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 0.760 A that flows for 50.0 min.

Answer 1

`m_(Ga)=0.550gGa`

Step-by-step explanation:

Hello!

In this case, since the applied current for the 50.0 mins provides the following charge to the system:

`q=0.760(C)/(s)*50.0min*(60s)/(1min)=2,280C`

As 1 mole of electrons carries a charge of 1 faraday, or 96,485 coulombs, we can compute the moles of electrons involved during the reduction:

`n_(e^-)=2,280C*(1mole^-)/(96,485C)=0.0236mole^-`

Then the reduction of Ga³⁺ to Ga involves the transference of three electrons, we are able to compute the moles and therefore the mass of deposited gallium:

`m_(Ga)=0.0236mole^-*(1molGa)/(3mole^-)*(69.72gGa)/(1molGa) \\\\m_(Ga)=0.550gGa`

Best regards!