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Chapter 18
Electrochemistry
2007, Prentice Hall
Chemistry: A Molecular Approach, 1stEd.
Nivaldo Tro
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
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Redox Reaction one or more elements change oxidation number
all single displacement, and combustion,some synthesis and decomposition
always have both oxidation and reductionsplit reaction into oxidation half-reaction and a
reduction half-reaction aka electron transfer reactions
half-reactions include electrons
oxidizing agentis reactant molecule that causes oxidationcontains element reduced reducing agentis reactant molecule that causes reduction
contains the element oxidized
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Oxidation & Reduction
oxidationis the process that occurs whenoxidation number of an element increases
element loses electrons
compound adds oxygen
compound loses hydrogenhalf-reaction has electrons as products
reductionis the process that occurs whenoxidation number of an element decreases
element gains electronscompound loses oxygen
compound gains hydrogen
half-reactions have electrons as reactants
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Rules for Assigning Oxidation States
rules are in order of priority1. free elements have an oxidation state = 0 Na = 0 and Cl2= 0 in 2 Na(s) + Cl2(g)
2. monatomic ions have an oxidation state equalto their charge
Na = +1 and Cl = -1 in NaCl
3. (a) the sum of the oxidation states of all theatoms in a compound is 0 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0
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Rules for Assigning Oxidation States
5. in their compounds, nonmetals have oxidationstates according to the table below nonmetals higher on the table take priority
Nonmetal Oxidation State Example
F -1 CF4
H +1 CH4
O -2 CO2
Group 7A -1 CCl4
Group 6A -2 CS2
Group 5A -3 NH3
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Oxidation and Reduction
oxidation occurs when an atoms oxidation stateincreases during a reaction
reduction occurs when an atoms oxidation statedecreases during a reaction
CH4 + 2 O2 CO2+ 2 H2O-4+1 0 +42 +1 -2
oxidation
reduction
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OxidationReduction oxidation and reduction must occur simultaneously
if an atom loses electrons another atom must take them
the reactant that reduces an element in another reactantis called the reducing agent
the reducing agent contains the element that is oxidized
the reactant that oxidizes an element in another reactantis called the oxidizing agent
the oxidizing agent contains the element that is reduced
2 Na(s) + Cl2(g) 2 Na+Cl(s)
Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2is the oxidizing agent
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Identify the Oxidizing and Reducing Agents
in Each of the Following3 H2S + 2 NO3
+ 2 H+ 3S + 2 NO + 4 H2O
MnO2+ 4 HBr MnBr2+ Br2+ 2 H2O
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Identify the Oxidizing and Reducing Agents
in Each of the Following3 H2S + 2 NO3
+ 2 H+ 3S + 2 NO + 4 H2O
MnO2+ 4 HBr MnBr2+ Br2+ 2 H2O
+1 -2 +5 -2 +1 0 +2 -2 +1 -2
ox agred ag
+4 -2 +1 -1 +2 -1 0 +1 -2
oxidationreduction
oxidation
reduction
red agox ag
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12
Common Reducing AgentsReducing Agent Product when Oxidized
H2 H+1
H2O2 O2
I-1
I2
NH3, N2H4 N2
S
-2
, H2S SSO3
-2SO4
-2
NO2-1
NO3-1
C (as coke or charcoal) CO or CO2
Fe+2(acid) Fe+3
Cr+2
Cr+3
Sn+2
Sn+4
metals metal ions
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Balancing Redox Reactions1) assign oxidation numbers
a) determine element oxidized and element reduced2) write ox. & red. half-reactions, including electrons
a) ox. electrons on right, red. electrons on left of arrow
3) balance half-reactions by massa) first balance elements other than H and O
b) add H2O where need O
c) add H+1where need H
d) neutralize H+with OH-in base
4) balance half-reactions by chargea) balance charge by adjusting electrons
5) balance electrons between half-reactions6) add half-reactions7) check
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Ex 18.3Balance the equation:
I(aq)+ MnO4
(aq)I2(aq)+ MnO2(s)in basic solution
Assign
Oxidation
States
I(aq)+ MnO4
(aq)I2(aq)+ MnO2(s)
Separate
into half-
reactions
ox:
red:
Assign
Oxidation
States
Separate
into half-
reactions
ox: I(aq)I2(aq)red: MnO4
(aq)MnO2(s)
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Balance
half-
reactions
by mass
then H by
adding H+
ox: 2 I(aq)I2(aq)red: 4 H+(aq)+ MnO4
(aq)MnO2(s)+ 2 H2O(l)
Balance half-
reactions by
mass
then O byadding H2O
ox: 2 I(aq)I2(aq)red: MnO4
(aq)MnO2(s)+ 2 H2O(l)
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Balancehalf-
reactions
by mass
in base,
neutralize
the H+
with OH-
ox: 2 I
(aq)I2(aq)red: 4 H+(aq)+ MnO4
(aq)MnO2(s)+ 2 H2O(l)
4 H+(aq)+ 4 OH
(aq)+ MnO4
(aq)MnO2(s)+ 2 H2O(l) + 4 OH(aq)
4 H2O(aq)+ MnO4
(aq)MnO2(s)+ 2 H2O(l) + 4 OH(aq)
MnO4
(aq)+ 2 H2O(l)MnO2(s)+ 4 OH(aq)
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Ex 18.3Balance the equation:
I(aq)+ MnO4
(aq)I2(aq)+ MnO2(s)in basic solution
Balance
Half-
reactions
bycharge
ox: 2 I(aq)I2(aq)+ 2 e
red: MnO4
(aq)+ 2 H2O(l) + 3 eMnO2(s) + 4 OH(aq)
Balance
electrons
between
half-
reactions
ox: 2 I(aq)I2(aq)+ 2 e}x3red: MnO
4
(aq)
+ 2 H2
O(l)
+ 3 eMnO2(s)
+ 4 OH(aq)
}x2
ox: 6 I(aq)3 I2(aq)+ 6 ered: 2 MnO4
(aq)+ 4 H2O(l) + 6 e
2 MnO2(s) + 8 OH(aq)
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Ex 18.3Balance the equation:
I(aq)+ MnO4
(aq)I2(aq)+ MnO2(s)in basic solution
Add the
Half-
reactions
ox: 6 I(aq)3 I2(aq)+ 6 ered: 2 MnO4
(aq)+ 4 H2O(l) + 6 e
2 MnO2(s) + 8 OH(aq)tot: 6 I(aq)+ 2 MnO4
(aq)+ 4 H2O(l)3 I2(aq)+ 2 MnO2(s) + 8 OH(aq)
Check ReactantCount Element
ProductCount
6 I 6
2 Mn 2
12 O 12
8 H 8
2 charge 2
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Practice - Balance the Equation
H2O2+ KI + H2SO4K2SO4+ I2+ H2O
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Practice - Balance the Equation
H2O2+ KI + H2SO4K2SO4+ I2+ H2O+1 -1 +1 -1 +1 +6 -2 +1 +6 -2 0 +1 -2
oxidationreduction
ox: 2 I-1
I2+ 2e-1
red: H2O2+ 2e
-1+ 2 H+2 H2Otot 2 I-1+ H2O2+ 2 H
+ I2+ 2 H2O
1 H2O
2+ 2 KI + H
2SO
4K
2SO
4+ 1 I
2+ 2 H
2O
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Practice - Balance the Equation
ClO3-1+ Cl-1 Cl2 (in acid)
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Practice - Balance the Equation
ClO3-1+ Cl-1 Cl2 (in acid)
+5 -2 -1 0
oxidationreduction
ox: 2 Cl-1
Cl2+ 2 e-1
} x5red: 2 ClO3
-1+ 10 e-1+ 12 H+Cl2+ 6 H2O} x1tot 10 Cl-1+ 2ClO3
-1+ 12 H+ 6 Cl2 + 6 H2O
1 ClO3-1+ 5 Cl-1 + 6 H+1 3 Cl2+ 3 H2O
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Tro, Chemistry: A Molecular Approach 24
Electrical Current
when we talk about the currentof a liquid in a stream, we arediscussing the amount of waterthat passes by in a given periodof time
when we discuss electriccurrent, we are discussing theamount of electric charge that
passes a point in a given period
of timewhether as electrons flowingthrough a wire or ions flowingthrough a solution
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Redox Reactions & Current
redox reactions involve the transfer of electronsfrom one substance to another
therefore, redox reactions have the potential togenerate an electric current
in order to use that current, we need to separate
the place where oxidation is occurring from theplace that reduction is occurring
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Electric Current Flowing
Indirectly Between Atoms
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Electrochemical Cells oxidation and reduction reactions kept separatehalf-cells
electron flow through a wire along with ion flowthrough a solution constitutes an electric circuit
requires a conductive solid (metal or graphite)electrodeto allow the transfer of electrons
through external circuit
ion exchange between the two halves of the system electrolyte
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Electrodes Anode
electrode where oxidation occursanions attracted to itconnected to positive end of battery in electrolytic
cell
loses weight in electrolytic cell Cathodeelectrode where reduction occurscations attracted to it
connected to negative end of battery in electrolyticcellgains weight in electrolytic cellelectrode where plating takes place in electroplating
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Voltaic Cell
the salt bridge is
required to complete
the circuit and
maintain charge
balance
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Current and Voltage the number of electrons that flow through the system per
second is the current unit = Ampere
1 A of current = 1 Coulomb of charge flowing by each second
1 A = 6.242 x 1018electrons/second
Electrode surface area dictates the number of electrons that canflow
the difference in potential energy between the reactantsand products is the potential difference unit = Volt
1 V of force = 1 J of energy/Coulomb of charge the voltage needed to drive electrons through the external circuit
amount of force pushing the electrons through the wire is calledthe electromotive force, emf
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Cell Potential the difference in potential energy between the
anode the cathode in a voltaic cell is called thecell potential
the cell potential depends on the relative ease
with which the oxidizing agent is reduced at thecathode and the reducing agent is oxidized at theanode
the cell potential under standard conditions iscalled the standard emf, Ecell25C, 1 atm for gases, 1 M concentration of solution
sum of the cell potentials for the half-reactions
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Cell Notation
shorthand description of Voltaic cell electrode | electrolyte || electrolyte | electrode oxidation half-cell on left, reduction half-cell on
the right single | = phase barrierif multiple electrolytes in same phase, a comma is
used rather than |
often use an inert electrode
double line || = salt bridge
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Fe(s) | Fe2+(aq) || MnO4(aq), Mn2+(aq), H+(aq) | Pt(s)
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Standard Reduction Potential a half-reaction with a strong tendency to
occur has a large + half-cell potential when two half-cells are connected, theelectrons will flow so that the half-reactionwith the stronger tendency will occur
we cannot measure the absolute tendencyof a half-reaction, we can only measure itrelative to another half-reaction
we select as a standard half-reaction thereduction of H+to H2under standard
conditions, which we assign a potentialdifference = 0 v standard hydrogen electrode, SHE
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Half-Cell Potentials
SHE reduction potential is defined to be exactly 0 v half-reactions with a stronger tendency towardreduction than the SHE have a + value for Ered
half-reactions with a stronger tendency towardoxidation than the SHE have a value for Ered
Ecell= Eoxidation+ EreductionEoxidation= Ereduction
when adding Evalues for the half-cells, do not multiplythehalf-cell E values, even if you need to multiply the half-
reactions to balance the equation
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Ex 18.4Calculate Ecellfor the reaction at 25CAl(s)+ NO3
(aq) + 4 H
+(aq)Al3+(aq)+ NO(g)+ 2 H2O(l)
Separate thereaction into
the oxidation
and reduction
half-reactions
ox: Al(s)Al3+(aq)+ 3 e
red: NO3
(aq)+ 4 H+
(aq) + 3 eNO(g) + 2 H2O(l)
find the Eforeach half-
reaction andsum to get
Ecell
Eox= Ered= +1.66 v
Ered= +0.96 v
Ecell= (+1.66 v) + (+0.96 v) = +2.62 v
E 18 4 P edi t if the f ll i e ti i
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Ex 18.4aPredict if the following reaction is
spontaneous under standard conditions
Fe(s)+ Mg2+
(aq)Fe2+(aq)+ Mg(s)
Separate thereaction into
the oxidation
and reduction
half-reactions
ox: Fe(s)Fe2+(aq)+ 2 e
red: Mg2+(aq)+ 2 eMg(s)
look up the
relative
positions of thereduction half-
reactions
red: Mg2+(aq)+ 2 eMg(s)
red: Fe2+(aq)+ 2 eFe(s)
since Mg2+reduction is below Fe2+reduction, the reaction is NOT spontaneous
as written
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Tro, Chemistry: A Molecular Approach 43
the reaction is
spontaneous in the
reverse direction
Mg(s)+ Fe2+
(aq)Mg2+(aq)+ Fe(s)ox: Mg(s)Mg2+(aq)+ 2 e
red: Fe2+(aq)+ 2 eFe(s)
sketch the cell and
label the parts
oxidation occurs atthe anode; electrons
flow from anode to
cathode
P ti Sk t h d L b l th V lt i C ll
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Practice - Sketch and Label the Voltaic Cell
Fe(s) Fe2+(aq) Pb2+(aq) Pb(s) , Write theHalf-Reactions and Overall Reaction, and Determine
the Cell Potential under Standard Conditions.
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ox: Fe(s) Fe2+(aq) + 2 e E= +0.45 V
red: Pb2+(aq) + 2 e Pb(s) E= 0.13 V
tot: Pb2+(aq) + Fe(s) Fe2+(aq) + Pb(s) E= +0.32 V
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Predicting Whether a Metal Will
Dissolve in an Acid acids dissolve in metals if the
reduction of the metal ion is
easier than the reduction ofH+(aq)
metals whose ion reduction
reaction lies below H+
reductionon the table will dissolve in acid
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Ecell,DG and K
for a spontaneous reactionone the proceeds in the forward direction with the
chemicals in their standard states
DG < 1 (negative)
E > 1 (positive)
K > 1
DG = RTlnK = nFEcell
nis the number of electrons
F = Faradays Constant = 96,485 C/mol e
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E l 18 7 C l l t K t 25C f th ti
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125.11 102.310
5.11
V0592.0
mol2V34.0log
K
eK
Example 18.7- Calculate K at 25C for the reactionCu(s)+ 2 H
+(aq)H2(g)+ Cu
2+(aq)
since K< 1, the position of equilibrium lies far tothe left under standard conditions
Answer:
Solve:
Concept Plan:
Relationships:
Cu(s)+ 2 H+
(aq)H2(g)+ Cu2+
(aq)K
Given:Find:
Eox, Ered Ecell K
redoxcell
EEE Kn
logV0592.0
Ecell
ox: Cu(s) Cu2+
(aq)+ 2 e E= 0.34 v
red: 2 H+(aq)+ 2 e H2(aq) E= +0.00 v
tot: Cu(s)+ 2H+
(aq) Cu2+
(aq)+ H2(g)E= 0.34 v
Kn
logV0592.0
Ecell
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Nonstandard Conditions -
the Nernst Equation DG = DG + RT ln Q
E = E - (0.0592/n) log Q at 25C
when Q = K, E = 0
use to calculate E when concentrations not 1 M
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l l l f h i
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Example 18.8- Calculate Ecellat 25C for the reaction3 Cu(s)+ 2 MnO4
(aq)+ 8 H
+(aq)2 MnO2(s)+ Cu
2+(aq)+ 4 H2O(l)
units are correct, Ecell> Ecell as expected because
[MnO4] > 1 M and [Cu2+] < 1 M
Check:
Solve:
Concept Plan:
Relationships:
3 Cu(s)+ 2 MnO4
(aq)+ 8 H
+
(aq)2 MnO2(s)+ Cu
2+
(aq)+ 4 H2O(l)[Cu2+] = 0.010 M, [MnO4
] = 2.0 M, [H+] = 1.0 M
Ecell
Given:
Find:
Eox, Ered Ecell Ecell
redoxcell EEE Qn
logV0592.0EE cellcell
red: MnO4
(aq)+ 4 H+
(aq)+ 3 e MnO2(s)+ 2
H2O(l)}x2 E= +1.68 v
tot: 3 Cu(s)+ 2 MnO4(aq)+ 8 H+(aq) 2 MnO2(s)+ Cu2+(aq)+ 4 H2O(l))E= +1.34 v
]C[V05920 32
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V41.1E
.0]1[.0]2[
]010.0[log
6
V0592.0V34.1E
][H][MnO
]Cu[log
V0592.0EE
cell
83
3
cell
834
32
cellcell
n
ox: Cu(s) Cu2+
(aq)+ 2 e}x3E= 0.34 v
Qn
logV0592.0
EE cellcell
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when the cell
concentrations
are equal there isno difference in
energy between
the half-cells and
no electrons flow
Concentration Cell
Cu(s)Cu2+(aq)(0.010 M) Cu2+(aq)(2.0 M)Cu(s)
when the cell concentrations
are different, electrons flow
from the side with the less
concentrated solution(anode) to the side with the
more concentrated solution
(cathode)
L Cl h A idi D C ll
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LeClanche Acidic Dry Cell electrolyte in paste form
ZnCl2+ NH4Clor MgBr2
anode = Zn (or Mg)Zn(s) Zn2+(aq) + 2 e-
cathode = graphite rod MnO2 is reduced
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e
-
2 NH4OH(aq) + 2 Mn(O)OH(s)
cell voltage = 1.5 v expensive, nonrechargeable, heavy,
easily corroded
Alk li D C ll
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Alkaline Dry Cell same basic cell as acidic dry cell, except
electrolyte is alkaline KOH paste anode = Zn (or Mg)
Zn(s) Zn2+(aq) + 2 e-
cathode = brass rod
MnO2 is reduced2 MnO2(s) + 2 NH4
+(aq) + 2 H2O(l) + 2 e-
2 NH4OH(aq) + 2 Mn(O)OH(s)
cell voltage = 1.54 v
longer shelf life than acidic dry cells andrechargeable, little corrosion of zinc
d S
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Lead Storage Battery
6 cells in series electrolyte = 30% H2SO4
anode = PbPb(s) + SO
4
2-(aq) PbSO4(s) + 2 e-
cathode = Pb coated with PbO2
PbO2 is reducedPbO2(s) + 4 H
+(aq) + SO42-(aq) + 2 e-
PbSO4(s) + 2 H2O(l)
cell voltage = 2.09 v
rechargeable, heavy
NiC d B tt
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NiCad Battery
electrolyte is concentrated KOH solution anode = CdCd(s) + 2 OH-1(aq) Cd(OH)2(s) + 2 e-1 E0 = 0.81 v cathode = Ni coated with NiO2
NiO2 is reducedNiO2(s) + 2 H2O(l) + 2 e
-1Ni(OH)2(s) + 2OH-1 E0= 0.49 v
cell voltage = 1.30 v rechargeable, long life, lighthowever
recharging incorrectly can lead to batterybreakdown
Ni MH B tt
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Ni-MH Battery
electrolyte is concentrated KOH solution anode = metal alloy with dissolved hydrogen oxidation of H from H0to H+1
MH(s) + OH-1(aq) M(s) + H2O(l) + e-1 E = 0.89 v
cathode = Ni coated with NiO2 NiO2 is reducedNiO2(s) + 2 H2O(l) + 2 e
-1Ni(OH)2(s) + 2OH-1 E0= 0.49 v
cell voltage = 1.30 v
rechargeable, long life, light, more environmentallyfriendly than NiCad, greater energy density than
NiCad
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Lithium Ion Battery electrolyte is concentrated KOH
solution
anode = graphite impregnated with Liions
cathode = Li - transition metal oxide reduction of transition metal
work on Li ion migration from anodeto cathode causing a correspondingmigration of electrons from anode tocathode
rechargeable, long life, very light,more environmentally friendly,greater energy density
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F l C ll
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Fuel Cells like batteries in which
reactants are constantlybeing added so it never runs down!
Anode and Cathodeboth Pt coated metal
Electrolyte is OHsolution
Anode Reaction:2 H2+ 4 OH
4 H2O(l) + 4 e
-
Cathode Reaction:O2+ 4 H2O + 4 e
- 4 OH
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electroplating
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electroplating
In electroplating, the work
piece is the cathode.
Cations are reduced at
cathode and plate to
the surface of the workpiece.
The anode is made of
the plate metal. The
anode oxidizes andreplaces the metal
cations in the solution
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Electrochemical Cells
in all electrochemical cells, oxidation occurs at theanode, reduction occurs at the cathode
in voltaic cells, anode is the source of electrons and has a () charge
cathode draws electrons and has a (+) charge
in electrolytic cells electrons are drawn off the anode, so it must have a place to
release the electrons, the + terminal of the battery
electrons are forced toward the anode, so it must have a
source of electrons, the terminal of the battery
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Electrolysis of Water
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Electrolysis of Water
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Electrolysis of Pure Compounds
must be in molten (liquid) state
electrodes normally graphite
cations are reduced at the cathode to metalelement
anions oxidized at anode to nonmetal element
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Tro, Chemistry: A Molecular Approach 72
Mixtures of Ions
when more than one cation is present, the cationthat is easiest to reduce will be reduced first atthe cathode
least negative or most positive Ered
when more than one anion is present, the anionthat is easiest to oxidize will be oxidized first at
the anodeleast negative or most positive Eox
Electrolysis of Aqueous Solutions
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Electrolysis of Aqueous Solutions Complicated by more than one possible oxidation and reduction possible cathode reactions reduction of cation to metal reduction of water to H2
2 H2O + 2 e-1H2+ 2 OH-1 E = -0.83 v @ stand. cond.
E = -0.41 v @ pH 7
possible anode reactionsoxidation of anion to elementoxidation of H2O to O2
2 H2O O2+ 4e-1+ 4H+1 E = -1.23 v @ stand. cond.E = -0.82 v @ pH 7
oxidation of electrodeparticularly Cugraphite doesnt oxidize
half-reactions that lead to least negative Etotwill occur unless overvoltage changes the conditions
Electrolysis of NaI
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Electrolysis of NaI(aq)with Inert Electrodes
possible oxidations
2 I-1 I2+ 2 e-1 E = 0.54 v2 H2O O2+ 4e-1+ 4H+1 E = 0.82 v
possible reductionsNa+1+ 1e-1Na0 E = 2.71 v2 H2O + 2 e
-1H2+ 2 OH-1 E = 0.41 v
possible oxidations
2 I-1 I2+ 2 e-1 E = 0.54 v2 H2O O2+ 4e-1+ 4H+1 E = 0.82 v
possible reductionsNa+1+ 1e-1Na0 E = 2.71 v2 H2O + 2 e
-1H2+ 2 OH-1 E = 0.41 v
overall reaction
2 I(aq)+ 2 H2O(l)I2(aq)+H2(g) + 2 OH-1(aq)
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Tro, Chemistry: A Molecular Approach 75
Faradays Law
the amount of metal deposited duringelectrolysis is directly proportional to the charge
on the cation, the current, and the length of timethe cell runs
charge that flows through the cell = current xtime
Example 18.10- Calculate the mass of Au that can be plated in
25 i i 5 5 A f h h lf i
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25 min using 5.5 A for the half-reaction
Au3+(aq) + 3 e Au(s)
units are correct, answer is reasonable since 10 A
running for 1 hr ~ 1/3 mol e
Check:
Solve:
Concept Plan:
Relationships:
3 mol e
: 1 mol Au, current = 5.5 amps, time = 25 minmass Au, g
Given:
Find:
s1
C5.5
Aug6.5
Aumol1
g196.97
mol3
Aumol1
C96,485
mol1
s1
C5.5
min1
s60min25
e
e
t(s), amp charge (C) mol e mol Au g Au
C6,4859mol1
e emol3
Aumol1
Aumol1
g196.97
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Tro, Chemistry: A Molecular Approach 77
Corrosion
corrosionis the spontaneous oxidation of ametal by chemicals in the environment
since many materials we use are activemetals, corrosion can be a very big problem
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Rusting
rust is hydrated iron(III) oxide moisture must be presentwater is a reactant
required for flow between cathode and anode electrolytes promote rustingenhances current flow
acids promote rustinglower pH = lower Ered
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Tro, Chemistry: A Molecular Approach 79
P i C i
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Preventing Corrosion
one way to reduce or slow corrosion is to coatthe metal surface to keep it from contactingcorrosive chemicals in the environment
paint
some metals, like Al, form an oxide that stronglyattaches to the metal surface, preventing the restfrom corroding
another method to protect one metal is to attachit to a more reactive metal that is cheap
sacrificial electrode
galvanized nails
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Sacrificial Anode