How do you balance oxidation reduction reactions in basic solution?

1 Answer
Sep 18, 2016


The same way as you balance redox equations in acid media; but there is a useful trick.


A general method for balancing redox reactions is given here. But let's suppose that you were asked to represent the reduction of #MnO_4^(-)# to manganous hydroxide, #Mn(OH)_2#.

We could first represent the reduction in acidic media:

#MnO_4^(-)+8H^++5e^−rarr Mn^(2+) +4H_2O#

Now this reduction reaction is balanced with respect to mass and charge, as it must be. However, basic conditions were specified, so we have to try to remove the #H^+#. How do we do this? We could add #8xxHO^-# to BOTH SIDES and preserve stoichiometry:

#MnO_4^(-)+8H^++8HO^(-)+5e^−rarr Mn^(2+) +4H_2O +8HO^(-)#

#MnO_4^(-)+8H_2O+5e^−rarr Mn^(2+) +4H_2O +8HO^(-)#, and remove the common reagents:

#MnO_4^(-)+4H_2O+5e^−rarr Mn^(2+) +cancel(4H_2O) +8HO^(-)#

#MnO_4^(-)+4H_2O+5e^−rarr Mn^(2+) +8HO^(-)#

Which is clearly balanced with respect to mass and charge as required. Possibly the following equation makes more chemical sense:

#MnO_4^(-)+4H_2O+5e^−rarr Mn(OH)_2(s) +6HO^(-)#

In that most hydroxides are insoluble. Is charge still balanced here?

So I have made a meal of this, but my advice is to balance the redox equation using acid conditions, and at the end to represent redox in basic conditions add appropriate equivalents of hydroxide, and remember to cancel the waters out on each side.