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

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Answer 1 · 2017-12-12T05:56:38

$2Li + FeBr_2 rarr Fe + 2LiBr$ is the stoichiometric equation....

$Fe(II)$ is reduced....

$FeBr_2 +2e^(-) rarr Fe + 2Br^(-)$ $(i)$

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

$Li rarr Li^+ + e^(-)$ $(ii)$

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

$2Li +FeBr_2 +2e^(-)rarr 2Li^+ +Fe + 2Br^(-)+ 2e^(-)$

To give finally:

$2Li +FeBr_2 rarr 2LiBr +Fe$

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

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