Question

3 attempts left Check my work Enter your answer in the provided box. A construction company is installing an iron culvert (a long cylindrical tube) that is 43.0 m long with a radius of 0.600 m. To prevent corrosion, the culvert must be galvanized. This process is carried out by first passing an iron sheet of appropriate dimensions through an electrolytic cell containing Zn2+ ions, using graphite as the anode and the iron sheet as the cathode. If the voltage is 3.26 V, what is the cost of electricity for depositing a layer 0.200 mm thick if the efficiency of the process is 95.0 percent? The electricity rate is $ 0.120 per kilowatt hour (kWh), where 1 W=1J/s and the density of Zn is 7.14 g/cm? References

Answer 1

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Given that,

Length of the cylindrical iron culvert, L= 43 m.

Radius of the culvert, R=0.6 m.

Therefore, surface area of the culvert is given by,

A = 27RL = 27 x (0.6 m) x (43 m)

:. A= 163.106 m2

Also, given that, Thickness of the Zn layer is, d=0.2 mm = 2 x 10^-4 m.

Therefore, Volume of the Zn layer is given by,

:: Volume = area x thickness = Axd = 163.206 m2 x 2 x 10-4 m

:: Volume = V = 0.03242 m3

Given that, Density of Zn, d= 7.14 g/cm³

Therefore, mass of the Zn layer deposited is given by,

1 cm 21 21 cm :: Mass = Volumex density = Vxd = 0.03242 m ( x7.14 g/cm 9/ * 10-6 m3)

:: Mass = 231488. g

Given that, Molar mass of Zn, M= 65.38 g/mol

Therefore, no.of moles of Zn deposited is given by,

in = Mass M 231488 g 65.38 g/mol

:in = 3.5406 x 103 mol

The Zn ions reduced in the electrolytic cell is given by the reaction:

Znt + 2e + Zn(s)

From this reaction we have, 2 electrons are transferred per mol of Zn.

From Faraday's law, we have

Total charge = nF

:: Total charge = C = nF = 2 x 3.5406 x 103 mol x 96500 C/mol

::C = 6.833452 x 1080

Therefore, Electric energy consumption is given by,

E = Volts x Charge = V XC = 3.26 V x 6.833452 x 108 C

:: E = 2.2277 x 106 kJ

Given that efficiency is 95%.

:. E' = E x 95% = 2.2277 x 106 kJ x 0.95

:. E' = 2.11632 x 106 kJ

Therefore the cost of the process is given by,

:. Cost = Energy x rate = E' < $0.12/kWh

We have,

1 kWh = 3.6 x 103 kJ

:: Cost = 2.11632 x 106 kJ x $0.12/3600 J

:: Cost = $70.54

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