Question

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

Answer 1

`5Fe^(+2)` + `MnO_4^(-) + 8H^+` => `Mn^(+2) + 5Fe^(+3) + 4H_2O`

Explanation:

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

`Fe^(+2) + MnO_4^-` => `Mn^(+2) + Fe^(+3)`

Oxidation Half Rxn: `Fe^(+2)` => `Fe^(+3) + e^-`
Reduction Half Rxn: `MnO_4^-` + `5e^-` => `Mn^(+2)`

Balancing with respect to charge transfer ...
`5Fe^(+2)` =>`5Fe^(+3) + 5e^-`
`MnO_4^-` + `5e^ -` => `Mn^(+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.

`5Fe^(+2)` + `MnO_4^(-) + 8H^+` => `Mn^(+2) + 5Fe^(+3) + 4H_2O`