Question #1f84a

1 Answer
Mar 1, 2017

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

Explanation:

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

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

Fe^(2+) rarr Fe^(3+) + e^-Fe2+Fe3++e (i)(i)

And, as is typical, "permanganate ion"permanganate ion is reduced to (almost) colourless Mn^(2+)Mn2+:

stackrel("+VII")(Mn)O_4^(-) +8H^(+) + 5e^(-)rarr stackrel("+II")Mn^(2+) + 4H_2O(l)+VIIMnO4+8H++5e+IIMn2++4H2O(l) (ii)(ii)

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

5Fe^(2+) + stackrel("+VII")(Mn)O_4^(-) + 8H^+ → 5Fe^(3+) + Mn^(2+) + 4H_2O5Fe2+++VIIMnO4+8H+5Fe3++Mn2++4H2O

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^-MnO4 solution to give the almost colourless Mn^(2+)Mn2+ ion.