Question #3fbeb

Provided that this is the complete problem given to you, you don't actually need to know the concentration of the aluminium chloride solution.

A concentration of #"1.0 M"# is usually given to let students know that they're dealing with standard conditions, which are defined as

• concentrations of #" 1.0 M"#
• a pressure of #" 1 atm"#
• a temperature of #" "25^@"C"#

This is important when using the standard reduction potentials for the oxidation and reduction half-reactions

#E_"reduction"^@ -># for the reduction half-reaction

#E_"oxidation"^@ = -E_"reduction"^@ -># for the oxidation half-reaction

These are then used to calculate the standard cell potential, #E_"cell"^@#

#color(purple)(|bar(ul(color(white)(a/a)color(black)(E_"cell"^@ = E_"reduction"^@ + E_"oxidation"^@)color(white)(a/a)|)))#

So, to sum this up, the concentration of the solution is not a factor provided that you're working under standard conditions, i.e. it will either be given as #"1.0 M"# or not given at all.

As the solution shows, all you have to do here is write the reduction half-reaction

#"Al"_ ((aq))^(3+) + 3"e"^(-) -> "Al"_ ((s))#

Since it takes #3# moles of electrons to reduce #1# mole of aluminium cations and produce #1# mole of aluminium metal, and since you need to get

#54 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(27color(red)(cancel(color(black)("g")))) = "2 moles Al"#

it follows that you will need

#2 color(red)(cancel(color(black)("moles Al"))) * "3 mole e"^(-)/(1color(red)(cancel(color(black)("mole Al")))) = "6 moles e"^(-)#