Aamc 7 ps # 49

[letaps]

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Hey,

I really don't understand this question. I know that a faraday is the charge per mole of electron. So, the question stem is saying there is 0.1 C/mole of electron (is that right?)...from here i don't where to go...


Approximately how many moles of Al3+ are reduced when 0.1 faraday of charge passes through a cell during the production of Al? (Note: Assume there is excess Al3+ available and that Al3+ is reduced to Al metal only.)

A) 0.033 mol
A faraday is equal to one mole of electric charge. Because each aluminum ion gains 3 electrons, 0.1 faraday of charge will reduce 0.1/3 moles of aluminum, or 0.033 moles of aluminum.

B) 0.050 mol

C) 0.067 mol

D) 0.10 mol

 

[DamienPro]

Hey,

I really don't understand this question. I know that a faraday is the charge per mole of electron. So, the question stem is saying there is 0.1 C/mole of electron (is that right?)...from here i don't where to go...


Approximately how many moles of Al3+ are reduced when 0.1 faraday of charge passes through a cell during the production of Al? (Note: Assume there is excess Al3+ available and that Al3+ is reduced to Al metal only.)

A) 0.033 mol
A faraday is equal to one mole of electric charge. Because each aluminum ion gains 3 electrons, 0.1 faraday of charge will reduce 0.1/3 moles of aluminum, or 0.033 moles of aluminum.

B) 0.050 mol

C) 0.067 mol

D) 0.10 mol

Look at the explanation in this way:

1 faraday is associated with 1 mole of electrons.
==> 0.1 faraday is equivalent to 0.1 mole of electrons.

Since 3 moles of electrons are needed to reduce 1 mole of Al3+, 0.1 mole of electrons will therefore reduce:

(0.1 mole of electrons x 1 mole of Al3+)/3 mole of electrons
= 0.033 mole of Al3+