4c Redox Part 2 Battery Lesson

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Transcript 4c Redox Part 2 Battery Lesson

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OXIDATION-REDUCTION
REACTIONS
Indirect Redox Reaction
A battery functions by transferring electrons
through an external wire from the reducer to
the oxidizer. The spontaneous flow of
electrons from one substance to another is
the driving force for this reaction.
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Direct Redox Reaction
With time, Cu plates out onto Zn
metal strip, and Zn strip
“disappears” as it becomes ions.
Electrons are transferred from Zn to
Cu2+, but there is no useful electric
current.
Oxidation: Zn(s) ---> Zn2+(aq) + 2eReduction: Cu2+(aq) + 2e- ---> Cu(s)
-------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
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•To obtain a useful
current, we separate the
oxidizer and reducer so
that
occurs through an
external wire.
This is accomplished in a GALVANIC or
VOLTAIC cell. A group of such cells is
called a battery.
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Fe --> Fe2+ + 2e-
Oxidation
Anode:
Cu2+ + 2e- --> Cu
Reduction
Cathode:
Fe
Fe gets
oxidized
<--Anions
Cations-->
Cu2+ gets
reduced
Fe
•Electrons travel through external wire.
•Salt bridge allows anions and cations to flow
into anode and cathode to keep charges balanced.
ANATOMY OF A BATTERY
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The
Cu|Cu2+
and
Ag|Ag+
Cell
Write the 2
half
reactions for
this battery.
Show how we
can find the
voltage of this
cell.
What species is
getting oxidized?
What is getting
reduced?
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Which is the oxidizer?
Which is the reducer?
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Electrons move
from anode to
cathode in the wire.
Anions & cations
move thru the salt
bridge.
Electrochemical
Cell: Describe
what particles are
doing…
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CELL POTENTIAL, E (volts)
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1.10 V
Zn and
anode
Cu and Cu2+,
cathode
Zn2+,
1.0 M
1.0 M
• Electrons are “driven” from anode to cathode by an electromotive
force.
• For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 ˚C and
when [Zn2+] and [Cu2+] = 1.0 M.
• Write the 2 half reactions. Show how the voltage is calculated.
• Standard reduction potentials are measured at standard conditions (1
M, 25oC)
• Positive voltage indicates which way the cell runs spontaneously!
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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.
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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
Use standard reduction potential table. Which two half
reactions occur? Calculate Eo for this cell. Label the oxidation
reaction and reduction reaction. Label anode and cathode.
Show direction of electron flow. Indicate movement of anions
and cations in salt bridge.
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