Question

Question #3fbeb

Answer 1

Here's what's going on here.

Explanation:

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"^(-)`