# Oxidation and Reduction Equation

**Example 1. Balance the following equations:**

**Solution:**

(i) Oxidising agent is

gain of electron = 5

Reducing agent is

Now add on right hand side to balance oxygen atoms and ions on left hand side to balance hydrogen atoms, we get

……..(I)

Reducing agent is

……..(II)

Multiply eq. (II) by 3 and add in eq. (I)

………(III)

Now balance the eq. (III) for oxygen by adding on right hand side and then add ion on left hand side. We get

Oxidising agent is

……..(I)

Reducing agent is KCl

…………(II)

Multiply eq. (I) by 2 and eq. (II) by 5 and add:

Adjusting for

** **

**Example 2.** *Balance the following equations:*

**Solution: **Two half reactions may be:

**Oxidation:** ……..(i)

**Reduction: ** ………….(ii)

Multiply eq. (ii) by 2 and on adding in eq. (i) we get:

**Example 3. Balance the following equations:**

(i)

(ii)

(iii)

**Solution:**

(i)

Multiplying eq. (a) by 3 and (b) by 2, and add

Charge -2 +12

Now add on left side to equalize the charge and on right side. We get

(ii)

In this case C of ion is reduced to CO(change in Ox. No. +4 to +2) and N of ion is oxidised to ion (change in Ox. N0. -5 to -3). Since both oxidation and reduction take place in the same compound hence it is written two times as:

On adding eq. (I) and (II), we get

Or,

Now balancing C and N atoms,

Now we can write,

(iii)

In this case is reduced to , therefore

……..(I)

In carbon atom is oxidised to and (Ox. No. from -2 to +2)

……..(II)

Now multiply eq. (I) by 4 and add in eq. (II)

Since three iodine atoms are in molecule hence on adjusting these atoms.

Since this reaction takes place in alkaline medium hence on balancing charge are added on left side then balance oxygen atoms by adding water molecules, we get

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