Question

In the electroplating of a silver spoon, the spoon acts as the cathode and a piece of pure silver as the anode. Both dip into a solution of silver cyanide $\left(AgCN\right)$. Suppose that a current of 1.5 A is passed through such a cell for 22 minutes and that the spoon has a surface area of $\mathbf{16}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{2}}$. Calculate the average thickness of the silver layer deposited on the spoon, taking the density of silver to be $\mathbf{10}\mathbf{.}\mathbf{5}\mathbf{}\mathit{g}\mathit{c}{\mathit{m}}^{\mathbf{-}\mathbf{3}}$.

Short answer

The average thickness of the silver layer deposited on the spoon is 0.0132 cm.

Step-by-step solution

Calculation of charge

A charge is the product of current and time.The following formula calculates it:

Charge, $\mathrm{Q}=\mathrm{It}$, I is the current, t is the time.

$\begin{array}{c}\mathrm{Q}=1.5\mathrm{A}\times 22\min \times 60\mathrm{s}\\ =1.98\times {10}^3\mathrm{C}\end{array}$

Charge is $1.98\times {10}^3\mathrm{C}$.

Step 2: Calculation of the number of moles of electrons

The number of moles of electrons is calculated as follows:

$\mathrm{n}=\frac{\mathrm{Q}}{\mathrm{F}}$, where F is Faraday constant, $\left(\mathrm{F}=96487\right)$

The number of moles,

data-custom-editor="chemistry" $\begin{array}{c}\mathrm{n}=\frac{1.98\times {10}^3\mathrm{C}}{96487}\\ =0.0205\mathrm{mol}\end{array}$

There are 0.205 moles of electrons.

Calculation of the mass of silver

Since one mole of electrons deposits one mole of silver on the spoon, we can calculate the mass of silver from this value.

The molar mass of silver, ${\mathrm{M}}_{\mathrm{Ag}}=107.87{\mathrm{gmol}}^{-1}$

The number of moles of electrons,

$\begin{array}{c}\mathrm{n}=\frac{{\mathrm{m}}_{\mathrm{Ag}}}{{\mathrm{M}}_{\mathrm{Ag}}}\\ 0.0205\mathrm{mol}=\frac{{\mathrm{m}}_{\mathrm{Ag}}}{107.87{\mathrm{gmol}}^{-1}}\\ {\mathrm{m}}_{\mathrm{Ag}}=0.0205\mathrm{mol}\times 107.87{\mathrm{gmol}}^{-1}\\ =2.21\mathrm{g}\end{array}$

The mass of silver is 2.21 g.

Calculate the average thickness

Given the density of silver, ${\rho }_{\mathrm{Ag}}=10.5{\mathrm{gcm}}^{-3}$

Mass of silver,${\mathrm{m}}_{\mathrm{Ag}}=2.21\mathrm{g}$

The volume of silver,

$\begin{array}{c}{\mathrm{V}}_{\mathrm{Ag}}=\frac{{\mathrm{m}}_{\mathrm{Ag}}}{{\rho }_{\mathrm{Ag}}}\\ =\frac{2.21\mathrm{g}}{10.5{\mathrm{gcm}}^{-3}}\\ =0.2105{\mathrm{cm}}^3\end{array}$

The volume is the product of surface area and the average thickness.

The volume of silver,

$\begin{array}{c}{\mathrm{V}}_{\mathrm{Ag}}={\mathrm{S}}_{\mathrm{Ag}}\times {\mathrm{d}}_{\mathrm{Ag}}\\ 0.2105{\mathrm{cm}}^3=16{\mathrm{cm}}^2\times {\mathrm{d}}_{\mathrm{Ag}}\\ {\mathrm{d}}_{\mathrm{Ag}}=\frac{0.2105{\mathrm{cm}}^3}{16{\mathrm{cm}}^2}\\ =0.0132\mathrm{cm}\end{array}$

The average thickness is 0.0132 cm.