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

1 Answer
Dec 12, 2017

Answer:

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

Explanation:

#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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