Transcript 2 I

ELECTROCHEMISTRY
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Electron Transfer Reactions
• Electron transfer reactions are oxidationreduction or redox reactions.
• Results in the generation of an electric
current (electricity) or be caused by
imposing an electric current.
• Therefore, this field of chemistry is often
called ELECTROCHEMISTRY.
Terminology for Redox
Reactions
• OXIDATION—loss of electron(s) by a species;
increase in oxidation number; increase in
oxygen.
• REDUCTION—gain of electron(s);
decrease in
©
oxidation number; decrease in oxygen;
increase in hydrogen.
• OXIDIZING AGENT—electron acceptor;
species is reduced.
• REDUCING AGENT—electron donor; species
is oxidized.
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Oxidation and reduction
happen at the same time
There is no net gain or loss of
electrons.
e
You can’t just create
them or destroy them!
Example:
Fe2O3 + Al  Fe + Al2O3
Fe3+
Al0
Fe0
Al3+
Fe3+ + 3e-  Fe
Reduction
Al  Al2O3 + 3e-
Oxidation
Ionic half-equations
Remember:
OIL - Oxidation Is Loss of electrons
RIG - Reduction Is Gain of electrons
And often... (as a quick and simple way to tell):
Oxidation is gain in oxygen or loss of hydrogen
Reduction is loss of oxygen or gain of hydrogen
Oxidation state
Oxidation
Reduction
Oxidation and
reduction can be
seen as movement
up or down a scale
of oxidation states
+7
+6
+5
+4
+3
+2
+1
0
-1
-2
-3
-4
-5
-6
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OXIDATION & REDUCTION
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• OXIDATION:
Zn(s)  Zn+2(aq) + 2 eMetallic zinc is oxidized to zinc ion. Metallic zinc is
serving as a reducing agent.(electron loser)
• REDUCTION:
Cu+2(aq) + 2 e- Cu(s)
Copper ion is reduced to copper metal. Copper ion
is serving as an oxidizing agent (electron gainer)
• In the overall reaction two electrons are transferred
from the zinc metal to the copper ion.
Zn(s) + Cu+2(aq) Zn+2(aq) + Cu(s)
You can’t have one… without
the other!
• Reduction (gaining electrons) can’t happen without an
oxidation to provide the electrons.
• You can’t have 2 oxidations or 2 reductions in the same
equation. Reduction has to occur at the cost of oxidation
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OXIDATION NUMBERS
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• Before discussing the mechanics of oxidation
numbers it is important to realize that:
• (1) oxidation numbers have no physical
significance. They are merely a way to tell which
substances gain or lose electrons and how many.
• (2) oxidation numbers are assigned to atoms never
molecules. Molecules contain atoms with oxidation
numbers but they themselves cannot be assigned
oxidation numbers.
• (3) there a several rules used to assign oxidation
numbers. They must be observed carefully.
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OXIDATION NUMBERS RULES
• (1) Elements have oxidation numbers of zero. For
example Cu has an oxidation number of zero. In
Cl2 each atom of chlorine has an oxidation number
of zero.
• (2) The oxidation number of a monatomic ion
(consisting of one atom) is the charge on the ion.
In Cu+2 the oxidation number is +2. In Cl- the
oxidation number is –1.
• (3) The oxidation number of combined oxygen is
always –2 except in peroxides such as hydrogen
peroxide, H2O2. In CO2 each oxygen atom has an
oxidation number of –2. In SO3 each oxygen is –2.
FINDING UNKNOWN OXIDATION NUMBERS
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• Problem: What is the oxidation number of Mn in MnO4-, the
permanganate ion?
• Solution: Mn is unknown
Each O is –2
All of the oxidation numbers must add up to –1
Mn + 4(-2) = -1, Mn = +7 in permanganate
• Problem: What is the oxidation number of Cr in K2Cr2O7,
potassium dichromate?
• Solution: Cr is unknown
Each K from Column I is +1
Each O is –2
All the oxidation numbers add up to 0 (no charge is shown for
K2Cr2O7)
2(+1) + 2 Cr + 7(-2) = 0,
2 Cr = 12,
Cr = 12 / 2 = +6
Cr = +6 in potassium dichromate
USING OXIDATION NUMBERS TO DETERMINE ELECTRON
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TRANSFERS
• Using oxidation numbers we will find the oxidizing agent, the
reducing agent, the number of electrons lost and gained, the
oxidation half-reaction, the reduction half-reaction and the final
balanced equation for:
H+ + MnO4- + Cl-  Mn+2 + Cl2 + H2O
• Step I – Find the oxidation number of each atom.
H+ + MnO4- + Cl-  Mn+2 + Cl2 + H2O
+1
(+7, -2)
-1
+2
0 (+1, -2)
• Step II – Determine which oxidation numbers change
MnO4-  Mn+2
2 Cl-  Cl2
+7
 +2
2(-1)  0
If oxidization # s decrease,
Reduction occurs
5 electrons gained
2 electrons lost
If oxidization # s increase,
Oxidation occurs
(MnO4- is the oxidizing agent. It gained electrons from Cl- )
(Cl- is the reducing agent. It lost electrons to MnO4- )
USING OXIDATION NUMBERS TO DETERMINE
ELECTRON TRANSFERS
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• Step III - Electrons lost by a reducing agent must always equal electrons
gained by the oxidizing agent!
• Therefore: 2 (MnO4- + 5e-  Mn+2 )
2MnO4- + 10e-  2Mn+2 (reduction half-reaction*)
• And
5(2Cl-  Cl2 + 2e-)
10 Cl-  5Cl2 + 10e- (oxidation half-reaction)
• Step IV -Completing the first half-reaction with H+ ions and H2O molecules:
16 H+ + 2MnO4- + 10e-  2Mn+2 + 8 H2O
• Step V - Adding the oxidation half-reaction and reduction half-reaction the 10
electrons gained and lost cancel to give the overall reaction:
16 H+ + 2MnO4- + 10 Cl-  2 Mn+2 + 5 Cl2 + 8 H2O
• * Half-reaction refers to the reaction showing either the electron gain
(reduction) or the electron loss (oxidation) step of the reaction.
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OXIDATION-REDUCTION
REACTIONS
Direct Redox Reaction
Oxidizing and
reducing agents in
direct contact.
Cu(s) + 2 Ag+(aq) --->
Cu2+(aq) + 2 Ag(s)
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OXIDATION-REDUCTION
REACTIONS
Indirect Redox Reaction
A battery functions by transferring electrons
through an external wire from the reducing
agent to the oxidizing agent.
Why Study Electrochemistry?
• Batteries
• Corrosion
• Industrial production
of chemicals such as
Cl2, NaOH, F2 and Al
• Biological redox
reactions
The heme group
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Balancing Equations
for Redox Reactions
Some redox reactions have equations that must be balanced by
special techniques.
MnO4- + 5 Fe2+ + 8 H+
---> Mn2+ + 5 Fe3+ + 4 H2O
Mn = +7
Fe = +2
Mn = +2 Fe = +3
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Balancing Equations
Consider the
reduction of Ag+
ions with copper
metal.
Cu + Ag+
--give--> Cu2+ + Ag
Balancing Equations
Step 1:
Divide the reaction into half-reactions, one
for oxidation and the other for reduction.
Ox
Cu ---> Cu2+
Red
Ag+ ---> Ag
Step 2:
Balance each element for mass. Already
done in this case.
Step 3:
Balance each half-reaction for charge by
adding electrons.
Ox
Cu ---> Cu2+ + 2eRed
Ag+ + e- ---> Ag
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Balancing Equations
Step 4:
Multiply each half-reaction by a factor so
that the reducing agent supplies as many electrons
as the oxidizing agent requires.
Reducing agent
Cu ---> Cu2+ + 2eOxidizing agent
2 Ag+ + 2 e- ---> 2 Ag
Step 5:
Add half-reactions to give the overall
equation.
Cu + 2 Ag+ ---> Cu2+ + 2Ag
The equation is now balanced for both
charge and mass.
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Redox in presence of acid
a) Identify, as oxidation or reduction,
the formation of NO2 from NO3- in the
presence of H+ and deduce the halfequation for the reaction.
NO3
+5
 NO2
+4
Reduction
Steps to take
1. Write a balanced equation for the species
2. Work out “before and after” oxidation states
3. Balance oxidation states with electrons
NO3-  NO2
+5
+4
NO3- + e-  NO2
4. If all the charges don’t balance,
- + e- + 2H+  NO
NO
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add H+ ions to one of the sides
to balance them
5. If the equation still doesn’t
balance, add enough water NO3- + e- + 2H+  NO2 + H2O
to one side so it balances
Beautiful!
Give these a go
Balance the half equations
Na

Na+
Fe2+

Fe3+
I2

I¯
C2O42-

CO2
H2O2

O2
H2O2

H2O
NO3-

NO
NO2

NO3-
SO42-

SO2
All good?
Na

Na+ +
e-
Fe2+

Fe3+ +
e-

2I¯
C2O42-

2CO2 +
H2O2

I2
+
2e-
H2O2 + 2H+ + 2e- 
O2
2e-
+ 2H+ + 2e-
2H2O
NO3- + 4H+ + 3e- 
NO +
2H2O

NO3- +
2H+ + e-
SO42- + 4H+ + 2e 
SO2 +
2H2O
NO2 + H2O
Combining half equations
Just a mashing together
of two half-equations!
... followed by some
satisfying cancelling-out.
b) Deduce the overall equation for the
reaction of copper with NO3- in acidic
conditions to give Cu2+.
NO3- + e- + 2H+  NO2 + H2O
Cu  Cu2+ + 2e-
Reduction
Oxidation
Now what?
Steps to take to combine equations
1. Multiply the equations so that the number of
electrons in each is the same
Cu  Cu2+ + 2eX2
2NO3- + 2e- + 4H+  2NO2 + 2H2O
2. Add the two equations and cancel out the electrons
on either side of the equation
Cu + 2NO3- + 2e- + 4H+  Cu2+ + 2e- + 2NO2 + 2H2O
3. If necessary, cancel out any other species which
appear on both sides of the equation
Cu + 2NO3- + 4H+  Cu2+ + 2NO2 + 2H2O
Give these a go
1. Fe2+ ions are oxidised to Fe3+ ions by ClO3- ions in
acidic conditions. The ClO3- ions are reduced to Clions. Write the overall reaction.
Fe2+  Fe3+ + e-
+ ClO3- + 6e- + 6H+  Cl- + 3H2O
6Fe2+ + ClO3- + 6H+  Fe3+ + Cl- + 3H2O
2. Write an overall reaction for MnO4- reducing H2O2
to O2 and creating Mn2+ .
2MnO4¯ + 5H2O2 + 6H+  2Mn2+ + 5O2 + 8H2O
Complete the half-equation
Na  Na+
Na  Na++ e-
Complete the half-equation
Pb4+  Pb2+
Pb4+ + 2e-  Pb2+
Complete the half-equation
H2  H+
H2  2H+ + 2e-
Complete the half-equation
Cr2O72-  Cr3+
Cr2O72- + 6e-  2Cr3+
What’s wrong with this equation?
Ce3+ + e-  Ce4+
Electron is on the wrong side!
What’s wrong with this equation?
Mg + 2H+ + e-  Mg+ + H2 + e-
Should be Mg2+
Electrons should be cancelled out
Balancing Equations
Balance the following in acid solution—
VO2+ + Zn ---> VO2+ + Zn2+
Step 1:
Write the half-reactions
Ox
Zn ---> Zn2+
Red
VO2+ ---> VO2+
Step 2:
Balance each half-reaction for
mass.
Ox
Zn ---> Zn2+
Red
2 H+ + VO2+ ---> VO2+ + H2O
Add H2O on O-deficient side and add H+
on other side for H-balance.
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Balancing Equations
Step 3:
Ox
Red
Step 4:
Ox
Red
Balance half-reactions for charge.
Zn ---> Zn2+ + 2ee- + 2 H+ + VO2+ ---> VO2+ + H2O
Multiply by an appropriate factor.
Zn ---> Zn2+ + 2e-
2e- + 4 H+ + 2 VO2+ ---> 2 VO2+ + 2 H2O
Step 5:
Add balanced half-reactions
Zn + 4 H+ + 2 VO2+ ---> Zn2+ + 2 VO2+ + 2 H2O
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Tips on Balancing Equations
• Never add O2, O atoms, or
O2- to balance oxygen.
• Never add H2 or H atoms to
balance hydrogen.
• Be sure to write the correct
charges on all the ions.
• Check your work at the end
to make sure mass and
charge are balanced.
• PRACTICE!
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Electrochemical Cells
• An apparatus that allows a
redox reaction to occur by
transferring electrons
through an external
connector.
• Product favoured reaction --> voltaic or galvanic cell ---> electric current
• Reactant favoured reaction
---> electrolytic cell --->
electric current used to
cause chemical change.
Batteries are voltaic
cells
Basic Concepts
of Electrochemical Cells
Anode
Cathode
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CHEMICAL CHANGE --->
ELECTRIC CURRENT
With time, Cu plates out
onto Zn metal strip, and
Zn strip “disappears.”
•Zn is oxidized and is the reducing agent
Zn(s) ---> Zn2+(aq) + 2e•Cu2+ is reduced and is the oxidizing agent
Cu2+(aq) + 2e- ---> Cu(s)
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CHEMICAL CHANGE --->
ELECTRIC CURRENT
•To obtain a useful
current, we separate the
oxidizing and reducing
agents so that electron
transfer occurs thru an
external wire.
This is accomplished in a GALVANIC or
VOLTAIC cell. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf
A group of such cells is called a battery.
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Zn --> Zn2+ + 2e-
Cu2+ + 2e- --> Cu
Oxidation
Anode
Negative
Reduction
Cathode
Positive
<--Anions
Cations-->
RED CAT
•Electrons travel through external wire.
•Salt bridge allows anions and cations to move
between electrode compartments.
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Terms Used for Voltaic Cells
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CELL POTENTIAL, E
• For Zn/Cu cell, potential is +1.10 V at 25 ˚C
and when [Zn2+] and [Cu2+] = 1.0 M.
• This is the STANDARD CELL
POTENTIAL, Eo
• —a quantitative measure of the tendency of
reactants to proceed to products when all
are in their standard states at 25 ˚C.
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Calculating Cell Voltage
• Balanced half-reactions can be added
together to get overall, balanced
equation.
Zn(s) ---> Zn2+(aq) + 2eCu2+(aq) + 2e- ---> Cu(s)
-------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
If we know Eo for each half-reaction, we
could get Eo for net reaction.
TABLE OF STANDARD
REDUCTION POTENTIALS
oxidizing
ability of ion
Eo (V)
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e-
H2
0.00
Zn2+ + 2e-
Zn
-0.76
To determine an oxidation
from a reduction table, just
take the opposite sign of the
reduction!
reducing ability
of element
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Zn/Cu Electrochemical Cell
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+
Anode,
negative,
source of
electrons
Cathode,
positive,
sink for
electrons
Zn(s) ---> Zn2+(aq) + 2eEo = +0.76 V
Cu2+(aq) + 2e- ---> Cu(s)
Eo = +0.34 V
--------------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Eo = +1.10 V
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Eo
for a Voltaic Cell
Cd --> Cd2+ + 2eor
Cd2+ + 2e- --> Cd
Fe --> Fe2+ + 2eor
Fe2+ + 2e- --> Fe
All ingredients are present. Which way does
reaction proceed?
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Eo for a Voltaic Cell
From the table, you see
• Fe is a better reducing
agent than Cd
• Cd2+ is a better
oxidizing agent than
Fe2+
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More About
Calculating Cell Voltage
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Assume I- ion can reduce water.
2 H2O + 2e- ---> H2 + 2 OHCathode
2 I- ---> I2 + 2eAnode
------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
Assuming reaction occurs as written,
E˚ = E˚red+ E˚ox= (-0.828 V) - (- +0.535 V) = -1.363 V
Minus E˚ means rxn. occurs in opposite direction
(the connection is backwards or you are
recharging the battery)
Charging a Battery
When you charge a battery, you are
forcing the electrons backwards (from
the + to the -). To do this, you will
need a higher voltage backwards than
forwards. This is why the ammeter in
your car often goes slightly higher
while your battery is charging, and
then returns to normal.
In your car, the battery charger is
called an alternator. If you have a
dead battery, it could be the
battery needs to be replaced OR
the alternator is not charging the
battery properly.
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Dry Cell Battery
Anode (-)
Zn ---> Zn2+ + 2eCathode (+)
2 NH4+ + 2e- --->
2 NH3 + H2
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Alkaline Battery
Nearly same reactions as
in common dry cell, but
under basic conditions.
Anode (-): Zn + 2 OH- ---> ZnO + H2O + 2eCathode (+): 2 MnO2 + H2O + 2e- --->
Mn2O3 + 2 OH-
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Mercury Battery
Anode:
Zn is reducing agent under basic conditions
Cathode:
HgO + H2O + 2e- ---> Hg + 2 OH-
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Lead Storage Battery
Anode (-) Eo = +0.36 V
Pb + HSO4- ---> PbSO4 + H+ + 2eCathode (+) Eo = +1.68 V
PbO2 + HSO4- + 3 H+ + 2e---> PbSO4 + 2 H2O
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Ni-Cad Battery
Anode (-)
Cd + 2 OH- ---> Cd(OH)2 + 2eCathode (+)
NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-
H2 as a Fuel
Cars can use electricity generated by H2/O2
fuel cells.
H2 carried in tanks or generated from
hydrocarbons
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