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

Jul 10, 2016

$2 {\text{Al"_ ((s)) + 3"Cl"_ (2(g)) -> 2"AlCl}}_{3 \left(s\right)}$

#### Explanation:

Let's balance this equation by using oxidation numbers.

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

${\stackrel{\textcolor{b l u e}{0}}{\text{Al") _ ((s)) + stackrel(color(blue)(0))("Cl")_ (2(g)) -> stackrel(color(blue)(+3))("Al") stackrel(color(blue)(-1))("Cl}}}_{3 \left(s\right)}$

Notice that the oxidation number of aluminium increases from $\textcolor{b l u e}{0}$ on the reactants' side to $\textcolor{b l u e}{+ 3}$ on the products' side, which means that it's being oxidized.

On the other hand, the oxidation number of chlorine decreases from $\textcolor{b l u e}{0}$ 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) :}
$\frac{\textcolor{w h i t e}{a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}$

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

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{2 {\text{Al"_ ((s)) + 3"Cl"_ (2(g)) -> 2"AlCl}}_{3 \left(s\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$