Question

The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hours of electricity required to produce 3.00kg of aluminum from electrolysis of compounds from bauxite is ________ when the applied emf is 4.50V.

Answer 1

Explanation:

According to Faraday's law, the amount of a substance deposited or liberated in electrolysis process is proportional to the quantity of electric charge passed and to the equivalent weight of the substance.

Formula to calculate the mass of substance liberated according to Faraday's law is as follows.

             m = $(\frac{Q}{F})(\frac{M}{Z})$

where,          m = mass of substance liberated at electrode

                     Q = electric charge passing through the substance

                     F = Faraday constant = 96,487 C $mol^{-1}$

                     M = molar mass of the substance

                     Z = valency number of ions of the substance

Since, it is given that mass is 3 kg or 3000 g (as 1 kg = 1000 g), molar mass of Al is 27, Z is 3.

Therefore, putting the values in the above formula as follows.

                      m = $(\frac{Q}{F})(\frac{M}{Z})$

                     3000 g = $(\frac{Q}{96,487 C mol^{-1}})(\frac{27}{3})$

                        Q = 32162333.33 C

As it is given that V = 4.50 Volt. Also, it is known that

                       Energy = $V \times Q$

Therefore, calculate the energy as follows.

                      Energy = $V \times Q$

                                   = $4.50 V \times 32162333.33 C$

                                   = 144730500 J

As it is known that $3.6 \times 10^{6}$ J = 1 KW Hr

So, convert 144730500 J into KW Hr as follows.

                   $\frac{144730500 J \times 1 KW Hr}{3600000 J}$

                          = 40.202 KW Hr

Thus, we can conclude that the number of kilowatt-hours of electricity required to produce 3.00 kg of aluminum from electrolysis of compounds from bauxite is 40.202 KW Hr when the applied emf is 4.50V.