The thickness of a pure Ag plate on a base metal is to be determined by controll

Question

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.

Explanation / Answer

96,485 Coulomb can strip 107.9 gm of Ag (mol wt of Ag)
0.600 Coulomb can strip ? gm of Ag
Cross multiply,
Therefore 0.600 C can strip 6.709851*10-4 gm of Ag
Area of the unmasked section = r2 = (0.5cm /2)2 = 0.196349 cm2
mass / volume = density
volume = (6.709851*10-4 gm) / 10.50 gm/cm3 = 6.39033*10-5cm3
Volume = Area * thickness
thickness = (6.39033*10-5cm3) /  0.196349 cm2 = 3.25457*10-4 cm = 3.2545 µm
[1 cm = 10000 µm]

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