Hope this helps. :)-separate into I and NO3^1- reactions2I- ---> I2NO3^1- --->NO2balance the Os by adding water2I- ---> I2NO3^1- --->NO2 + H2Obalance the Hs by adding H+2I- ---> I2NO3^1- + 2H+--->NO2 + H2Obalance the charges by adding electrons2I- ---> I2 +2e-NO3^1- + 2H+ +1e----> NO2 + H2Oequalize the number of electrons exchanged by multiplying the second equation by 22I- ---> I2 +2e-2NO3^1- + 4H+ +2e----> 2NO2 + 2H2Oadd the equations together2I- + 2NO3^1- + 4H+ + 2e- ---> I2 + 2e- + 2NO2 + 2H2Oadd out any common items (usually e-, H+ and H2O)2I- + 2NO3^1- + 4H+ ---> I2 + 2NO2 + 2H2OCheck mass balance: 2I, 2N, 6O, 4HCheck charge balance: 2- + 2- + 4+ = 0 + 0 + 0,......
Final Reaction:
2I-(aq) + 4H+ + 2NO3-(aq) ---> 2NO2(g) + 2H2O + I2(aq)
this should be it. if the medium is basic then simply add same no. of OH- ions as that of H+ on both sides. On the side already containing H+, H+ will combine with OH- to form H2O.
Hope this helps. :)
