# How can you balance oxidation reduction reactions?

Feb 4, 2017

You have to balance mass and charge.

#### Explanation:

See this old answer. The method of half equations, where electrons are introduced as virtual particles, is the standard model. The conservation of mass and charge is fundamental to ALL chemical reactivity.

For a complicated example, let us presume that ammonia is oxidized up to nitrate ion, by say potassium permanganate.

Now the oxidation reaction is standard:

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$

Are mass and charge conserved? They should be. We have introduced electrons are virtual particles; they are presumed to come from somewhere, and here we presume they come from (for instance) a corresponding oxidation reaction of ammonia to nitrate anion:

${\text{^(-III)NH_3 + 3H_2O rarr }}^{+ V} N {O}_{3}^{-} + 9 {H}^{+} + 8 {e}^{-}$

Mass and charge are again balanced. To represent the reaction, we cross-multiply the half-equations so that the electrons are removed from the final equation.

$5 N {H}_{4}^{+} + 8 M n {O}_{4}^{-} + 14 {H}^{+} \rightarrow 5 N {O}_{3}^{-} + 8 M {n}^{2 +} + 17 {H}_{2} O$

As improbable as it might seem, this equation is balanced with respect to mass and charge. What would we see in the reaction? We would see the deep purple colour of permanganate dissipating to give the (almost) colourless $M {n}^{2 +}$ ion.