Question

How do you balance `Al+Cl_2->AlCl_3`?

Answer 1

`2"Al"_ ((s)) + 3"Cl"_ (2(g)) -> 2"AlCl"_ (3(s))`

Explanation:

Let's balance this equation by using oxidation numbers.

Start by assigning oxidation numbers to the atoms that take part in the reaction

`stackrel(color(blue)(0))("Al") _ ((s)) + stackrel(color(blue)(0))("Cl")_ (2(g)) -> stackrel(color(blue)(+3))("Al") stackrel(color(blue)(-1))("Cl")_ (3(s))`

Notice that the oxidation number of aluminium increases from `color(blue)(0)` on the reactants' side to `color(blue)(+3)` on the products' side, which means that it's being oxidized.

On the other hand, the oxidation number of chlorine decreases from `color(blue)(0)` on the reactants' side to `-1` on the products' side, which means that chlorine is being reduced.

The oxidation half-reaction looks like this

`stackrel(color(blue)(0))("Al") -> stackrel(color(blue)(+3))("Al")""^(3+) + 3"e"^(-)`

The reduction half-reaction looks like this

`stackrel(color(blue)(0))("Cl")_ 2 + "e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-)`

Balance the chlorine atoms by multiplying the product by `2`. Since one atom of chlorine gains `1` electron, it follows that two atoms will gain `2` electrons

`stackrel(color(blue)(0))("Cl")_ 2 + 2"e"^(-) -> 2stackrel(color(blue)(-1))("Cl")""^(-)`

Now, in any redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

To get them to balance out, multiply the oxidation half-reaction by `2` and the reduction half-reaction by `3`., then add the two half-reactions to get

`{(color(white)(aaaaaa)stackrel(color(blue)(0))("Al") -> stackrel(color(blue)(+3))("Al")""^(3+) + 3"e"^(-)" "| xx 2), (stackrel(color(blue)(0))("Cl")_ 2 + 2"e"^(-) -> 2stackrel(color(blue)(-1))("Cl")""^(-)" "color(white)(aaaaa)| xx 3) :}`
`color(white)(a)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)`

`2"Al" + 3"Cl"_ 2 + color(red)(cancel(color(black)(6"e"^(-)))) -> 2"Al"^(3+) + 6"Cl"^(-) + color(red)(cancel(color(black)(6"e"^(-))))`

This will be equivalent to

`2"Al" + 3"Cl"_ 2 -> 2 xx ["Al"^(3+) + 3"Cl"^(-)]`

which of course gets you

`color(green)(|bar(ul(color(white)(a/a)color(black)(2"Al"_ ((s)) + 3"Cl"_ (2(g)) -> 2"AlCl"_ (3(s)))color(white)(a/a)|)))`