Balancing redox reactions

The changes in oxidation state must balance so that the totals on the two sides are similar, in any complete redox reaction. Problems can arise with ions in solution, like the ionic charges may not be similar as the oxidation states. Refer the unbalanced redox reaction in acidified aqueous solution:

It is very easy to balance the redox changes by first dividing this into two half reactions, one involving oxidation and the other reduction. The oxidation step is

                                                                                              (3)

with electrons (e^-) being separated.  The conversion of  to Mn²⁺   involves a transform of oxidation state from MnVII

to Mn^II  and so is a reduction  requiring  five electrons.  To balance half reaction

four oxygen atoms are needed on the right-hand side, that (in aqueous solution) will be in the form of H₂O. The reaction

                                                                    (4)

is then finished  by balancing hydrogen with 8H⁺  on the left-hand side, like this reaction  occurs in acid. The entire redox reaction is now written  by combining  the two half reactions in such type of a way that the free electrons are removed. This needs 5 moles of Equation 3 to every 1 mole of Equation 4, giving

In alkaline solution  it is more suitable  to use OH^- rather than H⁺. The other species exist may also be dissimilar  from  those in acid,  as several  metal  cations form  insoluble  hydroxides  or even oxoanions . As an instance, refer the reaction of aluminum  metal with water to form [Al^III(OH)₄]^-  and H₂. The balanced half reactions are

and

which may be combined in the suitable proportions (two to three) to give

A specific benefit of the half-reaction approach is that it leads naturally to the discussion of the thermodynamics of redox reactions in terms of electrode potentials.