Answer:
2.51 g of Ca.
Step-by-step 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.