Question

(a) Seawater contains approximately 0.055 M MgCl2. How many liters of seawater are required to produce 1.00 kg of magnesium metal? (b) The final step of the extraction of magnesium involves the electrolysis of molten MgCl2. How many liters of Cl2 (measured at STP) are produced for every kilogram of Mg extracted? (c) If the electrolysis is carried out using cells that operate at 1.0 x 104 A and 3.0 V, how long will it take to produce 1.0 kg of Mg and how much electrical energy (in kJ) is consumed?

Answer 1

let x = Volume of sea water. mole of MgCl2 in sea water = x*0.055

2MgCl2---->2 Mg + 2Cl2

2 moles of MgCl2 gives 2 moles of Mg

x*0.055 moles of MgCl2 gives x*0.055 moles of Mg

Atomic weight of Mg= 24, mass of Mg= 1kg= 1000 gm

Moles of Mg= 1000/24= x*0.055

x=758 L

2. Moles of Cl2 to be produced = 1000/24 ( same as Mg) =42 moles

1 mole of Cl2 correspond to 22.4 L of Cl2

42 moles of Cl2 correspond to 42*22.4 L=941 L

c) P = VI = 1*10⁴*3 joules/sec

Reactions are 2Cl- -------->Cl2+2e   Eo = -1.36 V and Mg+2 + 2e- ------> Mg Eo= -2.38 V

2Cl- + Mg+2 --------->Cl2 + Mg EO= -1.38-2.38= -3.76V

deltaG = -nFE= 2*96500*3.76=725680 joules/mole. for 1 kg ,elenctrical enrgy consumed= 42*725680 joules=3.047*107 joules

Time required = 3.04*10⁷ joules/3*104 joules/sec=1016 seconds