Electrolysis of molten MgCl2 is the final production step in the isolation of ma

Question

Electrolysis of molten MgCl2 is the final production step in the isolation of magnesium from seawater by the Dow process 2. Write the cell reaction for this electrolysis cell. b. a. Assuming that 752 g of Mg metal forms, calculate the number of moles of electrons required (MsMg 24.3 g/mol. c. Calculate the coulombs of charge required. d. How many amps of current will produce this amount in 5.75 hours? e. What volume of Cl2(g) measured at STP will be produced at the anode during this electrolysis?

Explanation / Answer

a)

Mg2+ + 2e- = Mg(s)

2Cl- --> Cl2(g) + 2e-

cell notation

Cl2(g)/Cl-(aq) // Mg2+(aq) / Mg(s)

b)

mol of Mg = mass/MW = 752/24.3 = 30.9465

1 mol of Mg requires = 2 mol of e-

30.9465 mol of Mg = 2*30.9465 = 61.893 mol of e-

c)

1 mol of e- = 96500 C

61.893 mol of e---> 61.893*96500 = 5972674.5 C

d)

find Amps for

t = 5.75 h = 5.75*36500 s = 209875 s

A = Charge/time=5972674.5 / 209875

A = 28.458 A or C/s

e)

1 mol of Mg = 1 mol of Cl2

30.9465 mol of Mg = 30.9465 mol of Cl2

at STP = 1 mol = 22.4 L

30.9465 mol --> 22.4*30.9465 = 693.2016 Liter of Cl2

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