Question

Question #1f84a

Answer 1

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.