Answer:
[tex]t=74.5hr[/tex]
Explanation:
Hello!
In this case, given the mass of aluminum metal which can undergo the following reduction from aluminum oxide:
[tex]Al^{3+}+3e^-\rightarrow Al[/tex]
We can see three moles of electrons are transferred per mole of aluminum, thus, we obtain:
[tex]mole^-=50.0gAl*\frac{1molAl}{26.98gAl}*\frac{3mole^-}{1molAl} =5.56mole^-[/tex]
Next, we compute the charge carried by those electrons via the faraday's constant:
[tex]q=5.56mole^-*\frac{96,485C}{1mole^-} =536,425C[/tex]
Finally, considering the applied current, we compute the elapsed time in hours to achieve such electrolysis:
[tex]t=536,425C*\frac{s}{2.00C} *\frac{1hr}{3600s} \\\\t=74.5hr[/tex]
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