Answer to Question #249657 in General Chemistry for Badeth

Question

. The galvanizing of iron sheet can be carried electrolytically using a bath containing zinc sulfate solution. The sheet

is made the cathode, and a graphite anode is used. Calculate the cost of electricity required to deposit 0.49 mm

layer of zinc on both sides of an iron sheet 2.0 m wide and 80m long if the current is 30 A, the voltage is 3.5 V, and

the energy efficiency of the process is 90%. Assume the cost of electricity is Php 12.50 per kilowatt-hr. The density

of zinc is 7.1g/cm3.

Answer 1

Galvanizing is a process where iron is coated with zinc which provide protection against corrosion of iron or steel. The galvanizing f iron sheet can be carried electrolytically using a bath containing zinc sulfate solution. The sheet is made the cathode, and a graphite anode is used.

1. Given that the dimensions of the iron sheet as follows:

Wide = 2.0 m = 2 × 10² cm

Length = 80 × 10² cm

Thickness = 0.49 mm = 49 × 10^{- 3} cm

Therefore, calculate the volume:

V o l u m e = 2 × w i d e × l e n g t h × t h i c k n e s s

V o l u m e = 2 × 2 × 10² cm × 80 × 10² cm × 49 × 10^{- 3} cm

V o l u m e = 156.8 × 10³ cm³

Next find the weight of zinc:

Weight of zinc = volume × density

Weight of zinc = 156.8 × 10³ cm³ × 7.1 g/cm³

Weight of zinc = 1118.5 × 10³ g

Then, find out the number of moles of zinc:

$n = \frac{Weight \; of\; zinc}{molecular \;weight} \\ n = \frac{1118.5 \times 10^3}{65.409} = 1.71 \times 10^4 \;mol$

Given that,

E = 3.5 V

F = 96,485 C/mol

∆G = -nFE

The given reaction,

Zn²⁺ + 2e^- → Zn( s )

One mole of zinc requires 2 moles of electrons.

Therefore,

∆G = -2 × 1.71 × 10⁴ mol × 96,485 C/mol × 3.5 V

∆G = -11550 MJ or 1.155×10⁴ MJ

The energy efficiency of the process is 90%.

Thus the total energy is calculated as follows:

$11550\; MJ \times \frac{100}{90} = 12833.3 \;MJ$

1kW h = 3.6 MJ

$\frac{12833.3}{3.6} = 3564.8 \;kWh$

Therefore, for Php 12.50 per kilowatt-hr, the cost of electricity is as follows:

3564.8 kWh × Php 12.50 = Php 44560

Hence, the cost of electricity per 3564.8 kWh is Php 44560 .