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

Answer: 0.256 g

Explanation:

The deposition of Ga from Ga(III) solution

$Ga^{3+}+3e^-\rightarrow Ga$

According to mole concept:

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

We know that:

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

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

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

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:

$0.590A=\frac{q}{1800}\\\\q=1062C$

$Ga^{3+}+3e^-\rightarrow Ga$

$96500\times 3=289500C$ will deposit = 70 g of Gallium

$1062C$ will deposit = $\frac{70}{289500}\times 1062=0.256 g$ 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.