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 Uses higher voltage than anodizing, e.g. 100 to 1000 V. 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.

Faraday’s laws of electrolysis 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 n = amount of material q = electrical charge z = number of electrons F = Faraday constant

Faraday’s Laws of Electrolysis: Units check

Faraday’s Laws of Electrolysis: Expanded Relationship 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 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 So, the rate of change of mass is given by:

Answers on Rate of Removal

Answers on Rate of Cu Removal How long does it take to deposit 1 mm over 100 cm2?)

Answers on Time to Deposit 1 mm Copper How long does it take to deposit 1 mm over 100 cm2?) Methods – either: (a) rearrange earlier equation or (b) use w/t expression

Time to Deposit 1 mm Cu on 100 cm2 (a) Rearrange earlier equation to give 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 So and

Time to Deposit 1 mm Cu on 100 cm2 (b) From our earlier result So, in one day, over a 100 cm2 area This mass of deposit has a thickness of 317.5 mm are deposited in 1 day so 1 mm is deposited in a much shorter time: