We are familiar with the balancing of chemical equations by inspection method. However, inspection method may not be useful for balancing the redox equations because in these equations, we have to keep in mind the conservation of charge as well as conservation of mass. The redox equations, are therefore, are balanced by using the concept of half equations and following certain set of rules. One of the methods used for balancing redox reactions is called ion-electron method.
31.1 BALANCING BY ION ELECTRON OR HALF REACTION METHOD
We know that during redox reactions there is a change in oxidation number of the elements due to the transference of electrons. The basic principle involved in balancing the redox equation is that the number of electrons lost during oxidation is equal to the number of electrons gained during reduction.
STEPS INVOLVED IN BALANCING REDOX EQUATIONS
The various steps involved in the balancing of redox equations ion-electron method are as follows:
1. Indicate the oxidation number of each atom involved in the reaction. Identify the elements which undergo a change in the oxidation number.
2. Divide the skeleton redox equation into two half reactions; oxidation half and reduction half. In each half reaction balance the atoms which undergo the change in oxidation number.
3. In order to make up for the difference in O.N. add electrons to left hand side or right hand side of the arrow in each half reaction.
4. Balance oxygen atoms by addition of proper number of H2O molecules to the side which is falling short of 0 atoms in each half reaction.
5. This step is meant for only ionic equations. It involves the balancing of H atoms in each half reaction as follows:
(i) For acidic medium Add proper number of H+ ions to the side falling short of H atoms.
( ii) For basic medium. Add proper number of H2O molecules to the side falling short of H atoms and equal number of OH- ions to the other side.
6. Equalise the number of electrons lost and gained by multiplying the half reactions with suitable integer and add them to get the final equation. The application of various steps described above has been illustrated as follows by balancing the redox equation representing the reaction between iodine and nitric acid.
HNO3 + I2 à HIO3 + NO2 + H2O
Step 1. Indication of oxidation numbers of each atom
Thus, only nitrogen and iodine undergo change in oxidation number.
Step 2. Division into two half reactions and balancing the atoms undergoing change in O.N.
Step 3. Addition of electrons to make up the difference in O.N.
(Each I atom loses 5e- therefore, two iodine atoms would lose 10e-)
(Each N atom gains 1 electron).
Step 4. Balancing of 0 by adding proper number of H2O molecule to the side falling short of 0 atoms.
Step 5. Not required because the equation is not ionic.
Step 6. To equalise the electrons multiply reduction half reaction by 10 and add the two.
Let us now proceed to balance the ionic equations in acidic and basic mediums.
Consider the chemical reaction between zinc and copper(II) tetraoxosulphate(V) solution.
Zn(s) + CuSO4(aq) à ZnSO(aq) + Cu(s)
Writing ionic form of each species we have
Zn(s) + Cu2+(aq) + SO4(aq) à Zn2+(aq) + SO4(aq) + Cu(s)
Now, it is evident that Cu atoms and Zn2+ ions undergo oxidation and reduction respectively and are active ions. SO4 lion, on the other hand, do not participate in the reaction and remain as such. They are inactive towards the reaction and are called spectator ions. In the similar way K+ ions are spectator ions in the following reaction.