# Why are oxidizing agents acidified?

Dec 23, 2016

Usually because oxidation reactions tend to specify acidic conditions.

#### Explanation:

And basic conditions tend to lead to the precipitation of insoluble metal hydroxides.

Metal oxidants, say permanganate, $M n {O}_{4}^{-}$, or dichromate, $C {r}_{2} {O}_{7}^{2 -}$ tend to be deployed in acidic media:

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

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O$

So we use ${H}^{+}$ or ${H}_{3} {O}^{+}$ in the reaction to represent the actual acidic condtions of reaction. Had the reaction been conducted in A BASIC medium, heavy metal oxides and hydroxides would precipitate out pdq.

If a redox reaction is specified to be in a BASIC medium, it is often easiest to balance the redox reaction in acidic conditions using ${H}^{+}$, and then add $H {O}^{-}$ ion to each side of the reaction.

For instance, aluminum metal can be oxidized up to aluminate ion, $A l {\left(O H\right)}_{4}^{-}$ in strong base. We could represent this as normal:

$A l \rightarrow A {l}^{3 +} + 3 {e}^{-}$, but then we could add $4 \times H {O}^{-}$ to BOTH sides of the reaction:

$A l + 4 H {O}^{-} \rightarrow {\left[A l {\left(O H\right)}_{4}\right]}^{-} + 3 {e}^{-}$

Charge and mass are balanced in this reaction as required.