Al³⁺ is reduced to Al(s)at an electrode.If a current of 2.75 ampere is passed for 36 hours,what mass of aluminum is deposited at the electrode? Assume 100% current efficiency. A) 9.2 $\times$ 10^-3 g B) 3.3 $\times$ 10¹ g C) 9.9 $\times$ 10¹ g D) 1.0 $\times$ 10² g E) 3.0 $\times$ 10² g

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of charge passed through the circuit using the formula:
Q = I*t
where I is the current in amperes and t is the time in seconds.

Since we are given the current in amperes and time in hours, we need to convert the time to seconds:
36 hours x 3600 seconds/hour = 129600 seconds

Now, we can calculate the charge:
Q = 2.75 A x 129600 s = 355200 C

Next, we can use Faraday's law to calculate the amount of aluminum deposited:
m = (Q/F) x (M/charge)
where Q is the charge in coulombs, F is Faraday's constant (96,485 C/mol), M is the molar mass of aluminum (27 g/mol), and charge is the number of electrons transferred per aluminum atom (3 in this case).

Plugging in the values, we get:
m = (355200 C/96485 C/mol) x (27 g/mol/3) = 101.79 g

This is equivalent to 3.3 x 10^10^-3 g, which matches option B.

Al3+ Is Reduced to Al(s)at an Electrode ×\times× 10-3 G B) 3

Al³⁺ Is Reduced to Al(s)at an Electrode $\times$ 10^-3 G
B) 3