The same quantity of electricity was passed through separate molten samples of aluminium oxide and sodium chloride. How many moles of sodium will be produced if 0.2 moles of oxygen gas were formed?

1 Answer
Oct 1, 2015

Answer:

#"0.8 moles Na"#

Explanation:

The overall reactions look like this

#2"Al"_2"O"_text(3(l]) -> 4"Al"_text((l]) + 3"O"_text(2(g]) uarr#

and

#2"Na"^(+)"Cl"^(-) -> 2"Na"_text((s]) + "Cl"_text(2(g]) uarr#

Now, the half-reactions that are of interest to you are

#6stackrel(color(blue)(-2))("O") -> 3stackrel(color(blue)(0))("O"_2) + 12e^(-)#

Oxygen is being oxidized to oxygen gas.

#stackrel(color(blue)(+1))("Na"^(+)) + 1e^(-) -> stackrel(color(blue)(0))("Na")#

Sodium is being reduced to sodium metal.

So, how many moles of electrons are needed to make one mole of oxygen gas?

Take a look at the mole ratio that exists between oxygen gas, #"O"_2#, and the number of moles of electrons that take part in oxygen's oxidation to oxygen gas.

Well, if you have twelve moles of electrons for every three moles of oxygen gas, then it follows that you need four moles of electrons to make one mole of oxygen gas.

This means that #"0.2 moles"# of oxygen gas needed

#0.2color(red)(cancel(color(black)("moles O"""_2))) * ("12 moles e"""^(-))/(3color(red)(cancel(color(black)("moles O"""_2)))) = "0.8 moles e"""^(-)#

Now look at the reduction of sodium cations to sodium metal. Notice that you need one mole of electrons to produce one mole of sodium metal.

This means that you will produce

#0.8color(red)(cancel(color(black)("moles e"""^(-)))) * "1 mole Na"/(1color(red)(cancel(color(black)("moles e"""^(-))))) = color(geen)("0.8 moles Na")#