Oxidation and Reduction Reactions 4
IB Topic 9.4: Voltaic Cells
https://www.heathscientific.net/p/509/potato_clock
Text Reference: Higher Level Chemistry p. 332-337
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IB Assessment Statements
9.4.1
Explain how a redox reaction is used to produce
electricity in a voltaic cell.
This should include a diagram to show how two half-cells
are used.
9.4.2
State that oxidation occurs at the negative electrode
(anode) and reduction occurs at the positive
electrode (cathode).
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Starter Activity - A Battery for Cents?
Turn coins into electricity! Make a “voltaic pile” with
different coins and see how much voltage you can generate.
You need: copper plated coins
nickel plated coins
filter paper soaked in salt water / vinegar
2 electrical leads
voltmeter / multimeter set to read voltage
1. Prepare a single “cell” by stacking a nickel (or equivalent),
filter paper soaked in salt and a copper penny.
2. Connect the leads to the voltmeter or multimeter as instructed. Hold the
electrode of each lead on each side of the cell and record the voltage.
3. Add another piece of filter paper on top of this cell, and then
add one more cell. Test the voltage. Keep adding more cells.
4. Repeat using different coins.
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Starter Activity - A Battery for Cents?
Results
voltage (V)
team
cell composition
1 cell
2 cells
3 cells
4 cells
5 cells
Q: Which “battery” gives the best voltage for the money?
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Single Displacement Reactions
Look at the reaction between
metallic zinc immersed in aqueous copper (II) ions:
Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)
What a waste of
potential energy!
Electrons are directly transferred from Zn atoms to
Cu2+ ions at the point of contact.
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Single Displacement Reactions
Is there another way to move the electrons
from zinc metal to the copper ion?
Yes!
Move the electrons from one substance to the other
via conducting materials!
Why?
The flow of electrons through a conductor generates electricity.
This is a VOLTAIC CELL.
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Voltaic Cells
Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)
How is this different than the
“direct” reaction between zinc
metal and copper ions?
Zn
http://en.wikipedia.org/wiki/Galvanic_cell
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Voltaic Cells
Components of a Voltaic Cell
1. two “half cells”
oxidation half cell + reduction half cell
Each half cell has two components:
electrode
(metal)
electrolyte
(solution with
the metallic ion)
Examples:
Zn(s) / ZnSO4(aq)
Electrode terms:
RED CAT
AN OX
Cu(s) / CuSO4(aq)
Ni(s) / Ni(NO3)2(aq)
cathode = electrode in reduction half cell
anode
= electrode in oxidation half cell
+ electrode
– electrode
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Voltaic Cells
Components of a Voltaic Cell
2. conducting material between the electrodes
allows electrons to move from
one half cell to the other through
an external circuit
(i.e. transfers electrons from the
reducing agent to the oxidizing agent)
allows electrons to flow
through a conductor
∴ electrical current is generated
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Voltaic Cells
Components of a Voltaic Cell
3. salt bridge between the electrolytes
allows ions to move from one
half cell to the other
(this prevents the build up of charge
in the half cell that would stop the
reaction from proceeding)
A salt bridge contains a solution of a soluble ionic compound
such as saturated KNO3 or KCl.
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Voltaic Cells
Putting a Voltaic Cell Together
Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)
oxidation:
Zn → Zn2+ + 2e–
–
anode = Zn(s)
metal in the
oxidation half
reaction
electrolyte = Zn(NO3)2(aq)
electrons flow
from the anode
to the cathode
reduction:
Cu2+ + 2e– → Cu(s)
cathode = Cu(s)
+
metal in the
reduction half
reaction
electrolyte = Cu(NO3)2(aq)
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Voltaic Cells
Role of the Salt Bridge
Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)
oxidation:
Zn → Zn2+ + 2e–
reduction:
Cu2+ + 2e– → Cu(s)
NO3– K+
anode = Zn(s)
Zn2+ ion
concentration
builds up.
(+ > – )
cathode = Cu(s)
Zn2+
NO3–
Zn2+
Anions from the salt bridge
move to balance the charge.
Cu2+ NO –
3
–
NO3
Cu2+ ion
concentration
decreases.
(– > +)
Cations from the salt bridge
move to balance the charge.
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Voltaic Cells
Operation of a Voltaic Cell
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf
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Voltaic Cells
Summary of a Voltaic Cell
oxidation half cell
reduction half cell
half reaction equation
X(s) → X2+(aq) + 2e–
Y2+(aq) + 2e– → Y(s)
name of electrode
anode
cathode
charge on electrode
negative
positive
more reactive
less reactive
decreases
increases
anions move to the
anode
cations move to the
cathode
nature of electrode
(more or less reactive metal?)
change in electrode
mass
salt bridge ion
movement
electron flow
anode to cathode (oxidation to reduction)
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Voltaic Cells
Example: Magnesium-Silver Voltaic Cell
oxidation half cell
reduction half cell
half reaction equation
Mg(s) → Mg2+(aq) + 2e–
Ag+(aq) + e– → Ag(s)
name of electrode
anode = Mg metal
cathode = Ag metal
charge on electrode
negative
positive
more reactive (Mg)
less reactive (Ag)
decreases
increases
anions move to the
magnesium half cell
cations move to the
silver half cell
nature of electrode
(more or less reactive metal?)
change in electrode
mass
salt bridge ion
movement
electron flow
from Mg to Ag
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Voltaic Cells
Example: Magnesium-Silver Voltaic Cell
Annotate this diagram.
overall equation: ______________________________
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