Outline Curriculum (5 lectures)
Each lecture  45 minutes
• Lecture 1: An introduction in electrochemical coating
• Lecture 2: Electrodeposition of coating
• Lecture 3: Anodizing of valve metal
• Lecture 4: Electroless deposition of coating
• Lecture 5: Revision in electrochemical coating
Lecture 5 of 5
Revision in electrochemical
coating
Type of electrochemical processes for the
production of surface coatings
• Electroplating
– Uses external power sources to reduce metal ions to metal deposit.
– Requires the substrate to be electrically conductive.
• Electroless deposition
– Uses chemical reaction to reduce metal ions.
– Requires substrate to be catalytically active.
– Does not requires an external power sources to initiate deposition.
• Immersion deposition
– Metal ion is reduced from the solution by exchange with metallic
substrate.
– Type of metal that can be deposited depends on the metal substrate
and metal ions in solution. (Electromotive Series Table).
– Displacement Reaction.
Type of electrochemical processes for the
production of surface coatings
• Anodizing
– Anodic oxidation of metal to form metal oxide.
– Metal oxide forms at the anode.
– Requires the external supply of high voltage, e.g. 10 to 100 V
to form oxide layer.
– Oxide layer thickness, e.g. 10 to 100 m.
• Plasma electrolytic oxidation
– Anodic oxidation of metal to form metal oxide.
– Uses higher voltage than anodizing, e.g. 100 to 1000 V.
– Metal oxide forms at the anode.
– Thicker oxide layer than anodizing, e.g. 100 to 500 m.
Current efficiency
• The percentage of the current which goes to the
useful electroplating reaction.
• Is the ratio between the actual amount of metal
deposit, Ma to that calculated theoretically from
Faradays Law, Mt.
Ma
Current efficiency 
 100%
Mt
Faraday’s laws of electrolysis
Faradays Law:
The mass of metal electroplated is directly related to the number of
coulombs passed through the electrochemical reactions.
amount of material = amount of electrical energy
q
n
zF
n = amount of material
q = electrical charge
z = number of electrons
F = Faraday constant
Faraday’s Laws of Electrolysis:
Units check
q
n
zF
[C]
[mol] 
1
[C mol ]
Faraday’s Laws of Electrolysis:
Expanded Relationship
q
n
zF
w
It

M zF
n = amount of material, mol
w = mass of material, g
M = molar mass of material, g mol-1
I = current, A
t = time, s
z = number of electrons
F = Faraday constant, 96 485 C mol-1
A Worked Example: Facts
• 200 cm3 of an acidic solution of 0.1M copper
sulphate pentahydrate (CuSO4.5H2O) is
electrolysed using inert electrodes.
• Molar mass of copper, M = 63.54 g mol-1
• Faraday constant, F = 96 485 C mol-1
A Worked Example:
Questions on reactions
• A sketch is useful.
• What is the cathode reaction?
• What is the anode reaction?
• What is the cell reaction?
• At which electrode does reduction occur?
Answers on reactions
• Cathode reaction:
Cu2+ + 2e- = Cu
• Anode reaction:
H2O - 2e- = 2H+ + 1/2O2
• Cell reaction:
Cu2+(l) + H2O(l) = Cu(s) + 2H+(l) + 1/2O2(g)
• Reduction (electron gain) occurs at cathode
A Worked Example:
Questions on Concentration
• What is the concentration of dissolved
metal in units of
– mol dm-3
– g dm-3
– ppm
Answers on Concentration
• The concentration of dissolved metal is the same
as that of the compound, i.e.:
c = 0.1M = 0.1 mol dm-3
• The concentration on a mass basis is:
c = (0.1 mol dm-3)(63.54 g mol-1)
c = 6.354 g dm-3
• Expressed as parts per million (= mg dm-3):
c = 6.354 x 103 mg dm-3
c = 6354 ppm
A Worked Example:
Questions on mass
• What is the mass of dissolved metal in g?
Answers on Mass
• The mass of dissolved copper, w, is given
by the product of concentration (on a mass
basis) and solution volume:
w = (6.354 g dm-3)(200 cm3)
w = (6.354 g dm-3)(0.200 dm3)
w = 1.271 g
A Worked Example:
Questions on Electrical Charge
• Calculate the electrical charge needed to
remove all of the copper from solution.
Answers on
Electrical Charge, q
q = nzF
n = (0.1 mol dm-3)(0.200 dm3)
n = 0.0200 mol
-1
q = (0.0200 mol)(2)(96485 C mol )
q = 3859.4 C
A Worked Example:
Questions on Rate of Removal
• If the current used is 1.0 A, calculate the
rate of copper deposition in
– g s-1
– kg day-1
Answers on Rate of Removal
w
It

M zF
So, the rate of change of mass is given by:
w MI

t
zF
Answers on Rate of Removal
1
w (63.54 g mol )(1.0 A )

1
t
( 2)(96485 C mol )
w
4
1
 3.293 x 10 g s
t
Answers on Rate of Cu Removal
w
4
3
1
 3.293 x 10 x 10 kg s
t
w
 3.293 x 10 7 x 60 x 60 x 24 kg day 1
t
w
 0.02845 kg day 1
t
How long does it take to deposit 1 m over 100 cm2?)
Answers on
Time to Deposit 1 m Copper
w
1
 0.02845 kg day
t
How long does it take to deposit 1 m over 100 cm2?)
Methods – either:
(a) rearrange earlier equation or
(b) use w/t expression
Time to Deposit 1 m Cu on 100 cm2
(a) Rearrange earlier equation
w M .I

t
z .F
to give
w.z .F
t
M .I
But the density of metal is the ratio of mass to volume

w
V
and the volume of metal deposited is the product of thickness and area
So
w   .x.A
t
and
t
 .x.A.z .F
V  x .A
M .I
 .x.A.z .F
M .I
( 8.96 g cm3 )(1 x 10 4 cm )(100cm2 )( 2 )( 96 485C m ol1 )
t
( 63.53 g m ol1 )(1 A )
t  272s  4 m in
Time to Deposit 1 m Cu on 100 cm2
(b) From our earlier result
w
 0.02845 kg day 1
t
So, in one day, over a 100 cm2 area
w
 28.45 g / 100 cm 2
A
This mass of deposit has a thickness of
2
0.2845g cm
x
 0.03175cm  317.5 m
3
8.96 g cm
317.5 m are deposited in 1 day so
1 m is deposited in a much shorter time:
1 x 24 x 60 x 60
t
s  272 s
317 .5