outline curriculum (5 lectures) each lecture 45 minutes

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.

%100M

MefficiencyCurrent

t

a

Faraday’s laws of electrolysis

amount of material = amount of electrical energy

zF

qn

n = amount of materialq = electrical chargez = number of electronsF = Faraday constant

Faradays Law:

The mass of metal electroplated is directly related to the number of coulombs passed through the electrochemical reactions.

Faraday’s Laws of Electrolysis: Units check

zF

qn

]molC[

]C[]mol[ 1

Faraday’s Laws of Electrolysis: Expanded Relationship

zF

qn

zF

It

M

w

n = amount of material, molw = mass of material, gM = molar mass of material, g mol-1

I = current, At = time, sz = number of electronsF = 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

q = (0.0200 mol)(2)(96485 C mol-1)

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

zF

It

M

w

zF

MI

t

w

So, the rate of change of mass is given by:

Answers on Rate of Removal

)molC96485)(2(

)A0.1)(molg54.63(

t

w1

1

14 sg10x293.3t

w

Answers on Rate of Cu Removal

134 1010293.3 skgxxt

w

17 24606010293.3 daykgxxxxt

w

102845.0 daykgt

w

How long does it take to deposit 1 m over 100 cm2?)

Answers on Time to Deposit 1 m Copper

102845.0 daykgt

w

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

F.z

I.M

t

w

V

w

I.M

F.z.wt (a) Rearrange earlier equation

But the density of metal is the ratio of mass to volume

and the volume of metal deposited is the product of thickness and area

A.xV So A.x.w and

to give

I.M

F.z.A.x.t

minst

)A)(molg.(

)molC)()(cm)(cmx)(cmg.(t

I.M

F.z.A.x.t

4272

15363

4859621001019681

1243

Time to Deposit 1 m Cu on 100 cm2

1028450 daykg.t

w

21004528 cm/g.A

w

m.cm.cmg.

cmg.x 5317031750

968

28450 2

3

(b) From our earlier result

So, in one day, over a 100 cm2 area

This mass of deposit has a thickness of

1 m is deposited in a much shorter time: ss

.

xxxt 272

5317

6060241

317.5 m are deposited in 1 day so