An object whose surface area is is to be plated with an even layer of gold thick. The density of gol — Electrochemistry Chemistry Question
Question
An object whose surface area is $80 \text{ cm}^2$ is to be plated with an even layer of gold $8.0 \times 10^{-4} \text{ cm}$ thick. The density of gold is $19.7 \text{ g/cm}^3$. The object is placed in a solution of $Au(NO_3)_3$ and a current of 2.4 A is applied. The time (in seconds) required for the electroplating to be completed, assuming that the layer of gold builds up evenly, is (Au = 197)
💡 Solution & Explanation
Volume of gold required = Area $\times$ thickness = $80 \text{ cm}^2 \times 8.0 \times 10^{-4} \text{ cm} = 0.064 \text{ cm}^3$. Mass of gold = Volume $\times$ Density = $0.064 \times 19.7 = 1.2608$ g. Moles of gold = $1.2608 / 197 = 0.0064$ mol. In $Au(NO_3)_3$, gold is in the +3 oxidation state, so 3 Faradays are required per mole of Au. Total charge $Q = 0.0064 \text{ mol} \times 3 \times 96500 \text{ C/mol} = 1852.8$ C. Time $t = Q / I = 1852.8 / 2.4 = 772$ s. Therefore, correct answer is 0772.