Please balance the following chemical equation by oxidation number method (redox reaction).

### Asked by Balbir | 26th Jan, 2018, 09:06: PM

In most situations of balancing an equation, you are not told whether the reaction is redox or not.

In these circumstances, you can use a procedure called the oxidation number method.

**Step 1**

The skeleton equation is:

I_{2} + HNO_{3} → HIO_{3} + NO_{2} + H_{2}O

**Step 2**

The oxidation number of various atoms involved in the reaction.

_{ 0 } _{+1 +5 -2 +1 +5 -2 +4 -2 +1 -2 }

I_{2} + HNO_{3} → HIO_{3} + NO_{2} + H_{2}O

**Step 3**

For N oxidation number changes from +5 to +4 so it is reduced. For I oxidation number changes from 0 to +5 so it is oxidized. No change in oxidation number of O.

**Step 4**

Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced.

For I 0 to +5 Net change = +5 ......... Oxidation

For N +5 to +4 Net change = -1 ..........Reduction

**Step 5**

Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number.

**I** atoms would yield a net increase in oxidation number of +5. (five electrons would be lost by five **I **atoms.). 1 N atom would yield a net decrease of -1. (One N atom would gain three electrons.)

Thus the ratio of I atoms to N atoms is 5:1.But we have I_{2} as a reactant then the ratio becomes 10:1

**Step 6**

To get the ratio identified in Step 5, add coefficients to the formulas which contain the elements whose oxidation number is changing.

and we will get the balanced equation,

I_{2} + 10 HNO_{3} → 2HIO_{3} + 10NO_{2} + 4H_{2}O

### Answered by Ramandeep | 29th Jan, 2018, 11:50: AM

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