Please expalin the following question in detail

Asked by Balbir | 11th Jan, 2018, 04:03: PM

Expert Answer:

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

I +  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  + I2→  2 I+5  +   10 N+4           

Thus the original form of equation is written as, 


10 space H N O subscript 3 space plus space I subscript 2 space end subscript rightwards arrow with blank on top space 2 H I O subscript 3 space space plus space space 10 space N O subscript 2 space space plus space 4 H subscript 2 O

Answered by Ramandeep | 11th Jan, 2018, 06:55: PM

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