc
Cu|Cu2+||Fe3+|Fe Iron Copper
C|Cl-,Cl2||Cu2+|Cu Copper Chlorine
C|Br-,Br2||Cu2+|Cu Copper Bromine
C|I-,I2||Cu2+|Cu Copper Iodine
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Cell Notation Cathode Anode
Cu(s)|Cu2+
(aq) (1M)|| Cu2+
(aq)
(1M)|Cu(s)
Cu2+(aq)(1M)|Cu(s)
Cu(s)|Cu2+
(aq) (1M)
Zn(s)|Zn2+
(aq) (1M)|| Cu2+
(aq)
(1M)|Cu(s)
Cu2+(aq)
(1M)|Cu(s)
Zn(s)|Zn2+
(aq) (1M)
Cu(s)|Cu2+
(aq) (1M) ||Fe3+
(aq) (1M),
Fe2+(aq) (1M)|C(graphite)
Fe3+(aq) (1M),
Fe2+(aq)(1M)|C(graphite)
Cu(s)|Cu2+
(aq) (1M)
C(graphite)| Cl2(g) |Cl-(aq) (1M)||Cu
2+(aq)
(1M)|Cu(s)
Cu(s)|Cu2+
(aq)
(1M)
C(graphite)| Cl2(g)
|Cl-(aq) (1M)
C(graphite)| Br2(l),Br-(aq) (1M)||Cu
2+(aq)
(1M)|Cu(s)
Cu(s)|Cu2+
(aq)
(1M)
C(graphite)| Br2(l),Br-
(aq) (1M)
C(graphite)| I2(s),I-(aq) (1M)||Cu
2+(aq)
(1M)|Cu(s)
Cu(s)|Cu2+
(aq)
(1M)
C(graphite)| I2(s),I-(aq)
(1M)
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Variable Half-
Cell Notation
Calculated
Volts
Book
Value
% Error
Cu|Cu2+ 0.2504 0.34 26 %
Zn|Zn2+ - 0. 4 - 0. 63 3.0 %
Fe3+|Fe 0. 54 0. 2.2 %
C|Cl-,Cl2 *0.808 .359 40.54 %
C|Br-,Br2 *0.609 .08 43.9 %
C|I-,I2 *0.46 0.6 5 25.04 %
* assuming concentration did not change with electrolysis
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More (-) Ered Higher tendency to be oxidized
Higher reducing power
When placed as anode, reaction is spontaneous
More (+) Ered
Higher tendency to be reduced Higher oxidizing power
When placed as cathode, reaction is spontaneous
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Take Electrolysis for 4th set-up:
C|Cl-, Cl2
Voltage is applied
Vapplied = Eanode - Ecathode
2Cl- (aq) Cl2+ 2e-
Reduction potential = 1.359 V spontaneous
Oxidation potential = -1.359 V non-spontaneous
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2Cl- (aq) Cl2+ 2e-
At equilibrium,Eanode = EcathodeV = 0Oxidation and reduction at same rate
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If Eanode = Ecathode system at equilibriumIf Eanode > Ecathode Ecathode,EanodeIf Eanode < Ecathode Ecathode,Eanode
2Cl- (aq) Cl2+ 2e-
In the experiment, we want to produce more Cl2
oxidation should predominate
Eanode should increase
should be Eanode > Ecathode
Vapplied = Eanode - Ecathode
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Ecathode = E - 0.0592 log [Cl-]2
n [Cl2]
2Cl- (aq) Cl2+ 2e-
Eanode
= E - 0.0592 log [Cl2
]n [Cl-]2
Cl2 = Ecathode = Ered = spontaneity
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Concentration changed as the halogen
solutions were electrolyzed
Electrolysis did not produce enough
halogen, X2 that will be reduced
Slightly different concentrations (not both
exactly 1 M)
increase in voltage reading(concentration difference can also cause
potential diff)
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Example: Calculate the mass ofcopper metal produced at the
cathode during the passage of 2.50
amperes of current through a solution
of copper(II) sulfate for 50.0 minutes.
Cu2+(aq) + 2e- Cu(s)
Current X
Time
no. of
Coulombs
Useful
Conversions:
1 J = 1 CV
1 A = 1 C/s
Mass of
substanceMol. of e-
passed
Cug47.2emol2
Cug5.63
C485,96
emol1
s
C50.2
min1
s60min0.50Cug !
!
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A galvanic cell is to be constructed using a Fe3+/Fe2+ halfcell and a Br2/Br
- half-cell. The Br2/Br- half-cell was
constructed by electrolyzing 50 mL of 0.5 M KBr using a 2.5A current for 3 minutes. The Fe3+/Fe2+ half cell wasconstructed by mixing 25 mL of 2 M FeSO4 and 25 mL of 2 M
FeCl3. (MW Br2: 159.8 g/mol)
Fe3+ (aq) + e- Fe2+
(aq) Eo
red = +0.771 VBr2(aq) + 2e
- 2Br-(aq) Eo
red = +1.087 V
a) Assuming that the bromide ions are converted to Br2(aq),
crcalculate [Br2] and [Br
-
] upon electrolysis.b) Write the cell notation (with the correct phases andconcentrations of ions) for the galvanic cell
c) Calculate the Ecell for the galvanic cell.
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Answers:
a) [Br2] = 0.046638 M, [Br-] = 0.406722 M
b) Fe2+(aq) (1M), Fe3+
(aq)
(1M)||Br2(aq)(0.046638 M), Br-(aq)
(0.406722 M)
c) 0.2997243 V
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Belcher, R. Quantitative Inorganic Analysis,1970
Christian, G.D. Analytical Chemistry, 1986
Day, Underwood, et al. Quantitative Analysis,
1967Haenisch, Pierce, et al. Quantitative Analysis,
1958
Skoog, et al., Fundamentals of Analytical
Chemistry, Eighth edition, 2004 http://academic.pgcc.edu/psc/chm103/103_m
anual.pdf