Question

Gallium is produced by the electrolysis of a solution obtained by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(III) solution by a current of 0.490 A that flows for 50.0 min.

Answer 1

Answer: The mass of gallium produced by the electrolysis is 0.0354 grams.

Explanation:

The equation for the deposition of Ga(s) from Ga(III) solution follows:

$Ga^{3+}(aq.)+3e^-\rightarrow Ga(s)$

• To calculate the total charge, we use the equation:

$C=I\times t$

where,

C = charge

I = current = 0.490 A

t = time required (in seconds) = $50\times 60=300s$    (Conversion factor: 1 min = 60 s)

Putting values in above equation, we get:

$C=0.490\times 300=147C$

• To calculate the moles of electrons, we use the equation:

$\text{Moles of electrons}=\frac{C}{F}$

where,

C = charge = 147 C

F = Faradays constant = 96500

$\text{Moles of electrons}=\frac{147}{96500}=1.52\times 10^{-3}mol$

• Now, to calculate the moles of gallium, we use the equation:

$\text{Moles of Gallium}=\frac{\text{Moles of electrons}}{n}$

where,

n = number of electrons transferred = 3

Putting values in above equation, we get:

$\text{Moles of Gallium}=\frac{1.52\times 10^{-3}}{3}=5.077\times 10^{-4}mol$

• To calculate the mass of gallium, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of Gallium = $5.077\times 10^{-4}mol$

Molar mass of Gallium = 69.72 g/mol

Putting values in above equation, we get:

$5.077\times 10^{-4}mol=\frac{\text{Mass of Gallium}}{69.72g/mol}\\\\\text{Mass of Gallium}=0.0354g$

Hence, the mass of gallium produced by the electrolysis is 0.0354 grams.