Question

the reaction between aluminum and bromine is an oxidation - reduction reaction. what is the oxidation number of aluminum as a reactant and in the product? al as reactant al in al2br6 (9 of 12) check

Explanation:

Step1: Recall oxidation - number rules for elements in compounds

In a free element, the oxidation number of an element is 0. Aluminum as a reactant is in its elemental form (Al), so its oxidation number is 0.

Step2: Recall oxidation - number rules for elements in compounds

In a compound, for binary compounds with a non - metal and a metal, the non - metal has a negative oxidation number. Bromine in compounds usually has an oxidation number of - 1. In $Al_2Br_6$, let the oxidation number of Al be $x$. Using the rule that the sum of oxidation numbers in a neutral compound is 0, we have $2x+6\times(- 1)=0$. Solving for $x$:
\[

$$\begin{align*} 2x-6&=0\\ 2x&=6\\ x& = + 3 \end{align*}$$

\]

Answer:

Al as a reactant: 0; Al in $Al_2Br_6$: + 3