Answer:
[tex]m_{Ga}=0.550gGa[/tex]
Explanation:
Hello!
In this case, since the applied current for the 50.0 mins provides the following charge to the system:
[tex]q=0.760\frac{C}{s}*50.0min*\frac{60s}{1min}=2,280C[/tex]
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:
[tex]n_{e^-}=2,280C*\frac{1mole^-}{96,485C}=0.0236mole^-[/tex]
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:
[tex]m_{Ga}=0.0236mole^-*\frac{1molGa}{3mole^-}*\frac{69.72gGa}{1molGa} \\\\m_{Ga}=0.550gGa[/tex]
Best regards!
