Q22.44P - Chemistry: Molecular Nature Of Matter And Change

Question

A piece of with a surface area of ${\mathbf{\text{2.5 m}}}^{\mathbf{\text{2}}}$ is anodized to produce a film of $\mathbf{\text{Al}}{}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{3 }}}\mathbf{\text{that is 23}}\mathbf{\mu }\mathbf{(}\mathbf{\text{23}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{-}\mathbf{6}}\mathbf{\text{ m}}\mathbf{)}$thick. 

(a) How many coulombs flow through the cell in this process (assume that the density of the $\mathbf{\text{Al}}{}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{3 }}}$layer is $\mathbf{\text{3.97 g/cm3 }}$)? 

(b) If it takes min to produce this film, what current must flow through the cell?

Step-by-Step Solution

Answer

a) $1.296\times {10}^6$Coulombs of charge are passed through the cell.

b) The current of $\text{1200 A}$ flows through the cell.

Step 1: Concept Introduction

Anodizing is an electrochemical technique that turns metal into an attractive, long-lasting, corrosion-resistant anodic oxide finish. Even if the aluminium is good for anodizing but the other non-ferrous metals such as magnesium as well as titanium can also be anodized.

Step 2: Calculating the Volume and Mass of the Film

a) We need to know the amount of substance in the cell to compute the charge (in Coulombs) that travelled through it. The mass and eventually, the amount of substance can be estimated using the volume and density of the rectangular film.

Let us calculate the volume of the film.

Given: The surface area $\left(\mathrm{A}\right) \mathrm{is} 2.5 {\mathrm{m}}^2$, and the thickness (or height ) is $23 \mathrm{\mu m}$, the volume of the cuboid (also known as a rectangular prism),

$$\begin{gathered} \mathrm{V}=\mathrm{A}\times \mathrm{h} \\ \mathrm{V}=2.5 {\mathrm{m}}^2\times 23\mathrm{\mu m} \end{gathered}$$

Now for simplification, all the units are converted to :

$$\begin{gathered} \mathrm{V}=2.5 {\mathrm{m}}^2\times \frac{{10}^4 {\mathrm{cm}}^2}{{\mathrm{m}}^2}\times 23\mathrm{\mu m}\times \frac{{10}^{-4} \mathrm{cm}}{\mathrm{\mu m}} \\ \mathrm{V}=57.5 {\mathrm{cm}}^3 \end{gathered}$$

Next, let us calculate the mass of the film.

Given the volume and density of the ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ layer, the mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ is then calculated,

$\begin{array}{rcl}\mathrm{m}& =& \mathrm{d}\times \mathrm{V}\\ \mathrm{m}& =& 3.97 \mathrm{g}/{\mathrm{cm}}^3\times 57.5 {\mathrm{cm}}^3\\ \mathrm{m}\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)& =& 228.275 \mathrm{g}\end{array}$

Therefore, the volume of the film is $57.5 {\mathrm{cm}}^3$, and the mass is $228.275 \mathrm{g}$.

Step 3: Calculating the Amount of Substance and Charge Passed through the Cell

Let us calculate amount of substance,

$\begin{array}{rcl}\mathrm{n}& =& \frac{\mathrm{m}}{{\mathrm{M}}_{\mathrm{R}}}\\ \mathrm{n}\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)& =& \frac{228.275 \mathrm{g}}{101.96 \mathrm{g}/\mathrm{mol}}\\ \mathrm{n}\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)& =& 2.239 \mathrm{mol}\end{array}$

The on anode is oxidized,

${\text{Al - 3e}}^{\text{ - }}\to {\text{Al}}^{\text{3 + }}$

In order for $\text{1 mol of Al}$ to be converted to oxide, ${\text{3 mol of e}}^{\text{ - }}$must flow through the cell. Because there are two $\text{Al}$ per molecule in ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{,}$ are required.

Now, doing the Coulomb calculation

The Faraday constant $\left(\text{F = 96485}\frac{\text{C}}{\text{mol}}\right)$ is used to define the amount of electric charge that is generally passed per mole of electrons. The charge (Coulombs) that is then passed through the cell is:

$\begin{array}{rcl}\mathrm{Charge}& =& \mathrm{n}\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)\times \frac{6 \mathrm{mol}\mathrm{of}{\mathrm{e}}^{-}}{1 \mathrm{mol}\mathrm{of}{\mathrm{Al}}_2{\mathrm{O}}_3}\times \mathrm{F}\\ \mathrm{Charge}& =& 2.239 \mathrm{mol}\times \frac{6 \mathrm{mol}\mathrm{of}{\mathrm{e}}^{-}}{1 \mathrm{mol}\mathrm{of}{\mathrm{Al}}_2{\mathrm{O}}_3}\times 96485\frac{\mathrm{C}}{\mathrm{mol}\mathrm{of}{\mathrm{e}}^{-}}\\ \mathrm{Charge}& =& 1296179.49\\ \mathrm{C}& =& 1.296\times {10}^6\end{array}$

In order to produce that amount of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ film, $1.296\times {10}^6$Coulombs of charge are passed through the cell.

Therefore, the amount of substance is $2.239 \mathrm{mol}$ and the coulombs of charge passed through the cell is $1.296\times {10}^6$.

Step 4: Calculating the Current

b) Currentis defined as the charge$\left(\text{Q}\right)$passed per time (T , second),

$\text{I = }\frac{\text{Q}}{\text{T}}$
and usually given in theunit, where

$1 \mathrm{A}=1\frac{\mathrm{C}}{\mathrm{s}}$

Given that it took 18 min to produce the film, and knowing the charge passing through the cell found in a), the current can be found:

$\begin{array}{rcl}\mathrm{I}& =& \frac{1.296\times {10}^6\mathrm{C}}{18 \min }\\ \mathrm{I}& =& \frac{1.296\times {10}^6\mathrm{C}}{1080 \mathrm{s}}\\ \mathrm{I}& =& 1200 \end{array}$

The cell operates at $\text{1200 A}$ current.