Balancing Of Redox Reactions

BALANCING OF REDOX REACTIONS

Some examples of the 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 oxidized as well as reduced. Such type of redox reaction is called Disproportionation reaction. Examples are

We know that during redox reactions there is change in O.N. of the elements due to transference of electrons. The number of electrons lost during oxidation is equal to the number of electrons gained during reduction. It is the basic principle of balancing of redox equations.

The two methods which are frequently employed are O.N. method and ion electron method.

Oxidation number method

Step – 1. Write the skeletal equation of all the reactants and products of the reaction.

Step – 2. Indicate the oxidation number of each element above its symbol and identify the elements which undergo a change in the oxidation number (O.N.).

Step – 3. Calculate the increase or decrease in O.N. and identify the oxidizing and reducing agents. If more than one atom of the same element is involved. Find out total increase or decrease in O.N. by multiplying this increase or decrease in O.N per atom by the number of atoms undergoing that change.

Step – 4. Multiplying the formula of the oxidizing and the reducing agents by suitable integers so as to equalize the total increase or decrease in O.N. as calculated in step 3.

Step – 5. Balance all atoms other than H and O.

Step – 6. Finally balance H and O atoms by adding $H_{2}O$molecules using hit and trial method.

Step – 7. In case of ionic reactions

(a) For acidic medium – First balance O atoms by adding $H_{2}O$ molecules on O deficient side and then balance H atom by adding H⁺ ions to whatever side deficient in H atoms.

(b) For basic medium – First balance O atom by adding $H_{2}O$ molecules on O excess side and then twice H₂O ions on opposite side.

Illustration 1. Balance the equation

P + HNO₃ $\to HP{O}_3$ + NO + H₂O

Step 1: The skeleton equation for the given reaction is P+HNO₃ $\to HP{O}_3$+NO+H₂O

Step 2:

Step 3:

Step 4: Equalise the increase / decrease in oxidation number Le. multiplying P and HPO₃ by 3 and HNO₃ and NO by 5.

3P+$5HN{O}_3\to 3HP{O}_3$+5NO

Step 5: Balance other atoms except Hand O. Here they are all balanced.

Step 6: Balance O atoms by adding Hp molecules to the side falling short of 0 atom.

3P + 5HNO₃ $\sim$3HPO3 + 5NO + Hp

Here, the H - atom is automatically balanced. So, the balanced equation is 3P + 5HNO₃ $\sim$3HPO3 + 5NO + H₂O

Ion Electron Method

This method involves the following steps:

$\bullet$ Divide the complete equation into two half reactions, one representing oxidation and the other reduction.

$\bullet$ 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.

$\bullet$ 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.

$\bullet$ 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 2. (a) $Mg(aq)+HN{O}_3(aq)\to Mg{(N{O}_3)}_2(aq)+N{O}_2(g)+H_{2}0(l)$

(b) $Fe+N_2{H}_4\underset{\text{alkaline medium}}{\stackrel{O{H}^{-}}{\rightleftharpoons }}Fe{\left(OH\right)}_2+N{H}_3$

Solution: (a) Step 1: Mg+HNO₃ I$\to Mg{(N{O}_3)}_2$ +N₂O+H₂O

Step 2:

Step 3:

Since there two N atoms in Np on RHS and only one in HN0₃ on LHS. Therefore multiply HN0₃ by 2. Mg on both sides is 1.

Step 4: Equalise the increase / decrease in oxidation number. Since the total increase in· oxidation number is 2 and decreases is 8. Therefore multiply Mg and $Mg{(N{O}_3)}_2$ by 4 and HN0₃ and Np by 1.

4Mg + 2HN0₃ $\to$4$Mg{(N{O}_3)}_2$ + N₂0 + H₂0

Step 5: Balance atoms other than O and H. N atom on RHS is 10 and only 2 on LHS. $\therefore$ The coefficient of HN0₃ is changed from 2 to 10.

Step 6: Balance '0' atoms by adding Hp molecules to the side short of ° atoms. 4Mg+10HN0₃ $\sim$ 4Mg(N03)2 +N₂0+5H₂0

The above equation is balanced equation.

Ion - electron method or Half reaction method. The various steps involved are -

(b) Step 1: $\text{Fe }\to {\text{ Fe}}_{\text{2}}^{\text{+}}\text{ + 2e}$

${\text{N}}_{\text{2}}^{\text{2-}}\text{ + 2e }\to \text{ 2}{\text{N}}^{\text{3-}}$

Step 2: ${\text{Fe + N}}_{\text{2}}^{\text{2-}}\to \text{ F}{\text{e}}^{\text{2+}}\text{ + 2}{\text{N}}^{\text{3-}}$

Step 3: $\text{Fe + }{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\to \text{ Fe(OH}{\text{)}}_{\text{2}}\text{ + 2N}{\text{H}}_{\text{3}}$

Step 4: No other atom (except H and O) is unbalanced and therefore no need for this step.

Step 5: $\text{Fe + }{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{ + 4O}{\text{H}}^{\text{-}}\to \text{ Fe(OH}{\text{)}}_{\text{2}}\text{ + 2N}{\text{H}}_{\text{3}}\text{ + 2}{\text{H}}_{\text{2}}\text{O}$

Step 6: Balance charge by H⁺

$\therefore \text{Fe + }{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{ + 4O}{\text{H}}^{\text{-}}\text{ + 4}{\text{H}}^{\text{+}}\to \text{ Fe(OH}{\text{)}}_{\text{2}}\text{ + 2N}{\text{H}}_{\text{3}}\text{ + 2}{\text{H}}_{\text{2}}\text{O}$

Finally $\text{Fe + }{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{ + 2}{\text{H}}_{\text{2}}\text{O }\to \text{ Fe(OH}{\text{)}}_{\text{2}}\text{ + 2N}{\text{H}}_{\text{3}}$