# How do we solve a redox equation when BASIC conditions are specified?

Dec 3, 2015

I only speak for myself, but I always solved the redox equation under acidic conditions. If basic conditions were specified, I could simply add the appropriate number of hydroxide ions to BOTH sides of the equation.

#### Explanation:

What the devil do I mean by the above spray? I will try an example. Consider the oxidation of chromium(III) hydroxide, $C r {\left(O H\right)}_{3}$ to chromate ion, $C r {O}_{4}^{2 -}$ under BASIC conditions. Now the metal has been oxidized from $C {r}^{I I I +}$ to $C {r}^{V I +}$, so I include the three electrons to get the oxidation equation (i):

$C r {\left(O H\right)}_{3} + {H}_{2} O \rightarrow C r {O}_{4}^{2 -} + 5 {H}^{+} + 3 {e}^{-}$ $\left(i\right)$

Now clearly, both charge and mass are balanced (and they MUST be!), however, basic conditions were specified, so how can I include the $5 {H}^{+}$ as a product? Well, I can remove the problem by adding $5 \times O {H}^{-}$ to EACH side of the equation to give:

$C r {\left(O H\right)}_{3} + {H}_{2} O + 5 O {H}^{-} \rightarrow C r {O}_{4}^{2 -} + 5 {H}_{2} O + 3 {e}^{-}$ $\left(i i\right)$ And, after cancelling common reagents (${H}_{2} O$) we get finally:

$C r {\left(O H\right)}_{3} + 5 O {H}^{-} \rightarrow C r {O}_{4}^{2 -} + 4 {H}_{2} O + 3 {e}^{-}$ $\left(i i i\right)$

On each side of the equation there is 1 metal atom, $8 \times O$, $8 \times H$, and there is a ${5}^{-}$ charge on each side. So the equation is balanced with respect to mass and charge, as required. See if you can manage to represent the oxidation of $M n {O}_{2}$ to manganate ion, $M n {O}_{4}^{2 -}$, which is normally performed in basic conditions.