# Question 1f84a

Mar 1, 2017

Given the equation (below), one equiv $\text{permanganate}$ will oxidize 5 equiv $\text{ferrous ion}$.

#### Explanation:

Well, we could conceive that $\text{permanganate ion}$ oxidized $\text{ferrous ion}$ up to $\text{ferric ion............}$

$\text{Ferrous ion}$ is oxidized up to $\text{ferric ion}$:

$F {e}^{2 +} \rightarrow F {e}^{3 +} + {e}^{-}$ $\left(i\right)$

And, as is typical, $\text{permanganate ion}$ is reduced to (almost) colourless $M {n}^{2 +}$:

$\stackrel{\text{+VII")(Mn)O_4^(-) +8H^(+) + 5e^(-)rarr stackrel("+II}}{M} {n}^{2 +} + 4 {H}_{2} O \left(l\right)$ $\left(i i\right)$

We cross multiply to eliminate the electrons from the final equation, i.e. $5 \times \left(i\right) + \left(i i\right) :$

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 $M n {O}_{4}^{-}$ solution to give the almost colourless $M {n}^{2 +}$ ion.