Some examples of redox reactions are
If one of the half reactions does not take place, other half will also not take place. We can say oxidation and reduction go side by side.
In this we find that Cl2 has been oxidised as well as reduced. Such type of redox reaction is called Disproportionation reaction. Examples are
How to Balance a Redox Reaction :
Ion Electron Method :
This method involves the following steps :
(i) Divide the complete equation into two half reactions, one representing oxidation and the other reduction.
(ii) Balance the atoms in each half reaction separately according to the following steps:
(a) First of all balance the atoms other than H and O.
(b) In a reaction taking place in acidic or neutral medium, oxygen atoms are balanced by adding molecules of water to the side deficient in oxygen atoms while hydrogen atoms are balanced by adding H+ ions to the other side deficient in hydrogen atoms. On the other hand, in alkaline medium (OH-), for every excess of oxygen atom on one side is balanced by adding one H2O to the same side and 2OH- to the other side. In case hydrogen is still unbalanced, then balance by adding one OH-, for every excess of H atom on the same side and one H2O on the other side.
(c) Equalize the charge on both sides by adding suitable number of electrons to the side deficient in negative charge.
(iii) Multiply the two half reactions by suitable integers so that the total number of electrons gained in one half reaction is equal to the number of electrons lost in the other half reaction.
(iv) Add the two balanced half equations and cancel any term common to both sides. There have been the common practices to balance the redox reaction by different methods like O.N. method and electron balance method. In the entrance examination it is never mentioned what method is to be used. We adopt here “quick” method that will certainly be a time-saving method.
Illustration : (a) NO3− + H2S HSO4− + NH4+
(b) Fe + N2H4 Fe(OH)2 + NH3
(a) Step 1 : N5+ + 8e → N3−
S2− → S6+ + 8e
Step 2: N5+ + S2− → N3− + S6+
Step 3 : NO3−+ H2S → NH4++ HSO4−
Step 4 : NO other atom (except H and O) is unbalanced therefore no need for this step.
Step 5: Balancing O atom. This is made by using H2O and H+ ions. Add desired molecules of H2O on the side deficient in O atom and double H+ on opposite side. Therefore
NO3− + H2S + H2O → NH4+ + HSO4− + 2H+
Step 6 : Balance charge by H+
NO3– + H2S + H2O + 3H+ → NH4+ + HSO4− + 2H+
∴ Balanced equation is
NO3– + H2S + H2O + H+ → NH4+ + HSO4−
(b) Step 1 : Fe → Fe2+ + 2e
N22− + 2e → 2N3−
Step 2 : Fe + N22− → Fe2+ + 2N3−
Step 3 : Fe + N2H4 → Fe(OH)2 + 2NH3
Step 4 : No other atom (except H and O) is unbalanced and therefore no need for this step.
Step 5: Fe + N2H4 + 4OH− → Fe(OH)2 + 2NH3 + 2H2O
Step 6 : Balance charge by H+
∴ Fe + N2H4 + 4OH− + 4H+ → Fe(OH)2 + 2NH3 + 2H2O
Finally Fe + N2H4 + 2H2O → Fe(OH)2 + 2NH3
Oxidation Number Method :
This method is based on the principle that the number of electrons lost in oxidation must be equal to the number of electrons gained in reduction. The steps to be followed are :
(i) Write the equation (if it is not complete, then complete it) representing the chemical changes.
(ii) By knowing oxidation number of elements, identify which atom(s) is(are) undergoing oxidation and reduction. Write down separate equations for oxidation and reduction.
(iii) Add required electrons on the right hand side of oxidation reaction and on the left hand side of reduction reaction. Care must be taken to ensure that the net charge on both the sides of the equation is same.
(iv) Multiply the oxidation and reduction reactions by suitable numbers to make the number of electrons lost in oxidation reactions equal to the number of electrons gained in reduction reactions.
(v) Transfer the coefficients of the oxidizing and reducing agents and their products to the main equation.
By inspection, arrive at the co-efficients of the species not undergoing oxidation or reduction.