Question

A thin circular-disk earring$\text{4.00 cm}$in diameter is plated with a coating of gold $\text{0.25 mm}$thick from an localid="1663324101600" ${\text{Au}}^{\text{3 +}}$bath.

(a) How many days does it take to deposit the gold on one side of one earring if the current is$\text{0.013 A}$(d of gold${\text{= 19.3 g/cm}}^{\text{3}}$)?

(b) How many days does it take to deposit the gold on both sides of the pair of earrings?

(c) If the price of gold is$\text{\$ 320}$per troy ounce ($\text{31.10 g}$), what is the total cost of the gold plating?

Short answer

(a) The number of days required to deposit the gold on one side of one earring is 8 days .

(b) The number of days required to deposit the gold on both sides of the pair of earrings is 32 days.

(c) The total cost of the gold plating is $250.

Step-by-step solution

Concept Introduction

The process of dissolving ionic compounds into their constituent components by delivering a direct electric current through the complex in a fluid form is known as electrolysis. At the cathode, cations are reduced, whereas anions are oxidised.

Information Provided

1. The reaction is:${\text{Au}}^{\text{3 +}}{\text{(aq) + 3e}}^{\text{-}}\rightleftharpoons \text{Au(s)}$
2. Diameter of earring is $\text{4.00 cm}$(which means that the radius is $\text{r =}\frac{\text{4 cm}}{\text{2}}\text{= 2 cm}$)
3. Thickness of coating is$\text{0.25 mm = 0.025 cm}$
4. The current is$\text{0.013A(A = C/s)}$
5. d of gold${\text{= 19.3 g/cm}}^{\text{3}}$
6. Faraday constant: charge of $\text{1 mole}$of electrons$\left(\text{F = 96485 C/mole}\right)$

(a) Calculation for Number of Days

The volume of gold coating is –

$$\begin{gathered} {\text{V =r}}^{\text{2}}\pi ·\text{thickness of coating} \\ {\text{= (2cm)}}^{\text{2}}·\pi ·\text{0.025cm} \\ {\text{= 0.31416cm}}^{\text{3}} \end{gathered}$$

The mass of gold coating is –

$$\begin{gathered} \text{m = V}·\text{d} \\ {\text{= 0.31416cm}}^{\text{3}}·{\text{19.3g/cm}}^{\text{3}} \\ \text{= 6.06g} \end{gathered}$$

The number of moles of gold is (molar mass of gold is $\text{196.96657g/mol}$) –

$$\begin{gathered} {\text{n}}_{\text{Au}}\text{=}\frac{\text{6.06g}}{\text{196.96657g/mol}} \\ \text{= 3.0767}·{\text{10}}^{\text{- 2}}\text{mol Au} \end{gathered}$$

Since 3 of electrons produces 1 mole of Au , the number of moles of electrons is –

$$\begin{gathered} {\text{n}}_{{\text{e}}^{\text{-}}}\text{= 3}·{\text{n}}_{\text{Au}} \\ \text{= 3}·\text{3.0767}·{\text{10}}^{\text{- 2}}\text{mol} \\ \text{= 9.23}·{\text{10}}^{\text{- 2}}{\text{mole}}^{\text{-}} \end{gathered}$$

Calculate the charge using Faraday constant –

$$\begin{gathered} {\text{Charge = Moles e}}^{-}\text{.F} \\ {\text{= 9.23×10}}^{-2}{\text{mol e}}^{-}·{\text{96485 C/mol e}}^{-} \\ \text{= 8905.57 C} \end{gathered}$$

The time it takes to deposit the gold on one side of one earring:

$$\begin{gathered} \text{Current =}\frac{\text{Charge}}{\text{Time}} \\ \text{Time =}\frac{\text{Charge}}{\text{Current}} \\ \text{=}\frac{\text{8905.57 C}}{\text{0.013}\frac{\text{C}}{\text{s}}} \\ \text{= 6.85}·{\text{10}}^{\text{5}}\text{s} \\ \text{= 6.85}·{\text{10}}^{\text{5}}\text{s}·\frac{\text{1h}}{\text{3600s}}·\frac{\text{1d}}{\text{24h}} \\ \text{= 7.93 days}\approx \text{8 days} \end{gathered}$$

Therefore, the vale for number of days is obtained as 8 days.

(b) Calculation for Number of Days

If it is needed to coat a pair of earrings, on both sides, then it needs to be coated total of 4 sides.

Since, 8 days are needed to coat 1 side of earring, to coat 4 sides –

$$\begin{gathered} \text{Time = 4 sides}·\text{8 days/side} \\ \text{= 32 days} \end{gathered}$$

Therefore, the vale for number of days is obtained as 32 days.

Calculation for Cost of Gold Plating

The price of gold is$\text{\$ 320}$per ounce ($\text{31.10 g}$).

The mass of gold on one side of earring is$\text{6.06 g}$ .

A pair of earring has 4 sides.

The mass of gold on a pair of earrings is –

$\text{m = 4}·\text{6.06 g = 24.24 g}$

Hence, the price of gold plating of a pair of earrings is –

$$\begin{gathered} \text{Price = 24.24 g}·\frac{\text{320 dollars}}{\text{31.10 g}} \\ \text{= 250 dollars} \end{gathered}$$

Therefore, the vale for cost is obtained as $250.