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-.

Question

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.

Answer 1

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