Question

Consider the electrolysis of molten barium chloride, BaCl2. (a) Write the half-reactions. (b) How many grams of barium metal can be produced by supplying 0.50 A for 30 minutes?

Answer 1

Answer: a)  Cathode(-): $Ba^+^2+2e^-\rightarrow Ba$

Anode(+): $2Cl^-\rightarrow Cl_2+2e^-$

b) 0.640 grams of Ba will be deposited.

Explanation:  a) The problem is based on Faraday law of electrolysis. Molten barium chloride has $Ba^+^2$ ion and $Cl^-$ ion. Barium ion is reduced and chloride is oxidized. Reduction takes place at cathode and oxidation at anode. So, the half reactions will be:

Cathode(-): $Ba^+^2+2e^-\rightarrow Ba$

Anode(+): $2Cl^-\rightarrow Cl_2+2e^-$

b) The question asks, how many grams of barium metal can be produced by supplying 0.50 ampere for 30 minutes.

From the Cathode half-reaction, 1 mole of Ba is deposited by 2 moles of electrons and we know that 1 mole of electron carries one Faraday that is 96485 Coulomb.

Coulombs for 2 moles of electrons will be = 2*96485 C = 192970 C

So, we can say that, 192970 C will deposit 1 mole of Ba metal.

Total available coulombs can be calculated using the formula:

q=i*t

where, q is electric charge in coulomb, i is current in ampere and t is time in seconds.

$q=q=0.50A*30min(\frac{60sec}{1min})$

q = 900 C         (note: 1 C = 1 A*sec)

Let's calculate how many moles of Ba will get deposited by 900 C.

$900C(\frac{1molBa}{192970C})$

= 0.00466 mole Ba

Convert the moles of Ba to grams and for this we multiply by molar mass of Ba which is 137.33 gram per mol.

$0.00466molBa(\frac{137.33g}{1mol})$

= 0.640 g Ba

So, 0.50 A for 30 minutes will deposit 0.640 grams of Ba metal.