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

Answer:

$m_{Ga}=0.550gGa$

Explanation:

Hello!

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

$q=0.760\frac{C}{s}*50.0min*\frac{60s}{1min}=2,280C$

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:

$n_{e^-}=2,280C*\frac{1mole^-}{96,485C}=0.0236mole^-$

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:

$m_{Ga}=0.0236mole^-*\frac{1molGa}{3mole^-}*\frac{69.72gGa}{1molGa} \\\\m_{Ga}=0.550gGa$

Best regards!