Question

In the production of a printed circuit boards for the electronics industry, a 0.60 mm layer of copper is laminated onto an insulating plastic board. Next a circular pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is

$\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathbf{Cl}}_{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{NH}}_{\mathbf{3}}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{}\mathbf{Cu}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }\mathbf{2}\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]\mathbf{Cl}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

a. Is this reaction an oxidation-reduction process? Explain.

b. A plant needs to manufacture 10,000 printed circuit boards each 8.0 cm × 16.0 cm in area. An average of 80 % of the copper is removed from each board (density of copper = 8.96 g/cm³). What masses of$\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathbf{Cl}}_{\mathbf{2}}$and ${\mathbf{NH}}_{\mathbf{3}}$are needed to accomplish this task? Assume 100 % yield.

Short answer

a. Yes, the reaction is an oxidation - reduction process

b. 1,745 kg of $\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{Cl}}_2$and 584 kg of ${\mathrm{NH}}_3$are needed to accomplish this task

Step-by-step solution

Oxidation-reduction process

(a)

The chemical etching is both oxidation and reduction reactions. The waste copper is oxidised from 0 to +1 state whereas Cu(II) complex used for etching is reduced to Cu(I) complex

Calculation regarding mass needed

(b)

The volume of copper laminated on each board 7.68 cm3 (8.0 cm × 16.0 cm × 0.06 cm)

From density, the weight of copper on each board is calculated as 69 g

The amount of copper removed from each board by etching is 55 g (80 % of 69 g)

The amount of copper removed from 10,000 boards is 550 kg

From etching reaction,

1 mole of copper = 1 mole of $\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{Cl}}_2$= 4 moles of${\mathrm{NH}}_3$

64 g of copper = 203 g of$\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{Cl}}_2$

55 g of copper = $\frac{203}{64}$× 55 g of$\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{Cl}}_2$

= 174.45 g of$\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{Cl}}_2$

550 kg of copper = 1745 kg of$\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{Cl}}_2$

64 g of copper = 68 g of${\mathrm{NH}}_3$

55 g of copper =$\frac{68}{64}$ × 55 g of${\mathrm{NH}}_3$

= 58.44 g of${\mathrm{NH}}_3$

550 kg of copper requires 584 kg of${\mathrm{NH}}_3$