Please expalin the following question in detail
Asked by Balbir | 11th Jan, 2018, 04:03: PM
It’s a simple redox reaction in which oxidation number of nitrogen is decreasing and oxidation number of iodine is increasing.
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)
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,
Answered by Ramandeep | 11th Jan, 2018, 06:55: PM
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