Q21.126 CP - Chemistry: Molecular Nature Of Matter And Change

Question

To improve conductivity in the electroplating of automobile bumpers, a thin coating of copper separates the steel from a heavy coating of chromium. 

1. What mass of $\text{Cu}$ is deposited on an automobile trim piece if plating continues for $\text{1.25 h}$at a current of $\text{5.0 A}$ ?
2. If the area of the trim piece is ${\text{50.0 cm}}^{\text{2}}$, what is the thickness of the $\text{Cu}$ coating ( d of ${\text{Cu = 8.95 g/cm}}^{\text{3}}$ )?

Step-by-Step Solution

Answer

(a) The mass of Cu that is deposited on an automobile trim piece is 7.41g .

(b) The thickness of the Cu coating is 0.017 cm .

Step 1: 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.

Step 2: (a) Calculation for Mass

To get the mass of Cu , get the charge by multiplying the time by the current.

$$\begin{gathered} \text{Current = }\frac{\text{Charge}}{Time} \\ \text{Charge = Current×Time} \\ \text{Charge = }\frac{\text{5.0 C}}{\text{s}}\text{×1.25 hr×}\frac{\text{60min}}{\text{1 hr}}\frac{\text{60s}}{\text{1min}} \\ \text{ = 22,500 C} \end{gathered}$$

Convert the charge to moles of electron by dividing by the Faraday's constant.

$\text{22,500 C×}\frac{{\text{1mol e}}^{\text{ - }}}{\text{96500 C}}{\text{ = 0.233mol e}}^{\text{ - }}$

Convert the moles of electrons to moles of copper. There are 2 electrons involved in the reduction of ${\text{Cu}}^{\text{2 + }}$ to Cu .

${\text{0.233mol e}}^{\text{ - }}\text{×}\frac{\text{1mol Cu}}{{\text{2mol e}}^{\text{ - }}}\text{ = 0.117mol Cu}$

Multiply by the molar mass of copper to obtain the mass deposited.

 $\text{0.117mol Cu×}\frac{\text{63.55g}}{{\text{1mol e}}^{\text{ - }}}\text{ = 7.41 g Cu}$

Therefore, the value of mass is obtained as $\text{7.41 g}$.

Step 3: (b) Calculation for Thickness

Convert the mass of Cu to its volume by dividing by its density.

$\text{7.41g×}\frac{{\text{cm}}^{\text{3}}}{\text{8.95g}}{\text{ = 0.828cm}}^{\text{3}}$

Get the thickness by dividing the volume by the given surface area.

$\frac{{\text{0.828cm}}^{\text{3}}}{{\text{50.0cm}}^{\text{2}}}\text{ = 0.017cm}$

 

Therefore, the value of thickness is obtained as $0.017\mathrm{cm}$.