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Semmy [17]
3 years ago
6

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the a

mount 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.
Chemistry
1 answer:
UkoKoshka [18]3 years ago
5 0

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!

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