# What is the balanced net ionic equation for the reaction in acidic solution of Fe^(2+) and KMnO_4?

Nov 19, 2017

$5 F {e}^{+ 2}$ + $M n {O}_{4}^{-} + 8 {H}^{+}$ => $M {n}^{+ 2} + 5 F {e}^{+ 3} + 4 {H}_{2} O$

#### Explanation:

Potassium is a spectator ion so assume the following reaction ...

$F {e}^{+ 2} + M n {O}_{4}^{-}$ => $M {n}^{+ 2} + F {e}^{+ 3}$

Oxidation Half Rxn: $F {e}^{+ 2}$ => $F {e}^{+ 3} + {e}^{-}$
Reduction Half Rxn: $M n {O}_{4}^{-}$ + $5 {e}^{-}$ => $M {n}^{+ 2}$

Balancing with respect to charge transfer ...
$5 F {e}^{+ 2}$ =>$5 F {e}^{+ 3} + 5 {e}^{-}$
$M n {O}_{4}^{-}$ + $5 {e}^{-}$ => $M {n}^{+ 2}$

Balancing with respect to mass ... (Rxn in acidic medium, add one water for each oxygen needed to the side needing oxygen and follow with the appropriate number of ${H}^{+}$ needed to the side needing hydrogen.

$5 F {e}^{+ 2}$ + $M n {O}_{4}^{-} + 8 {H}^{+}$ => $M {n}^{+ 2} + 5 F {e}^{+ 3} + 4 {H}_{2} O$