Question

When add nitrous acid to potassium iodide solution?

Answer 1

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Reaction: 
HNO2 + I- ---> NO + I2 

The two half-reactions are: 
HNO2 ---> NO 
I- ---> I2 

---------------------------- 
For the first half-reaction, 
in an acid solution : 

Add water on the right, 
to balance with O on the left. 
Then add H+ on the left to 
balance with H on the right: 
HNO2 + 1H+ ---> NO + H2O 

Next, add one electron to the left, 
to balance with zero charge on the right. 
HNO2 + 1H+ + 1e- ---> NO + H2O 

---------------------------- 
For the second half-reaction, 
in an acid solution: 

Balancing I: 
2I- ---> I2 

Next, add one electron to the right, 
to balance with -2 charge on the left. 
2I- ---> I2 + 2e- 

---------------------------- 
Multiplying the first half-reaction by 2, to 
balance electrons with the second half-reaction: 
2 x (HNO2 + 1H+ + 1e- ---> NO + H2O) 
2I- ---> I2 + 2e- 

Adding both half-reactions together: 
2HNO2 + 2H+ + 2e- ---> 2NO + 2H2O 
+ 2I- ---> I2 + 2e- 
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
** 2HNO2 + 2H+ + 2I- ----> 2NO + 2H2O + I2 ** 

--------------------------------------... 
If it's a base solution, add OH- to both sides 
of the acid based solution result, to cancel H+: 
2HNO2 + 2H+ + (2OH-) + 2I- ---> 2NO + 2H2O + I2 + (2OH-) 

Adding 2H+ + 2OH- ---> 2H2O: 
2HNO2 + 2H2O + 2I- ---> 2NO + 2H2O + I2 + 2OH- 

Subtracting 2H2O from both sides: 
** 2HNO2 + 2I- ---> 2NO + I2 + 2OH- ** 

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