Please expalin the following question in detail

It’s a simple redox reaction in which oxidation number of nitrogen is decreasing and oxidation number of iodine is increasing.

1. Nitrogen:

O.S. of N in HNO₃ is +5 which on reduction become +4

HNO3  → H⁺  +  NO₃^-

NO₃^- + 2H⁺ + e^-   →    NO₂ + H₂O

i.e., N⁺⁵  +  e^-   →    N⁺⁴ ………………(1)

2. Iodine:

O.S. of I in I₂ is 0 which on oxidation become +5

I₂   +  6H₂O   →  2IO₃^-  +  12 H⁺  + 10 e^- 

 i.e., I₂⁰      →   2 I⁺⁵  +  10 e^-     …………..(2)

Multiply eq (1) by 10, we get,

10N⁺⁵  +  10e^-   →   10 N⁺⁴           ….(3)

By adding  eq(2) by (3) we get,

10N⁺⁵  + I₂⁰  →  2 I⁺⁵  +   10 N⁺⁴           

Thus the original form of equation is written as,