Question

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

Answer 1

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, `Cr(OH)_3` to chromate ion, `CrO_4^(2-)` under BASIC conditions. Now the metal has been oxidized from `Cr^(III+)` to `Cr^(VI+)`, so I include the three electrons to get the oxidation equation (i):

`Cr(OH)_3 + H_2O rarr CrO_4^(2-) + 5H^+ + 3e^-` `(i)`

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

`Cr(OH)_3 + H_2O + 5OH^(-) rarr CrO_4^(2-) + 5H_2O + 3e^-` `(ii)` And, after cancelling common reagents (`H_2O`) we get finally:

`Cr(OH)_3 + 5OH^(-) rarr CrO_4^(2-) + 4H_2O + 3e^-` `(iii)`

On each side of the equation there is 1 metal atom, `8xxO`, `8xxH`, 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 `MnO_2` to manganate ion, `MnO_4^(2-)`, which is normally performed in basic conditions.