In the electrolysis of molten CaCl2, a current of 1.12 A is passed through the cell for 3.0 h. What is the mass of Ca produced at the cathode

2.51 g of Ca.

Explanation:

From the question given above, the following data were obtained:

Current (I) = 1.12 A

Time = 3 h

Mass of Ca =?

Next, we shall convert 3 h to s. This can be obtained as follow:

1 h = 3600 s

Therefore,

3 h = 3h × 3600 s / 1 h

3 h = 10800 s

Next, we shall determine the quantity of electricity flowing through the solution. This can be obtained as follow:

Current (I) = 1.12 A

Time = 10800 s

Quantity of electricity (Q) =?

Q = It

Q = 1.12 × 10800

Q = 12096 C

Next, we shall determine the quantity of electricity needed to deposit 1 mole (i.e 40 g) of Ca. This can be obtained as follow:

In solution, CaCl₂ will dissociate as follow:

CaCl₂(aq) <—> Ca²⁺(aq) + 2Cl¯(aq)

Ca²⁺(aq) + 2e –> Ca(s)

From the equation above, 2 moles of electrons is required to deposit 1 mole (i.e 40 g) of Ca.

Recall:

1 e = 1 faraday = 96500 C

1 e = 96500 C

Therefore,

2 e = 2 × 96500 C

2 e = 193000 C

Thus, 193000 C of electricity is needed to deposit 40 g of Ca.

Finally, we shall determine the mass of Ca deposited by 12096 C. This can be obtained as follow:

193000 C of electricity is needed to deposit 40 g of Ca.

Therefore, 12096 C of electricity will deposit = (12096 × 40)/193000 = 2.51 g of Ca.

Thus, 2.51 g of Ca is deposited at the cathode.

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