Question

Sheet iron can be galvanised bypassing a direct current through a cell containing a solution of zinc sulphate between a graphite anode and the iron sheet. Zinc plates out on the iron. The process can be made continuous if the iron sheet is a coil that unwinds as it passes through the electrolysis cell and coils up again after it emerges from a rinse bath. Calculate the cost of the electricity required to deposit a 0.250 mm thick layer of zinc on both sides of an iron sheet that is 1.0 m wide and 100 m long, if a current of 25 A at a voltage of 3.5 V Used and the energy efficiency off process is 90%. The cost of electricity is $0.10 per kilowatt hour (1KWh=3.6 MJ). Consult appendix F for data on zinc.

Short answer

Total cost for the electricity required is 11363.5 cents.

Step-by-step solution

Step 1: Total volume of sheet

Dimensions of the iron sheet are

$\text{Width = 1} \text{m = 100} \text{cm}$

$\text{Length = 100} \text{m = 10000} \text{cm}$

$\text{thickness} \text{of the Zn layer = 0.25} \text{mm = 0.025} \text{cm}$

Area ${\text{= 1m×100m = 100m}}^{\text{2}}$

Therfore, total volume of zinc required to cover both the surface $\text{= 2×100} {\text{m}}^{\text{2}}{\text{×0.25×10}}^{-3}\text{m = 0.05} {\text{m}}^3=5\times {10}^4 {\text{cm}}^{\text{3}}$

Step 2: calculate number of moles of zinc

• Density of zinc =$\text{7.133} {\text{g/cm}}^{\text{3}}$

$\begin{array}{rcl}\text{Mass}& =& \text{Volume×Density}\\ & =& {\text{7.133g/cm}}^{\text{3}}{\text{×5×10}}^{\text{4}}{\text{cm}}^{\text{3}}\\ & =& {\text{35.665×10}}^{\text{4}}\text{g}\end{array}$

• $\begin{array}{rcl}\text{number} \text{of} \text{moles}& =& \frac{\text{mass}}{\text{molar} \text{mass}}\\ & =& \frac{{\text{35.665×10}}^{\text{4}}\text{g}}{\text{65.409} \text{g/moles}}\\ & =& \text{5452.6} \text{moles}\end{array}$
  

Step 3: calculation of  work done

Zn is reduced by following equation:

$Z{n}^{2+}+2e^{-}=Zn$

Hence, 1 mole of Zn formation required 2 moles of electrons

So, 5452.6135 moles of Zn required $2\times \text{5452.6} =10905.2 \text{moles}$ of electrons.

So, Gibbs free energy and potential is

$$\begin{gathered} \Delta G=-nFE \\ \Delta G=-\left(10905.227\right)mol.\left(96485Cmo{l}^{-1}\right)\left(3.5V\right) \\ \Delta G=-\left(3682680872.0526\right)J \\ \Delta G=-3682.680 MJ \end{gathered}$$

$\text{workdone = - 3682.60} \text{MJ}$

But energy efficiency is only 90%

So, total required energy is :

$\begin{array}{rcl}Total energy& =& 3682.68\times \frac{100}{90}\\ & =& 4090.866MJ\end{array}$

required electrical energy

As, we know, 1 KWh = 3.6 MJ

Required electrical energy:

$$\begin{gathered} =4090.86MJ\times \frac{1KWh}{3.6MJ} \\ =1136.35KWh \end{gathered}$$

Step 5: cost for electricity

1KWh energy costs 10 cents

1136.35 KWh energy costs=1136.35*10

=11363.5 cents

Hence, total cost is 11363.5 cents