Question

The surface area of an object to be gold plated is 49.8 cm2 and the density of gold is 19.3 g/cm-3. A current of 3.15. A is applied to a solution that contains gold in the +3 oxidation state. Calculate the time required to deposit an even layer of gold 1.2 x 10 -3cm thick on the object.

Answer 1

Answer: Time required to deposit an even layer of gold with given thickness is $5.3 \times 10^{2}$ sec.

Explanation:

The given data is as follows.

     Surface area = 49.8 $cm^{2}$,

     Density of gold = 19.3 $g/cm^{3}$,

     Current = 3.15 A,       thickness of gold layer = $1.2 \times 10^{-3} cm$

It is known that relation between volume, area and thickness is as follows.

           V = Surface area × Thickness

               = $49.8 \times 1.2 \times 10^{-3} cm$

               = 0.05988 $cm^{3}$

Therefore, we will calculate the time required to deposit an even layer of gold with given thickness is calculated as follows.

  $0.05988 \times cm^{3} \times \frac{19.3 g Au}{1 cm^{3}} \times \frac{1 mol Au}{197 g Au} \times \frac{3 mol e^{-}}{1 mol Au} \times \frac{96485}{1 mol e^{-}} \times \frac{1 As}{1 C} \times \frac{1}{3.20 A}$

        = $5.3 \times 10^{2}$ sec

Thus, we can conclude that time required to deposit an even layer of gold with given thickness is $5.3 \times 10^{2}$ sec.