Aluminum(III)ion (Al³⁺)is reduced to solid aluminum at an electrode.If a current of 2.75 amperes is passed for 36 hours,calculate the mass of aluminum deposited at the electrode.(Assume 100% current efficiency.)

Question

Quizplus · Accepted Answer

To calculate the mass of aluminum deposited, we need to use Faraday's Law of Electrolysis which states that the amount of substance produced at an electrode is directly proportional to the amount of electrical charge passing through the electrode.

We first need to calculate the amount of charge (Q) passing through the electrode by using the formula:

Q = I x t

Where I is the current and t is the time.

Substituting the given values, we get:

Q = 2.75 A x 36 hours x 3600 seconds/hour
Q = 351360 C

Next, we need to calculate the number of moles of aluminum produced using the formula:

n = Q / F

Where F is Faraday's constant (96485 C/mol).

Substituting the values, we get:

n = 351360 C / 96485 C/mol
n = 3.64 mol

Finally, to get the mass of aluminum produced, we need to multiply the number of moles by the molar mass of aluminum (27 g/mol).

m = n x M
m = 3.64 mol x 27 g/mol
m = 98.3 g

Therefore, the mass of aluminum deposited at the electrode is 3.3 x 10^11 g (option B).

Aluminum(III)ion (Al3+)is Reduced to Solid Aluminum at an Electrode

Aluminum(III)ion (Al³⁺)is Reduced to Solid Aluminum at an Electrode