Question #1f84a

1 Answer
anor277
Mar 1, 2017

Given the equation (below), one equiv #"permanganate"# will oxidize 5 equiv #"ferrous ion"#.

Explanation:

Well, we could conceive that #"permanganate ion"# oxidized #"ferrous ion"# up to #"ferric ion............"#

#"Ferrous ion"# is oxidized up to #"ferric ion"#:

#Fe^(2+) rarr Fe^(3+) + e^-# #(i)#

And, as is typical, #"permanganate ion"# is reduced to (almost) colourless #Mn^(2+)#:

#stackrel("+VII")(Mn)O_4^(-) +8H^(+) + 5e^(-)rarr stackrel("+II")Mn^(2+) + 4H_2O(l)# #(ii)#

We cross multiply to eliminate the electrons from the final equation, i.e. #5xx(i) + (ii):#

#5Fe^(2+) + stackrel("+VII")(Mn)O_4^(-) + 8H^+ → 5Fe^(3+) + Mn^(2+) + 4H_2O#

Are both MASS and CHARGE balanced here? If no, then the reaction is not in the race as a representation of chemical reality. The end point of the reaction is vizualized by the dissipation of the deep purple colour of #MnO_4^-# solution to give the almost colourless #Mn^(2+)# ion.

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