# How could "lithium metal" reduce "ferrous bromide" to give "iron metal"?

##### 1 Answer
Dec 12, 2017

$2 L i + F e B {r}_{2} \rightarrow F e + 2 L i B r$ is the stoichiometric equation....

#### Explanation:

$F e \left(I I\right)$ is reduced....

$F e B {r}_{2} + 2 {e}^{-} \rightarrow F e + 2 B {r}^{-}$ $\left(i\right)$

$\text{Lithium metal}$ is oxidized....

$L i \rightarrow L {i}^{+} + {e}^{-}$ $\left(i i\right)$

We add $\left(i\right)$ and $\left(i i\right)$ together in such a way that the electrons are retired from the equation....$\left(i\right) + 2 \times \left(i i\right)$

$2 L i + F e B {r}_{2} + 2 {e}^{-} \rightarrow 2 L {i}^{+} + F e + 2 B {r}^{-} + 2 {e}^{-}$

To give finally:

$2 L i + F e B {r}_{2} \rightarrow 2 L i B r + F e$

Both charge and mass are balanced as is absolutely required (and this is why we employ the oxidation number method).

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