• What is
powering
this clock?
How much Voltage
• You can see the
battery is missing and
the clips are attached
to the terminals.
• What is the voltage
required to run the
clock?
Make a Battery
Place a drop of Copper nitrate on one end
of the paper. Place a piece of copper in
the center of the wet spot
Make a Battery
On the other end place a drop of zinc nitrate and place a
piece of zinc in the wet solution.
Make a Battery
Add a couple of drops of KNO3 in the
middle of the two solutions to make a salt
bridge.
Make a Battery
V
Touch the probe leads
to the two metals as
pictured here.
Record the voltage.
What’s the sign?
• If the reading is negative, switch the leads
to the other metals. You want to get a
positive voltage reading.
• Record the metal that is at the red lead
and the metal at the black lead.
REDUCE RED CATS
• This is the way I remember that reduction
occurs at the cathode and it is at the red
lead.
• Reduction ?
• Oxidation ?
Look at the Standard Reduction
Potential Table
•
•
•
•
Cu2+
Cu
Zn2+
Zn
Find the voltage for each pair of
metals you have.
?
• Cu2+ + 2 e-  Cu
•
Zn  Zn2+ + 2e-
•
0.34 volts
0.76 volts
1.10 volts
1.1
volts
•
Zn
Cu
•
Zn(NO3)2
Cu(NO3)2
What is the purpose of the salt
bridge?
1.1
volts
Zn
Zn(NO3)2
Cu
Cu(NO3)2
• What is
powering
this clock?
How much Voltage
• You can see the
battery is missing and
the clips are attached
to the terminals.
• What is the voltage
required to run the
clock?
• After adding the
phenolphthalein
around the strip of
magnesium a pink
color is observed.
• Also there are tiny
bubbles all along the
sides of the
magnesium
Lead Battery
Pb(s) + HSO4- PbSO4(s) +H+(aq) 2 e-
Anode:
cathode:
+ +HSO + 2e-  PbSO + H O
+
3
H
2(s)
4
4
2
0.296 V
1.628 V
1.924 V
PbO
Mercury Battery
STEEL
cathode
HgO in KOH
Zn
container
Zn(OH)2
anode
Watches, pacemakers, calculators
Rechargeable Nickel-cadmium
anode Cd + OH-  Cd(OH)2 + 2 ecathode NiO(OH)S + H2O Ni(OH)2 + OH-
Recharge many times because the
solid products adhere to the surface
of the electrode renewing the
battery.
Corrosion
• Corrosion is the oxidative deterioration of a
metal such at rust.
Drop of water
O2 from the air
O2 + 4H+ + 4 e-  2 H2O
cathode
Fe --> Fe2+ + 2 eanode
Rust
How can you prevent corrosion?
• Look at the equation and prevent the
reaction from happening. What can you do?
Electrochemical Cells
• There are 2 types of cells
– Galvanic also called voltaic is a
spontaneous reaction that produces an
electric current
- Electrolytic requires an outside source
to supply the current such as a battery
or electrical outlet
Electroplating
• Example of an Electrolytic cell
– Silverplated dinnerware - Silver is
a soft metal what would happen if
you used a solid silver fork?
Electrolysis
Electrolysis
Red lead +
Black lead
Battery
cathode -
anode
Graphite
electrodes
Na2SO4(aq)
What is happening??
• Reduction:
2 H2O(l) + 2 -  H2(g) + 2 H-(aq) -0.83 V
2 H+(aq) 3 e-  H2(g)
0.00 V
Na+(aq) + e-  Na(s)
-2.71 V
• Oxidation:
2 H2O  O2(g) + 4H+(aq) + 4e- -1.23 V
2 SO42-  S2O8 + 2 e-2.00 V
See bubbles? What is the clue?
• Look at the data table again and see
which reactions you think took place
What is happening??
• Reduction:
2 H2O(l) + 2 -  H2(g) + 2 H-(aq) -0.83 V
2 H+(aq) 3 e-  H2(g)
0.00 V
Na+(aq) + e-  Na(s)
-2.71 V
• Oxidation:
2 H2O  O2(g) + 4H+(aq) + 4e- -1.23 V
2 SO42-  S2O8 + 2 e-2.00 V
Answer
• Reduction was water or Na+
• We know is must be water for 3
reasons – 1. a gas was produced
– 2. sodium reacts with water violently
– 3. It became more basic
2H2O + 2 e- --> H2(g) + 2 OH- -.83V
• Oxidation was either water of sulfate
ion
Oxidation of water produces H+ and a
gas. Do we have evidence of that?
H2O --> O2(g) + 4 H+ + 4 e-
-1.23 V
Energy Involved
2H2O + 2 e- --> H2(g) + 2 OH- -.83V
H2O --> O2(g) + 4 H+ + 4 e-1.23 V
- 2.06 V
What does the negative sign mean?
Change the electrodes to Copper
Black lead
Red lead +
Battery
cathode -
anode
Copper
electrodes
Na2SO4(aq)
What is happening??
• Reduction:
2 H2O(l) + 2 -  H2(g) + 2 H-(aq) -0.83 V
2 H+(aq) 3 e-  H2(g)
0.00 V
Na+(aq) + e-  Na(s)
-2.71 V
Cu2+ + 2 e- -  Cu
- 0.34 V
Oxidation:
2 H2O  O2(g) + 4H+(aq) + 4e- -1.23 V
2 SO42-  S2O8 + 2 e-2.00 V
Cu(s) -  Cu2+ + 2 e+0.34 V