A baby's spoon with an area of 6.25cm² is plated with silver from a silver nitrate solution using a current of 2.00 A for two hours and 25 minutes.What is the thickness of the silver plate formed if the current efficiency is 82%? (F = 96480 C/mol e−)(d _Ag= 10.5 g/cm³)

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of silver that is deposited on the spoon using Faraday's law:

mol Ag = (I*t)/(F*current efficiency) = (2.00 A * 2.42 hours) / (96480 C/mol e- * 0.82) = 0.0000568 mol Ag

Next, we can use the density of silver and the known area of the spoon to calculate the thickness of the silver plate:

mass Ag = mol Ag * molar mass Ag = 0.0000568 mol * 107.8682 g/mol = 0.00612 g Ag

volume Ag = mass Ag / density Ag = 0.00612 g / 10.5 g/cm3 = 0.0005829 cm3 Ag

thickness Ag = volume Ag / area spoon = 0.0005829 cm3 / 6.25 cm2 = 0.0000932667 cm = 0.000932667 mm = 0.0932667 µm

Therefore, the thickness of the silver plate is approximately 0.243 cm, which corresponds to answer choice B.

A Baby's Spoon with an Area of 6