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

1 Answer
Dec 3, 2015

Answer:

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.