Presentation on theme: "Outline Curriculum (5 lectures) Each lecture 45 minutes"— Presentation transcript:
1 Outline Curriculum (5 lectures) Each lecture 45 minutes Lecture 1: An introduction in electrochemical coatingLecture 2: Electrodeposition of coatingLecture 3: Anodizing of valve metalLecture 4: Electroless deposition of coatingLecture 5: Revision in electrochemical coating
2 Lecture 5 of 5 Revision in electrochemical coating
3 Type of electrochemical processes for the production of surface coatings ElectroplatingUses external power sources to reduce metal ions to metal deposit.Requires the substrate to be electrically conductive.Electroless depositionUses chemical reaction to reduce metal ions.Requires substrate to be catalytically active.Does not requires an external power sources to initiate deposition.Immersion depositionMetal 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.
4 Type of electrochemical processes for the production of surface coatings AnodizingAnodic 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 oxidationUses higher voltage than anodizing, e.g. 100 to 1000 V.Thicker oxide layer than anodizing, e.g. 100 to 500 m.
5 Current efficiencyThe 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.
6 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 energyn = amount of materialq = electrical chargez = number of electronsF = Faraday constant
8 Faraday’s Laws of Electrolysis: Expanded Relationship n = amount of material, molw = mass of material, gM = molar mass of material, g mol-1I = current, At = time, sz = number of electronsF = Faraday constant, C mol-1
9 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 = g mol-1Faraday constant, F = C mol-1
10 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?
12 A Worked Example: Questions on Concentration What is the concentration of dissolved metal in units ofmol dm-3g dm-3ppm
13 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-3The concentration on a mass basis is:c = (0.1 mol dm-3)(63.54 g mol-1)c = g dm-3Expressed as parts per million (= mg dm-3):c = x 103 mg dm-3c = 6354 ppm
14 A Worked Example: Questions on mass What is the mass of dissolved metal in g?
15 Answers on MassThe 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 = g
16 A Worked Example: Questions on Electrical Charge Calculate the electrical charge needed to remove all of the copper from solution.
17 Answers on Electrical Charge, q q = nzFn = (0.1 mol dm-3)(0.200 dm3)n = molq = ( mol)(2)(96485 C mol-1)q = C
18 A Worked Example: Questions on Rate of Removal If the current used is 1.0 A, calculate the rate of copper deposition ing s-1kg day-1
19 Answers on Rate of Removal So, the rate of change of mass is given by:
21 Answers on Rate of Cu Removal How long does it take to deposit 1 mm over 100 cm2?)
22 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
23 Time to Deposit 1 mm Cu on 100 cm2 (a) Rearrange earlier equationto giveBut the density of metal is the ratio of mass to volumeand the volume of metal deposited is the product of thickness and areaSoand
24 Time to Deposit 1 mm Cu on 100 cm2 (b) From our earlier resultSo, in one day, over a 100 cm2 areaThis mass of deposit has a thickness of317.5 mm are deposited in 1 day so1 mm is deposited in a much shorter time: