Electrolytic cells

Introduction to and mathematics of electrolytic cells, including electrolysis of water, sodium chloride solution and liquid sodium chloride.

1. Electrolytic Cells
Electrolysis, electroplating: similar processes
From Pink Monkey

2. Electrochemistry
The Two Types
▪Electrochemical Cells:
▸Spontaneous flow of electrons from anode through
a meter or “work” to the cathode. Energy is
produced by the cell and used to do work
▪Electrolytic Cells:
▸Forced flow of electrons from anode to the
cathode by an external battery or generator. Energy
is used by the electrochemical cell to make high-
energy elements or compounds

3. Example
The Lead-Acid Battery (Pb/H
2
SO
4
/PbO
2
)
▪Charged battery:
Pb in contact with H
2
SO
4
: Pb oxidized to form
PbSO
4
. This oxidation generates electrons (cathode)
PbO
2
in contact with H
2
SO
4
: PbO
2
is reduced by
adding electrons to form PbSO
4
.
Discharged battery: PbSO
4
coats both
electrodes, surrounded by depleted sulfuric acid

4. Example
The Lead-Acid Battery (Pb/H
2
SO
4
/PbO
2
)
▪Discharged battery is recharged:
PbSO
4
coats both electrodes, surrounded by depleted sulfuric
acid
▸Electrons are forced in the opposite direction by the automobile
generator
PbSO
4
is oxidized to PbO
2
(+2 to +4)
PbSO
4
is reduced to Pb at the other electrode
Now the Pb and PbO
2
are restored at their proper electrodes
so they can start your car the next time you need some “juice”

5. Summary
Lead-Acid Battery
▪Discharging: producing of electrons
Pb + H
2
SO
4
→ PbSO
4
+ 2H
+
+ 2e
-
PbO
2
+ 4H
+
+ SO
4
2-
+ 2e
-
→ PbSO
4
+ 2H
2
O
▪Charging: restoring original situation
PbSO
4
+ 2H
+
+ 2e
-
→ Pb + H
2
SO
4
PbSO
4
+ 2H
2
O → PbO
2
+ 4H
+
+ SO
4
2-
+ 2e
-
Batteries “die” when enough PbSO
4
falls off one or the
other electrode so that it can’t “hold a charge” in the
electrolysis cycle

6. What Reaction Happens?
If a voltage is supplied to a mixture of cations,
which one will plate out?
Since this would cause reduction, look at reduction
potentials
Ag
+
+ e
-
→ Ag (s) E°= 0.80 V
Cu
2+
+ 2e
-
→ 2Cu (s) E°= 0.34 V
Zn
2+
+ 2e
-
→ 2Zn (s) E°= -0.76 V
The more positive voltage has a more
negativeG
Thus Ag
+
>Cu
2+
>Zn
2+

7. Chlorine & NaOH
Electrolysis of Brine (concentrated NaCl)
▪Cathode: reduction--electrons provided
2 H
2
O + 2e
-
→ H
2
(g) + 2 OH
-
(aq)
▪Anode: oxidation--electrons taken away
2 Cl
-
(aq) → Cl
2
(g) + 2e
-
▪Valuable stuff produced:
▸Hydrogen gas
▸Sodium hydroxide (lye)
▸Chlorine gas

8. Hydrogen & Oxygen
Electrolysis of dilute sulfuric acid
▪Cathode: reduction--electrons provided
2 H
2
O + 2e
-
→ H
2
(g) + 2 OH
-
(aq)
▪Anode: oxidation--electrons taken away
2 H
2
O → O
2
(g) + 4 H
-
(aq) + 4e
-
▪Valuable stuff produced:
▸Hydrogen gas
▸Oxygen gas
▪Oxygen is more cheaply produced by liquefying air

9. Downs Cell
Electrolysis of moltensodium chloride (m.p. = 1074K)
From voltaicpower.com

10. Sodium & Chlorine
Electrolysis of moltensodium chloride (m.p. = 1074K)
▪Anode: oxidation
2 Cl
-
(l) → Cl
2
(g) + 2e
-
▪Cathode: reduction
2 Na
+
(l) + 2e
-
→ 2 Na (l)
▪This process can also be used to produce
K from KCl, Li from LiCl, etc.

11. Electroplating
Commercial use for electrolysis
▪Silver plating
Ag
+
+ e
-
→ Ag (s)
▸Should we put the item to be plated at
–the cathode or the anode?
▪Electrorefining:
▸Impure metal at the anode; pure metal appears at the
cathode
▪How would we produce the selectively plated Toyota
logo at top right?

12. Electrolysis Stoichiometry
Commercial use for electrolysis
We can predict amount of product at either electrode
Current x time = coulombs of charge
Coulombsx 1 mol e
-
= mol electrons
96,485 Coulombs
Moles product = mol e
-
x Stoichiometric ratio
Grams product = mol product x molar mass
1 mol electrons = 96,485 C = 1 Faraday

13. Electrolysis Stoichiometry
Problem to work
Current x time = coulombs of charge
Coulombsx 1 mol e
-
= mol electrons
96,485 Coulombs
Mol product = mol e
-
x Stoichiometric ratio
Grams product = mol product x molar mass
How much silver can we
plate out with a 12.0 A
current running 3.5 hours?
Coulombs = 12.0 A x 3.5 hr x 60 min/hr = 2520 C
Mol e-= 2520 C x 1 mol e-= 2.6 x 10
-2
mol e-
96,485 C
Mol Ag = 2.6 x 10
-2
mol e-x 1 mol Ag = 2.6 x 10
-2
mol Ag
1 mol e-
Mass Ag = 2.6 x 10
-2
mol Ag x 107.9 g Ag
1 mol Ag
= 2.8 g Ag