Question

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.590 A that flows for 30.0 min.

Answer 1

To calculate the mass of gallium deposited from a Ga(III) solution by electrolysis with a current of 0.590 A for 30.0 minutes, the total charge passed is first found and then used to determine the moles of electrons involved in the reduction of Ga(III) to Ga metal. Finally, the mass of gallium produced is calculated to be 0.0851 g.

Step-by-step explanation:

To calculate the amount of Ga(s) deposited from a Ga(III) solution, we'll use Faraday's laws of electrolysis. The first step is to calculate the total charge passed through the solution during electrolysis. Since current (I) is the rate of charge flow and it is given as 0.590 A, and time (t) is 30.0 minutes, we must convert minutes to seconds to use the correct units (1 minute = 60 seconds).

Charge (Q) = Current (I) × Time (t) = 0.590 A × (30.0 min × 60 s/min) = 1062 C

Next, we apply Faraday's laws, which state that the amount of a substance produced at an electrode during electrolysis is proportional to the quantity of electricity that passes through the electrolyte. The molar mass of gallium (Ga) is 69.72 g/mol and the valency of Ga(III) ions is 3, as they lose three electrons to form gallium metal during electrolysis.

The number of moles of electrons required to deposit one mole of gallium can be calculated using the equation: n = Q / (F × z), where n is the number of moles of electrons, Q is the charge, F is Faraday's constant (96485 C/mol), and z is the valency of the ion.

n = 1062 C / (96485 C/mol × 3) = 0.00366 mol

Since one mole of Ga(III) ions requires three moles of electrons for complete reduction to gallium metal, the number of moles of Ga produced is equal to the number of moles of electrons divided by 3 (because each Ga(III) ion requires 3 electrons to reduce to Ga metal).

Number of moles of Ga = 0.00366 mol / 3 = 0.00122 mol

Finally, the mass of Ga produced is found by multiplying the number of moles of Ga by the molar mass of Ga.

Mass of Ga = Number of moles of Ga × Molar mass of Ga = 0.00122 mol × 69.72 g/mol = 0.0851 g

Therefore, the mass of gallium deposited is 0.0851 g.

Answer 2

Answer: 0.256 g

Step-by-step explanation:

The deposition of Ga from Ga(III) solution

`Ga^(3+)+3e^-\rightarrow Ga`

According to mole concept:

1 mole of an atom contains
`6.022* 10^(23)` number of particles.

We know that:

Charge on 1 electron =
`1.6* 10^(-19)C`

Charge on 1 mole of electrons =
`1.6* 10^(-19)* 6.022* 10^(23)=96500C`

`I=(q)/(t)`

where,

I = current passed = 0.590 A

q = total charge = ?

t = time required = 30.0 min = 1800 s (1min=60s)

Putting values in above equation, we get:

`0.590A=(q)/(1800)\\\\q=1062C`

`Ga^(3+)+3e^-\rightarrow Ga`

`96500* 3=289500C` will deposit = 70 g of Gallium

`1062C` will deposit =
`(70)/(289500)* 1062=0.256 g` of Gallium

Thus 0.256 g of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.590 A that flows for 30.0 min.