The thickness of a pure Ag plate on a base metal is to be determined by controlled potential coulometry. The metal sheet is masked except for a circular area, 0.5cm in diameter. Electrical connection is made to the metal and the sheet is clamped in a cell so that the unmasked area is covered with electrolyte. the Ag plate was then anodically stripped according to the reaction Ag-->Ag+ + e-.
Calculate the average thickness of the Ag plating in um (micrometer, if the stripping required 0.600 C. the density of Ag is 10.50 gcm-3. I have no notes on this question as I missed the class due to family issues, if anyone could help on how to approach this question that would be great.
96,485 C will stip 107.9 g Ag.
107.9 x 0.600/96.485 = g Ag stipped
g = v x d. You know g and d, solve for volume.
Area of the unmasked section is A = pi*r^2 = pi*(0.5/2)^2 = ?
volume = A * thickness. You know volume and you know A, solve for thickness in cm and convert to um.
You know volume
To calculate the average thickness of the Ag plating, we can use the information given and the principles of controlled potential coulometry.
1. Convert the given diameter of the circular unmasked area to radius:
Radius = Diameter/2 = 0.5 cm / 2 = 0.25 cm
2. Calculate the surface area of the circular unmasked area:
Surface Area = π * Radius^2
3. Convert the surface area to cm^2:
Surface Area = Surface Area in cm^2 / 1 = Surface Area in cm^2
4. Determine the moles of Ag deposited during the anodic stripping using Faraday’s Law:
Moles of Ag = Charge (Coulombs) / Faraday's Constant
where Faraday's Constant = 96,485 C/mol
5. Convert the moles of Ag deposited to grams:
Mass of Ag = Moles of Ag * Atomic Mass of Ag
where Atomic Mass of Ag = 107.87 g/mol
6. Calculate the volume of the Ag deposited using the density of Ag:
Volume of Ag = Mass of Ag / Density of Ag
7. Convert the volume of Ag deposited to cubic centimeters (cm^3):
Volume of Ag = Volume of Ag in cm^3 / 1 = Volume of Ag in cm^3
8. Calculate the average thickness of the Ag plating:
Thickness of Ag = Volume of Ag / Surface Area
9. Convert the thickness of Ag to micrometers (μm):
Thickness of Ag = Thickness of Ag in cm / 0.0001 = Thickness of Ag in μm
Now, plug in the given values and calculate the answer.
Charge (Coulombs) = 0.600 C
Density of Ag = 10.50 g/cm^3
Radius = 0.25 cm
Surface Area = π * (0.25 cm)^2
Surface Area = π * 0.0625 cm^2
Faraday's Constant = 96,485 C/mol
Moles of Ag = 0.600 C / 96485 C/mol
Mass of Ag = Moles of Ag * 107.87 g/mol
Volume of Ag = Mass of Ag / Density of Ag
Volume of Ag = (Moles of Ag * 107.87 g/mol) / 10.50 g/cm^3
Thickness of Ag = Volume of Ag / Surface Area
Thickness of Ag = [(Moles of Ag * 107.87 g/mol) / 10.50 g/cm^3] / Surface Area
Finally, convert the thickness of Ag from cm to μm:
Thickness of Ag (in μm) = Thickness of Ag (in cm) / 0.0001