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.

Step-by-step 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* t`

where,

C = charge

I = current = 0.490 A

t = time required (in seconds) =
`50* 60=300s` (Conversion factor: 1 min = 60 s)

Putting values in above equation, we get:

`C=0.490* 300=147C`

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

`\text{Moles of electrons}=(C)/(F)`

where,

C = charge = 147 C

F = Faradays constant = 96500

`\text{Moles of electrons}=(147)/(96500)=1.52* 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}=(1.52* 10^(-3))/(3)=5.077* 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* 10^(-4)mol`

Molar mass of Gallium = 69.72 g/mol

Putting values in above equation, we get:

`5.077* 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.