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

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

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Answer 1 · 2017-12-12T05:56:38

$2 L i + F e B r_{2} \to F e + 2 L i B r$ $2Li + FeBr_2 rarr Fe + 2LiBr$ is the stoichiometric equation....

Explanation:

$F e (I I)$ $Fe(II)$ is reduced....

$F e B r_{2} + 2 e^{-} \to F e + 2 B r^{-}$ $FeBr_2 +2e^(-) rarr Fe + 2Br^(-)$ $(i)$ $(i)$

$Lithium metal$ $"Lithium metal"$ is oxidized....

$L i \to L i^{+} + e^{-}$ $Li rarr Li^+ + e^(-)$ $(i i)$ $(ii)$

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

$2 L i + F e B r_{2} + 2 e^{-} \to 2 L i^{+} + F e + 2 B r^{-} + 2 e^{-}$ $2Li +FeBr_2 +2e^(-)rarr 2Li^+ +Fe + 2Br^(-)+ 2e^(-)$

To give finally:

$2 L i + F e B r_{2} \to 2 L i B r + F e$ $2Li +FeBr_2 rarr 2LiBr +Fe$

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

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

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

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