Question

How many seconds are required to produce 4.00 g of aluminum metal from the electrolysis of molten alcl3 with an electrical current of 12.0 a?

Answer 1

Answer: 3618 seconds

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{4g}{27g/mol}=0.15moles$

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

$AlCl_3\rightarrow Al^{3+}+3Cl^-$

At cathode: $Al^{3+}+3e^-\rightarrow Al$

1 mole of aluminium is deposited by = $3\times 96500=289500C$

Thus 0.15 moles of aluminium is deposited by = $\frac{289500C}{1}\times 0.15=43425C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed =12.0 A

q = total charge = $43425C$

t = time required in seconds = ?

Putting values in above equation, we get:

$12.0A=\frac{43425C}{t}\\\\t=\frac{43425C}{12.0A}=3618s$

Hence, the amount of time required to produce 4.00 g of aluminum metal from the electrolysis of molten $AlCl_3$ with an electrical current of 12.0 A is 3618 seconds